VQucBCnNguynVnThchNguynPhiHngPhanHngSnVThnhVnCollectedproblemsAboutinequalityNgy 19 thng 5 nm 2007www.VNMATH.comiiwww.VNMATH.comMclc1 Problems 12 Solution 172.1 Li gii cc bi ton . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 172.2 Tc gi cc bi ton . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 164iiiwww.VNMATH.comiv MCLCwww.VNMATH.comChng 1Problems1. Chox, y, z l cc s dng thaxy + yz + zx = 1, chng minh1_1 + (2x y)2+1_1 + (2y z)2+1_1 + (2z x)23322. Cho cc s dnga, b, c thaabc = 1, chng minh rngab + cb + c + 1 +bc + ac + a + 1 +ca + ba + b + 1 23. Vi mi s khng ma, b, c, ta c_a4a + 4b + c +_b4b + 4c + a +_c4c + 4a + b 14. Cho cc s dnga, b, c, chng minh1a2+ bc +1b2+ ca +1c2+ ab a + b + cab + bc + ca_1a + b +1b + c +1c + a_5. Chng minh rng vi mi s dnga, b, c ta lun ca32a2ab + 2b2+b32b2bc + 2c2+c32c2ca + 2a2 a + b + c36. Cho cc s khng ma, b, c thaa + b + c = 1. Chng minh bt ng thc_a + (b c)24+_b + (c a)24+_c + (a b)243 +_1 32_(|a b| +|b c| +|c a|)7. Cho cc s dnga, b, c thaa + b + c = 3, chng minh bt ng thca3/2b + b3/2c + c3/2a 38. Chng minh rng vi mi s thca, b, c, ta cab4a2+ b2+ 4c2+bc4b2+ c2+ 4a2+ca4c2+ a2+ 4b2 131www.VNMATH.com2 CHNG1. PROBLEMS9. Cho cc s khng ma, b, c thaa + b + c = 3, chng minha2+ b2(a + 1)(b + 1) +b2+ c2(b + 1)(c + 1) +c2+ a2(c + 1)(a + 1) 3210. Vi mia b c 0, tP=ab + c +bc + a +ca + bQ =2(b + c) a4a + b + c+ 2(c + a) b4b + c + a+ 2(a + b) c4c + a + bChng minh rng(a) Nua + c 2b thP Q.(b) Nua + c 2b thP Q.11. Cho cc s khng ma, b, c thaa + b + c = 1, tx = a2+ b2+ c2, chng minh bt ng thc_1 + 2a2x +_1 + 2b2x +_1 + 2c2x 11 9x12. Chng minh rng vi mia, b, c > 0, ta c1a(a + b) +1b(b + c) +1c(c + a) 32(abc)2/313. Chng minh rng nua, b, c > 0 th1aa + b+1bb + c+1cc + a 32abc14. Cho cc s dngx, y, z thax2+ y2+ z2 3, chng minh rngx5x2x5+ y2+ z2+y5y2y5+ z2+ x2+z5z2z5+ x2+ y2 015. Chon 3 va1, a2, . . . , anl cc s khng m thaa21 + a22 + + a2n = 1, chng minh bt ngthc13(a1 + a2 + + an) a1a2 + a2a3 + + ana116. Cho cc s dnga, b, c, chng minh bt ng thc_ab+bc +ca +_ab + bc + caa2+ b2+ c2 3 + 117. Chng minh rng vi mia, b, c > 0, ta ca2b2+b2c2+c2a2+ 8(ab + bc + ca)a2+ b2+ c2 1118. Chng minh rng vi mi s dnga1, a2, . . . , an, b1, b2, . . . , bn, ta c_n

i=1a2i__n

i=1b2i__n

i=1bi(ai + bi)__n

i=1a2ibiai + bi_www.VNMATH.com319. Chng minh rng vi cc s thca, b, c i mt khc nhau, ta c(a2+ b2+ c2ab bc ca)_1(a b)2+1(b c)2+1(c a)2_27420. Cho cc s khng ma, b, c, d thaa2+ b2+ c2+ d2= 4, chng minh bt ng thc13 abc +13 bcd +13 cda +13 dab 221. Cho cc s dnga, b, c, chng minh bt ng thcab+bc +ca 3_a2+ b2+ c2ab + bc + ca22. Cho cc s khng ma, b, c, chng minh bt ng thc7_3(a2+ b2+ c2)a + b + c+a2b + b2c + c2aa3+ b3+ c3 823. Chng minh rng vi mi s dnga, b, c ta ca3a3+ abc + b3+b3b3+ abc + c3+c3c3+ abc + a3 124. Cho cc s dnga, b, c, d, chng minh rngabc(d + a)(d + b)(d + c) +abd(c + a)(c + b)(c + d) +acd(b + a)(b + c)(b + d) +bcd(a + b)(a + c)(a + d) 1225. Chng minh rng vi mia, b, c > 0, ta cab+c+ bc+a+ ca+b 126. Chon 3, n Nvx1, x2, . . . , xnlccskhngmctngbng1.Tmgitrlnnhtcabiu thcP(x1, x2, . . . , xn) = x31x22 + x32x23 + + x3nx21 + n2(n1)x31x32 x3n27. Cho cc s thca1, a2, . . . , anthaa1a2 an = 1, tm cc hng s tt nhtm, Msao cho_a21 + n21 +_a22 + n21 + +_a2n + n21 m(a1 + a2 + + an) + M28. Chng minh rng vi mi s dnga, b, c, d, ta ca3a2+ 2b2+ c2+b3b2+ 2c2+ d2+c3c2+ 2d2+ a2+d3d2+ 2a2+ b2 16_1a + 1b + 1c + 1d_29. Cho cc s dngx, y, z, chng minh bt ng thcx(y + z)x2+ yz+y(z + x)y2+ zx+z(x + y)z2+ xyx + y + z3xyzx2+ yzx(y + z) +y2+ zxy(z + x) +z2+ xyz(x + y)30. Vi mi s dnga, b, c thaa + b + c = 3, ta cab2+ c +bc2+ a +ca2+ b 32www.VNMATH.com4 CHNG1. PROBLEMS31. Vi mi s khng ma, b, c thaa + b + c = 3, ta ca_b3+ 1 + b_c3+ 1 + c_a3+ 1 532. Tm hng sk tt nht sao cho bt ng thc sau ng vi mia, b, c > 0(a + b + c)_1a + 1b + 1c_ 9 +k max{(a b)2, (b c)2, (c a)2}(a + b + c)233. Cho cc s dngx, y, z c tch bng 1, chng minh rng vi mik 0, ta c3_xy + k +3_yz + k +3_zx + k 33k + 134. Cho cc s dnga, b, c, chng minh bt ng thcb2+ c2a(b + c) +c2+ a2b(c + a) +a2+ b2c(a + b) (a2+ b2+ c2)3abc(a + b + c)35. Cho cc s dnga, b, c, chng minh bt ng thc2_a2b+b2c+c2a_+ 3(a + b + c) 15(a2+ b2+ c2)a + b + c36. Chng minh rng vi mi s thc dngx, y, z c tch bng 1 v vi mik 0, ta c4_xy + k +4_yz + k +4_zx + k 34k + 137. Chng minh rng vi mi s khng ma, b, c v vi mik 3, ta ca(bk+ ck)a2+ bc+b(ck+ ak)b2+ ca+c(ak+ bk)c2+ ab ak1+ bk1+ ck138. Cho cc s khng ma, b, c, chng minh bt ng thca4a3+ abc + b3+b4b3+ abc + c3+c4c3+ abc + a3 a3+ b3+ c3a2+ b2+ c239. Cho cc s dngx, y, z, t tha1x + 1 +1y + 1 +1z + 1 +1t + 1= 1Chng minh rngmin_1x + 1y + 1z, 1y + 1z+ 1t, 1z+ 1t+ 1x, 1t+ 1x + 1y_ 1 max_1x + 1y + 1z, 1y + 1z+ 1t, 1z+ 1t+ 1x, 1t+ 1x + 1y_40. Cho cc s khng ma, b, c, chng minh bt ng thca24a2+ ab + 4b2+b24b2+ bc + 4c2+c24c2+ ca + 4a2a + b + c3www.VNMATH.com541. Cho cc s dnga, b, c, chng minh bt ng thca(b + c)a2+ bc+b(c + a)b2+ ca+c(a + b)c2+ ab12(a + b + c)_1a + 1b + 1c_+ 2742. Cho cc s khng ma, b, c thaa + b + c = 1, chng minh bt ng thcaa + 2b+bb + 2c+cc + 2a_3243. Cho cc s khng ma, b, c, tm hng sk tt nht bt ng thc sau ngab + c +bc + a +ca + b 32 +k max{(a b)2, (b c)2, (c a)2}ab + bc + ca44. Cho cc s khng ma, b, c, chng minh bt ng thc_aa + b_3+_bb + c_3+_cc + a_338 _a2+ b2+ c2ab + bc + ca_245. Choa, b, c, d l cc s dng tha mna, b, c 1 vabcd = 1, chng minh rng1(a2a + 1)2+1(b2b + 1)2+1(c2c + 1)2+1(d2d + 1)2 446. Vi mi s khng ma, b, c, chng minh rng_a2+ 4bcb2+ c2+_b2+ 4cac2+ a2+_c2+ 4aba2+ b2 2 +247. Cho cc s khng ma, b, c, chng minh bt ng thc(a b)(13a + 5b)a2+ b2+ (b c)(13b + 5c)b2+ c2+ (c a)(13c + 5a)c2+ a2 048. Chng minh rng vi mi s dnga, b, c, n, ta c_a2+ bcb + c_n+_b2+ cac + a_n+_c2+ aba + b_n an+ bn+ cn49. Cho cc s khng ma, b, c thaa +b +c = 1. Ty theo gi tr can N, hy tm gi tr ln nhtv gi tr nh nht ca biu thcP(a, b, c) = a(b c)n+ b(c a)n+ c(a b)n50. Cho cc s dnga, b, c thaa + b + c = 3, tm hng sk ln nht sao choa5+ b5+ c53a3+ b3+ c33 k51. Cho cc s khng ma, b, c thaa2+ b2+ c2= 8, chng minh bt ng thc4(a + b + c 4) abcwww.VNMATH.com6 CHNG1. PROBLEMS52. Chom, n(3n2>m2)lccsthcchotrcva, b, clccsthcthamna + b + c=m, a2+ b2+ c2= n2. Tm gi tr ln nht v gi tr nh nht ca biu thc sauP= a2b + b2c + c2a53. Tm hng sk nh nht sao cho vi mia, b, c 0 tha3ka2+ (b + c)2+b3kb2+ (c + a)2+c3kc2+ (a + b)2 _3(a + b + c)k + 454. Chng minh rng nua, b, c > 0 va + b + c = 3 th(ab + bc + ca)_ab2+ 9 +bc2+ 9 +ca2+ 9_91055. Cho cc s dnga, b, c thaa + b + c = 3, chng minh bt ng thcabc2+ 3 +bca2+ 3 +cab2+ 3 3256. Chng minh rng vi mia, b, c dng th_b + ca+_c + ab+_a + bc16(a + b + c)33(a + b)(b + c)(c + a)57. Tm hng sk ln nht sao cho bt ng thc sau ng1a(1 + bc)2+1b(1 + ca)2+1c(1 + ab)2 k(1 + ab)(1 + bc)(1 + ca) + 34 k8trong a, b, c l cc s dng thaabc = 1.58. Cho cc s khng ma, b, c, chng minh bt ng thc sau vik =ln 3ln 3ln 2_a2b2+ bc + c2_1/k+_b2c2+ ca + a2_1/k+_c2a2+ ab + b2_1/k 259. Cho cc s khng ma, b, c chng minh bt ng thc_a2+ bcb2+ bc + c2+_b2+ cac2+ ca + a2+_c2+ aba2+ ab + b2 660. Chng minh rng vi mix, y [0, 1], ta c1x2x + 1 +1y2y + 1 1 +1x2y2xy + 161. Cho cc s dnga, b, c, chng minh bt ng thc_aa + b +_bb + c +_cc + a 32 _ab + bc + caa2+ b2+ c262. Chng minh rng vi mia, b, c 0, ta c bt ng thca2(b + c)(b2+ c2)(2a + b + c) +b2(c + a)(c2+ a2)(2b + c + a) +c2(a + b)(a2+ b2)(2c + a + b) 23www.VNMATH.com763. Cho cc s dnga, b, c, chng minh rng vi mik 2, ta c bt ng thca + b + c3abck_a + cb + c+k_c + ba + b +k_b + ac + a64. Cho cc s khng ma, b, c, chng minh bt ng thc3_ab + c +3_bc + a +3_ca + b 2abc(a + b)(b + c)(c + a) + 165. Cho cc s thca, b, c, d thaa2+ b2+ c2+ d2= 4, chng minh bt ng thc9(a + b + c + d) 4abcd + 3266. Cho cc s khng ma, b, c, chng minh bt ng thc_a2+ 256bcb2+ c2+_b2+ 256cac2+ a2+_c2+ 256aba2+ b2 1267. Cho cc s dngx, y, z c tch bng 1, chng minh rngxy4+ 2 +yz4+ 2 +zx4+ 2 168. Chng minh rng vi mi s dnga, b, c, d ta c bt ng thc_1a + 1b + 1c + 1d__1a + b +1b + c +1c + d +1d + a_16abcd + 169. Cho cc s dnga, b, c, d thaa2+ b2+ c2+ d2= 4, chng minh bt ng thca + b + c + d23(abcd + 1)_1a + 1b + 1c + 1d_70. Cho cc s dnga1, a2, . . . , anthaa1a2 an = 1. Khi , vi mik R, ta c1(1 + a1)k+1(1 + a2)k+ +1(1 + an)k min_1,n2k_71. Choa, b, c l cc s dng, chng minh rng(a)a9bc+b9ca +c9ab +2abc a5+ b5+ c5+ 2(b)a9bc+b9ca +c9ab +3abc a4+ b4+ c4+ 372. Chox, y, z, t l cc s dng thaxyzt = 1, chng minh rng1xy + yz + zx + 1 +1yz + zt + ty + 1 +1zt + tx + xz + 1 +1tx + xy + yt + 1 173. Chng minh rng vi mix, y, z, t > 0 th(x + y)(x + z)(x + t)(y + z)(y + t)(z + t) 4xyzt(x + y + z + t)2www.VNMATH.com8 CHNG1. PROBLEMS74. Chng minh rng vi mi s dnga1, a2, . . . , anthaa1a2 an = 1 ta c bt ng thc_a21 + 1 +_a22 + 1 + +_a2n + 1 2(a1 + a2 + + an)75. Chng minh rng vi mi s dnga, b, c ta c bt ng thca +ab +3abc33_a a + b2a + b + c376. Cho cc s khng ma, b, c, chng minh bt ng thca3b2bc + c2+b3c2ca + a2+c3a2ab + b2 a2+ b2+ c277. Chng minh rng vi mia, b, c khng m_a2a2+ 6ab + 2b2+_b2b2+ 6bc + 2c2+_c2c2+ 6ca + 2a2 178. Cho cc s khng ma, b, c, chng minh bt ng thc_ab + c +_bc + a +_ca + b + 3_3(ab + bc + ca)a2+ b2+ c272279. Cho cc s khng ma, b, c, chng minh bt ng thcab + c +bc + a +ca + b + 16(ab + bc + ca)a2+ b2+ c2 880. Cho cc s khng ma, b, c, chng minh bt ng thc3(a3+ b3+ c3) + 2abc 11_a2+ b2+ c23_3/281. Cho cc s khng ma, b, c, d thaa2+ b2+ c2+ d2= 1, chng minh bt ng thca31 bcd +b31 cda +c31 dab +d31 abc 4782. Cho cc s khng ma, b, c, d thaa3+ b3+ c3+ d3= 1, chng minh bt ng thc1 a31 bcd +b31 cda +c31 dab +d31 abc 4383. Cho cc s dnga, b, c, d thaa + b + c + d = 4, chng minh rng1ab +1bc +1cd +1da a2+ b2+ c2+ d284. Cho cc s dngx, y, z, tm hng sk ln nht sao choxy+yz+zx + 3k (k + 1) x + y + z3xyzwww.VNMATH.com985. Cho cc s khng ma, b, c, d, chng minh bt ng thc_aa + b + c +_bb + c + d +_cc + d + a +_dd + a + b 4386. Chng minh rng vi mia, b, c, d [1, 2], ta ca + bc + d +c + da + b a + cb + d 3287. Chng minh rng vi mia, b, c > 0, ta lun ca2bc(b + c) +b2ca(c + a) +c2ab(a + b) 32a2+ b2+ c2a + b + c88. Cho cc s khng ma, b, c, thaa2+ b2+ c2= 3, chng minh rng1 + 4abc 5 min{a, b, c}89. Vi mia, b, c 0 vab + bc + ca = 1, ta c12a2+ 3bc+12b2+ 3ca+12c2+ 3ab26390. Choa, b, c l cc s thc khc 0 thaa2+ b2+ c2= (a b)2+ (b c)2+ (c a)2, chng minh btng thc1.ab+bc +ca 5 2.112 a2b + b2c + c2a(a + b + c)353691. Tm hng sk > 0 nh nht sao cho bt ng thc_a + k(b c)2+_b + k(c a)2+_c + k(a b)23ng vi mia, b, c 0 va + b + c = 1.92. Chng minh rng vi mia, b, c 0 tha3+ abc(b + c)3+b3+ abc(c + a)3+c3+ abc(a + b)3 ab + c +bc + a +ca + b93. Cho cc s dnga, b, c, chng minh rngab2c2+bc2a2+ca2b2+ a + b + c 6(a2+ b2+ c2)a + b + c94. Tm gi tr ln nht ca biu thcP= (a b)(b c)(c a)(a + b + c)via, b, c 0 thaa2+ b2+ c2= 1.95. Vi mi s dnga, b, c, d,b(a + c)c(a + b) +c(b + d)d(b + c) +d(c + a)a(c + d) +a(d + b)b(d + a) 4www.VNMATH.com10 CHNG1. PROBLEMS96. Chng mnh rng vi mi s thca, b, c tha2bca2+ 2b2+ 3c2+b2cab2+ 2c2+ 3a2+c2cac2+ 2a2+ 3b2 097. Cho cc s khng mx, y, z, chng minh bt ng thcx4x4+ x2yz + y2z2+y4y4+ y2zx + z2x2+z4z4+ z2xy + x2y2 198. Cho cc s dnga, b, c thaabc = 1, chng minh rng1a2a + 1 +1b2b + 1 +1c2c + 1 399. Chng minh rng vi mi s dnga, b, c,3a22ab b23a2+ 2ab + 3b2+3b22bc c23b2+ 2bc + 3c2+3c22ca a23c2+ 2ca + 3a2 0100. Cho cc s dnga, b, c thaa4+ b4+ c4= 3, chng minh bt ng thca2b3+ 1 +b2c3+ 1 +c2a3+ 1 32101. Cho cc s dnga, b, c, chng minh bt ng thc92 (a2+ b2+ c2)3(a + b + c)4a3a + b +b3b + c +c3c + a102. Cho cc s dnga, b, c, d thaa + b + c + d = 4, tm hng sk tt nht sao cho1a + 1b + 1c + 1d 4 k(a2+ b2+ c2+ d24)103. Cho cc s dngx, y, z thaxy + yz + zx = 1, chng minh bt ng thcx(y + z)2(1 + yz)2+y(z + x)2(1 + zx)2+z(x + y)2(1 + xy)2 334104. Cho cc s khng ma, b, c thaa + b + c = 3, chng minh bt ng thc_a +_b2+ c2+_b +_c2+ a2+_c +_a2+ b2 3_2 + 1105. Choa, b, c l di ba cnh ca mt tam gic, chng minh rnga3a + b c +b3b + c a +c3c + a b 1106. Cho cc s dnga, b, c thaa2+ b2+ c2= 3, chng minh bt ng thcaab + 3 +bbc + 3 +cca + 3 34www.VNMATH.com11107. Cho cc s khng ma, b, c, chng minh bt ng thca2b2+ (c + a)2+b2c2+ (a + b)2+c2a2+ (b + c)2 35108. Choa, b, c l di ba cnh ca mt tam gic, chng minh bt ng thca(a b)a2+ 2bc +b(b c)b2+ 2ca +c(c a)c2+ 2ab 0109. Cho cc s dnga, b, c, chng minh_a2a2+ 7ab + b2+_b2b2+ 7bc + c2+_c2c2+ 7ca + a2 1110. Cho cc s khng ma, b, c, chng minh bt ng thc1a2+ bc+1b2+ ca+1c2+ ab2_1a + b +1b + c +1c + a_111. Choa, b, c l di 3 cnh ca 1 tam gic, chng minh rng3_ab+bc +ca 3_ 2_ba +cb +ac 3_112. Chng minh rng nua, b, c l di 3 cnh ca 1 tam gic tha2bc+b2ca+c2ab a2+ b2+ c2113. Cho cc s khng ma, b, c chng minh bt ng thca2b2+b2c2+c2a2+ 9(ab + bc + ca)a2+ b2+ c2 12114. Cho cc s khng ma, b, c, chng minh bt ng thcab+bc +ca 3_a2+ b2+ c2ab + bc + ca_2/3115. Cho cc s khng ma, b, c, chng minh bt ng thcab+bc +ca 239(a3+ b3+ c3)(a + b)(b + c)(c + a)116. Cho cc s khng mx, y, z thax + y2+ z2= 1, chng minh bt ng thcx3x2+ xy + y2+y3y2+ yz + z2+z3z2+ zx + x2 12117. Choa, b, c l di 3 cnh ca mt tam gic, chng minh bt ng thca2+ b2a2+ c2+b2+ c2b2+ a2+c2+ a2c2+ b2 a + ba + c +b + cb + a +c + ac + bwww.VNMATH.com12 CHNG1. PROBLEMS118. Choa, b, c l di 3 cnh ca mt tam gic, chng minh rng3(a3b + b3c + c3a) (a2+ b2+ c2)(ab + bc + ca)119. Cho cc s thca, b, c, chng minh bt ng thc15a2b2c2+ 12(a4+ b4+ c4)(a2+ b2+ c2) 11(a6+ b6+ c6) + 30abc(a3+ b3+ c3)120. Cho cc s khng ma, b, c, d thaa + b + c + d = 3, chng minh bt ng thcab(b + c) + bc(c + d) + cd(d + a) + da(a + b) 4121. Choa, b, c l cc s khn m thaa2+ b2+ c2= 1, chng minh rng_1 _a + b2_2__1 _b + c2_2__1 _c + a2_2_827122. Cho cc s khng ma, b, c, d, chng minh bt ng thcaba + b +bcb + c +cdc + d +dad + a _(a + c)(b + d)123. Chng minh rng vi mi s dnga, b, c ta c bt ng thcab+bc +ca _a2+ c2b2+ c2+_c2+ b2a2+ b2+_b2+ a2c2+ a2124. Cho cc s khng ma, b, c thaa + b + c = 5, chng minh bt ng thc16(a3b + b3c + c3a) + 640 11(ab3+ bc3+ ca3)125. Cho cc s dnga, b, c, chng minh bt ng thc1a + b + c _1a + b +1b + c +1c + a_1ab + bc + ca +12(a2+ b2+ c2)126. Chng minh rng vi mi s khng ma, b, c, d ta c1a3+ b3+1a3+ c3+1a3+ d3+1b3+ c3+1b3+ d3+1c3+ d3 2432(a + b + c + d)3127. Chng minh rng vi mi s khng ma, b, c, d ta c1a2+ b2+ c2+1b2+ c2+ d2+1c2+ d2+ a2+1d2+ a2+ b2 12(a + b + c + d)2128. Cho cc s dnga, b, c, chng minh bt ng thc_a(b + c)a2+ bc+_b(c + a)b2+ ca+_c(a + b)c2+ ab_a +b +c__1a +1b+1c_129. Chng minh rng vi mi s dnga, b, c tha2bca2+ 2b2+ 3c2+b2cab2+ 2c2+ 3a2+c2abc2+ 2a2+ 3b2 0www.VNMATH.com13130. Cho cc s dnga, b, c thaa + b + c = 1, chng minh bt ng thc_1a 2_2+_1b 2_2+_1c 2_28(a2+ b2+ c2)2(1 a)(1 b)(1 c)131. Cho cc s khng ma, b, c, d thaa + b + c + d = 1, chng minh bt ng thca4b4+ c4d42a2c2+ 2b2d2+ 4ab2c + 4cd2a 4bc2d 4da2b 1132. Cho cc s dnga, b, c, chng minh bt ng thcab(a2+ bc)b + c+bc(b2+ ca)c + a+ca(c2+ ab)a + b_3abc(ab2+ bc2+ ca2)133. Tm hng sa nh nht sao cho bt ng thc sau_x + y + z3_a_xy + yz + zx3_3a2(x + y)(y + z)(z + x)8ng vi mi s thc dngx, y, z.134. Cho cc s khng ma, b, c thaa2+ b2+ c2= 1, chng minh bt ng thc1 a1 + bc+b1 + ca+c1 + ab32135. Choa, b, c l cc s khng m, chng minh bt ng thc_a(b + c)b2+ c2+_b(c + a)c2+ a2+_c(a + b)a2+ b2_2 + 2_1 + 4abc(a + b)(b + c)(c + a)(a2+ b2)(b2+ c2)(c2+ a2)136. Choa, b, c l cc s thc dng, chng minh rnga2ab + b2a + b+b2bc + c2b + c+c2ca + a2c + a32 a3+ b3+ c3a2+ b2+ c2137. Chng minh rng vi mi s dnga, b, c > 0 thaabc = 1, ta c bt ng thc1(1 + a)2+1(1 + b)2+1(1 + c)2+1a + b + c + 1 1138. Cho cc s dngx, y, x thax + y + z = 1. Chng minh rng_x2+ xyz +_y2+ xyz +_z2+ xyz _x2+ y2+ z2+ xy + yz + zx + 2_3xyz139. Chng minh rng nux, y, z l cc s khng m thax2+ y2+ z2= 1 th9318 13_1 _x+y2_2+13_1 _y+z2_2+13_1 _z+x2_2 1 +436140. Chng minh rng vi mi s khng ma, b, c thaa + b + c = 1,a4a + 5b2+b4b + 5c2+c4c + 5a2317www.VNMATH.com14 CHNG1. PROBLEMS141. Tm hng sk = k(n) ln nht sao cho bt ng thc sau ng vi mi s thca1, a2, . . . , ana21 + a22 + + a2n k(n)(a1a2 + a2a3 + + an1an)142. Vi mi s dnga, b, c, ta c3_a2+ bcb + c+3_b2+ cac + a+3_c2+ aba + b3_9(a + b + c)143. Cho cc s khng ma, b, c, chng minh bt ng thc_a +b2c_2+_b +c2a_2+_c +a2b_212(a3+ b3+ c3)a + b + c144. Cho cc s khng ma, b, c thaab + bc + ca = 1, chng minh bt ng thc1a + bc+1b + ca+1c + ab 22145. Cho cc s dnga, b, c thaa + b + c =1a +1b +1c, chng minh_a + bb + 1 +_b + cc + 1 +_c + aa + 1 3146. Choa1, a2, . . . , a5l cc s dng thaa1a2 a5 = a1(1 + a2) + a2(1 + a3) + + a5(1 + a1) + 2Tm gi tr nh nht ca biu thcP=1a1+1a2+ +1a5.147. Vi mi s dnga, b, c, ta ca(a + c)b(b + c)+b(b + a)c(c + a) +c(c + b)a(a + b) 3(a2+ b2+ c2)ab + bc + ca148. Chng minh rng vi mia, b, c dng,a(b + c)a2+ bc+b(c + a)b2+ ca+c(a + b)c2+ ab_6(a2+ b2+ c2)149. Choa, b, c l cc s dng, chng minh rng3 +ab+bc +ca 2(a + b + c)_1a + 1b + 1c_150. Choa, b, c l cc s khng m tha mnab + bc + ca = 1, chng minha2b+b2c+c2a 2(a2+ b2+ c2) 3 2151. Tm hng sk ln nht sao cho bt ng thc sau nga + b + c + kabc k + 3vi mi s khng ma, b, c tha mnab + bc + ca + 6abc = 9.www.VNMATH.com15152. Cho cc s khng ma, b, c thaa2+ b2+ c2= 1. Chng minh rnga3b2bc + c2+b3c2ca + a2+c3a2ab + b2 2153. Cho cc s khng mx, y, z tha 6 x + y + z 3, chng minh rng1 + x +_1 + y +1 + z _xy + yz + zx + 15154. Cho cc s dngx, y, z thaxyz = 1, chng minh bt ng thcy + zx3+ yz+z + xy3+ zx +x + yz3+ xy 1x2+1y2+1z2155. Cho cc s dnga, b, c, chng minh bt ng thc399a(a + b)2(a + b + c)2+36bc(a + b)(a + b + c) 4156. Cho cc s khng ma, b, c, chng minh bt ng thc1(a + 2b)2+1(b + 2c)2+1(c + 2a)2 1ab + bc + ca157. Cho cc s khng ma, b, c, chng minh bt ng thca2a2+ ab + b2+b2b2+ bc + c2+c2c2+ ca + a2+ab + bc + caa2+ b2+ c2 2158. Cho cc s khng mx, y, z thax + y + z = 3, chng minh bt ng thcx2y + y2z + 32xyz 4159. Cho cc s khng ma, b, c, chng minh bt ng thc1a2+ bc +1b2+ ca +1c2+ ab 3(a + b + c)22(a2+ b2+ c2)(ab + bc + ca)160. Cho cc s khng ma, b, c, chng minh bt ng thc43(ab2+ bc2+ ca2) + a2+ b2+ c2+ 2 3(ab + bc + ca)161. Cho cc s khng ma, b, c, chng minh bt ng thc14a2+ bc+14b2+ ca+14c2+ ab4a + b + c162. Cho cc s thca, b, c, chng minh bt ng thc1 + a2b2(a b)2+ 1 + b2c2(b c)2+ 1 + c2a2(c a)2 32163. Cho cc s khng ma, b, c, chng minh rnga2b+b2c+c2a 3_a4+ b4+ c4a2+ b2+ c2www.VNMATH.com16 CHNG1. PROBLEMS164. Cho cc s dnga, b, c, chng minh rng_ab+bc +ca 2 +8abc(a + b)(b + c)(c + a) 2165. Cho cc s thca, b, c, chng minh bt ng thc_a(b + c)(a + b)(a + c)_2+_b(c + a)(b + c)(b + a)_2+_c(a + b)(c + a)(c + b)_212166. Cho cc s khng mx, y, z thax + y + z = 1. Chng minh bt ng thc_x + y2+_y + z2+_z + x2115167. Cho cc s khng ma, b, c, d thaa + b + c + d = 4, tm hng sk>6427nh nht bt ngthc sau ng1k abc +1k bcd +1k cda +1k dab 4k 1168. Cho cc s khng ma, b, c, chng minh bt ng thc3(a + b + c) 2__a2+ bc +_b2+ ca +_c2+ ab_169. Cho dy dng {xn} thak

i=1xi k vi mik = 1, 2, . . . , n, chng minh bt ng thcx21 + x22 + + x2n 14_1 + 12 + 13 + +1n_170. Cho cc s khng ma, b, c tha 6 a + b + c 3, chng minh bt ng thca + 1 +b + 1 +c + 1 15 + ab + bc + cawww.VNMATH.comChng 2Solution2.1.Ligiiccbiton1Chox, y, zl cc s dng thaxy + yz + zx = 1, chng minh1_1 + (2x y)2+1_1 + (2y z)2+1_1 + (2z x)2332Li gii. t a = 2x y, b = 2y z, c = 2z x, do a +b +c = x +y +z> 0 v txy +yz +zx = 1,ta c14(a2+ b2+ c2) + 35(ab + bc + ca) = 49Li c 3(14(a2+ b2+ c2) + 35(ab + bc + ca)) 49(a + b + c)2, nna + b + c 3Ta s chng minh vi mi s thca, b, c tha mna + b + c 3, thP(a, b, c) =1a2+ 1 +1b2+ 1 +1c2+ 1 332Nu c 0, thay c bi c

= c, th ta cng c a+b+c

3, v gi tr ca biu thc Pvn khng i,do , khng mt tnh tng qut, ta c th gi sa, b, c > 0, khi , ta = ka1, b = kb1, c = kc1vik 1, a1, b1, c1> 0 sao choa1 + b1 + c1 =3, thP(a, b, c) =

cyc1_k2a21 + 1

cyc1_a21 + 1= P(a1, b1, c1)Nh vy, ta c th gi sa, b, c > 0 va + b + c =3. Xt hm sf(x) =1x2+1, ta cf

(x) =2x21(x2+ 1)5/2T y, ta c th d dng kim tra cflm trn_0,12_ v li trn_12,3_.Khng mt tnh tng qut, gi sa b c > 0, t y suy rac 13, Xt 2 trng hpTrnghp1.b 12, s dng bt ng thc Jensenf(b) + f(c) 2f_b + c2_=2__b+c2_2+ 1=4__3 a_2+ 4Ta cn chng minh4__3 a_2+ 4+1a2+ 1 332(2.1)17www.VNMATH.com18 CHNG2. SOLUTIONTht vy, ta =t3th 3 t 1 v ta cn chng minh4t26t + 21 +1t2+ 3 32Hay16t26t + 21 +1t2+ 3 + 8_(t2+ 3)(t26t + 21)(t2+ 3)(t26t + 21)94S dng bt ng thc AMGM, ta c_t2+ 3 t2+ 74,_t26t + 21 t26t + 378Nh vy, ta ch cn chng minh16t26t + 21 +1t2+ 3 +(t2+ 7)(t26t + 37)4(t2+ 3)(t26t + 21) 94Hay(t 1)2(t 2)2 0Bt ng thc ny hin nhin ng.Trnghp2.b 12, ta cf(a) + f(b) f_12_+ f_a + b 12_S dng bt ng thc Jensen,f_12_+ f(c) 2f_c +122_= 2f__3 _a + b 12_2__Nh vy, ta cn chng minh2f__3 _a + b 12_2__+ f_a + b 12_332Bt ng thc ny ng theo (2.1). Bt ng thc c chng minh xong. ng thc xy ra khiv ch khix = y = z =13.Nhnxt. Btngthctrnvnngvimix, y, z Rthamnxy + yz + zx = 1.2Cho cc s dnga, b, c thaabc = 1, chng minh rngab + cb + c + 1 +bc + ac + a + 1 +ca + ba + b + 1 2Li gii. S dng bt ng thc Holder, ta c_

cycab + cb + c + 1_2_

cyca(b + c + 1)2b + c_ (a + b + c)3www.VNMATH.com19Do , ta cn chng minh(a + b + c)3 2

cyca(b + c + 1)2b + chay

cyca3+ 3

cycab+ 3

cycba + 6 4

cycab + 4

cyca + 2

cycab + cS dng bt ng thc AMGM, ta li c

cycab

cycab,

cycba

cycab, 2

cycab + c 12

cycab+ 12

cycbaDo ,V T V P

cyca3+ 52

cycab+ 52

cycba 4

cycab 4

cyca + 6

cyca3+

cycab 4

cyca + 6 =

cyc_a34a + 1a + 2_Xt hm sf(x) = x34x +1x + 2 + 2 ln x vix > 0, ta cf

(x) = (x 1)_3x + 3 +1x2 1x_Nux 1 th1x2 1x, nux 1 th 1 1x, do f

(x) = 0 x = 1T y, ta d dng kim tra cf(x) f(1) = 0 x > 0Hayx34x + 1x + 2 2 ln x x > 0Vy

cyc_a34a + 1a + 2_ 2

cycln a = 0Bt ng thc c chng minh xong. ng thc xy ra khi v ch khia = b = c = 1.3Vi mi s khng ma, b, c, ta c_a4a + 4b + c +_b4b + 4c + a +_c4c + 4a + b 1Li gii. Cch1. S dng bt ng thc Cauchy Schwarz, ta c

cyc_a4a + 4b + c 3

cyca4a + 4b + cKhng mt tnh tng qut, gi s a +b +c = 3 v b l s hng nm gia a v c, ta cn chng minh

cycaa + b + 1 1www.VNMATH.com20 CHNG2. SOLUTIONhaya2b + b2c + c2a + abc 4Vb l s hng nm giaa vc nnc(b a)(b c) 0Suy rab2c + c2a abc + bc2Do a2b + b2c + c2a + abc b(a + c)212_2b + (a + c) + (a + c)3_3= 4Bt ng thc c chng minh xong. ng thc xy ra khi v ch khia = b = c.Cch2. S dng bt ng thc Cauchy Schwarz, ta c_

cyc_a4a + 4b + c_2=_

cyc_a(4a + 4b + c)(4a + b + 4c) 4a + b + 4c_2_

cyca(4a + 4b + c)(4a + b + 4c)__

cyc(4a + b + 4c)_=9(a + b + c)(a2+ b2+ c2+ 8(ab + bc + ca))(4a + 4b + c)(4b + 4c + a)(4c + 4a + b)Ta cn chng minh9(a + b + c)(a2+ b2+ c2+ 8(ab + bc + ca)) (4a + 4b + c)(4b + 4c + a)(4c + 4a + b)Hay7

cyca3+ 3

cycab(a + b) 39abcTheo bt ng thc AMGM th

cyca3 3abc,

cycab(a + b) 6abcDo ta c pcm.4Cho cc s dnga, b, c, chng minh1a2+ bc +1b2+ ca +1c2+ ab a + b + cab + bc + ca_1a + b +1b + c +1c + a_Li gii. Ta c bt ng thc cn chng minh tng ng vi

cycab + bc + caa2+ bc

cyca + b + cb + cHay

cyca(a2b2c2+ ab + ac bc)(b + c)(a2+ bc) 0

cyca(a + 2b + c)(a b) + a(a + b + 2c)(a c)(b + c)(a2+ bc) 0www.VNMATH.com21

cyc(a b)_a(a + 2b + c)(b + c)(a2+ bc) b(2a + b + c)(a + c)(b2+ ca)_ 0

cycz(a2b2)(a b) 0Vix = (a(b + c)(b2+ c2) + 2a2(b2+ c2) + 3a2bc + a3(b + c) b2c2)(a2+ bc)y = (b(c + a)(c2+ a2) + 2b2(c2+ a2) + 2b2ca + b3(c + a) c2a2)(b2+ ca)z = (c(a + b)(a2+ b2) + 2c2(a2+ b2) + 2c2ab + c3(a + b) a2b2)(c2+ ab)Khng mt tnh tng qut, gi sa b c > 0, khi d thyx, y 0. Li cy + z b(c + a)(c2+ a2)(b2+ ca) a2b2(c2+ ab) a3b(b2+ ca) a2b2(c2+ ab) = a2bc(a2bc) 0Ch rnga b c > 0 nn (c2a2)(c a) (a2b2)(a b). T y, ta c pcm. ng thcxy ra khi v ch khia = b = c hoca = t > 0, b = c 0 v cc hon v.Cch2. Ta c2

cyc1(b + c)2

cyca + b + cab + bc + ca1b + c=

cyc1b + c_2b + c a + b + cab + bc + ca_=

cycb(a b) + c(a c)(b + c)2(ab + bc + ca)=

cyca bab + bc + ca_b(b + c)2 a(c + a)2_=1ab + bc + ca

cyc(ab c2)(a b)2(a + c)2(b + c)2Ch rng2

cyc1(a + c)2

cyc1a2+ bc=

cyc_1(a + c)2+1(b + c)2 1c2+ ab_=

cycab(a b)2+ (c2ab)2(a + c)2(b + c)2(c2+ ab)Do bt ng thc tng ng0

cyc(a b)2(a + c)2(b + c)2_abc2+ ab ab c2ab + bc + ca_+

cyc(c2ab)2(a + c)2(b + c)2(c2+ ab)=

cycc(c3+ a2b + b2a)(a b)2(ab + bc + ca)(a + c)2(b + c)2(c2+ ab) +

cyc(c2ab)2(a + c)2(b + c)2(c2+ ab)Bt ng thc ny hin nhin ng. Vy ta c pcm.Nhnxt.Tbtngthcny,tac1a2+ bc+1b2+ ca+1c2+ ab 32(a + b + c)2(ab + bc + ca)25Chng minh rng vi mi s dnga, b, c ta lun ca32a2ab + 2b2+b32b2bc + 2c2+c32c2ca + 2a2 a + b + c3www.VNMATH.com22 CHNG2. SOLUTIONLi gii. Ta c bt ng thc cn chng minh tng ng vi

cycSc(a b)2 0trong Sa =2c b2b2bc + 2c2, Sb =2a c2c2ca + 2a2, Sc =2b a2a2ab + 2b2Xt 2 trng hpTrnghp1.a b c > 0, khi , d thySb 0. Ta s chng minhSb + 2Sc 0 (1)a2Sb + 2b2Sa 0 (2)Tht vy, ta c(1) 6a2b + 4ab23abc + 8bc24ac22b2c 0 (ng doa b c)(2) f(a) =a2(2a c)2c2ca + 2a2+2b2(2c b)2b2bc + 2c2 0Li cf

(a) =a(4a34a2c + 13ac24c3)(2c2ca + 2a2)2 0Do ,f(a) l hm ng bin. Suy ra,f(a) f(b) =3b2c2b2bc + 2c2 0Cc bt ng thc (1) v (2) c chng minh.T cc bt ng thc ny v vi ch rng a b c > 0 nn (ac)2 max_a2b2(b c)2, (a b)2_,ta c2

cycSc(a b)2=_Sb(c a)2+ 2Sa(b c)2_+_Sb(c a)2+ 2Sc(a b)2_(b c)2b2(a2Sb + 2b2Sa) + (a b)2(Sb + 2Sc) 0Trnghp2.c b a > 0, d thySc, Sa 0. Nu 2a c th bt ng thc cn chng minhhin nhin ng. Xt trng hp ngc lic 2a, tc lSb 0. Xt 2 trng hp nhTrnghp2.1. 2b c + a, ta cSb(c a)2+ Sc(a b)2 0 (3)m(b) =(a b)2(2b a)2a2ab + 2b2+ (c a)2(2a c)2c2ca + 2a2 0m

(b) =(b a)(4b3+ 9a2b 7a3)(2a2ab + 2b2)2 0Do ,m(b) l hm ng bin. Suy ra,m(b) m_a + c2_=a(a c)2(16a22ac + c2)2(4a2+ ac + c2)(2a2ac + 2c2) 0www.VNMATH.com23Vy (3) ng. Do ,

cycSc(a b)2 Sb(c a)2+ Sc(a b)2 0Trnghp2.2.c + a 2b.Trnghp2.2.1. 2b a 4a, ta s chng minhSc + 3Sb 0 (4)Sa + 32Sb 0 (5)Tht vy(4) g(c) =2b a2a2ab + 2b2+3(2a c)2c2ca + 2a2 0Ta cg

(c) =6c(c 4a)(2c2ca + 2a2)2 0Do ,g(c) l hm ng bin. Suy ra,g(c) g(2b a) =4b34ab2a2b + 13a3(2a2ab + 2b2)(5a210ab + 8b2) 0(5) h(a) =4c 2b2b2bc + 2c2+6a 3c2c2ca + 2a2 0h

(a) =3(3c2+ 4ca 4a2)(2c2ca + 2a2)2 0Do ,h(a) l hm ng bin. Suy ra,++, Nuc 2b thh(a) h(0) =(c 2b)(2c + 3b)2c(2c2bc + 2b2) 0++, Nu 2b c thh(a) h(2b c) =(2b c)(4b2+ 13bc 2c2)(2b2bc + 2c2)(8b210bc + 5c2) 0Tm li, ta lun ch(a) 0.T (4) v (5) vi ch rng (c a)2 3(b a)2+32(b c)2, ta c

cycSc(a b)2 (Sc + 3Sb)(a b)2+_Sa + 32Sb_(b c)2 0Trnghp2.2.2. 2b a 4a a 25b, ta cSa + Sb + Sc 0 (6)SaSb + SbSc + ScSa 0 (7)(6) hin nhin ng v theo (5), ta cSa + Sb + Sc = Sa +_32Sb + Sc_ 12Sb 0www.VNMATH.com24 CHNG2. SOLUTIONBy gi ta s chng minh (7), ta c(7) k(c) = 4(ab3+ bc3+ ca3) + 7abc(a + b + c) 2(a3b + b3c + c3a)6(a2b2+ b2c2+ c2a2) 0k

(c) = 12bc2+ 4a3+ 14abc + 7ab(a + b) 2b36ac212c(a2+ b2)k

(c) = 24bc 12ac + 14ab 12a212b2 24b212ab + 14ab 12a212b2= 12b2+ 2ab 12a2 0Do ,k

(c) l hm ng bin. Suy ra,k

(c) k

(b) = 4a35a2b + 15ab22b3 0 (doa 25b)Suy ra,k(c) l hm ng bin. Do ,k(c) k(b) = b(2a35a2b + 16ab24b3) 0 (doa 25b)T y, ta c pcm. ng thc xy ra khi v ch khia = b = c.6Cho cc s khng ma, b, c thaa + b + c = 1. Chng minh bt ng thc_a + (b c)24+_b + (c a)24+_c + (a b)243 +_1 32_(|a b| +|b c| +|c a|)Li gii. Khng mt tnh tng qut, gi sa b c 0. ta + b = 2t, a b = 2m, k =14th do githit, ta ct m t c 0. Khi , bt ng thc cn chng minh tr thnhf(m) =_t + m + k(m + c t)2+_t m + k(m + t c)2+_c + 4km2_2 3_(t + mc) 3Ta cf

(m) =4k(2t c) 14 (t + m + k(m + c t)2)3/2+4k(2t c) 14 (t m + k(m + t c)2)3/2+4kc(c + 4km2)3/2= c2_a +14(b c)2_3/2 c2_b +14(c a)2_3/2+c_c +14(a b)2_3/2Ch rnga b c 0 nna + 14(b c)2c 14(a b)2=34(a c)(b + 1) 0b + 14(c a)2c 14(a b)2=34(b c)(a + 1) 0Suy ra_a + 14(b c)2_3/2_c + 14(a b)2_3/2 0_b + 14(c a)2_3/2_c + 14(a b)2_3/2 0www.VNMATH.com25Do ,f

(m) c2_c +14(a b)2_3/2 c2_c +14(a b)2_3/2+c_c +14(a b)2_3/2= 0Suy ra,f(m) l hm li. Do ,f(m) max {f(0), f(t c)}Nh vy, ta ch cn chng minhmax {f(0), f(t c)} 3iu ny c ngha l ta ch cn chng minh bt ng thc cho trong trng hp 3 sa, b, c c2 s bng nhau, khng mt tnh tng qut, gi sb = c. Ta cn chng minha + 2_b + (a b)243 +_2 3_|a b|Hay_3a 12_2+ 2(1 a) 3 +_1 32_|3a 1| att =3a th ta ct 3, ta cn chng minh_3t414t2+ 27 6 2t +_23 2_|t21|Xt 2 trng hpTrnghp1.t 1, ta c bt ng thc tng ng_3t414t2+ 27 6 2t +_23 2_(t21)Hay2(t 1)_t 3___63 9_t2+_3 +3_t + 18 113_ 0Bt ng thc ny hin nhin ng do 3 t 1.Trnghp2.t 1, bt ng thc tr thnh_3t414t2+ 27 6 2t _23 2_(t21)Hay2(t 1)__63 9_t3+_23 3_t2+_23 9_t + 63 3_ 0Bt ng thc ny cng ng do 1 t 0. Bi ton c gii quyt hon ton. ng thc xy rakhi v ch khia = b = c =13hoca = 1, b = c = 0 v cc hon v.Nhnxt.Tac1ktqu"yu"hnnhngkh"p"l_a +(b c)24+_b +(c a)24+_c +(a b)24 2vimia, b, c 0, a + b + c = 1.7Cho cc s dnga, b, c thaa + b + c = 3, chng minh bt ng thca3/2b + b3/2c + c3/2a 3www.VNMATH.com26 CHNG2. SOLUTIONLi gii. S dng bt ng thc Cauchy Schwarz, ta ca3/2b + b3/2c + c3/2a _(ab + bc + ca)(a2b + b2c + c2a)Nh vy, ta ch cn chng minh(ab + bc + ca)(a2b + b2c + c2a) 9Hay(ab + bc + ca)(a + b + c)(a2b + b2c + c2a) 27Hay(ab + bc + ca)_

cyca3b +

cyca2b2+ 3abc_ 27Ch rng12

cyc(a2c22ab + bc + ca)2 0 nn

cyca3b 13_

cyca2_2Ta cn chng minh(ab + bc + ca)___

cyca2_2+ 3

cyca2b2+ 9abc__ 81tx =ab + bc + cath theobtngthcAMGMv Schur, tacx 3, 3abc 4x 9, btng thc tr thnhx((9 2x)2+ 3x29abc) 81Nh vy, ta ch cn chng minhx((9 2x)2+ 3x23(4x 9)) 81Hay(x 3)(7x227x + 27) 0Btngthcnyhinnhinngdo3 x 0.Btngthccchngminhxong.ngthc xy ra khi v ch khia = b = c = 1.8Chng minh rng vi mi s thca, b, c, ta cab4a2+ b2+ 4c2+bc4b2+ c2+ 4a2+ca4c2+ a2+ 4b2 13Li gii. D thy trong 3 sa, b, c lun tn ti t nht 2 s cng du, gi sbc 0, nuab 0, ac 0th

cycab4a2+ b2+ 4c2 bc4b2+ c2+ 4a2 14 0Do P(a, b, c) P_a,_b2+ c22,_b2+ c22_Nh vy, ta ch cn chng minhP(a, t, t) 0via t 0, a2+ 2t2= 3.Hay(a2+ 2t2)3+ 12t2(a2+ 2t2)(2a2+ t2) 27a2t4+ 4t(2a + t)(a2+ 2t2)2Hay(a t)2(a2(a 3t)2+ 4a2t2+ 16t4) 0Bt ng thc ny hin nhin ng. Vy ta c pcm. ng thc xy ra khi v ch khia = b = c.www.VNMATH.com28 CHNG2. SOLUTION9Cho cc s khng ma, b, c thaa + b + c = 3, chng minha2+ b2(a + 1)(b + 1) +b2+ c2(b + 1)(c + 1) +c2+ a2(c + 1)(a + 1) 32Li gii. Bnh phng 2 v ri nhn 2 v vi (a + 1)(b + 1)(c + 1), ta c bt ng thc tng ng

cyc(a2+ b2)(1 + c) + 2

cyc_(a2+ c2)(b2+ c2)(a + 1)(b + 1) 92(a + 1)(b + 1)(c + 1)Theo bt ng thc Cauchy Schwarz th

cyc_(a2+ c2)(b2+ c2)(a + 1)(b + 1)

cyc(c2+ ab)_1 +ab_Ta cn chng minh

cyc(a2+ b2)(1 + c) + 2

cyc(c2+ ab)_1 +ab_92(a + 1)(b + 1)(c + 1)Hay8

cyca2+

cycab + 4

cycab(c2+ ab) 36 + 15abcS dng bt ng thc AMGM v Schur, ta c4

cycab(c2+ ab) 15abc 9abc 12

cycab 27Nh vy, ta ch cn chng minh8

cyca2+

cycab + 12

cycab 27 36Hayab + bc + ca 3Bt ng thc ny ng theo bt ng thc AMGM. Vy ta c pcm. ng thc xy ra khi vch khia = b = c = 1.10Vi mia b c 0, tP=ab + c +bc + a +ca + bQ =2(b + c) a4a + b + c+ 2(c + a) b4b + c + a+ 2(a + b) c4c + a + bChng minh rng1. Nua + c 2b thP Q.2. Nua + c 2b thP Q.Li gii. Khng mt tnh tng qut, gi sa + b + c = 1.(1) Bt ng thc cn chng minh tng ng vi

cyc3a 1(3a + 1)(1 a) 0www.VNMATH.com29Hay

cycz(a b)2 0vix = (1 9a2)(1 a), y = (1 9b2)(1 b), z = (1 9c2)(1 c).Ch rnga b c, a + c 2b nnb 13, y, z 0, a c 2(b c) 0, a b b c 0Do , ta ch cn chng minhx + 4y + z 0HayF(a, b, c) = 9(a3+ c3) 9(a2+ c2) + 36b336b23b + 5 0Ta cF(a, b, c) =(1 3b)(11 + 30b 45b2+ 9(a c)2)4 0(2) Bng bin i tng t, ta c bt ng thc tng ng

cycSc(a b)2 0viSa = (9a21)(1 a), Sb = (9b21)(1 b), Sc = (9c21)(1 c).Doa b c, 2b a + c nn12 b 13, Sa, Sb 0, a c 2(a b) 0, b c a b 0Nh vy, ta ch cn chng minhSa + 4Sb + Sc 0HayG(a, b, c) = 9(a3+ c3) + 9(a2+ c2) 36b3+ 36b2+ 3b 5 0Ta cG(a, b, c) =(3b 1)(11 + 30b 45b2+ 9(a c)2)4 0Bi ton c gii quyt hon ton.ng thc c 2 bt ng thc xy ra khi v ch khi 2b = a + c.11Cho cc s khng ma, b, c thaa + b + c = 1, tx = a2+ b2+ c2, chng minh bt ng thc_1 + 2a2x +_1 + 2b2x +_1 + 2c2x 11 9xLi gii. Bnh phng 2 v ri thu gn, ta c th vit li bt ng thc nh sau

cyc_(1 + a2b2c2)(1 + b2c2a2) 8

cycabS dng bt ng thc GM-HM, ta c_(1 + a2b2c2)(1 + b2c2a2) (1 + a2b2c2)(1 + b2c2a2)1 c2www.VNMATH.com30 CHNG2. SOLUTIONTa cn chng minh

cyc(1 + a2b2c2)(1 + b2c2a2)1 c2 8

cycabHay2

cycc(a b)21 + c 0Bt ng thc ny hin nhin ng, vy ta c pcm. ng thc xy ra khi v ch khi a = b = c =13hoca = 1, b = c = 0 v cc hon v.12Chng minh rng vi mia, b, c > 0, ta c1a(a + b) +1b(b + c) +1c(c + a) 32(abc)2/3Li gii. Khng mt tnh tng qut, ta c th gi sabc = 1. Khi , tn ti cc s dngx, y, zsaochoa =xy, b =zx, c =yz, bt ng thc tr thnh

cycy2x2+ yz 32S dng bt ng thc Cauchy Schwarz, ta c

cycy2x2+ yz (x2+ y2+ z2)2x2y2+ y2z2+ z2x2+ x3y + y3z + z3xMt khc, ta li c(x2+ y2+ z2)23(x3y + y3z + z3x) =12

cyc(x2z22xy + yz + zx)2 0(x2+ y2+ z2)23(x2y2+ y2z2+ z2x2) =12

cyc(x2y2)2 0Nn t y, ta d dng suy ra pcm. ng thc xy ra khi v ch khia = b = c.13Chng minh rng nua, b, c > 0 th1aa + b+1bb + c+1cc + a 32abcLi gii. Tng t bi trn, ta cng a bi ton v chng minh rng vi mix, y, z> 0 th

cycyy_x(x2+ yz) 32S dng bt ng thc Cauchy Schwarz, ta cV T (x + y + z)2_xy(x2+ yz) +_yz(y2+ zx) +_zx(z2+ xy)(x + y + z)2_(xy + yz + zx)(x2+ y2+ z2+ xy + yz + zx)www.VNMATH.com31Mt khc, theo bt ng thc AMGM th8(xy + yz + zx)(x2+ y2+ z2+ xy + yz + zx) (x2+ y2+ z2+ 3(xy + yz + zx))2169 (x + y + z)2T y, ta c pcm. ng thc xy ra khi v ch khia = b = c.14Cho cc s dngx, y, zthax2+ y2+ z2 3, chng minh rngx5x2x5+ y2+ z2+y5y2y5+ z2+ x2+z5z2z5+ x2+ y2 0Li gii. Bt ng thc cho c vit li nh sau

cyc1x5+ y2+ z2 3x2+ y2+ z2T y, ta suy ra c ch cn xt trng hpx2+y2+z2= 3 l , khi , bt ng thc tngng

cyc1x5x2+ 3 1S dng bt ng thc AMGM, ta cx5=x6x2x6x2+ 1ta = x2, b = y2, c = z2th ta ca + b + c = 3 v ta cn chng minh

cyc12a3a+1 a + 3 1Hay

cyca + 12a3a2+ 2a + 3 1Hay

cyc(a 1)2(2a2+ 3a + 3)2a3a2+ 2a + 3 0Khng mt tnh tng qut, gi sa b c, suy raa 1 c. Xt 2 trng hpTrnghp1.b + c 1, suy raa 2, khi , ta c2a2+ 3a + 3 > 0, 2b2+ 3b + 3 > 0, 2c2+ 3c + 3 > 0Nn kt qu bi ton l hin nhin.Trnghp2.b + c 1, suy raa 2, ta c(2a3a2+ 2a + 3) 5(a + 1) = 2a3a23a 2 = a3_2 1a 3a2 2a3_ a3_2 12 322 223_=12a3> 0www.VNMATH.com32 CHNG2. SOLUTIONSuy raa+12a3a2+2a+3 15. Nh vy, ta ch cn chng minhb + 12b3b2+ 2b + 3 +c + 12c3c2+ 2c + 3 45iu ny lun ng v vi mi 1 x 0, ta cx + 12x3x2+ 2x + 3 25Tht vy, bt ng thc tng ng4x3 (x + 1)(2x 1)Nux 12th ta c ngay pcm, nux 12th4x3(x + 1)(2x 1) 4x32(2x 1) = 2(2x32x + 1) 2(x22x + 1) = 2(x 1)2 0Bt ng thc c chng minh xong. ng thc xy ra khi v ch khix = y = z = 1.15Chon 3va1, a2, . . . , anlccskhngmthaa21 + a22 + + a2n= 1,chngminhbtngthc13(a1 + a2 + + an) a1a2 + a2a3 + + ana1Li gii. tfn(a1, a2, . . . , an) =13(a1 +a2 + +an) (a1a2 +a2a3 + +ana1). Khng mt tnhtng qut, gi sa1 = max{a1, a2, . . . , an}. Nuan 13thfn(a1, a2, . . . , an) fn1_a1, a2, . . . , an2,_a2n1 + a2n_=13_an1 + an_a2n1 + a2n_+ (an2 + a1)_a2n1 + a2nan1(an2 + an) ana1_13 an__an1 + an_a2n1 + a2n_ 0Suy rafn(a1, a2, . . . , an) fn1_a1, a2, . . . , an2,_a2n1 + a2n_Nuan 13th ta ca1 13, suy raan1 13, do fn(a1, a2, . . . , an) fn1_a1, a2, . . . , an3,_a2n2 + a2n1, an_=13_an2 + an1_a2n2 + a2n1_+ an3__a2n2 + a2n1an2_+ an__a2n2 + a2n1an1_an2an1 an2_13 an1_ 0Suy rafn(a1, a2, . . . , an) fn1_a1, a2, . . . , an3,_a2n2 + a2n1, an_T y, ta suy ra c ta ch cn chng minh bt ng thc trong trng hpn = 3 l nhngtrong trng hp ny, bt ng thc l hin nhin nn ta c pcm. ng thc xy ra khi v chkhin = 3 va1 = a2 = a3 =13.www.VNMATH.com3316Cho cc s dnga, b, c, chng minh bt ng thc_ab+bc +ca +_ab + bc + caa2+ b2+ c2 3 + 1Li gii. Trc ht, ta chng minh kt qu sau vi mia, b, c > 0(a + b + c)2_ab+bc +ca_ 9(a2+ b2+ c2)Tht vy, bt ng thc tng ng

cycSc(a b)2 0trong Sa =bc +ab+ 2ac 52, Sb =ca +bc + 2ba 52, Sc =ab+ca + 2cb 52Khng mt tnh tng qut, gi sa = max{a, b, c}. Nuc b th ta cab +bc +ca ac +cb +bannkhng mt tnh tng qut, ta ch cn xta b c > 0 l , khi , d thySa 0. Ta s chngminhSa + 2Sb 0, Sc + 2Sb 0, Sb + Sc 0Tht vy, ta cSa + 2Sb = 2_ac+ca_+_ab+ 4ba_+ 3bc 152 4 + 4 + 3 152> 0Sc + 2Sb = 2_bc +cb_+_ab+ 4ba_+ 3ca 152 4 + 4 152> 0Sb + Sc =_ab+ 2ba_+_b2c + 2cb_+_b2c + 2ca_5 ab+ 2_ba + 2ba+ 2 5=_a2b + 2ba_+_a2b +_ba +_ba_3 2 +332 3 > 0T y, ta c+, NuSb 0 th

cycSc(a b)2 (Sa + 2Sb)(b c)2+ (Sc + 2Sb)(a b)2 0+, NuSb 0 th

cycSc(a b)2 (Sc + Sb)(a b)2 0Bt ng thc trn c chng minh, s dng bt ng thc ny, ta suy ra c, ta ch cn chngminh3a2+ b2+ c2(a + b + c)2+_ab + bc + caa2+ b2+ c2 3 + 1tx =_ab+bc+caa2+b2+c2 1, ta cn chng minh32x2+ 1 + x 3 + 1D dng kim tra c bt ng thc ny ng vi mi 1 x 0, vy ta c pcm. ng thc xyra khi v ch khia = b = c.www.VNMATH.com34 CHNG2. SOLUTION17Chng minh rng vi mia, b, c > 0, ta ca2b2+b2c2+c2a2+ 8(ab + bc + ca)a2+ b2+ c2 11Li gii. Trc ht, ta s chng minh kt qu sau vi mix, y, z> 0 thaxyz = 1x2+ y2+ z2+ 6 32_x + y + z + 1x + 1y + 1z_Tht vy, khng mt tnh tng qut, gi sx = min{x, y, z}. tt =yz vP(x, y, z) = x2+ y2+ z2+ 6 32_x + y + z + 1x + 1y + 1z_Ta cP(x, y, z) P(x, t, t) =12_y z_2_2_y +z_23 3bc_12_y z_2(8 3 3) 0Li cP(x, t, t) = P_1t2, t, t_=(t 1)2((t22t 1)2+ t2+ 1)2t4 0Bt ng thc c chng minh. Tr li bi ton ca ta, s dng bt ng thc trn vi x =ab, y =bc, z =ca, ta suy ra c ta ch cn chng minh32

cyca2+ b2ab+ 8(ab + bc + ca)a2+ b2+ c2 17Hay

cycSc(a b)2 0trong Sa =3bc 8a2+ b2+ c2, Sb =3ca 8a2+ b2+ c2, Sc =3ab 8a2+ b2+ c2Khng mt tnh tng qut, gi sa b c, khi , d thySa Sb Sc, li cSb + Sc =3(b + c)abc16a2+ b2+ c2 6abc16a2+ 2bc 6abc162a2bc> 0T y, ta d dng suy ra pcm. ng thc xy ra khi v ch khia = b = c.Nhnxt.Chrng_ab+bc+caa2+b2+c2_2+ 1 2(ab+bc+ca)a2+b2+c2,tasuyraa2b2+b2c2+c2a2+ 4_ab + bc + caa2+ b2+ c2_2 7KtqunyctmrabibnNguynAnhCngvcalnhttp://mathnfriend.org/18Chng minh rng vi mi s dnga1, a2, . . . , an, b1, b2, . . . , bn, ta c_n

i=1a2i__n

i=1b2i__n

i=1bi(ai + bi)__n

i=1a2ibiai + bi_www.VNMATH.com35Li gii. tfn(a1, a2, . . . , an) =V T V P.Taschngminhbtngthcchobngquynp.Vin := 1 th bt ng thc l hin nhin, gi s bt ng thc ng vin := n, khi , s dnggi thit quy np, ta cfn+1(a1, a2, . . . , an+1) fn+1(a1, a2, . . . , an+1) fn(a1, a2, . . . , an)=1an+1 + bn+1n

i=1(an+1bibn+1ai)2(an+1ai + bn+1ai + an+1bi)ai + bi 0Vy bt ng thc cho cng ng khin := n + 1 nn theo nguyn l quy np, n ng vi min.19Chng minh rng vi cc s thca, b, c i mt khc nhau, ta c(a2+ b2+ c2ab bc ca)_1(a b)2+1(b c)2+1(c a)2_274Li gii. Khngmt tnhtngqut, gisa=min{a, b, c}, t b =a +x, c =a +yth tacx, y> 0, x = y (doa, b, c phn bit nhau) v bt ng thc tr thnh(x2xy + y2)_ 1x2+1y2+1(x y)2_274Li tt =xy +yx 1, d thyt > 1, bt ng thc c vit li nh sau4t3t 1 27Hay(2t 3)2(t + 3) 0Bt ng thc ny hin nhin ng, vy ta c pcm.20Cho cc s khng ma, b, c, d thaa2+ b2+ c2+ d2= 4, chng minh bt ng thc13 abc +13 bcd +13 cda +13 dab 2Li gii. D dng chng minh c vi mi833 x 0, ta c23 x 5x23x + 1214S dng bt ng thc ny vi ch l max{abc, bcd, cda, dab} 833, ta suy ra c ta ch cnchng minh5(a2b2c2+ b2c2d2+ c2d2a2+ d2a2b2) 3(abc + bcd + cda + dab) 8Cnhiucchchngminhchobtngthcny, xincgii thiuvi ccbncchchngminhsaudavokthuthmli. t t2=a2+b22, k2=c2+d22vx=ab, y=cdth tact2 x 0, k2 y 0, bt ng thc c vit li nh sauf(x) = 10x2k2+ 10y2t23x_2y + 2k23y_2x + 2t28 0www.VNMATH.com36 CHNG2. SOLUTIONTa cf

(x) = 20k2+3y(2x + 2t2)3/2 0Suy raf(x) l hm li, do f(x) max{f(t2), f(0)}Ta cf(0) =_yt2 + 1__5yt2 8_(doyt2 833 0Bt ng thc c chng minh xong. ng thc xy ra khi v ch khia = b = c.22Cho cc s khng ma, b, c, chng minh bt ng thc7_3(a2+ b2+ c2)a + b + c+a2b + b2c + c2aa3+ b3+ c3 8Li gii. S dng kt qu bi ton 16, ta cba +ac+cb 9(a2+ b2+ c2)(a + b + c)2Suy raa2b + b2c + c2a 9abc(a2+ b2+ c2)(a + b + c)2Ta cn chng minh7_3(a2+ b2+ c2)a + b + c+9abc(a2+ b2+ c2)(a + b + c)2(a3+ b3+ c3) 8Khng mt tnh tng qut, gi s a +b +c = 1, t x = ab +bc +ca th ta c13 x 0. Hn na,theo bt ng thc Schur, ta suyabc 4x19, do 9abca3+ b3+ c3=9abc3abc + 1 3x 3(4x 1)2 5xNh th, ta phi chng minh7_3(1 2x) + 3(4x 1)(1 2x)2 5x 8Ta c147(1 2x) _8 3(4x 1)(1 2x)2 5x_2=(3x 1)2(227 550x 64x2)(2 5x)2(3x 1)2_227 55013 6419_(2 5x)2=329(3x 1)29(2 5x)2 0Bt ng thc c chng minh xong. ng thc xy ra khi v ch khia = b = c.23Chng minh rng vi mi s dnga, b, c ta ca3a3+ abc + b3+b3b3+ abc + c3+c3c3+ abc + a3 1Li gii. tx =ba, y =ac, z =cbth ta cx, y, z> 0, xyz = 1 v bt ng thc tr thnh

cyc1x3+xy + 1 1www.VNMATH.com38 CHNG2. SOLUTIONHay

cyc1x3+ x2z + 1 1Hay

cycyzx2+ yz + zx 1S dng bt ng thc Cauchy Schwarz, ta c

cycyzx2+ yz + zx (yz + zx + xy)2yz(x2+ yz + zx) + zx(y2+ zx + xy) + xy(z2+ xy + yz)= 1Bt ng thc c chng minh xong. ng thc xy ra khi v ch khi a = b = c hocab 0,bc 0v cc hon v.24Cho cc s dnga, b, c, d, chng minh rngabc(d + a)(d + b)(d + c) +abd(c + a)(c + b)(c + d) +acd(b + a)(b + c)(b + d) +bcd(a + b)(a + c)(a + d) 12Li gii. tx =1a, y =1b, z =1c, t =1dth ta cx, y, z, t > 0 v bt ng thc tr thnh

cycx3(x + y)(x + z)(x + t) 12S dng bt ng thc Cauchy Schwarz, ta suy ra c ta ch cn chng minh2_

cycx2_2

cycx(x + y)(x + z)(x + t)Hay2_

cycx2_2

cycx4+

cyc(x3y + y3z + z3x) + (x + y + z + t)(xyz + yzt + zxt + txy)S dng kt qu bi ton trc, ta c 3(x3y + y3z + z3x) (x2+ y2+ z2)2, ta cn chng minh2_

cycx2_2

cycx4+ 13

cyc(x2+ y2+ z2)2+ (x + y + z + t)(xyz + yzt + zxt + txy)Hay43

cyc(x2y2+ y2z2+ z2x2) (x + y + z + t)(xyz + yzt + zxt + txy)S dng bt ng thc AMGM, ta cV T

cycxyz(x + y + z) + 13

cyc(x2y2+ y2z2+ z2x2)

cycxyz(x + y + z) + 4xyzt= (x + y + z + t)(xyz + yzt + zxt + txy) = V PBt ng thc c chng minh xong. ng thc xy ra khi v ch khia = b = c = d.www.VNMATH.com3925Chng minh rng vi mia, b, c > 0, ta cab+c+ bc+a+ ca+b 1Li gii. Nu 1trong ccsa, b, ckhng nhhn 1 thbt ngthcl hinnhin.Xt trng hpa, b, c 1, khi c 2 kh nngKh nng 1. Nu a+b+c 1, suy ra max{a+b, b+c, c+a} 1, s dng bt ng thc Bernoulli,ta c1ab+c=_1 + 1a 1_b+c 1 +_1a 1_(b + c) 1 +b + ca=a + b + caSuy raab+caa + b + cS dng tng t vib, c ri cng li, ta c pcm.Khnng2. Nua + b + c 1, li s dng bt ng thc Bernoulli, ta c1ac a + c(1 a)a,1ab a + b(1 a)aSuy raab+ca2(a + b(1 a))(a + c(1 a))S dng tng t vib, c, ta cn chng minh

cyca2(a + b(1 a))(a + c(1 a)) 1S dng bt ng thc Cauchy Schwarz, ta c

cyca2(a + b(1 a))(a + c(1 a)) (a + b + c)2

cyc(a + b(1 a))(a + c(1 a))Ta li c(a +b +c)2

cyc(a +b(1 a))(a +c(1 a)) = (ab +bc +ca)(a +b +c 1) +abc(3 a b c) 0Bt ng thc c chng minh xong.26Chon 3, n N vx1, x2, . . . , xnl cc s khng m c tng bng 1. Tm gi tr ln nht ca biuthcP(x1, x2, . . . , xn) = x31x22 + x32x23 + + x3nx21 + n2(n1)x31x32 x3nLi gii. Khng mt tnh tng qut, ta c th gi sx1 = max{x1, x2, . . . , xn}, ta s chng minhP(x1, x2, . . . , xn) P(x1, x2 + + xn, 0, . . . , 0)Tht vy, ta cP(x1, x2 + + xn, 0, . . . , 0) = x31(x2 + + xn)2 2(x31x2x3 + x31x3x4 + + x31xn1xn) + x31x22 + x31x2n (x31x2x3 + x31x3x4 + + x31xn1xn) + (x32x23 + + x3n1x2n) + x31x22 + x3nx21 x31x2x3 + x31x22 + x32x23 + + x3nx21www.VNMATH.com40 CHNG2. SOLUTIONTa cn chng minhx31x2x3 n2(n1)x31x32 x3nNun = 3, bt ng thc tr thnhx2x3 19. S dng bt ng thc AMGM, ta cx2x3 _x2 + x32_2_x1 + x2 + x33_2=19Nun > 3, bt ng thc tr thnhx22x23x34 . . . x3n 1n2(n1)S dng bt ng thc AMGM, ta cx22x23x34 x3n (x2x3 xn)2_x2 + x3 + + xnn 1_2(n1)_x1 + x2 + + xnn_2(n1)=1n2(n1)Bt ng thc c chng minh xong. Li cx31(x2 + + xn)2= 108_x13_3_x2 + . . . + xn2_2 108_x1 + x2 + + xn5_5=1083125Suyra P(x1, x2, . . . , xn)1083125. Mt khc, cho x1=35, x2=25, x3= =xn=0, tacP(x1, x2, . . . , xn) =1083125. Vymax P(x1, x2, . . . , xn) =1083125.27Cho cc s thca1, a2, . . . , anthaa1a2 an = 1, tm cc hng s tt nhtm, Msao cho_a21 + n21 +_a22 + n21 + +_a2n + n21 m(a1 + a2 + + an) + MLi gii. Choa1 =a2 = =an = 1, ta suy ra cmn + M n2. Li choa1 =a2 = =an1 =1x> 0, an = xn1, ta c_x2n2+ n21 + (n 1)_ 1x2+ n21 m_xn1+n 1x_+ M_1 +n21x2n2+ (n 1)_1x2n+n21x2n2 m_1 +n 1xn_+Mxn1Chox , ta suy ra cm 1, do mn

i=1ai + M mn

i=1ai + n2mn = m_n

i=1ain_+ n2n

i=1ai + n(n 1)T y, ta s chng minhm = 1, M= n(n 1) l cc hng s cn tm, tc ln

i=1_a2i+ n21 n

i=1ai + n(n 1)Ta s chng minh vi mix > 0 th_x2+ n21 x +n(n 1)x + n 1www.VNMATH.com41n + 1x +x2+ n21 nx + n 1S dng bt ng thc Cauchy Schwarz, ta cx +_x2+ n21 x +x + n21n=(n + 1)(x + n 1)nS dng bt ng thc ny ln lt choa1, a2, . . . , anri cng li, ta cn chng minhn

i=11ai + n 1 1tai = xni> 0, ta c theo bt ng thc AMGMn 1ai + n 1= 1aiai + n 1= 1xn1ixn1i+ (n 1)x1 xi1xi+1 xn 1xn1ixn11+ xn12+ + xn1nChoi = 1, 2, . . . , n, ri cng li ta c pcm. Vym = 1, M= n(n 1) l cc hng s tt nht cabt ng thc cho.28Chng minh rng vi mi s dnga, b, c, d, ta ca3a2+ 2b2+ c2+b3b2+ 2c2+ d2+c3c2+ 2d2+ a2+d3d2+ 2a2+ b2 16_1a + 1b + 1c + 1d_Li gii. S dng bt ng thc AMGM v bt ng thc Cauchy Schwarz, ta c18a3a2+ 2b2+ c2=18a2(a2+ b2) + a2+ c2 92b + c 2b + 1cTng t, ta c18b3b2+ 2c2+ d2 2c + 1d,18c3c2+ 2d2+ a2 2d + 1a,18d3d2+ 2a2+ b2 2a + 1bCng tng ng cc bt ng thc trn v theo v, ta c pcm. ng thc xy ra khi v ch khia = b = c = d.29Cho cc s dngx, y, z, chng minh bt ng thcx(y + z)x2+ yz+y(z + x)y2+ zx+z(x + y)z2+ xyx + y + z3xyzx2+ yzx(y + z) +y2+ zxy(z + x) +z2+ xyz(x + y)Li gii. 1. Trc ht, ta s chng minh

cycx(y + z)x2+ yzx + y + z3xyztx = a3, b = y3, z = c3(a, b, c > 0), ta c bt ng thc tng ng

cyca3(b3+ c3)a6+ b3c3a3+ b3+ c3abcS dng bt ng thcx3+ y3 xy(x + y) x, y> 0, ta c

cyca3(b3+ c3)a6+ b3c31abc

cyca2(b3+ c3)a2+ bcwww.VNMATH.com42 CHNG2. SOLUTIONTa cn chng minh

cyca2(b3+ c3)a2+ bc

cyca3Hay

cyca2(a3+ abc b3c3)a2+ bc 0

cyca3(a b)(a c)a2+ bc+

cycab(a b)2(a + b)(ac + bc ab)(a2+ bc)(b2+ ca) 0Khng mt tnh tng qut, gi sc = min{a, b, c}, ta cV T a3(a b)(a c)a2+ bc+b3(b a)(b c)b2+ caa2b2(a b)2(a + b)(a2+ bc)(b2+ ca)=c(a b)2(a + b)(a3+ b3a2c b2c)(a2+ bc)(b2+ ca) 02. Ta cn phi chng minh

cycx2+ yzx(y + z) x + y + z3xyzNuxy+yz+zxx+y+z3xyz, s dng bt ng thc Cauchy Schwarz v bt ng thc AMGM,ta cV T=

cycx2x(y + z) +

cycyzx(y + z) (x + y + z)22(xy + yz + zx) + (xy + yz + zx)22xyz(x + y + z)(x + y + z)(xy + yz + zx)xyzx + y + z3xyzNu3xyz xy+yz+zxx+y+z, s dng bt ng thc ny, ta cn chng minh(xy + yz + zx)

cycx2+ yzx(y + z) (x + y + z)2Hay

cycy2z2+ x2yzx(y + z) xy + yz + zxHay

cyc(yz + xy)(yz + zx)xy + zx (xy + yz) + (yz + zx) + (zx + xy)Bt ng thc ny ng theo bt ng thc AMGM, vy ta c pcm.30Vi mi s dnga, b, c thaa + b + c = 3, ta cab2+ c +bc2+ a +ca2+ b 32www.VNMATH.com43Li gii. Xt 2 trng hpTrnghp1.a b c, s dng bt ng thc Cauchy Schwarz, ta c

cycab2+ c (a + b + c)2

cyc ab2+

cyc ab=9

cyc ab2+

cyc abTa cn chng minh

cycab2+

cycab 6Hay2

cyca3+ 3

cyca2b + 3abc 6

cycab2Bt ng thc ny ng do

cyca3

cyca2b

cycab2

cyca3+ 3abc

cyca2b +

cycab2 2

cycab2Trnghp2.c b a, bt ng thc c vit li nh sau2

cyca4+ 2

cyca2b3+ 2

cyca2b2+ 3abc 3a2b2c2+ 3

cycab3+ 3

cyca3b2S dng kt qu bi ton trc v bt ng thc AMGM, ta c

cyca4+ 2

cyca2b2 3

cycab3, 1 abcTa cn phi chng minh

cyca4+ 2

cyca2b3 3

cyca3b2Hay

cyca5+

cycab(a3+ b3) + 6

cyca2b3 9

cyca3b2Bt ng thc ny ng do

cyca5

cyca2b3

cyca3b2

cycab(a3+ b3)

cyca2b2(a + b) =

cyca3b2+

cyca2b3 2

cyca3b2Bt ng thc c chng minh xong. ng thc xy ra khi v ch khia = b = c = 1.31Vi mi s khng ma, b, c thaa + b + c = 3, ta ca_b3+ 1 + b_c3+ 1 + c_a3+ 1 5Li gii. S dng bt ng thc AMGM, ta c

cyca_b3+ 1 =

cyca_(b + 1)(b2b + 1) 12

cyca(b2+ 2) =12

cycab2+ 3www.VNMATH.com44 CHNG2. SOLUTIONTa cn chng minh

cycab2 4Khn mt tnh tng qut, gi sc b a 0, ta ca(b a)(b c) 0Suy raab2+ a2c a2b + abc a2b + 2abcDo

cycab2 bc2+ a2b + 2abc = b(a + c)212_2b + (a + c) + (a + c)3_3= 4Bt ng thc c chng minh xong. ng thc xy ra khi v ch khi (a, b, c) = (0, 1, 2) v cchon v tng ng.32Tm hng sktt nht sao cho bt ng thc sau ng vi mia, b, c > 0(a + b + c)_1a + 1b + 1c_ 9 +k max{(a b)2, (b c)2, (c a)2}(a + b + c)2Li gii. Khng mt tnh tng qut, gi sa b c, ta cn tmk sao cho(a + b + c)_1a + 1b + 1c_ 9 +k(a c)2(a + b + c)2Chob =a+c2, bt ng thc tr thnh3(a c)22ac4k(a c)29(a + c)2Hayk 27(a + c)28acChoa = c, ta suy ra ck 272 , ta s chng minh y l gi tr cn tm, tc l(a + b + c)_1a + 1b + 1c_ 9 +27(a c)22(a + b + c)2ta = b + x, c = b y th ta cx 0, b y 0, bt ng thc tng ng vi9(xy)2b3+3(y x)(x2+16xy +y2)b2+(4x4+11x3y +78x2y2+11xy3+4y4)b +2xy(y x)3 0Nuy x th ta c ngay pcm, xtx y, khi , ta c2x4b + 2xy(y x)3 2x4y + 2xy(y x)3 2x4y 2x4y = 0Ta cn chng minhf(b) = 9(x y)2b2+ 3(y x)(x2+ 16xy + y2)b + 2x4+ 11x3y + 78x2y2+ 11xy3+ 4y4 0Nhng bt ng thc ny ng vf= 9(7x4+ 12x3y + 54x2y2+ 12xy3+ 15y4)(x y)2 0Vy ta c pcm, t ta i n kt lunkmax =272.www.VNMATH.com4533Cho cc s dngx, y, zc tch bng 1, chng minh rng vi mik 0, ta c3_xy + k +3_yz + k +3_zx + k 33k + 1Li gii. Dox, y, z> 0, xyz= 1 nn tn ti a, b, c> 0 sao chox =a4b4, y =c4a4, z=b4c4, bt ng thctr thnh

cyca8/3b4/33c4+ ka433k + 1S dng bt ng thc Holder, ta c_

cyca8/3b4/33c4+ ka4_3_

cyc(c4+ ka4)__

cyca2b_4Ta cn chng minh_

cyca2b_427k + 1

cyc(c4+ ka4)Hay

cyca2b4_27(a4+ b4+ c4)Khng mt tnh tng qut, gi sa4+ b4+ c4= 3, suy raa 43 < 2, do

cyc4(a3a2) =

cyc4(a3a2)

cyc(a41) =

cyc(a 1)2(1 + 2a a2) 0Suy raa3+ b3+ c3a2+ b2+ c2 1Do , ta ch cn chng minh

cyca2b3(a3+ b3+ c3)a2+ b2+ c2Hay

cycSa(b c)2trong Sa =a2+ b2cb, Sb =b2+ c2ac, Sc =c2+ a2baC 2 trng hp xy raTrnghp1.a b c, khi d thySa, Sc 0, ta li cSa + 2Sb =a2c+b2a 2c +b(b c)c+b2+ 2c2aa2c+b2a 2c a2b+b2a 2c a + b 2c 0Ta s chng minhSc + 2Sb 0. Tht vy, nua2 2b2, ta cSc + 2Sb =(a b)(a22b2)ab+c2b+ 2c2a+ 2(b c) 0Nu 2b2 a2va 2c, ta cSc + 2Sb =a(a b)b+c2b+ 2c2a+ 2b2a2c 2b2a2c a 2c 0www.VNMATH.com46 CHNG2. SOLUTIONNu 2b2 a2v 2c a, ta cSc + 2Sb =a(a b)b+c2b+ 2c2a+ 2b2a2c a(a b)b+a24b +a2 + 2b2a2c=(a b)(5a 4b)4ab+ 3a4+b2a+ b 2c 3a4+a2 c 0Trnghp2.c b a, khi d thySb, Sc 0, ta cSb + Sa =b(b a) + c(c a)a+a2+ b2c 0Bt ng thc c chng minh xong. ng thc xy ra khi v ch khix = y = z = 1.Nhnxt.Btngthcvi4svncnng3_xy + k+3_yz + k+3_zt + k+3_tx + k 43k + 1x, y, z, t, k> 0, xyzt = 134Cho cc s dnga, b, c, chng minh bt ng thcb2+ c2a(b + c) +c2+ a2b(c + a) +a2+ b2c(a + b) (a2+ b2+ c2)3abc(a + b + c)Li gii. Bt ng thc tng ng

cycb2+ c2a(b + c) (a2+ b2+ c2)_3abc(a + b + c)abc(a + b + c)S dng bt ng thc AMGM, ta cab + bc + ca _3abc(a + b + c), ta cn chng minhabc(a + b + c)

cycb2+ c2a(b + c) (ab + bc + ca)(a2+ b2+ c2)Hay12abc

cyc(b c)2b + c 0Bt ng thc c chng minh. ng thc xy ra khi v ch khia = b = c.35Cho cc s dnga, b, c, chng minh bt ng thc2_a2b+b2c+c2a_+ 3(a + b + c) 15(a2+ b2+ c2)a + b + cLi gii. Khngmttnhtngqut,gisa = min{a, b, c},tb =a + x, c =a + y(x, y 0),btng thc c th vit li nh sau(x2xy + y2)a3+ 3xy(2y x)a2+ (x45x3y + 6x2y2+ xy3+ y4)a + xy3(x + y) 0Ta s chng minhg(a) = (x2xy + y2)a2+ 3xy(2y x)a + x45x3y + 6x2y2+ xy3+ y4 0www.VNMATH.com47Tht vy, ta cg = (4x624x5y + 39x4y24x3y312x2y4+ 4y6) = f(x)Nu x 3y, ta c f

(x) = 12x(x2y)(2x2(x3y) +xy2+y3) 0 nn f(x) l hm ng bin, suyraf(x) f(3y) = 31y6 0. Nux 3y, ta cf(x) = (2x36x2y + xy2+ y3)2+ x3y2(3y x) + y3(x3+ 4y3y(x + y)2) y3(x3+ 4y3y(x + y)2) y3_14(x + y)3+ 3y3y(x + y)2_= y3_18(x + y)3+ 18(x + y)3+ 3y3y(x + y)2__33341_y4(x + y)2 0Nh th, ta lun cf(x) 0, suy rag(a) 0. Vy ta c pcm. ng thc xy ra khi v ch khia = b = c.36Chng minh rng vi mi s thc dngx, y, zc tch bng 1 v vi mik 0, ta c4_xy + k +4_yz + k +4_zx + k 34k + 1Li gii. Dox, y, z> 0, xyz = 1 nn tn ti cc s dnga, b, c sao chox =a5b5, y =c5a5, z =b5c5, khi bt ng thc tr thnh

cyca5/2b5/44c5+ ka534k + 1S dng bt ng thc Holder, ta c_

cyca5/2b5/44c5+ ka5_4_

cyc(c5+ ka5)__

cyca2b_5Ta cn chng minh_

cyca2b_581k + 1

cyc(c5+ ka5)Hay

cyca2b5_81(a5+ b5+ c5)S dng kt qu bi trn, ta cn chng minh15(a2+ b2+ c2)a + b + c3(a + b + c) 25_81(a5+ b5+ c5)Khng mt tnh tng qut, gi sa + b + c = 1, tab + bc + ca =1q23, r = abc (1 q 0), thth ta cr (1q)2(1+2q)27, khi , ta ca5+ b5+ c5=19(15(q2+ 2)r + 35q425q21)Do , bt ng thc tng ng5q2+ 1 5_9(15(q2+ 2)r + 35q425q21)www.VNMATH.com48 CHNG2. SOLUTIONDor (1q)2(1+2q)27nn ta ch cn chng minh5q2+ 1 5_5(q2+ 2)(1 q)2(1 + 2q) + 315q4225q29Hay5q2+ 1 5_10q5+ 300q4+ 20q3250q2+ 1Ta cV T5V P= 5q2(625q8+ 625q6+ 250q42q310q24q + 55) 0 (do 1 q 0)Bt ng thc c chng minh xong. ng thc xy ra khi v ch khix = y = z = 1.Nhnxt.Tktqubiny,tacthsuyracktqubi33.Mtcuhitnhintral:Vinhnggitrnocanthbtngthcsaungn_xy + k+n_yz + k+n_zx + k 3nk + 137Chng minh rng vi mi s khng ma, b, c v vi mik 3, ta ca(bk+ ck)a2+ bc+b(ck+ ak)b2+ ca+c(ak+ bk)c2+ ab ak1+ bk1+ ck1Li gii. Ta s ch ra rng ta ch cn xt bt ng thc trong trng hp k = 3 l , tht vy xt hmsf(k) =1ak1+ bk1+ ck1

cyca(bk+ ck)a2+ bcTa c(ak1+ bk1+ ck1)2f

(k) =

cycak1bk1(a b)(ln a ln b)_cc2+ ab +ab(a + b c)(a2+ bc)(b2+ ca)_

cyccak1bk1(a b)(ln a ln b)_1c2+ ab ab(a2+ bc)(b2+ ca)_=

cycc2ak1bk1(a b)(ln a ln b)(a3+ b3)(a2+ bc)(b2+ ca)(c2+ ab) 0Do f(k) l hm ng bin, v nh th nu bt ng thc trong trng hp k = 3 th cng ngcho mik 3. Ta cn phi chng minh vik = 3 th bt ng thc ng, hay

cyca(b3+ c3)a2+ bc

cyca2Hay

cyca(b3+ c3)a2+ bc

cyc(b2bc + c2)

cycab

cyca2

cycMa(a b)(a c) 0trong Ma =(a + b c)(a + c b)a2+ bc, Mb =(b + c a)(a + b c)b2+ ca, Mc =(c + a b)(b + c a)c2+ abwww.VNMATH.com49Khng mt tnh tng qut, gi sa b c, xt 2 trng hpTrnghp1.b + c a, khi , ta cMa, Mb, Mc 0, li caMabMb =(a + b c)(a b)(ab(a + b c) + c(a2+ b2) + c2(a + b))(a2+ bc)(b2+ ca) 0Trnghp2.a b + c, khi , ta cMa 0 Mb, Mc, vit li bt ng thc nh sau(a c)(Ma(a b) + Mc(b c)) Mb(a b)(b c) 0Ta cn chng minhMa(a b) + Mc(b c) 0Ta cMa(a b) + Mc(b c) = (a + c b)_(a b)(a + b c)a2+ bc (b c)(a b c)c2+ ab_ (a + c b)(a b c)_aa2+ bc b cc2+ ab_=c(a + c b)(a b c)(a2b2+ ac + bc)(a2+ bc)(c2+ ab) 0Bt ng thc c chng minh xong. ng thc xy ra khi v ch khia =b =c hoc (a, b, c) (1, 1, 0).38Cho cc s khng ma, b, c, chng minh bt ng thca4a3+ abc + b3+b4b3+ abc + c3+c4c3+ abc + a3 a3+ b3+ c3a2+ b2+ c2Li gii. Xt 2 trng hpTrnghp1. a3b2+ b3c2+ c3a2abc(a2+ b2+ c2), khi sdngbtngthcCauchySchwarz, ta c

cyca4a3+ abc + b3 (a3+ b3+ c3)2a5+ b5+ c5+ abc(a2+ b2+ c2) + a2b3+ b2c3+ c2a3Ta cn chng minh(a3+ b3+ c3)(a2+ b2+ c2) a5+ b5+ c5+ abc(a2+ b2+ c2) + a2b3+ b2c3+ c2a3Haya3b2+ b3c2+ c3a2 abc(a2+ b2+ c2) (ng)Trnghp2.a3b2+ b3c2+ c3a2 abc(a2+ b2+ c2), khi , ta chng minh c3(a4b3+ b4c3+ c4a3) 1abc(a3b2+ b3c2+ c3a2)2 (a3b2+ b3c2+ c3a2)(a2+ b2+ c2) abc(a2+ b2+ c2)2Do , s dng bt ng thc Cauchy Schwarz, ta c

cyca4a3+ abc + b3 (a4+ b4+ c4)2a7+ b7+ c7+ abc(a4+ b4+ c4) + a4b3+ b4c3+ c4a33(a4+ b4+ c4)23(a7+ b7+ c7) + 3abc(a4+ b4+ c4) + abc(a2+ b2+ c2)2www.VNMATH.com50 CHNG2. SOLUTIONTa cn chng minh3(a4+ b4+ c4)23(a7+ b7+ c7) + 3abc(a4+ b4+ c4) + abc(a2+ b2+ c2)2 a3+ b3+ c3a2+ b2+ c2Khng mt tnh tng qut, gi sa + b + c = 1, tab + bc + ca =1q23(1 q 0) vr =abc,th th ta c 0 r (1q)2(1+2q)27v bt ng thc tr thnhf(r) 0, vif(r) = 99r3+ 29 178q231q43r2(q2+ 5)(20q46q2+ 1)9r + (q21)2(8q6+ 21q43q2+ 1)81Theo bt ng thc Schur, ta c 27r 1 4q2, do f

(r) = 594r + 2(29 178q231q4)3 22(1 4q2) + 2(29 178q231q4)3= 2(31q4+ 46q2+ 4)3< 0Suy raf(r) l hm lm, do f(r) min_f(0)f,_(1 q)2(1 + 2q)27__Li cf(0) =(q21)2(8q6+ 21q43q2+ 1)81 0f_(1 q)2(1 + 2q)27_=2q2(q 1)2(46q6+ 54q5+ 102q4+ 13q3+ 36q212q + 4)2187 0 hon thnh chng minh ca bi ton, xin c nu mt li gii cho bt ng thc3abc(a4b3+ b4c3+ c4a3) (a3b2+ b3c2+ c3a2)2tx =1a, y =1b, z =1c, th th bt ng thc tr thnh_

cycx2y_2 3

cycx3yHay

cycSx(y z)2 0trong Sx =y2z2 yz+ 2xy 32, Sy =z2x2 zx + 2yz 32, Sz =x2y2 xy+ 2zx 32C 2 trng hp xy ra+, Nux y z, khi ta cSx 0, li cSy + Sz =z2x2+zx + 2yz+x2y2 xy 3 zx + 2yz+x2y2 xy 3yx +x2y2 xy 1 =(x + y)(x y)2xy2 0Sz + 2Sy =x2y2 xy+ 2z2x2+ 4yz 92 x2y2 xy+ 2y2x2 12 0www.VNMATH.com51Sx + 2Sy =y2z2+ 3yz+ 2xy+ 2z2x2 2zx 92 2xy+ 2y2x2 2yx 12 0+, Nuz y x, khi ta cSy =z2x2 zx + 2yz 32 z2x2 zx + 2xz 32 0Sy + Sz =z2x2+zx + 2yz+x2y2 xy 3 z2x2+ 2yz3 z2y2+ 2yz3 0Sx + Sy =z2x2+y2z2+yz+ 2xyzx 3 zx +y2z2+yz+ 2xy4=yz3+ 2(x22xy)z2+ xy2z + xy3xyz2y4+ 2(x22xy)y2+ xy3+ xy3xyz2=y(x2+ (x y)2)xz2 0Btngthccchngminhxong. ngthcxyrakhi vch khi (a, b, c) (1, 1, 1)hoc(a, b, c) (1, 0, 0).39Cho cc s dngx, y, z, t tha1x + 1 +1y + 1 +1z + 1 +1t + 1= 1Chng minh rngmin_1x + 1y + 1z, 1y + 1z+ 1t, 1z+ 1t+ 1x, 1t+ 1x + 1y_ 1 max_1x + 1y + 1z, 1y + 1z+ 1t, 1z+ 1t+ 1x, 1t+ 1x + 1y_Li gii. ta =1x+1, b =1y+1, c =1z+1, d =1t+1, th th ta c 1 a, b, c, d 0 va + b + c + d = 1.Khng mt tnh tng qut, gi sa b c d, bt ng thc tng ng vib1 b +c1 c +d1 d 1 a1 a +b1 b +c1 cHayba + c + d +ca + b + d +da + b + c 1 ab + c + d +bc + d + a +cd + a + bDoa b c d nnba + c + d bb + c + d,ca + b + d cb + c + d,da + b + c db + c + dSuy raba + c + d +ca + b + d +da + b + c 1Tng t, ta cab + c + d aa + b + c,bc + d + a ba + b + c,cd + a + b ca + b + cDo ab + c + d +bc + d + a +cd + a + b 1Bt ng thc c chng minh xong.www.VNMATH.com52 CHNG2. SOLUTION40Cho cc s khng ma, b, c, chng minh bt ng thca24a2+ ab + 4b2+b24b2+ bc + 4c2+c24c2+ ca + 4a2a + b + c3Li gii. Vi mix 0, ta c6x24x2+ x + 4 3x 1Tht vy, nux 13, bt ng thc l hin nhin. Nux 13, ta c36x44x2+ x + 4 (3x 1)2=(x 1)2(15x 4)4x2+ x + 4 0S dng bt ng thc trn, ln lt thayx biab,bc,ca, ta c6a24a2+ ab + 4b2 3a b,6b24b2+ bc + 4c2 3b c,6c24c2+ ca + 4a2 3c aCng ln lt v vi v 3 bt ng thc trn, ta c pcm. ng thc xy ra khi v ch khi a = b = c.41Cho cc s dnga, b, c, chng minh bt ng thca(b + c)a2+ bc+b(c + a)b2+ ca+c(a + b)c2+ ab12(a + b + c)_1a + 1b + 1c_+ 27Li gii. Bt ng thc tng ng4

cyca2(b + c)2(a2+ bc)2+ 8

cycab(a + c)(b + c)(a2+ bc)(b2+ ca) 27 + (a + b + c)_1a + 1b + 1c_Hay4

cyca2(b + c)2(a2+ bc)2+ 8

cycab(a + c)(b + c)(a2+ bc)(b2+ ca) 24 +

cyc(b + c)2bc

cyc(b + c)2(a2bc)2bc(a2+ bc)2+ 8

cycc(a b)2(a + b)(a2+ bc)(b2+ ca) 0Bt ng thc c chng minh xong. ng thc xy ra khi v ch khia = b = c.42Cho cc s khng ma, b, c thaa + b + c = 1, chng minh bt ng thcaa + 2b+bb + 2c+cc + 2a_32Li gii. Xt 2 trng hpTrnghp1.c b a, khi s dng bt ng thc Cauchy Schwarz, ta c_

cycaa + 2b_2 (a + b + c)

cycaa + 2b=

cycaa + 2bwww.VNMATH.com53Ta cn chng minh

cycaa + 2b 32tx =ba, y =cb, z =acth ta cx, y 1 z vxyz = 1, bt ng thc tng ng12x + 1 +12y + 1 +12z + 1 32Dox, y 1 nn12x + 1 +12y + 1=12xy + 1 + 13 2(x 1)(y 1)(4xy 1)3(2x + 1)(2y + 1)(2xy + 1) 12xy + 1 + 13=zz + 2 + 13Ta phi chng minhzz + 2 +12z + 1 76Hay2z2+ 23z + 26(z + 2)(2z + 1) 0 (ng)Trnghp2.a b c, xt 2 kh nngKhnng2.1.a 4b, khi ta s chng minhaa + 2ba + ca + c + 2b,bb + 2c+cc + 2abTht vy, ta c(a + c)2a + c + 2b a2a + 2b=c(a2+ 4ab + ac + 2bc)(a + 2b + c)(a + 2b) 0bb + 2c+cc + 2abb + 2c+cc + 8bTa phi chng minhbb + 2c+cc + 8bbHayb2b + 2c +c28b + c +2bc_(b + 2c)(8b + c) b2bb + 2c c8b + c 2b_(b + 2c)(8b + c)224b436b3c 71b2c24bc3+ 4c4(b + 2c)2(8b + c)2 0 (ng)Khnng2.2. 4b a, khi ta s chng minhaa + 2ba +c2_a + 2b +3c2,bb + 2c+cc + 2a_b +c2Tht vy, ta c_a +c2_2a + 2b +3c2a2a + 2b=c(2a(4b a) + c(a + 2b))2(a + 2b)(2a + 4b + 3c) 0bb + 2c+cc + 2abb + 2c+cc + 2bwww.VNMATH.com54 CHNG2. SOLUTIONTa phi chng minhbb + 2c+cc + 2b_b +c2Hayb2b + 2c +c22b + c +2bc_(2b + c)(b + 2c) b +c22bb + 2c + 12 2b_(2b + c)(b + 2c) +c2b + cDob c nnc2b + c 13 0, xyz = 1 th

cyc1(1 + x)3+5(1 + x)(1 + y)(1 + z) 1Thtvy, tm =1x1+x, n =1y1+y, p =1z1+zthtacm, n, p [1, 1]v(1 m)(1 n)(1 p) =(1 + m)(1 + n)(1 + p), suy ram + n + p + mnp = 0tq = mn + np + pm, r = mnp th |r| 1, ta cm2n2p2= (m + n + p)2= m2+ n2+ p2+ 2qSuy ra2q = m2(n2p21) n2p2 0Mt khc, ta li c2q = m2(n2p21) n2p2 (n2p21) n2p2= n2(p21) p21 (p21) p21 = 2Do q [1, 0], ta c bt ng thc tng ngr3+ 3r2q(1 + 3r) 0Nur 13th ta c pcm, xtr 13, khi , ta cr3+ 3r2q(1 + 3r) r3+ 3r2+ (1 + 3r) = (1 + r)3 0Bt ng thc c chng minh xong, s dng kt qu ny vi x =ab, y =bc, z =ca, ta suy ra c

cycb3(a + b)3 1 5abc(a + b)(b + c)(c + a)www.VNMATH.com56 CHNG2. SOLUTIONTr li bi ton ca ta, bt ng thc c vit li nh sau

cyc_1 a3(a + b)3_+ 3(a2+ b2+ c2)28(ab + bc + ca)2 3Hay

cycb3(a + b)3+ 3

cycab(a + b)2+ 3(a2+ b2+ c2)28(ab + bc + ca)2 3S dng kt qu trn, ta cn chng minh3

cycab(a + b)2+ 3(a2+ b2+ c2)28(ab + bc + ca)2 2 +5abc(a + b)(b + c)(c + a)Khng mt tnh tng qut, gi sa + b + c = 1, tab + bc + ca =1q23, r = abc (1 q 0), khi ta cr max_0,(1+q)2(12q)27_, bt ng thc tng ngf(r) =38_1 + 2q21 q2_2+ 108r2+ (15 + 20q2)r (1 q2)2(1 + q2)(1 q23r)2 0Ta cf

(r) =3((87 60q2)r + (1 q2)(2q4+ 4q2+ 3))(1 q23r)3 0Do f(r) l hm ng bin, suy raNu 2q 1 thf(r) f(0) =38_1+2q21q2_2 0, nu 1 2q thf(r) f_(1 + q)2(1 2q)27_=3q2(14(2 10q + 13q2) + 124q2(1 2q) + q2(1 + 52q2+ 10q3) + 18q5(1 q))8(1 q2)2(2 q)4 0Bt ng thc c chng minh xong. ng thc xy ra khi v ch khia = b = c.45Choa, b, c, d l cc s dng tha mna, b, c 1 vabcd = 1, chng minh rng1(a2a + 1)2+1(b2b + 1)2+1(c2c + 1)2+1(d2d + 1)2 4Li gii. Ta c1(x2x + 1)2+1(y2y + 1)2 1 +1(x2y2xy + 1)2vi mix, y 1. Tht vy, bt ng thc tng ng_1 1(x2x + 1)2__1 1(y2y + 1)2_1(x2x + 1)2(y2y + 1)2 1(x2y2xy + 1)2Hayxy(x 1)(y 1)(x2x + 2)(y2y + 2)(x2y2xy + 1)2 (x 1)(y 1)(x + y)[2x2y2xy(x + y) + x2+ y2x y + 2]www.VNMATH.com57xy(x2x + 2)(y2y + 2)(x2y2xy + 1)2 (x + y)[2x2y2xy(x + y) + x2+ y2x y + 2]Dox, y 1 nn(x2x + 2)(y2y + 2)(x2y2xy + 1) 4, 2xy x + yDo xy(x2x + 2)(y2y + 2)(x2y2xy + 1)2 2(x + y)(x2y2xy + 1)Ta cn chng minh2(x2y2xy + 1) 2x2y2xy(x + y) + x2+ y2x y + 2Hay(x 1)(y 1)(x + y) 0 (ng)S dng kt qu trn ln lt via, b, c, ta c1(a2a + 1)2+1(b2b + 1)2 1 +1(a2b2ab + 1)21(a2b2ab + 1)2+1(c2c + 1)2 1 +1(a2b2c2abc + 1)2= 1 +d4(d2d + 1)2Do 1(a2a + 1)2+1(b2b + 1)2+1(c2c + 1)2 2 +d4(d2d + 1)2Mt khc, ta li cd4(d2d + 1)2+1(d2d + 1)2= (d 1)4(d2d + 1)2+ 2 2Bt ng thc c chng minh xong. ng thc xy ra khi v ch khia = b = c = d = 1.46Vi mi s khng ma, b, c, chng minh rng_a2+ 4bcb2+ c2+_b2+ 4cac2+ a2+_c2+ 4aba2+ b2 2 +2Li gii. Khng mt tnh tng qut, gi sa b c 0. Xt 2 trng hpTrnghp1. 4b3 a2c, khi ta ca2+ 4bcb2+ c2a2b2=c(4b3a2c)b2(b2+ c2) 0,b2+ 4cac2+ a2b2a2=c(4a3b2c)a2(c2+ a2) 0Suy ra

cyc_a2+ 4bcb2+ c2ab+ba + 2_aba2+ b2S dng bt ng thc AMGM, ta cab+ba + 2_aba2+ b2=_1 12_a2+ b2ab+_a2+ b22ab+ 2_aba2+ b2_ 2_1 12_+ 24_2(a2+ b2)ab 2_1 12_+ 22 = 2 +2www.VNMATH.com58 CHNG2. SOLUTIONTrnghp2. 4b3 a2c, suy raa 2b, ta ca2+ 4bcb2+ c2a2+ 4b22b2=(b c)(a2(b + c) 4b2(b c))2b2(b2+ c2) 0Suy ra

cyc_a2+ 4bcb2+ c2_a2+ 4b22b2+ba + 2_aba2+ b2tx =ab 2, ta cn chng minhf(x) =_x22+ 2 + 1x + 2_xx2+ 1 2 +2Dox 2 nnx(x2+ 1) (x + 1)2= x3_1 1x 1x2 1x3_ x3_1 12 122 123_=18x3> 0, suyra _x(x2+ 1) > x + 1, do f

(x) =x2_x22+ 21x2 x21(x2+ 1)_x(x2+ 1)>12_12 +2x21x2 x21(x2+ 1)(x + 1)12 1x2 x 1x2+ 1=x244x2+ (x 2)2+ 14(x2+ 1)> 0Vyf(x) l hm ng bin trn [2, +), do f(x) f(2) =52 + 2_25> 2 +2Bt ng thc c chng minh. ng thc xy ra khi v ch khi (a, b, c) (1, 1, 0).47Cho cc s khng ma, b, c, chng minh bt ng thc(a b)(13a + 5b)a2+ b2+ (b c)(13b + 5c)b2+ c2+ (c a)(13c + 5a)c2+ a2 0Li gii. Bt ng thc c vit li nh sau

cyc13a28ab 5b2a2+ b2 0Hay4

cyc(a b)2a2+ b2+ 9

cyca2b2a2+ b2 0Ch rng_1 +a2b2a2+ b2__1 +b2c2b2+ c2__1 +c2a2c2+ a2_=_1 a2b2a2+ b2__1 b2c2b2+ c2__1 c2a2c2+ a2_Nn

cyca2b2a2+ b2= (a2b2)(b2c2)(c2a2)(a2+ b2)(b2+ c2)(c2+ a2)Ta cn chng minh4

cyc(a b)2a2+ b29(a2b2)(b2c2)(c2a2)(a2+ b2)(b2+ c2)(c2+ a2)www.VNMATH.com59S dng bt ng thc AMGM, ta c

cyc(a b)2a2+ b2 33(a b)2(b c)2(c a)2(a2+ b2)(b2+ c2)(c2+ a2)Ta phi chng minh43(a b)2(b c)2(c a)2(a2+ b2)(b2+ c2)(c2+ a2) 3(a2b2)(b2c2)(c2a2)(a2+ b2)(b2+ c2)(c2+ a2)Hay64

cyc(a2+ b2)2 27

cyc(a2b2)(a + b)2Bt ng thc ny l h qu ca bt ng thc sau vi mix y 04(x2+ y2)2 3(x2y2)(x + y)2Hayx46x3y + 8x2y2+ 6xy3+ 7y4 0+, Nux 6y th bt ng thc hin nhin ng.+, Nux 6y th ta cx46x3y + 8x2y2+ 6xy3+ 7y4= x2(x 3y)2+ xy2(6y x) + 7y4 0Bt ng thc c chng minh xong. ng thc xy ra khi v ch khia = b = c.48Chng minh rng vi mi s dnga, b, c, n, ta c_a2+ bcb + c_n+_b2+ cac + a_n+_c2+ aba + b_n an+ bn+ cnLi gii. Khng mt tnh tng qut, gi sa b c > 0. Ta c_a2+ bca(b + c)_n 1_a2+ bca(b + c)_n+_b2+ cab(c + a)_n 2_(a2+ bc)(b2+ ca)ab(a + c)(b + c)_n/2 2Do(a2+ bc)(b2+ ca) ab(a + c)(b + c) = c(a b)2(a + b) 0V do _a2+ bca(b + c)_n+_b2+ cab(c + a)_n+_c2+ abc(a + b)_n 3tx =_a2+bca(b+c)_n1, y =_b2+cab(c+a)_n1, z =_c2+abc(a+b)_n1 th ta cx 0, x + y 0, x + y + z 0Do

cyc_a2+ bcb + c_n

cycan= anx + bny + cnz= (anbn)x + (bncn)(x + y) + cn(x + y + z) 0Bt ng thc c chng minh xong. ng thc xy ra khi v ch khia = b = c hocn 0.www.VNMATH.com60 CHNG2. SOLUTION49Cho cc s khng ma, b, c thaa + b + c = 1. Ty theo gi tr can N, hy tm gi tr ln nhtv gi tr nh nht ca biu thcP(a, b, c) = a(b c)n+ b(c a)n+ c(a b)nLi gii. Trong trng hp n = 0 v n = 1 th ta c P= 1 v P= 0. Xt n 2, khi c 2 trng hpTrnghp1.n l, suy ran 3, vi gi thitb l s hng nm giaa vc, ta s chng minhP(a + c, b, 0) P(a, b, c) P(a + c, 0, b)C 2 kh nngKh nng 1.a b c 0, xt hm sg(a) = P(a +c, 0, b) P(a, b, c) = (a +c)nb (a +c)bna(b c)nb(c a)nc(a b)n, ta cg

(a) = (nb(a + c)n1bn(b c)n) + n(b(a c)n1c(a b)n1) 0Suy rag(a) l hm ng bin, do g(a) g(b) = b(b + c)((b + c)n1bn1) 0Xt tip hm s h(a) = P(a, b, c)P(a+c, b, 0) = (a+c)nb(a+c)bn+a(bc)n+b(ca)n+c(ab)n,ta ch

(a) = nb((a + c)n1(a c)n1) bn+ (b c)n+ nc(a b)n1 nb((b + c)n1(b c)n1) bn+ (b c)n= 2nn32

i=0C2i+1n1 bn2i1c2i+1n12

i=0C2i+1nbn2i1c2i+1+n12

i=1C2in bn2ic2i=n32

i=0bn2i1c2i+1(2nC2i+1n1 C2i+1n) +n12

i=1C2in bn2ic2icn 0Suy rah(a) l hm ng bin, do h(a) h(b) = b(b + c)((b + c)n1bn1) 0Khnng2.a b c, khi , ta cP(a, b, c) = a(b c)n+ b(c a)n+ c(a b)n= (c(b a)n+ b(a c)n+ a(c b)n) = P(c, b, a)Theo trn, ta cP(c + a, b, 0) P(c, b, a) P(c + a, 0, b)Do P(a + c, b, 0) = P(c + a, 0, b) P(a, b, c) P(a + c, b, 0) = P(a + c, 0, b)Vy trong mi kh nng, ta lun cP(a + c, b, 0) P(a, b, c) P(a + c, 0, b)Xt hm sf(x) =xnx(x+1)n+1vix 0, ta cf

(x) = xn+ nxn1+ nx 1(x + 1)n+2f

(x) = 0 (x) = xnnxn1nx + 1 = 0www.VNMATH.com61D thyx = 0, x = 1 khng l nghim ca(x) v nux> 0 (x = 1) l 1 nghim ca(x) th1xcng l nghim ca(x), do , ta ch cn xt nghim ca(x) trn [0, 1] l . Khi , ta c

(x) = (n(1 xn1) + n(n 1)xn2) 0Suyra(t)lhmnghchbin, li c(0)=1>0, (1)=2(1 n) 0,suy raa + b 2, ta c(4Ma + Mb + Mc) (4Mb + Mc + Ma) = 3(MaMb) = 3(a3b3) + 6(a b)_a + b + 19 1133 + 1_ 3(a3b3) + 6(a b)_2 + 19 1133 + 1_ 0www.VNMATH.com63Do ,

cyc(4Ma + Mb + Mc)(a b)(a c) =(a b)((4Ma + Mb + Mc)(a c) (4Mb + Mc + Ma)(b c))+ (4Mc + Ma + Mb)(a c)(b c) 0Bt ng thc c chng minh xong. Vykmax =5_53 7_3 + 1.51[NguynPhiHng] Cho cc s khng ma, b, c thaa2+ b2+ c2= 8, chng minh bt ng thc4(a + b + c 4) abcLi gii. tx = a +b +c, y = ab +bc +ca th ta cx22y = 8. S dng bt ng thc Schur bc 4,ta cabc (4y x2)(x2y)6x=(x216)(x2+ 8)12xTa cn chng minh(x216)(x2+ 8)12x 4(x 4)Hay(x 4)2(x2+ 8x 8)12x 0Bt ng thc ny hin nhin ng, vy ta c pcm. ng thc xy ra khi v ch khi a = b = 2, c = 0v cc hon v.52Chom, n(3n2>m2) lccsthcchotrcva, b, clccsthcthamna +b +c =m, a2+ b2+ c2= n2. Tm gi tr ln nht v gi tr nh nht ca biu thc sauP= a2b + b2c + c2aLi gii. ta=x +m3 , b=y +m3 , c=z +m3 ,ththiukinbitonchotax + y + z=0vx2+ y2+ z2=3n2m23. Biu thcPtr thnhP= x2y + y2z + z2x +m39Ta c

cyc_3x_23n2m2 183n2m2xy 1_2= 3 +183n2m2_

cycx_2+324(3n2m2)2

cycx2y26_23n2m2

cycx 54_23n2m2_3/2

cycx2y= 3 +324(3n2m2)2

cycx2y254_23n2m2_3/2

cycx2yDox + y + z = 0 nnxy + yz + zx = 12(x2+ y2+ z2) = 3n2m26, suy ra

cycx2y2=_

cycxy_22xyz

cycx =_

cycxy_2=(3n2m2)236www.VNMATH.com64 CHNG2. SOLUTIONDo 12 54_23n2m2_3/2

cycx2y 0Suy ra

cycx2y 29_3n2m22_3/2VyP 29_3n2m22_3/2+m39Mt khc, chox =2(3n2m2)3cos29, y =2(3n2m2)3cos49, z =2(3n2m2)3cos89 , ta cP=29_3n2m22_3/2+m39Vymax P=29_3n2m22_3/2+m39Hontontngt, bngcchxtbiuthc cyc_3x_23n2m2+183n2m2xy + 1_2, taddngsuy ra cmin P= 29_3n2m22_3/2m39Bi ton c gii quyt hon ton.53Tm hng sknh nht sao cho vi mia, b, c 0 tha3ka2+ (b + c)2+b3kb2+ (c + a)2+c3kc2+ (a + b)2 _3(a + b + c)k + 4Li gii. Choa = b = 1, c = 0, suy rak 5. Ta s chng minh y l gi tr cn tm, tc l

cyca35a2+ (b + c)2 _a + b + c3S dng bt ng thc Cauchy Schwarz, ta c_

cyca35a2+ (b + c)2_2_

cyca__

cyca25a2+ (b + c)2_Ta cn chng minh

cyca25a2+ (b + c)2 13Khng mt tnh tng qut, gi sa +b +c = 1 va b c 0, suy raa 13 c. Bt ng thctr thnh

cyca26a22a + 1 13www.VNMATH.com65Xt 2 trng hp+, Nuc 18, ta c9

cyc27a26a22a + 1=

cyc_12a 1 27a26a22a + 1_=

cyc(3a 1)2(8a 1)6a22a + 1 0+, Nuc 18, ta c6(V T V P) =2a 16a22a + 1 +2b 16b22b + 1 +6c26c22c + 1=a b c6a22a + 1 +b c a6b22b + 1 +6c26c22c + 1=2(a b)2(3c 2)(6a22a + 1)(6b22b + 1) + c_6c6c22c + 1 16a22a + 1 16b22b + 1_Ta phi chng minh16a22a + 1 +16b22b + 1 6c6c22c + 1Doc 18nn6c6c22c+1 1, suy ra ta ch cn chng minh16a22a + 1 +16b22b + 1 1++, Nub 13th16b22b + 1 1++, Nub 13, s dng bt ng thc Cauchy Schwarz, ta ch cn chng minh4 6(a2+ b2) 2(a + b) + 2Hay(2(a + b) + c)(a + b + c) 3(a2+ b2)Dob 13nn 3b a, do (2(a + b) + c)(a + b + c) 2(a + b)2= 3(a2+ b2) + 4ab a2b2 3(a2+ b2) + 3ab a2 3(a2+ b2)Bt ng thc c chng minh xong. Vykmin = 5.54Chng minh rng nua, b, c > 0 va + b + c = 3 th(ab + bc + ca)_ab2+ 9 +bc2+ 9 +ca2+ 9_910Li gii. Bt ng thc tng ng vi

cycab2+ 9 910(ab + bc + ca)www.VNMATH.com66 CHNG2. SOLUTIONHay910(ab + bc + ca) + 19

cycab2b2+ 9 13S dng bt ng thc Cauchy Schwarz, ta c

cycab2b2+ 9 (ab + bc + ca)2ab2+ bc2+ ca2+ 27Ta cn chng minh910(ab + bc + ca) +(ab + bc + ca)29(ab2+ bc2+ ca2+ 27) 13S dng bt ng thc AMGM,25(ab + bc + ca) +(ab + bc + ca)29(ab2+ bc2+ ca2+ 27) 5519 104(ab + bc + ca)2(ab2+ bc2+ ca2+ 27)Li s dng bt ng thc Cauchy Schwarz v bt ng thc AMGM, ta c(ab + bc + ca)4(ab2+ bc2+ ca2)2 (ab + bc + ca)4(a2b2+ b2c2+ c2a2)(a2+ b2+ c2) 27(ab + bc + ca)2(a2b2+ b2c2+ c2a2)tx = ab +bc +ca, theo bt ng thc AMGM v bt ng thc Schur,x 3, abc 4x93. Suyra(ab + bc + ca)2(a2b2+ b2c2+ c2a2) = x2(x26abc) x2(x28x + 18)= (x 3)3(x + 1) + 27 27Do (ab + bc + ca)2(ab2+ bc2+ ca2) 27Suy ra25(ab + bc + ca) +(ab + bc + ca)29(ab2+ bc2+ ca2+ 27) 16V nh th910(ab + bc + ca) +(ab + bc + ca)29(ab2+ bc2+ ca2+ 27) 13Bt ng thc c chng minh xong. ng thc xy ra khi v ch khia = b = c = 1.55Cho cc s dnga, b, c thaa + b + c = 3, chng minh bt ng thcabc2+ 3 +bca2+ 3 +cab2+ 3 32Li gii. S dng bt ng thc Cauchy Schwarz, ta cn chng minh_

cycab__

cycabc2+ 3_94Khng mt tnh tng qut, gi sa b c > 0. ta + b = 2t, a b = 2m, suy ra32> t 1, c =3 2t, xt hm sf(m) =t2m23 + c2+c(t + m)3 + (t m)2+c(t m)3 + (t + m)2www.VNMATH.com67Ta cf

(m) = 2c(t2m2)_1(3 + (t m)2)2 1(3 + (t + m)2)2_2m3 + c2+ c_13 + (t m)2 13 + (t + m)2_=c(a2b2)(3 + a2)(3 + b2) + 2abc(a2b2)(a2+ b2+ 6)(3 + a2)2(3 + b2)2a b3 + c2= (a b)_c(a + b)(3 + a2)(3 + b2) + 2abc(a + b)(a2+ b2+ 6)(3 + a2)2(3 + b2)213 + c2_Ta s chng minhf

(m) 0, hayc(a + b)(3 + a2)(3 + b2) + 2abc(a + b)(a2+ b2+ 6)(3 + a2)2(3 + b2)213 + c2Ch rng (3 +a2)(3 +b2) _3 + t2_2doa +b < 3 vab(a2+b2) 2t4. Do , ta ch cn chngminh2tc(3 + t2)2+8t3c(3 + t2)3 13 + c2Hay2t(3 2t)(3 + t2)2+ 8t3(3 2t)(3 + t2)313 + (3 2t)29(t 1)(9t531t4+ 42t322t2+ 21t 3) 0Bt ng thc ny ng do32> t 1. Do ,f(m) l hm khng tng, suy raf(m) f(0) =t23 + c2+2tc3 + t2Mt khc, d thyab + bc + ca t(t + 2c). Ta cn phi chng minht(t + 2c)_t23 + c2+2tc3 + t2_94Hayt(t + 2(3 2t))_t23 + (3 2t)2+ 2t(3 2t)3 + t2_94(t 1)2(5t424t3+ 33t29t 9) 0Bt ng thc ng do32> t 1.Vy ta c pcm. ng thc xy ra khi v ch khia = b = c = 1.56Chng minh rng vi mia, b, c dng th_b + ca+_c + ab+_a + bc16(a + b + c)33(a + b)(b + c)(c + a)Li gii. S dng bt ng thc Holder, ta c_

cyc_b + ca_2_

cyc1a2(b + c)__

cyc1a_3www.VNMATH.com68 CHNG2. SOLUTIONDo , ta cn chng minh_

cyc1a_316(a + b + c)33(a + b)(b + c)(c + a)

cyc1a2(b + c)tx =1a, y =1b, z =1c, bt ng thc tr thnh(x + y + z)316(xy + yz + zx)33(x + y)(y + z)(z + x)

cycxy + zS dng bt ng thc AMGM, ta c(x + y)(y + z)(z + x) = (x + y + z)(xy + yz + zx) xyz 89(x + y + z)(xy + yz + zx)Ta phi chng minh(x + y + z)4xy + yz + zx 6(xy + yz + zx)

cycxy + zHay(x + y + z)4xy + yz + zx 6(x2+ y2+ z2) + 6xyz

cyc1y + zLi s dng bt ng thc AMGM,4y + z 1y + 1z,4z + x 1z+ 1x,4x + y 1x + 1ySuy ra

cyc1y + z 12_1x + 1y + 1z_=xy + yz + zx2xyzTa cn chng minh(x + y + z)4xy + yz + zx 6(x2+ y2+ z2) + 3(xy + yz + zx)Hay(x + y + z)4xy + yz + zx 6(x + y + z)29(xy + yz + zx)((x + y + z)23(xy + yz + zx))2xy + yz + zx 0Bt ng thc c chng minh xong. ng thc xy ra khi v ch khia = b = c.57Tm hng skln nht sao cho bt ng thc sau ng1a(1 + bc)2+1b(1 + ca)2+1c(1 + ab)2 k(1 + ab)(1 + bc)(1 + ca) + 34 k8trong a, b, c l cc s dng thaabc = 1.www.VNMATH.com69Li gii. Choa = 2, b = 1, c =12, ta ck 4. Ta s chng minh4

cyc1a(1 + bc)2 1 +16(1 + ab)(1 + bc)(1 + ca)Hay4

cyca(a + 1)2 1 +16(1 + a)(1 + b)(1 + c)tx =1a1+a, y =1b1+b, z =1c1+c, thx, y, z [1, 1] v(1 x)(1 y)(1 z) = (1 + x)(1 + y)(1 + z)Suy rax + y + z + xyz = 0Bt ng thc tr thnh

cyc(1 x2) 1 + 2(1 + x)(1 + y)(1 + z)Hayx2+ y2+ z2+ 2(xy + yz + zx) + 2(x + y + z + xyz) 0(x + y + z)2 0Vy ta c pcm, do kmax = 4.58Cho cc s khng ma, b, c, chng minh bt ng thc sau vik =ln 3ln 3ln 2_a2b2+ bc + c2_1/k+_b2c2+ ca + a2_1/k+_c2a2+ ab + b2_1/k 2Li gii. S dng bt ng thc Holder, ta c_

cyc_a2b2+ bc + c2_1/k_k_

cyca(b2+ bc + c2)_ (a3/(k+1)+ b3/(k+1)+ c3/(k+1))k+1Ta phi chng minh(a3/(k+1)+ b3/(k+1)+ c3/(k+1))k+1 2k(a + b + c)(ab + bc + ca)Khng mt tnh tng qut, gi sa b c, ta =t + m, b =t m vi t m + c, m 0, xthm sf(m) = (k + 1) ln ((t + m)3/(k+1)+ (t m)3/(k+1)+ c3/(k+1)) ln (t2+ 2tc m2)Ta cf

(m) =3((t + m)(2k)/(k+1)(t m)(2k)/(k+1))(t + m)3/(k+1)+ (t m)3/(k+1)+ c3/(k+1)+2mt2+ 2tc m2=3(a(2k)/(k+1)b(2k)/(k+1))a3/(k+1)+ b3/(k+1)+ c3/(k+1)+a bab + bc + cawww.VNMATH.com70 CHNG2. SOLUTIONTaschngminhf

(m) 0,ta =xk+1, b =yk+1, c =zk+1(x y z 0),taphichngminhxk2yk2(xk+1yk+1)xk+1yk+1+ yk+1zk+1+ zk+1xk+1 3(xk2yk2)x3+ y3+ z3 0Dthyrngdox yvk> 2nnxk+1 yk+1k+1k2y3(xk2 yk2),nhthtacnchngminhk + 1k 2xk2yk+1(x3+ y3+ z3) 3(xk+1yk+1+ yk+1zk+1+ zk+1xk+1)Hay7 2kk 2xk+1yk+13xk+1zk+1+k + 1k 2xk2yk+4+_k + 1k 2xk23zk2_yk+1z3 0Nhth, nuk+1k2xk2yk+4xk+1zk+1th dox yv72kk2>2nnbt ngthcng.Xttrnghpngcli,k+1k2xk2yk+4xk+1zk+1, suyrax 3_k+1k2y 34y, khi tacxk+1yk+1 6y3(xk2yk2), vy nn ta cn chng minh2xk2yk+1(x3+ y3+ z3) (xk+1yk+1+ yk+1zk+1+ zk+1xk+1)Hayxk+1(yk+1zk+1) + 2xk2yk+4+ (2xk2zk2)yk+1z3 0 (ng)Vytaluncf

(m) 0, vnhvy, tach cnxtbtngthcchotrongtrnghpa = b c l , tc l ta phi chng minh(2a3/(k+1)+ c3/(k+1))k+1 2ka(a + 2c)(2a + c)tu =_ca_1/(k+1) 1, ta cn chng minhg(u) =(u3+ 2)k+1(uk+1+ 2)(2uk+1+ 1) 2kC th d dng kim tra c bt ng thc trn. Vy ta c pcm.59Cho cc s khng ma, b, c chng minh bt ng thc_a2+ bcb2+ bc + c2+_b2+ cac2+ ca + a2+_c2+ aba2+ ab + b2 6Li gii. S dng bt ng thc Holder, ta c_

cyc_a2+ bcb2+ bc + c2_2_

cyc(a2+ bc)2(b2+ bc + c2)_ (a2+ b2+ c2+ ab + bc + ca)3Ta cn chng minh(a2+ b2+ c2+ ab + bc + ca)3 6

cyc(a2+ bc)2(b2+ bc + c2)Hay(a2+b2+c2+ab+bc+ca)3 12

cyca2b2(a2+b2)+6

cyca3b3+6

cyca4bc+12

cyca2b2c(a+b)+36a2b2c2www.VNMATH.com71Khng mt tnh tng qut, gi s a+b +c = 1, t ab +bc +ca =1q23, r = abc th ta c 1 q 0vr max_0,(12q)(1+q)227_, bt ng thc tr thnh2(4q2+ 5)r + 1727q6 89q4 209q2+ 1027 0Nu 2q 1 th ta c1727q689q4209q2+1027 0 nn bt ng thc hin nhin ng, nu 2q 1th ta cV T 227(4q2+ 5)(1 2q)(q + 1)2+ 1727q6 89q4 209q2+ 1027=127q2(17q4+ 16q3+ 20q 38) 0Bt ng thc c chng minh xong. ng thc xy ra khi v ch khia = b = c.60Chng minh rng vi mix, y [0, 1], ta c1x2x + 1 +1y2y + 1 1 +1x2y2xy + 1Li gii. Bt ng thc tng ng vi_1 1x2x + 1__1 1y2y + 1_1(x2x + 1)(y2y + 1) 1x2y2xy + 1Hayxy(1 x)(1 y)(x2x + 1)(y2y + 1) (1 x)(1 y)(x + y)(x2x + 1)(y2y + 1)(x2y2xy + 1)x + y xy(x2y2xy + 1)Dox, y 1 nnx2y2xy + 1 1, do x + y xy(x2y2xy + 1) x + y xy = x(1 y) + y 0Bt ng thc c chng minh xong.61Cho cc s dnga, b, c, chng minh bt ng thc_aa + b +_bb + c +_cc + a 32 _ab + bc + caa2+ b2+ c2Li gii. Trc ht ta chng minh rng3(a2+ b2+ c2)a + b + c 2

cyca2a + cTht vy, ta c(V T V P)(a + b + c) = 3(a2+ b2+ c2) (a + b + c)

cyca2+ b2a + b= a2+ b2+ c2

cycc(a2+ b2)a + b=

cycab(a b)2(a + c)(b + c) 0www.VNMATH.com72 CHNG2. SOLUTIONS dng bt ng thc ny v bt ng thc Holder, ta c3(a2+ b2+ c2)a + b + c_

cyc_aa + b_2_

cyca(a + b)(a + c)_ 2_

cyca2a + c__

cyc_aa + b_2_

cyca(a + b)(a + c)_ 2(a + b + c)4Ta cn chng minh4(a + b + c)5 27(ab + bc + ca)

cyca(a + b)(a + c)Khng mt tnh tng qut, gi sa + b + c = 1, tab + bc + ca =1q23, r = abc (1 q 0), thth ta cr (1q)2(1+2q)27, bt ng thc tr thnh4 (1 q2)(6q2+ 3 + 27r)Ta c4 (1 q2)(6q2+3 +27r) 4 (1 q2)(6q2+3 +(1 q)2(1 +2q)) = q2(2q3+2q2+(q 1)2) 0Bt ng thc c chng minh xong. ng thc xy ra khi v ch khia = b = c.62Chng minh rng vi mia, b, c 0, ta c bt ng thca2(b + c)(b2+ c2)(2a + b + c) +b2(c + a)(c2+ a2)(2b + c + a) +c2(a + b)(a2+ b2)(2c + a + b) 23Li gii. S dng bt ng thc Cauchy Schwarz, ta c_

cyca2(b + c)(b2+ c2)(2a + b + c)__

cyca2(b2+ c2)(2a + b + c)b + c_ (a2+ b2+ c2)2Do , ta cn chng minh3(a2+ b2+ c2)2 2

cyca2(b2+ c2)(2a + b + c)b + cHay3

cyca4+ 2

cyca2b2 4

cyca3(b2+ c2)b + c3

cyc_a4a3(b2+ c2)b + c_+

cyc_a2(b2+ c2) a3(b2+ c2)b + c_ 03

cyca3b(a b) ca3(c a)b + c+

cyca2(b2+ c2)(b + c a)b + c 03

cycab(a b)2(a2+ b2+ ab + bc + ca)(a + c)(b + c)+

cyca2(b2+ c2)(b + c a)b + c 0www.VNMATH.com73Khng mt tnh tng qut, gi sa b c 0. Khi , ta cV T 2ab(a b)2(a2+ b2+ ab + bc + ca)(a + c)(b + c)+a2(b2+ c2)(b a)b + c+b2(a2+ c2)(a b)a + c=(a b)2(2ab(a2+ b2+ ab + bc + ca) (a2b2+ (a2+ b2+ ab)c2+ (a + b)c3))(a + c)(b + c)(a b)2(2ab(a2+ b2+ ab + bc + ca) (a2b2+ (a2+ b2+ ab)ab + (a + b)abc))(a + c)(b + c)=ab(a b)2(a2+ b2+ ac + bc)(a + c)(b + c) 0Bt ng thc c chng minh xong. ng thc xy ra khi v ch khi (a, b, c) (1, 1, 0).63Cho cc s dnga, b, c, chng minh rng vi mik 2, ta c bt ng thca + b + c3abck_a + cb + c+k_c + ba + b +k_b + ac + aLi gii. S dng bt ng thc Holder, ta c_

cyck_a + cb + c_k2 3k21

cyc_a + cb + cMt khc, theo bt ng thc AMGM th_

cyck_a + cb + c_k2=_

cyck_a + cb + c__

cyck_a + cb + c_k21 3k21

cyck_a + cb + cSuy ra

cyck_a + cb + c

cyc_a + cb + cDo , ta ch cn chng minh

cyc a3abc

cyc_a + cb + cS dng bt ng thc Cauchy Schwarz, ta c_

cyc_a + cb + c_2 2_

cyca__

cyc1a + b_Nh vy, ta ch cn chng minha + b + c3a2b2c22a + b +2b + c +2c + aHay(a + b + c)(a + b)(b + c)(c + a) 23a2b2c2(a2+ b2+ c2+ 3ab + 3bc + 3ca)S dng bt ng thc AMGM, ta cab + bc + ca 3a2b2c2, (a + b + c)(ab + bc + ca) 9abcwww.VNMATH.com74 CHNG2. SOLUTIONSuy ra(a + b)(b + c)(c + a) = (a + b + c)(ab + bc + ca) abc89(a + b + c)(ab + bc + ca)833a2b2c2(a + b + c)Do (a + b + c)(a + b)(b + c)(c + a) 833a2b2c2(a + b + c)2Nh vy, ta ch cn chng minh43(a + b + c)2 a2+ b2+ c2+ 3ab + 3bc + 3caHay(a + b + c)2 3(ab + bc + ca) (ng theo AMGM)Vy bt ng thc cn chng minh ng. ng thc xy ra khi v ch khia = b = c.64Cho cc s khng ma, b, c, chng minh bt ng thc3_ab + c +3_bc + a +3_ca + b 2abc(a + b)(b + c)(c + a) + 1Li gii. Trc ht, ta chng minh_ab + c +_bc + a +_ca + b 2abc(a + b)(b + c)(c + a) + 1Tht vy, bt ng thc tng ng

cyc_a(a + b)(a + c) 2_(a + b + c)(ab + bc + ca)Hay_

cyc_a(a + b)(a + c)_2 4(a + b + c)(ab + bc + ca)

cyca3+ 2

cyc(a + b)_ab(a + c)(b + c) 3

cycab(a + b) + 9abcS dng cc bt ng thc AMGM, Cauchy Schwarz v Schur, ta cV T

cyca3+ 2

cyc(a + b)_ab + cab_=

cyca3+ 2

cycab(a + b) + 2abc

cycc(a + b)

cyca3+ 2

cycab(a + b) + 12abc =_

cyca3+ 3abc_+ 2

cycab(a + b) + 9abc 3

cycab(a + b) + 9abcwww.VNMATH.com75Tip theo, ta s chng minh3_ab + c a2/3b2/3+ c2/3Tht vy, bt ng thc tng nga2(b + c)2 a2(b2/3+ c2/3)3Hay(b2/3+ c2/3)3 (b + c)23b2/3c2/3(b2/3+ c2/3) 2bc (ng)T bt ng thc ny, suy ra

cyc3_ab + c

cyca2/3b2/3+ c2/3Theo trn th

cyca2/3b2/3+ c2/3 2a2/3b2/3c2/3(a2/3+ b2/3)(b2/3+ c2/3)(c2/3+ a2/3) + 1Suy ra

cyc3_ab + c 2a2/3b2/3c2/3(a2/3+ b2/3)(b2/3+ c2/3)(c2/3+ a2/3) + 1Do , ta ch cn chng minha2/3b2/3c2/3(a2/3+ b2/3)(b2/3+ c2/3)(c2/3+ a2/3) + 1 abc(a + b)(b + c)(c + a) + 1Hay(a + b)(b + c)(c + a) a1/3b1/3c1/3(a2/3+ b2/3)(b2/3+ c2/3)(c2/3+ a2/3)(x3+ y3)(y3+ z3)(z3+ x3) xyz(x2+ y2)(y2+ z2)(z2+ x2)trong x = a1/3, y = b1/3, z = c1/3.Bt ng thc ny chnh l h qu ca cc bt ng thc hin nhin saux + y 2xy, x2xy + y212(x2+ y2) x, y 0Vy ta c pcm. ng thc xy ra khi v ch khi (a, b, c) (1, 1, 0).65Cho cc s thca, b, c, d thaa2+ b2+ c2+ d2= 4, chng minh bt ng thc9(a + b + c + d) 4abcd + 32Li gii. Ddngkimtracmax{abc, bcd, cda, dab} 0. Khng mt tnh tngqut, gi sd = min{a, b, c, d}, suy ra 1 d> 0, tP(a, b, c, d) =V T V P, x =_a2+b2+c23vp = a +b +c, th th ta c 23 3x p x3, ta s chng minh P(a, b, c, d) P(x, x, x, d), thtvy bt ng thc tng ng9(3x p) 4d(x3abc)T bt ng thc Schur bc 4

cyc a2(ab)(ac) 0, ta suy ra c abc (p26x2)(p2+3x2)12p, nhvy, ta ch cn chng minh9(3x p) 4d_x3 (p26x2)(p2+ 3x2)12p_Hay(3x p)_27 d(p3+ 3p2x + 6px2+ 6x3)p_ 0Ch rng 3x p x3 nn81 3d(p3+ 3p2x + 6px2+ 6x3)p 81 78x2d = 81 26d(4 d2) = 3 + 26(1 d)(3 d d2) 0Nh vy bt ng thc ng v ta cn phi chng minh9(3x + d) 4x3d 32Hay(9 4x3)_4 3x2= d(9 4x3) 32 27xf(x) =32 27x(9 4x3)4 3x2 1Ta cf

(x) =12(x 1)(81x447x3119x2+ 9x + 81)(9 4x3)2(4 3x2)3/2 0Suy raf(x) l hm ng bin, do f(x) f(1) = 1, vy ta c pcm. ng thc xy ra khi vch khia = b = c = d = 1.66Cho cc s khng ma, b, c, chng minh bt ng thc_a2+ 256bcb2+ c2+_b2+ 256cac2+ a2+_c2+ 256aba2+ b2 12Li gii. Khng mt tnh tng qut, gi sa b c 0, xt cc trng hp sauTrnghp1. 256b3 a2c, suy raa2 256b2, do _a2+ 256bcb2+ c2 16_b2+ bcb2+ c2 16 > 12Trnghp2. 256b3 a2c, khi ta ca2+ 256bcb2+ c2=a2b2+c(256b3a2c)b2(b2+ c2)a2b2 ,b2+ 256cac2+ a2=b2a2+c(256a3b2c)a2(a2+ c2)b2a2Do V T ab+ba + 16_aba2+ b2=a2+ b2ab+ 2 8_aba2+ b2 3382= 12Bt ng thc c chng minh xong. ng thc xy ra khi v ch khi (a, b, c) _2 +3, 1, 0_.www.VNMATH.com7767Cho cc s dngx, y, zc tch bng 1, chng minh rngxy4+ 2 +yz4+ 2 +zx4+ 2 1Li gii. Dox, y, z> 0, xyz = 1 nn tn ti cc s dnga, b, c sao chox =ab, y =ca, z =bc, khi btng thc tr thnh

cyca5b(c4+ 2a4) 1S dng bt ng thc Cauchy Schwarz, ta cV T (a3+ b3+ c3)22(a5b + b5c + c5a) + abc(a3+ b3+ c3)Ta cn chng minh(a3+ b3+ c3)2 2(a5b + b5c + c5a) + abc(a3+ b3+ c3)Hay(a3+ b3+ c3)2(ab + bc + ca)(a4+ b4+ c4)

cyca5b

cycab52_

cyca3_2_

cyca2__

cyca4_+_

cyca4__2

cyca22

cycab_ 2_

cyca5b

cycab5_

cyc(a4+ b4+ c42a2b2)(a b)2 2(a b)(b c)(a c)((a + b + c)(a2+ b2+ c2) + abc)T y, khng mt tnh tng qut, ta ch cn xt a b c l . t a = a1+t, b = b1+t, c = c1+tvit c1, xt hm sf(t) =

cyc((a1 + t)4+ (b1 + t)4+ (c1 + t)42(a1 + t)2(b1 + t)2)(a b)2+ 2

cyc(a b)__

cyc(a1 + t)__

cyc(a1 + t)2_+ (a1 + t)(b1 + t)(c1 + t)_Ta cf

(t) = 4

cyc((a1 + t)3+ (b1 + t)3+ (c1 + t)3(a1 + t)2(b1 + t) (a1 + t)(b1 + t)2)(a b)2+ 2

cyc(a b)__3

cyc(a1 + t)2+ 2_

cyc(a1 + t)_2+

cyc(a1 + t)(b1 + t)__f

(t) = 4

cyc(2(a1 + t)2+ 2(b1 + t)2+ 3(c1 + t)24(a1 + t)(b1 + t))(a b)2+ 24

cyc(a b)

cyc(a1 + t)= 4

cyc(2a2+ 2b2+ 3c24ab)(a b)224(a b)(b c)(a c)(a + b + c)= 2

cyc(2a22b2+ 5bc 3ca 2ab)2+ 6

cycc2(a b)2 0www.VNMATH.com78 CHNG2. SOLUTIONSuy raf

(t) l hm ng bin, do f

(t) f

(c1) = 2(4x511x4y + 6x3y2+ 6x2y3xy4+ 4y5) 0trong x = a1c1, y = b1c1. Nh vyf(t) l hm ng bin, suy raf(t) f(c1) = x62x5y + 2x3y3+ y6 0Bt ng thc c chng minh xong. ng thc xy ra khi v ch khix = y = z = 1.68Chng minh rng vi mi s dnga, b, c, d ta c bt ng thc_1a + 1b + 1c + 1d__1a + b +1b + c +1c + d +1d + a_16abcd + 1Li gii. S dng bt ng thc AMGM, ta cV T=_ 1ab +1cd_+_a + bab(c + d) +c + dcd(a + b)_+a + bab(d + a) +a + bab(b + c) +c + dcd(b + c) +c + dcd(d + a)4abcd+a + bab(d + a) +a + bab(b + c) +c + dcd(b + c) +c + dcd(d + a)Tng t, ta cV T 4abcd+b + cbc(a + b) +b + cbc(c + d) +a + dad(a + b) +a + dad(c + d)Do 2V T 8abcd+_a + bab(d + a) +a + dad(a + b)_+_a + bab(b + c) +b + cbc(a + b)_+_c + dcd(b + c) +b + cbc(c + d)_+_c + dcd(d + a) +a + dad(c + d)_8abcd+2abd+2bca +2cbd+2dca=8abcd+2bd_1a + 1c_+2ac_1b + 1d_16abcdSuy raV T 8abcd16abcd + 1Vy ta c pcm. ng thc xy ra khi v ch khia = b = c = d = 1.69Cho cc s dnga, b, c, d thaa2+ b2+ c2+ d2= 4, chng minh bt ng thca + b + c + d23(abcd + 1)_1a + 1b + 1c + 1d_Li gii. S dng kt qu bi ton 65, ta c9(a + b + c + d) 4abcd + 32www.VNMATH.com79Mt khc, theo bt ng thc AMGM thabcd 1,1a + 1b + 1c + 1d 44abcdNh vy, ta cn chng minh2(x4+ 8) 93_4(x4+ 1)xvix =4abcd 1.Hayf(x) =x(x4+ 8)3x4+ 17292Ta cf

(x) =(x4+8)2(9x811x4+8)(x4+1)2> 0, suy raf(x) l hm ng bin, do f(x) f(1) =7292.Bt ng thc c chng minh xong. ng thc xy ra khi v ch khia = b = c = d = 1.70Cho cc s dnga1, a2, . . . , anthaa1a2 an = 1. Khi , vi mik R, ta c1(1 + a1)k+1(1 + a2)k+ +1(1 + an)k min_1,n2k_Li gii. Nhn xt rng ta ch cn chng minh trong trng hpk> 0 l . tf(t) =1(t+1)k. GiMl trung bnh nhn caa1, a2, . . . , an. Khi , bt ng thc cn chng minh tng ng vif(a1) + f(a2) + + f(an) min {nf(M), 1}Ta c B sauB. Nu 0 < a b c d vad = bc thf(a) + f(d) min {f(b) + f(c), 1}Tht vy, tm =ad =bc vag(t) = f(mt) + f_mt_= (mt + 1)k+_mt+ 1_kvi mit > 0.Li tt1 =cm, t2 =dmtht2 t1 1. Ta cn chng minhg(t2) min {g(t1), 1}Xt tnh n iu cag trn [1, +), ta cg

(t) = mk_1t2_mt+ 1_k1(mt + 1)k1_g

(t) > 0 1t2_mt+ 1_k1> (mt + 1)k1h(t) = t2k+1mt + mt1k1+k1 < 0Li ch

(t) =2k + 1t1k1+km + 1 k1 + kmt2kk+1h

(t) =2(1 k)(k + 1)2t3k+1k+1(t mk)h(1) = 0, h

(1) =2(1 km)k + 1www.VNMATH.com80 CHNG2. SOLUTIONTy thuc vo cc gi tr cam vk, xt cc trng hp sau(i)k = 1, m 1, ta ch(t) = (1 m)(t 1) 0 t > 1, do h 0 trn (1, +).(ii)k = 1, m > 1, ta ch(t) = (1 m)(t 1) < 0 t > 1, do h < 0 trn (1, +).(iii)k< 1, m 1k, khi , ta ch

> 0 t > 1, vh

(1) 0 nnh

> 0 trn (1, +). Vh(1) = 0vh lin tc nnh > 0 trn (1, +).(iv)k< 1, m>1k, khi , ta ch

(1)< 0 vh

< 0 t (1, mk), suy rah

< 0 t (1, mk). Vh(1) = 0vhlintcnnh< 0 t (1, mk].Trn(mk, +),tach

> 0,tchlhmlmtrn (mk, +). Ta li ch(mk) < 0 v limt+h(t) = + nn tn ti duy nhtp > 1 sao choh < 0t (1, p) vh > 0 t (p, +).(v)k > 1, m 1k, khi , ta ch

< 0 t > 1, tch l hm li trn (1, +). Doh l hm lin tcnnh(t) min_h(1), limt+h(t)_= 0 t > 1.(vi)k> 1, m>1k,khi,tach

> 0 t (1, mk),tchlhmlmtrn(1, mk)vh

< 0t (mk, +), tchlhmlitrn(mk, +).Nuh(mk)mksaochohmk, doh l hm lm trn (1, mk) vh(1) = 0 nn tntiv [1, mk] sao choh 0 t (1, v] vh 0 t [v, mk].T cc trng hp ni trn+, Nu h(t2) 0 th h 0 t (t2, +), tc l g

0. Suy ra, g l hm khng tng trn [t2, +).Do g(t2) limt+g(t) = 1+, Nuh(t2) < 0 thh 0 t (1, t2), tc lg

0. Suy ra,g l hm khng gim trn (1, t2). Dog(t2) g(t1)Vy, ta cg(t2) min {g(t1), 1}B c chng minh hon ton.Tr li bi ton ca ta, ta s chng minh bng quy np theon. Trng hpn = 1, n = 2 th btng thc hin nhin ng. Gi s bt ng thc ng cho s bin b hnn (n 3). Ta s chngminh minh n cng ng cho s bin bngn. Ta cn chng minhf(a1) + f(a2) + + f(an) min {nf(M), 1}Dthyrngtrongdya1, a2, . . . , anluntnti tnhtmtskhnglnhnMvtnhtmtskhngnhhnM. Khngmttnhtngqut, tacthgisa1 M a2. Khiux1 = min_M,a1a2M_ vx2 = max_M,a1a2M_. Khi , ta ca1 x1 x2 a2vx1x2 = a1a2, do s dng B trn, ta cf(a1) + f(a2) min {f(x1) + f(x2), 1} = min_f(M) + f_a1a2M_, 1_Ch rnga1a2M, a3, a4, . . . , an cng c trung bnh nhn lMv s bin ln 1 < n nn theo githit quy np, ta cf_a1a2M_+ f(a3) + . . . + f(an) min {(n 1)f(M), 1}www.VNMATH.com81Do n

i=1f(ai) min_f(M) + f_a1a2M_, 1_+ f(a3) + + f(an) min_f(M) + f_a1a2M_+ f(a3) + + f(an), 1_ min{nf(M), 1}Vy bt ng thc trn cng ng cho s bin bngn. Theo nguyn l quy np, ta suy ra n ngvi min. Bi ton c gii quyt hon ton.71Choa, b, c l cc s dng, chng minh rng1.a9bc+b9ca +c9ab +2abc a5+ b5+ c5+ 22.a9bc+b9ca +c9ab +3abc a4+ b4+ c4+ 3Li gii. (1) S dng bt ng thc AMGM, ta ca9bc+ abc 2a5,b9ca + abc 2b5,c9ab + abc 2c5Suy raa9bc+b9ca +c9ab 2(a5+ b5+ c5) 3abc a5+ b5+ c5+ 33a5b5c53abcDo chng minh bt ng thc cho, ta ch cn chng minh33a5b5c53abc +2abc 2Hay3t53t3+2t3 2vit =3abc > 0.(t 1)2(3t6+ 6t5+ 6t4+ 6t3+ 6t2+ 4t + 2)t3 0 (ng)Vy bt ng thc cn chng minh ng. ng thc xy ra khi v ch khia = b = c = 1.(2) Tng t nh trn, p dng bt ng thc AMGM, ta cng ca9bc+ abc + a2 3a4,b9ca + abc + b2 3b4,c9ab + abc + c2 3c4Suy raa9bc+b9ca +c9ab 3(a4+ b4+ c4) (a2+ b2+ c2) 3abcLi p dng bt ng thc AMGM, ta c12(a4+ 1) a2,12(b4+ 1) b2,12(c4+ 1) c2Suy ra12(a4+ b4+ c4) a2+ b2+ c2 32www.VNMATH.com82 CHNG2. SOLUTIONDo t trn, ta ca9bc+b9ca +c9ab 52(a4+ b4+ c4) 3abc 32 923a4b4c4+ (a4+ b4+ c4) 3abc 32Nh vy, chng minh bt ng thc cho, ta ch cn chng minh923a4b4c43abc 32 +3abc 3Hay92t43t3+3t3 92vit =3abc > 0.32(t 1)2(t + 1)(3t4+ t3+ 4t2+ 2t + 2) 0 (ng)Vy bt ng thc cn chng minh ng. ng thc xy ra khi v ch khia = b = c = 1.72Chox, y, z, t l cc s dng thaxyzt = 1, chng minh rng1xy + yz + zx + 1 +1yz + zt + ty + 1 +1zt + tx + xz + 1 +1tx + xy + yt + 1 1Li gii. ta =1x, b =1y, c =1z, d =1t, th ta ca, b, c, d> 0 vabcd = 1. Bt ng thc cn chngminh tr thnh1a(b + c + d) + 1 +1b(c + d + a) + 1 +1c(d + a + b) + 1 +1d(a + b + c) + 1 1Khng mt tnh tng qut, ta c th gi sa b c d > 0, thcd 1. Khi , theo bt ngthc AMGM, ta c1c(d + a + b) + 1 +1d(a + b + c) + 1 1c_d + 2ab_+ 1+1d_2ab + c_+ 1Mt khc1a(b + c + d) + 1 +1b(c + d + a) + 1=(a + b)(c + d) + 2(ab + 1)(a + b)(c + d)(ab + 1) + ab(c + d)2+ (ab + 1)2=m + 2(ab + 1)m(ab + 1) + ab(c + d)2+ (ab + 1)2= f(m)trong m = (a + b)(c + d) 2ab(c + d) > 0. Ta cf

(m) =ab(c2+ d2+ cd ab 2)(m(ab + 1) + ab(c + d)2+ (ab + 1)2)2 0Do f(m) l hm ngch bin, suy raf(m) f_2ab(c + d)_=2ab_c + d +ab_+ 1tA =ab, th ta cA 1 vA2cd = 1. Do V T 2A(c + d + A) + 1 +1c (d + 2A) + 1 +1d (2A + c) + 1www.VNMATH.com83Nh vy, chng minh bt ng thc cho, ta ch cn chng minh2A(c + d + A) + 1 +1c (d + 2A) + 1 +1d (2A + c) + 1 1tp = A(c +d), khi sau mt vi tnh ton n gin (vi ch rngA2cd = 1), ta c bt ngthc trn tng ng vi2cdp2+ (cd 1)2p + 3A2c2d23cd 7 0Hay2A2cd(c + d)2+ A(c + d)(cd 1)2+ 3A2c2d23cd 7 02(c + d)2+ A(c + d)(cd 1)2+ 3A2c2d23cd 7 02(c d)2+ A(c + d)(cd 1)2+ 3A2c2d2+ 5cd 7 02(c d)2+ A(c + d)(cd 1)2+3cd c2d2+ 5cd 7 02(c d)2+ A(c + d)(cd 1)2+ (cd 1)2(3 cd)cd 0Bt ng thc ny hin nhin ng v cd 1. Vy bt ng thc cn chng minh ng. ng thcxy ra khi v ch khia = b = c = d = 1.73Chng minh rng vi mix, y, z, t > 0 th(x + y)(x + z)(x + t)(y + z)(y + t)(z + t) 4xyzt(x + y + z + t)2Li gii. ta =1x, b =1y, c =1z, d =1tth ta ca, b, c, d > 0. Khi , bt ng thc trn tr thnh(a + b)(a + c)(a + d)(b + c)(b + d)(c + d) 4(abc + abd + acd + bcd)2S dng bt ng thc Cauchy Schwa