[VNMATH.com]-Chuyen de BĐT 2015 Ôn Thi ĐH

Chuyn Bt ng thc LTH nm 2015 Bin son: Thy L Xun i cvp

1

L XUN I

(GV THPT Chuyn Vnh Phc)

Bt ng thc (BT) l mt trong nhng dng ton thng c trong cc thi H-

C. Cc th sinh ca chng ta u rt s v lng tng khi gp phi bi ton chng minh

BT hoc tm gi tr ln nht, nh nht. n gin l do cc bi ton v BT thng l bi

ton kh trong thi, nhm phn loi v chn c cc hc sinh kh gii. Thng th cc s

t khng bit bt u t u gii quyt cc bi ton v BT. Chuyn ny mun h

thng cho cc bn cc phng php c bn v mt s dng bi tp v BT. Hy vng s

gip cc em hc sinh lp 12 t kt qu cao trong k thi H- C sp ti.

c xong chuyn ny ti tin cc bn s khng cn cm gic s bt ng thc na

Khi chng ta ht i s s hi v ngi ngn th chng ta s am m v dnh tnh yu cho n.

Dnh tnh yu v s am m cho ton hc ni chung v BT ni ring l iu rt cn thit

ca mt ngi lm ton s cp chn chnh v s lng mn ca ton hc cng bt ngun t

Thnh cng ch n khi bn lm vic tn tm v lun ngh n nhng iu tt

p

Nhng li khuyn b ch khi hc v BT:

1. Nm chc cc tnh cht c bn ca BT.

2. Nm vng cc phng php c bn chng minh BT nh: PP bin i tng

ng; PP s dng BT C si; PP s dng o hm

3. c bit ch trng vo n tp cc k thut s dng BT C si, lun bit t v tr

li cc cu hi nh: khi no p dng; iu kin cho cc bin l g; du bng xy ra khi no;

nu p dng th th c xy ra du bng khng; ti sao li thm bt nh vy

4. Lun bt u vi cc BT c bn (iu ny v cng quan trng); hc thuc mt

s BT c bn c nhiu p dng nhng phi ch iu kin p dng c, chng hn nh:

* 2 2 2a b c ab bc ca (1) vi mi a,b,c

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* 2(a b c) 3(ab bc ca) (2) vi mi a,b,c

* 2 2 2 2(a b c) 3(a b c ) (3) vi mi a,b,c

* 1 1 4 1 1 1 9; (4)a b a b a b c a b c

vi mi a,b,c dng

* 2 2 2 2 2 2a x b y (a b) (x y) (5) vi mi a,b,x,y.

* 2 2 2x y (x y) (6)

a b a b

vi mi a,b dng v x,y bt k

* 2 2 2 2x y z (x y z) (7)

a b c a b c

vi mi a,b,c dng v x,y,z bt k

Du bng xy ra cc BT (1), (2), (3) v (4) l a=b=c.

Du bng xy ra BT (5) v (6) l x ya b ; (7) l x y z

a b c (vi mu khc 0).

(Cc em hy bt tay ngay vo vic chng minh cc BT c bn trn nh. Hy tm cho mnh

mt cch gii nht qun, n gin, nh n v khi lm bi thi u phi chng minh li, ri

mi c p dng).

Trc ht xin a ra 3 phng php thng dng nht chng minh BT

I. PHNG PHP BIN I TNG NG:

1. Phng php chung

chng minh A B ta thng thc hin theo mt trong hai cch sau:

Cch 1: Ta chng minh A B 0 . lm c iu ny ta thng s dng hng ng

thc phn tch A B thnh tng hoc tch ca nhng biu thc khng m.

Cch 2: Xut pht t mt BT ng no ta bin i n BT cn chng minh. i vi

cch ny thng cho ta li gii khng c t nhin cho lm v thng s dng khi cc

bin c nhng rng buc c bit.

Ch : Mt s kt qu hay s dng

* 2x 0 vi mi x v 2x 0 x 0

* x 0 vi mi x v x 0 x 0

2. Mt s v d

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V d 1: Chng minh rng vi mi a,b ta c: 2 2a b 2ab (1)

Gii: Ta c 2 2 2 2 2a b 2ab (a b) 0 a b 2ab (pcm).

ng thc xy ra khi v ch khi a=b.

Tht n gin phi khng cc bn, nu tinh thm mt cht thi cc bn s tm ra nhng

kt qu tng qut hn v nim tin vt qua bi BT trong thi H l hon ton kh

thi.

C th l vi ba s thc a,b,c bt k ta c 2 2a b 2ab ; 2 2b c 2bc v 2 2a c 2ac

Cng tng v ca 3 BT ta c kt qu sau: 2 2 2a b c ab bc ca (2)

C th thy ngay c hai BT tng ng vi (2) rt quen thuc l

2(a b c) 3(ab bc ca) (3) vi mi a,b,c

2 2 2 2(a b c) 3(a b c ) (4) vi mi a,b,c

Chng ta s ni thm ng dng tuyt vi ca ba BT (2), (3) v (4) nhng phn sau

V d 2: Chng minh rng vi mi a,b,c ta c: 4 4 4a b c abc(a b c)

Gii: p dng lin tip BT (2) trong v d 1 ta c:

4 4 4 2 2 2 2 2 2

2 2 2 2 2 2 2 2 2

a b c (a ) (b ) (c )a b b c c a (ab) (bc) (ac)ab.bc ab.ac bc.ac abc(a b c)

Nh vy nu thi hi cc bn mt bi nh sau:

Cho 3 s thc a,b,c tho mn a b c 1 . Chng minh rng: 4 4 4a b c abc th

chc cc bn c c hi cao t im 10 ri! (Hy c t tin ln nh th!)

V d 3: Chng minh rng vi mi a,b 0 ta c:

3 3 2 2a b a b ab

Gii: Ta bin i 3 3 2 2 2a b a b ab (a b) (a b) 0 , suy ra pcm.

Nhn xt: BT trn tht n gin nhng cng c kh nhiu ng dng vi cc bi ton kh

hn, chng hn ta xt 3 bi ton sau:

Bi 3.1. Cho a,b,c 0 . Chng minh rng:

3 3 3 3 3 31 1 1 1

a b abc b c abc a c abc abc

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Hng gii: Ta c 3 3 2 2 3 3a b a b ab ab(a b) a b abc ab(a b c)

Suy ra 3 31 1

a b abc ab(a b c)

.

Cng hai BT tng t ta c

1 1 1 1VTab(a b c) bc(a b c) ac(a b c) abc

(pcm).

Xin a ra thm hai h qu ca bi ton trn (coi nh bi tp cho cc bn luyn tp)

* Cho a,b,c 0 tho mn abc=1. Khi : 3 3 3 3 3 31 1 1 1

a b 1 b c 1 a c 1

* Cho a,b,c 0 tho mn abc=1. Khi : 1 1 1 1a b 1 b c 1 a c 1

(che du bn cht hn)

Bi 3.2. Cho a,b,c khng m tho mn a b c 2012 . Tm gi tr nh nht ca biu thc

3 3 3 3 3 33 3 3P 4(a b ) 4(b c ) 4(a c )

Hng gii: Mi nhn BT ta cm thy rt kh khn v c cn bc 3 v iu quan trng l

phi x l c biu thc trong du cn. Bt ng thc 3 3 2 2a b a b ab cho ta mt manh

mi tm ra li gii bi ton, nhng nu p dng nguyn xi nh vy th cha n. Ta bin

i mt cht BT ny

3 3 2 2 3 3 2 2 3 3 3a b a b ab 3(a b ) 3(a b ab ) 4(a b ) (a b)

Nh vy ta c thu c BT 3 3 34(a b ) (a b) .

Chc cc bn cng ng vi ti rng php bin i rt t nhin ch.

By gi p dng BT va tm c ta c

3 3 3 3 3 33 3 3P 4(a b ) 4(b c ) 4(a c ) (a b) (b c) (c a) 2(a b c) 4024

ng thc xy ra khi v ch khi 2012a b c3

.

Vy GTNN ca P bng 4024.

Bi ton tng qut: Cho a,b,c khng m tho mn a b c k . Tm gi tr nh nht ca

3 3 3 3 3 33 3 3P m(a b ) m(b c ) m(a c )

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( m,k l cc hng s dng cho trc)

Bi 3.3. K hiu A,B,C l ba gc ca mt tam gic bt k. Tm gi tr ln nht ca

3 3 3

3 3 3

sin A sin B sin CPA B Ccos cos cos2 2 2

Hng gii: y qu l mt bi ton kh, ta hy m mm theo cc u mi nh nh

* Th nht: Ta c mt nh gi rt quen thuc trong tam gic:

C A B Csin A sin B 2cos .cos 2cos2 2 2

* Th hai: Cc cn bc 3 gi ta ngh ti BT: 3 33a b 4(a b )

Nh vy, ta c 3 3 3 3 3C Csin A sin B 4(sin A sin B) 4.2cos 2. cos2 2

Tng t ta c 3 3 3 Asin B sin C 2. cos2

v 3 3 3 Bsin A sin C 2. cos2

Cng tng v 3 BT trn ta c

3 3 3 3 3 3A B Csin A sin B sin C cos cos cos2 2 2

Vy P 1 . Du bng xy ra khi v ch khi A=B=C

Do gi tr ln nht ca P bng 1 khi tam gic ABC u.

V d 4: Chng minh rng vi a,b,c l 3 cnh mt tam gic bt k ta c:

2 2 2ab bc ca a b c 2(ab bc ca)

Gii: BT bn tri chng minh, chng minh BT bn phi ta xut pht t mt BT

c bn trong tam gic l b c a b c .

* Nu s dng b c a th ta bin i nh sau:

2 2 2 2 2 2 2a b c a (b c) b c 2bc a b c 2bc

Tng t 2 2 2b a c 2ac ; 2 2 2c a b 2ab . Cng theo tng v ba BT ta c pcm.

* Nu s dng a b c th ta bin i nh sau:

2a b c a ab ac , cng hai BT tng t ta c pcm.

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ATH.

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V d 5: Chng minh rng vi mi a,b,x, y ta c BT sau (BT Mincpxki)

2 2 2 2 2 2a x b y (a b) (x y) (1)

Gii: Bnh phng hai v v bin i tng ng:

2 2 2 2 2 2 2 2 2 2 2 2

2 2 2 2

a x b y 2 (a x )(b y ) a x b y 2ab 2xy

(a x )(b y ) ab xy (*)

+ Nu ab xy 0 th hin nhin (*) ng

+ Nu ab xy 0 th 2 2 2 2 2 2(*) (a x )(b y ) (ab xy) (bx ay) 0 (lun ng)

Vy bi ton c chng minh. ng thc xy ra khi bx=ay.

Ch : C th chng minh BT trn bng cch s dng BT vc t rt n gin nh sau

(khi lm bi thi H cc bn phi chng minh BT ny trc khi dng n, lc cc bn

hy chn mt phng n chng minh m cc bn cho l hay v d nh nht. OK).

t u (a; x)

v v (b; y)

, khi u v (a b;x y)

.

T BT vc t u v u v

v cng thc di vc t ta c ngay pcm.

Nu p dng hai ln BT (1) ta c BT sau:

2 2 2 2 2 2 2 2a x b y c z (a b c) (x y z) vi mi a, b,c, x, y,z .

Nhn xt: BT Mincpxki c rt nhiu ng dng hay v c th gii quyt c nhiu bi

BT hc ba. Xin c minh ho iu ny qua 3 bi ton c bn sau y:

Bi 5.1. Cho a,b khng m tho mn a b 1 .

a) Chng minh rng: 2 21 a 1 b 5

b) Tm gi tr nh nht ca 2 22 21 1P a bb a

Hng gii:

a) Ta c 2 2 2 21 a 1 b (1 1) (a b) 5 . ng thc xy ra khi 1a b2

.

b) Ta c 2 2

2 2 2 22 2

1 1 1 1 4P a b (a b) (a b) 17b a a b a b

.

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ATH.

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ng thc xy ra khi 1a b2

. Vy GTNN ca P bng 17 .

Bi 5.2. Cho x,y,z dng tho mn x y z 1 . Chng minh rng:

2 2 22 2 2

1 1 1x y z 82x y z

Hng gii: p dng BT Mincpxki ta c

22 2 2 2

2 2 2

22

1 1 1 1 1 1P x y z (x y z)x y z x y z

9(x y z) 82 (*)x y z

Du bng xy ra khi v ch khi 1x y z3

.

Vi gi thit x y z 1 ta thay trc tip vo (*) v c kt qu l 82 . Tuy nhin nhiu

khi bi li cho gi thit kh i rt nhiu, mc d du bng vn xy ra khi 1x y z3

.

Chng hn H khi A nm 2003: Cho x,y,z dng tho mn x y z 1 .

Chng minh rng: 2 2 22 2 21 1 1x y z 82x y z

.

Vi bi ton ny ta khng th thay x y z 1 ra ngay kt qu nh bi trn c. ng

trc tnh hung ny ta c ngay hai hng gii quyt.

Hng 1: t 2t (x y z) 0 t 1 . Ta c 81P tt

.

Ta tch kho dng BT Csi: 81 1 80 1 80t t 2 t. 82 P 82t t t t 1

.

Hng 2: Vn t 2t (x y z) v xt hm 81f (t) t ; 0 t 1t

.

Ta c 2

2 2

81 t 81f '(t) 1 0 t 0;1t t

, suy ra hm f(t) nghch bin trn 0;1 .

Do f (t) f (1) 82 P 82 .

Hng gii quyt th hai s c cp phn sau ca chuyn .

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ATH.

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Bi 5.3. Cho x,y,z dng tho mn x y z 5 . Tm gi tr nh nht ca

2 2 2P 223 x 223 y 223 z

Hng gii: Ta c

2 2 2 2 2 2 2 2P ( 223) x ( 223) y ( 223) z (3 223) (x y z) 2012

ng thc xy ra khi 5x y z3

. Vy GTNN ca P bng 2012 .

C l khng phi ni g thm na th cc bn cng thy v p v sc mnh ca

BT Mincpxki. Nhng ti nhc li rng phi chng minh li BT ny trc khi p

dng nh!

3. Bi tp t luyn

Bi 1. Chng minh rng: , , , , ,a b c d e R ta c:

a) 2 2 2 2 2 ( )a b c d e a b c d e .

b)

33 3

( 0).2 2

a b a b a b

Bi 2. Chng minh rng:

a) 5 5 4 4 2 2( )( ) ( )( ), , : 0.a b a b a b a b a b ab

b) 2 21 1 2 , , 1.

1 1 1a b

a b ab

Bi 3. Cho ABC. Chng minh rng:

a) 2 2 2 3 3 3( ) ( ) ( )a b c b c a c a b a b c .

b) 2 2 2 2( ).a b c ab bc ca


a) (a c)(b d) ab cd , a,b,c,d 0

b) 2 2 2 2 2 2(a c) (b d) a b c d , a,b,c,d R

c) 1 1 1 1 1b (c a) (c a)c a b c a

0 a b c

d) b c a a b ca b c b c a 0 a b c

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ATH.

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Bi 5. Cho a, b > 0: a + b = 2. Chng minh rng a bab a b

Bi 6. Cho hai s thc a ,b tho mn a + b 2. Chng minh rng : a4 + b4 a3 + b3

Bi 7. Cho ba s a ,b ,c [0;1]. Chng minh rng : a + b + c ab bc ca 1

Bi 8. Cho a,b,c tho mn a b c 1 . Chng minh rng:

a b c a b c

1 1 1 a b c33 3 3 3 3 3

Bi 9. Cho a,b,c dng. Chng minh rng:

2 2 2 2 2 2a ab b b bc c a ac c a b c

II. PHNG PHP S DNG BT NG THC C SI

1. Bt ng thc Csi

a) Cho a 0, b 0 . Khi a b ab2

. ng thc xy ra khi a=b.

b) Cho a 0, b 0, c 0 . Khi 3a b c abc3

. ng thc xy ra khi a=b=c.

Cc dng tng ng l: a b 2 ab ; 2a bab

2

3a b c 3 abc ; 3a b cabc

3

c) Tng qut: Cho n s thc khng m 1 2, ,..., ( 2)na a a n . Khi ta c

1 2 1 2... ... nn na a a n a a a

ng thc xy ra khi v ch khi 1 2 ... na a a .

Ch : Vi cc bi thi H- C thng thng ch cn p dng BT Csi vi 2 hoc 3 s.

2. Mt s v d

V d 1: Chng minh rng:

a) a b 2 a,b 0b a b) a b 2 a,b 0

b a

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Gii. a) p dng BT Csi cho hai s dng ta c: a b a b2 . 2b a b a (pcm)

Du bng xy ra khi a=b.

b) Ta khng th p dng ngay BT Csi v ch c iu kin a,b 0 . Bin i tng

ng BT bng cch bnh phng hai v:

2 2 2

2 2a b a b a b2 4 2b a b a b a

n y theo BT csi th BT sau l ng, vy ta c pcm

Ch l du bng xy ra khi a b .

Cng c th thy ngay rng ab

v ba

cng du nn ta c

a b a b 2b a b a (lc ny li p dng BT Csi c)

V d 2: Cho a,b,c dng. Chng minh rng:

a) 1 1 4a b a b

(1) b) 1 1 1 9

a b c a b c

(2)

Gii. a) Nu vit li BT cn chng minh di dng 1 1(a b) 4a b

th hng gii

quyt l qu r rng. Tht vy, p dng BT Csi cho hai s dng ta c

a b 2 ab v 1 1 12a b ab .

Suy ra 1 1 1(a b) 4 ab. 4a b ab

. Du = xy ra khi v ch khi a=b

b) Hon ton tng t vi phn a) bng cch p dng BT Csi vi 3 s.

Nhn xt: Hai BT trong v d 1 c rt nhiu ng dng v cng l con ng sng to ra

v vn cc BT hay. C th ni phn ln cc BT trong thi H- C c gc tch ca hai

BT ny. Ni ra cc p dng hay ca hai BT ny th nhiu v k v khng bit s tn

km bao giy mc, ti xin ch dn chng ra vi bi ton in hnh.

Bi 2.1. Cho a,b,c dng. Chng minh rng:

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ATH.

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* Cho x,y,z dng tho mn x 2y 4z 12 . Chng minh rng:

2xy 8yz 4xz 6x 2y 2y 4z 4z x

.

Vi bi ton ny, cc bn ch cn coi a x;b 2y;c 4z th a b c 12 v BT cn

chng minh tr thnh: ab bc ac 6a b b c a c

(y chnh l h qu ca (7) ri. OK)

Bi 2.4. Gi a,b,c l ba cnh ca mt tam gic, p l na chu vi tam gic . Chng minh

rng:

1 1 1 1 1 12p a p b p c a b c

(8)

Hng gii: D thy p a 0;p b 0;p c 0 v nhn xt rng

(p a) (p b) 2p a b c

iu ny gi ta dng BT (1) cho hai s p-a v p-b. C th l:

1 1 4 4p a p b (p a) (p b) c

Cng hai BT tng t ta c BT (8) cn chng minh

Bi 2.5. Cho a,b,c dng. Chng minh rng:

2 2 2 1 1 1 3a b c a b ca b b c c a 2

(9)

Hng gii: Ta c 2 2 2

2 2 2 9 3 3(a b c ) 3VT(9) a b c . . (a b c)2(a b c) 2 a b c 2

( do 2 2 2 2(a b c) 3(a b c ) )

Bi 2.6. Cho x,y,z dng tho mn x y z 1 . Tm gi tr ln nht ca biu thc

x y zPx 1 y 1 z 1

.

Hng gii: c th p dng c BT (2) ta bin i P nh sau:

x 1 1 y 1 1 z 1 1 1 1 1P 3x 1 y 1 z 1 x 1 y 1 z 1

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Ta c 1 1 1 9 9x 1 y 1 z 1 x y z 3 4

, suy ra 9 3P 34 4

Du bng xy ra khi v ch khi 1x y z3

. Vy GTLN ca P bng 34

.

Vi li gii nh trn cc bn c th lm hon ton tng t vi bi ton tng qut hn

Bi 2.7. Cho x,y,z dng tho mn x y z 1 v k l hng s dng cho trc.

Tm gi tr ln nht ca biu thc x y zPkx 1 ky 1 kz 1

.

Bi 2.8. Cho a,b,c dng tho mn a b c 1 . Tm gi tr nh nht ca biu thc

2 2 21 1 1P

a 2bc b 2ac c 2ab

Hng gii: Ta c ngay 2 2 2 29 9P 9

a 2bc b 2ac c 2ab (a b c)

.

Du bng xy ra khi a b c 1a b ca b c 1 3

. Vy minP 9 .

Bi 2.9. Cho A,B,C l ba gc ca mt tam gic. Chng minh rng:

1 1 1 62 cos2A 2 cos2B 2 cos2C 5

Hng gii: Ta c 1 1 1 92 cos2A 2 cos2B 2 cos2C 6 cos2A cos2B cos2C

D chng minh c rng 3cos2A cos2B cos2C2

(cc bn hy t chng minh nh)

Suy ra 1 1 1 9 632 cos2A 2 cos2B 2 cos2C 562

(pcm)

Bi 2.10. Cho a,b,c dng tho mn a b c 1 . Chng minh rng:

2 2 21 1 1 1 30

ab bc aca b c

(10)

Hng gii: Ta nh gi v tri ca (10) mt cch rt t nhin nh sau:

2 2 2 2 2 21 1 1 1 1 9

ab bc ac ab bc caa b c a b c

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ATH.

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= 2 2 21 1 1 7

ab bc ca ab bc ca ab bc caa b c

29 7 9 7 301ab bc ca 1(a b c)

3

(do BT c bn 21 1ab bc ca (a b c)3 3

)

3. Bi tp t luyn

Bi 1. Cho a, b, c > 0. Chng minh rng:

a) 1 1 1( )( ) 9a b ca b c

. b) 33(1 a)(1 b )(1 c ) 1 abc

c) 2 2 2a b c a b c

b c c a a b 2

d) abc

acb

cbacba

222

222222

Bi 2. Tm gi tr nh nht ca mi biu thc sau:

21

2A aa

vi a > 0. 3 23

B xx

vi x > 0.

Bi 3. Cho a, b, c > 0 tho mn: a + b + c = 2. Tm gi tr nh nht ca: 3 3 3T a b c

Bi 4. Cho x, y, z > 0: x + y + z = 1. Tm Min: 4 4 4R x y z .

Bi 5. Tm gi tr ln nht ca mi biu thc sau:

(3 2 ); (0 3 / 2).M x x x

(1 )(2 )(4 ); (0 x 1, 0 2).N x y x y y

3(1 ) ; 0 1.P x x x

Bi 6. Chng minh rng vi a, b, c > 0 ta c:

a) 6a b b c c ac a b

b) 32

a b cb c c a a b

.

c) bc ca ab a b ca b c

d) 2 2 2 1 1 1 3a b c (a b c)a b b c c a 2

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ATH.

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e) 2 2 2 2 2 2

a b c 1 1 1 12 a b ca b b c c a

f) 2 2 2 2 2 2

a b b c c a 1 1 1a b ca b b c c a

Bi 7. Cho ABC vi ba cnh l a,b,c. CMR: 3a b cb c a c a b a b c

.

Bi 8. Cho , 1a b . Chng minh rng: 1 1a b b a ab .

Bi 9. Cho ABC. Chng minh rng:

a) 1( )( )( )8

p a p b p c abc .

b) 1 1 1 1 1 12( )p a p b p c a b c

.

Bi 10. Cho a,b,c 0 v a b c 1 . Chng minh rng:

2 2 2

1 1 1 9a 2bc b 2ca c 2ab


a) 1a 3b(a b)

, a b 0 b) 2

1a 2 2b(a b)

, a b 0

c) 2

4a 3(a b)(b 1)

, a b 0 d) 2

2

a 2 2a 1

, a R

e) 2 2

4 4

x y 141 16x 1 16y

x,y R f) 22

1 2(x 1) 1 16

x x

0 x

Bi 12. Cho a,b,c 0 v a b c 1 . CMR: 8abc(a b)(b c)(c a)729

Bi 13. Cho a,b,c 0 v 2 2 2a b c 1 . CMR: 2 2 2 2 2 2

a b c 3 32b c c a a b

Bi 14. Cho a,b,c 0 : 1 1 2a c b . CMR: a b c b 4

2a b 2c b

Bi 15. a) Cho a , b , c > 0 v 1 1 1 21 a 1 b 1 c

. CMR: 18

abc .

b) Cho a,b,c,d 0 tho mn 1 1 1 1 31 a 1 b 1 c 1 d

. CMR: 1abcd81

Bi 16. Cho a, b , c R v a + b + c = 0. CMR: a b c a b c8 8 8 2 2 2 .

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ATH.

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Bi 17. Chng minh rng 2 2( 2) 3 ( 0)2

x xx

Bi 18. Cho a > 0, b > 0 v a + b = 1. Chng minh rng 2 4ab27

.

Bi 19. Chng minh rng nu x > - 3 th

13

93

22

x

x

Bi 20. Chng minh rng nu a > b > 0 th 24a 3

(a b)(b 1)

Bi 21. Tm gi tr nh nht ca biu thc Q = x + y bit x > 0, y > 0 tho mn: 132 yx

Bi 22. Vi xyz = 1, x, y, z > 0. CMR: 23222

yx

zzx

yyz

x

Bi 23. Tm gi tr nh nht ca P = a

bb

a

11

33

vi a, b l cc s dng tho mn iu

kin ab = 1.

Bi 24. Tm gi tr nh nht ca 2 3 Px y

vi x, y l cc s dng tha mn x+y=1.

Bi 25. Cho x, y, z > 0. Chng minh rng

26)(16)(9)(4 z

xyy

zxx

zy

Bi 26. Cho x + y = 1, x, y > 0. Tm gi tr nh nht ca biu thc: xyyx

A 11 22

Bi 27. Cho x,y 0 , x+ y= 1. CMR: 121 22 yx

Bi 28. Cho a + b = 5, a, b > 0. Tm gi tr nh nht ba

P 11

Bi 29. Cho x,y,z dng tho mn xyz=1. Tm gi tr nh nht ca

a) P=x+y+z b) P= x y z c) 1 1 1x y z

Bi 30. Cho 3 s a,b,c > 0. CMR: 2 a

a3 + b2 + 2 b

b3 + c2 + 2 c

c3 + a2 1a2 +

1b2 +

1c2

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ATH.

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Bi 31. Cho x ,y ,z [0;1]. CMR: (2x + 2y + 2z)(2 x + 2 y + 2 z) 818

Bi 32. Cho a 3 ; b 4 ; c 2 . Tm GTLN ca A = ab c 2 + bc a 3 + ca b 4

abc

Bi 33. Cho a,b,c dng . Chng minh rng:

1 1 4 16 64a b c d a b c d

Bi 34. Cho a,b,c dng tho mn a b c 1 . Chng minh rng:

3 3 3 3a b b c a c 18

Bi 35. Cho a,b,c dng tho mn ab bc ca 5 . Chng minh rng:

2 2 23a 3b c 10

Bi 36. Cho a,b,c dng tho mn abc=1. Chng minh rng:

a) 3 3 3a b c a b c

b) 3 3 3 2 2 2a b c a b c

Bi 37. Cho a,b,c dng tho mn abc=1. Chng minh rng:

1 1 1a 1 . b 1 . c 1 1b c a

Bi 38. Gi s a,b,c l ba cnh ca mt tam gic. Chng minh rng:

a b c 3a b c b c a c a b

Bi 39. Cho a,b,c dng tho mn a b c 1 . Chng minh rng:

(1 a)(1 b)(1 c) 8(1 a)(1 b)(1 c)

Bi 40. Cho a,b,c dng tho mn a b c abc . Chng minh rng:

3 3 3a b c 1b c a

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ATH.

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cc bn c thm k thut khi p dng BT Csi ti xin gii thiu mt cht v

phng php chn im ri csi. y c th ni l mt tuyt chiu c o gip cc em

nhanh chng tm ra li gii bi ton.

III. PHNG PHP THM HNG T V CHN IM RI CSI

T vic d on c du bng xy ra (im ri Csi), thm bt cc s hng cho ph

hp v s dng kho lo bt ng thc Csi ta c th t c nhng kt qu khng ng.

c c mt nh hng ng n chng ta thc hin cc bc phn tch bi ton nh

sau:

1. D on du bng xy ra hay cc im m ti t c GTLN, GTNN.

2. T d on du bng, kt hp vi cc BT quen bit, d on cch nh gi (tt

nhin l thm mt cht nhy cm v kh nng ton hc ca mi ngi) cho mi

bi ton. Ch rng mi php nh gi phi m bo nguyn tc du bng xy ra

mi bc ny phi ging nh du bng m ta d on ban u.

lm r iu ny ti xin phn tch cch suy ngh tm ra li gii trong cc v d sau:

V d 1. Chng minh rng vi a,b,c 0 ta c:

2 2 2a b c a b cb c a

Phn tch bi ton:

* Trc ht ta nhn thy nu p dng ngay bt ng thc C si cho 3 s th khng ra

c kt qu mong mun.

* D nhn thy du bng xy ra khi a = b = c.

Khi 2a b

b . V vy ta thm b vo phn t i din

2ab

c chng minh sau:

Li gii.

p dng bt ng thc C si cho hai s dng ta c: 2 2 2

2 2 2 2 2 2

a b cb 2a; c 2b; a 2cb c a

a b c a b cb c a 2a 2b 2c a b cb c a b c a

V d 2. Chng minh rng vi x,y,z > 0 ta c 3 3 3

2 2 2x y z x y zy z x

Phn tch bi ton:

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ATH.

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Ta thy rng vi hng t 3x

y c th c hai hng sau:

Hng 1: Thm 3

2x xy 2xy , cng vi BT c bn 2 2 2x y z xy yz zx

cng cc bt ng thc li ta c iu phi chng minh.

Hng 2: Thm 3 3 3 3 3 3

2 2 2 2 2 2x x y y z zy 3x ; z 3y ; x 3zy y z z x x ri cng li

ta c iu phi chng minh.

V d 3. Cho x, y, z l cc s dng tha mn xyz = 1. Chng minh rng:

x3 + y3 +z3 x + y + z

Phn tch bi ton:

* D on du bng xy ra khi x = y = z = 1.

* Ta mun t hai mc ch l nh gi gim bc t bc 3 xung bc 1 v m bo

du bng khi x=1, nh vy phi s dng BT csi vi 3 s, l iu d hiu. Vy th phi

thm hng s no vo vi 3x . Chc cc bn u thng nht l s 1 ri.

Li gii: p dng BT Csi cho 3 s dng ta c 3x 1 1 3x ; 3y 1 1 3y ; 3z 1 1 3z

Cng tng v 3 BT ta c : 3 3 3x y z 3(x y z) 6

Mt khc 3x y z 3 xyz 3 nn 3(x y z) 6 x y z

Vy bi ton c chng minh.

Cng theo hng ny ta c cc kt qu sau:

Vi x, y, z l cc s dng tha mn xyz = 1 ta c: 3 3 3 2 2 2x y z x y z 2012 2012 2012 2011 2011 2011x y z x y z

(Cc bn hy chng minh cc kt qu ny nh)

V d 4. Cho a, b, c dng tho mn abc=1. Chng minh rng: 3 3 3a b c 3

(1 b)(1 c) (1 c)(1 a) (1 a)(1 b) 4

Phn tch bi ton:

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ATH.

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Ta s thm cho 3a

(1 b)(1 c) nhng hng t g? tr li c cu hi cc

bn ch l du bng xy ra khi a=b=c=1.

Lc th 3a 1 1 1 1 b 1 c

(1 b)(1 c) 4 8 8 8

V vy ta c cch chng minh sau:

Li gii: p dng BT Csi cho 3 s dng ta c 3 3

3a 1 b 1 c a 1 b 1 c 33. . . a

(1 b)(1 c) 8 8 (1 b)(1 c) 8 8 4

Cng hai BT tng t ta c: 3 3 3a b c 3 1 3(a b c)

(1 b)(1 c) (1 c)(1 a) (1 a)(1 b) 4 2 2

(pcm).

iu phi chng minh.

V d 5. Cho a, b, c dng. Chng minh rng: 3 3 3a b c 1 (a b c)

b(c a) c(a b) a(b c) 2

Phn tch bi ton:

* D on du bng xy ra khi a=b=c.

* Khi 3 3a a a c a b

b(c a) a(a a) 2 4 2

. Vit nh vy v dng ca ta l phi

kh c mu s v tri. Nh vy c th thc hin li gii n gin nh sau:

Li gii: p dng BT Csi cho 3 s dng ta c 3 3

3a c a b a c a b 33. . . a

b(c a) 4 2 b(c a) 4 2 2

Cng hai BT tng t ta c iu phi chng minh.

V d 6. Cho a, b, c dng tho mn a b c 3 . Tm gi tr ln nht ca 3 3 3P a 2010b b 2010c c 2010a

Phn tch bi ton:

* D on P t GTLN ti a b c 1 (tt nhin khng phi lc no iu d on

ca ta cng ng)

* Khi 3 3a 2010b 2011 v d on gi tr ln nht ca P bng 33 2011

(th ny m thi trc nghim th ngon qu...)

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ATH.

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* By gi vi mt tham s m>0 no , ta vit

3 32 23 3

1 1 (a 2010b) m ma 2010b . (a 2010b).m.m .3m m

.

Vn by gi l ta chn m bng bao nhiu th ph hp?

D thy du bng xy ra khi a b 1

m 2011a 2010b m

Li gii: p dng BT Csi cho 3 s dng ta c:

3 32 23 3

1 1 (a 2010b) 2011 2011a 2010b . (a 2010b).2011.2011 .32011 2011

Cng hai BT tng t v cng li ta c:

3

23

1 2011(a b c) 6.2011P . 3. 201132011

Du bng xy ra khi a b c 1 . Vy GTLN ca P bng 33 2011 .

V d 7. Cho a,b,c khng m tha mn a+b+c=3. Tm gi tr nh nht ca

3 3 3P a 64b c .

Phn tch bi ton:

y l bi ton m cc vai tr ca cc bin khng nh nhau. Tuy nhin ta vn d

on c P t GTNN khi a=c. Vn l bng bao nhiu th cha th ni ngay c.

bit iu ta xt hai tham s , 0 v vit P nh sau:

3 3 3 3 3 3 3 3 3 3 3P (a ) (64b ) (c ) 4 2

p dng BT Csi ta c:

2 2 2 3 3P 3 a 3.4 b 3 c 4 2 (*)

Du bng xy ra khi

a c

b / 4 2 3 (1)4

a b c 3

n y vn cha c th tm ra , .

rng gi thit cho a+b+c=3 nn t (*) ta s lm cho cc h s ng trc a,b,c bng

nhau. C th l 2 23 12 (2)

T (1) v (2) d tm ra 24 12,17 17

.

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ATH.

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Khi 24 3a c ,b17 17

v 3

2

12P

17

Mi th th l n. Cc bn hy t vit li li gii bi ton ny v nng nng trong nim

vui chin thng nh.

Nhn xt: Nu ta b gi thit a+b+c=3 th ta c th thu c BT sau:

Cho a,b,c khng m. Chng minh rng

3 3 3 3289(a 64b c ) 64(a b c) .

Li gii ca bi ton ny dnh cho bn c (gi l c th chun ho a v bi

ton trn).

V d 8. Cho x,y,z dng tha mn xy yz xz 1 . Tm gi tr nh nht ca

2 2 2P 3(x y ) z .

Phn tch bi ton:

Chc khng phi bnh lun g th cc bn u cng nhn vi ti rng bi ton ny

qu hay, cu trc p mt nhng khng h d dng. Tt nhin ai chng mong rng bi s

cho tm GTNN ca 2 2 2P x y z hoc vui hn l tm GTNN ca

2 2 2P x y z 2013 .... Ta tr li qu trnh phn tch v tm ti li gii cho bi ton:

iu kin rng buc gi thit l i xng vi x,y,z, nhng trong biu thc P ch

i xng vi x,y; vai tr ca z vi x,y l nh nhau. V vy ta d on P t GTNN khi x=y

v 2z x y

2 (vi 0 no ).

Ta a ra nh gi nh sau: 2

2 zx 2 .xz2 2

;

22 zy 2 .yz

2 2

v 2 2. x y 2. .xy2 2

Do : 2 2 2. x y z 2. . xy yz xz 22 2 2

.

Nh th ta chn 0 sao cho 32

(s 3 trong bi), c th thy ngay

mt s 2 .

Du bng xy ra khi 22 2

1x yxy yz xz 1 z 2x 2y 5z xy yz xz 1 2x y z2 5

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ATH.

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Lc GTNN ca P bng 2.

Cc bn hy bt tay t gii bi ton tng t sau nh:

Cho x,y,z dng tha mn x y z 1 . Tm gi tr nh nht ca

3

3 3 zP x y4

.

IV. PHNG PHP S DNG O HM

1. Ni dung phng php

a) Cc kin thc lin quan:

1. Hm f(x) ng bin trn D khi v ch khi f '(x) 0 x D .

2. Hm f(x) nghch bin trn D khi v ch khi f '(x) 0 x D .

3. Cho hm f(x) ng bin trn D, khi vi u,v D ta c: u v f (u) f (v)

4. Cho hm f(x) nghch bin trn D, khi vi u,v D ta c: u v f (u) f (v)

b) Phng php gii: chng minh BT bng PP o hm, ta kho st s bin thin

ca mt hm s f(x) no c lin quan ti cu trc ca BT cn chng minh. T s bin

thin ca hm s f(x) ta suy ra BT cn chng minh. Ch l cc bin b rng buc theo

gi thit ca bi ton.

cc bn c th hiu ngay t tng ca phng php ny ti xin a ra mt bi

ton n gin sau:

Cho a b . Chng minh rng: 2 21 1a b

a 1 b 1

Cc bn c th chng minh bi ton ny bng PP bin i tng ng, tuy nhin

nhn vo c im hai v ca BT ta xt hm s 21f (x) x

x 1

vi x

Ta c 4 2 4 2 2

2 2 2 2 2 22x x 2x 1 2x x x (x 1)f '(x) 1 0

(x 1) (x 1) (x 1)

vi mi x .

Suy ra hm f(x) ng bin trn .

M a b f (a) f (b) , hay 2 21 1a b

a 1 b 1

(pcm).

Nhn xt: Bi ton trn th hin kh r v PP s dng o hm trong bi ton BT.

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ATH.

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2. Cc dng ton c bn

Trong cc thi vo H- C thng xut hin hai dng bi ton sau:

Dng 1: Bt ng thc cn chng minh ch c mt bin.

V d 1: Chng minh rng vi mi x 0 ta c: xe 1 x (1)

Li gii. Xt hm xf (x) e x 1 vi x 0 . Ta c xf '(x) e 1 0 vi mi x 0 .

Suy ra hm f(x) ng bin trn 0; f (x) f (0) 0 . Vy xe 1 x

Du bng xy ra khi v ch khi x=0.

Nhn xt: Bng vic xt o hm hai ln v s dng v d 1 ta c kt qu sau:

2x xe 1 x

2 vi mi x 0 (2)

Hoc ta c kt qu tng qut hn: Cho n nguyn dng. Chng minh rng:

2 nx x xe 1 x ...

2 n! vi mi x 0 (3)

V d 2: Cho 0 x2

. Chng minh rng:

a) sin x x b) sinx3x

x6

(4) b) sinx 2x

(5)

Li gii. a) Xt hm f(x) x sin x vi x 0;

2. Ta c f '(x) 1 cosx 0 ,

x 0;2

Suy ra hm f(x) ng bin trn 0;

2. Do f(x) f(0) 0 (pcm).

b) Xt hm 3x

f(x) sinx x6

vi x 0;

2. Ta c

2x

f '(x) cosx 12

; f ''(x) sin x x 0 (theo phn a)

Do f '(x) f '(0) 0 f(x) f(0) 0 (pcm).

c) Xt hm

2xf(x) sin x , ta c

2

f '(x) cosx .

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ATH.

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n y kch bn khng nh hai phn (a) v (b) na v f(x) c nghim duy nht

x 0;

2. C l n y cc bn s lng tng. Hy tht bnh tnh v nh li rng khi

o hm f(x) c nghim trong on 0;

2 ta phi ngh ti bng bin thin ca n (hy v

ngay ra nhp BBT i).

T BBT ta suy ra ngay

f(x) f 02

. Vy

2xsinx (pcm).

V d 3: Cho x 2 . Chng minh rng 1 5xx 2

(6)

Li gii. Nu bn no cha tho v vic s dng BT Csi gii bi ton ny th PP s

dng hm s l mt v kh lp l hng . Tht n gin khi ta xt hm s

1

f(x) xx

vi x 2 . Ta c ngay 2

2 2

1 x 1f '(x) 1 0

x x, suy ra hm f(x) ng bin

trn 2; . Do 5f(x) f(2)2

(pcm).

Dng 2: Bt ng thc cn chng minh c nhiu bin

V d 1: Cho a,b,c dng. Chng minh rng:

3

3

a b c abc 10a b c 3abc

(7)

Li gii. t 3

a b ctabc

, khi theo BT Csi th t 3 .

Ta cn chng minh 1 10tt 3

vi t 3 .

n y th bi ton c gii hon ton tng t vi vic chng minh BT (6).

Du bng xy ra khi v ch khi a=b=c.

V d 2: Cho a,b,c dng tho mn 2 2 2a b c 1 . Chng minh rng:

2 2 2 2 2 2

a b c 3 3b c c a a b 2

(8)

Gii. T gi thit ta vit li (8) di dng:

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ATH.

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2 2 2

2 2 22 2 2 2 2 2

a b c 3 3 a b c 3 3 (a b c )1 a 1 b 1 c 2 a(1 a ) b(1 b ) c(1 c ) 2

Ch thm rng a,b,c (0;1)

iu ny cho ta ngh ti vic chng minh 2 2 3x(1 x )9

vi x (0;1)

Xt hm s 2f(x) x(1 x ) trn khong (0;1) ta c ngay kt qu trn.

Ch : C mt cch hi theo dng hnh hc kh th v ca bi ton trn nh sau:

Cho hnh hp ch nht c di ba cnh l a,b,c v di ng cho chnh l 1.

Tm gi tr nh nht ca 2 2 2 2 2 2a b cP

b c c a a b

.

V d 3. (H Khi A-2003): Cho x,y,z dng tho mn x y z 1 . Chng minh rng:

2 2 22 2 2

1 1 1x y z 82

x y z (9)

Gii. t v tri ca (9) l P. Theo cc kt qu quen bit ta c:

2 2

2 21 1 1 9P (x y z) (x y z)x y z x y z

Nu gi thit cho x y z 1 th qu n ri, nhng y gi thit li cho x y z 1 .

Gi thit ny thng lm cho cc em HS bi ri. Ti sao khng ngh ti hm s nh. Cc

bn ch cn t 2(x y z) t th 0 t 1 v BT cn chng minh tr thnh 81t 82t

vi 0 t 1 . iu ny th qu n gin ri.

Ch l du bng vn xy ra khi x y z 1 v x=y=z, hay 1x y z3

.

Nh vy sc mnh ca PP hm s tht kinh khng v c th xuyn thng bt k

hng phng ng no cho d gi thit v kt lun ca bi ton mi nhn chong. Ti

xin dn chng thm mt vi bi BT khng i xng na, c l cc bn s tm phc

khu phc iu ti ni thi.

V d 4. Cho a,b,c dng tha mn a+b+c=1. Tm gi tr nh nht ca

3 3 31P a b c4

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Gii. S dng nh gi 3

3 3 (a b)a b4

, suy ra

3 33 3 2(a b) 1 (1 c) 1 1P c c (3c 3c 1)

4 4 4 4 4

Ta thnh cng khi a BT vi 3 bin v ch cn 1 bin c, ch l c (0;1) .

Xt hm 21f (c) (3c 3c 1)4

, ta c 1 1f '(c) (6c 3) 0 c4 2

T 1 1f (c) f2 16

. Du bng xy ra khi

1a b a b4

1 1c c2 2

.

Vy GTNN ca P bng 116

, khi 1 1a b ;c4 2

.

V d 5. Cho a,b,c khng m tha mn iu kin a+b+c=1. Chng minh rng:

7ab bc ca 2abc27

(10)

Gii. W.L.O.G, gi s c l s nh nht trong 3 s a,b,c. Khi 1c 0;3

.

Ta s tp trung vo vic nh gi VT ca (10) theo bin c.

Ta c 2a bab bc ca 2abc c(1 c) ab(1 2c) c(1 c) (1 2c)

2

= 21 cc(1 c) (1 2c)

2

.

Xt hm s 21 cf (c) c(1 c) (1 2c)

2

vi 1c 0;

3

Ta c 1f '(c) c(1 3c) 02

, suy ra hm f(c) ng bin trn 10;3

Do 1 7f (c) f3 27

(pcm). Du bng xy ra khi v ch khi 1a b c3

.

3. Bi tp t luyn

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a) , 0x sinx x b) 2

1 , 02

x xe x x

c) tan 2 , 0;2

sinx x x x

Bi 2. Tm m : 2 4 22 0 m x x m x .

Bi 3. a) Cho x (0;1) . CMR: 2 2 3x(1 x )9

.

b) vi 2 2 2, , 0 : 1a b c a b c ta c: 2 2 2 2 2 23 3

2

a b c

b c c a a b

Bi 4. Cho ab0. CMR: 2

1a

a 1

2

1b

b 1

.

Bi 5. Cho >1. CMR vi mi x0 th : (1+x) 1+x.

Bi 6. a) CMR vi mi x th 5 5 1x (1 x)16

.

b) Tng qut: n nn 1

1a b

2 vi n nguyn dng; a,b>0 tho mn a+b=1

c) Tm GTNN ca f(x)= 10 10sin x cos x .

Bi 7. Chng minh rng vi mi x>0 th: lnx< x .

Bi 8. Cho 0 x2

. CMR:

3 12sin tan 22 2 2x

x x .

Bi 9. Cho 00. CMR: x y x y2 ln x ln y

.

Bi 11. Chng minh rng 2 2x x 1 x x 3 1 2 .

Bi 12. (Khi A-2003). Cho x,y,z>0 tho mn: x+y+z1. Chng minh rng:

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2 2 22 2 2

1 1 1x y z 82

x y z .

Bi 13. (Khi D-2001). CMR vi mi x0 v vi mi >1 ta lun c x 1 x .

T hy CMR: Vi ba s dng a,b,c bt k ta c:

3 3 3

3 3 3

a b c a b c

b c a b c a .

Bi 14. Chng minh rng vi mi x0 ta c: cosx2x

12

.

Bi15. Cho tam giac ABC nhn. Chng minh rng

sin A sin B sinC tan A tan B tanC 2

Bi 16. Chng minh rng mi x 0 th ta c: x x 2e e 2 ln(x 1 x ) .

Bi 17. Cho 2

cosxx 0; . CMR: 8

4 sin x(cosx-sinx).

Bi 18. Tm GTNN ca hm s 29

f(x) 4x sin x, x 0x

Bi 19. Cho 430 . Chng minh rng 31.2 2

Bi 20. Cho x y z 0 . Chng minh rng:

x z y x y zz y x y z x

Bi 21. Cho x,y,z khng m tha mn 2 2 2x y z 1 . Chng minh rng:

6(y z x) 27xyz 10

Bi 22. Cho x,y,z, dng tho mn x+y+z=4 v xyz=2. Tm GTLN v GTNN ca

4 4 4P x y z

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V. MT S THI I HC V LI GII

Bi gii

Vi a, b, c > 0 ta chng minh c: 3 3 2 2

3 3 2 2

3 3 2 2

a b a b abb c b c bcc a c a ca

VT (1) 2 2 2 2 2 2

1 1 1a b ab abc b c bc abc c a ca abc

= 1 1 1ab(a b c) bc(a b c) ca(a b c)

= 1 1 1 1 1a b c ab bc ca abc

.

Bi gii

* Ta c: 21 x 2x 2

x 121 x

. Tng t 2

y 121 y

v 2

z 121 z

Suy ra 2 2 2

x y z 321 x 1 y 1 z

* Ta c: 1 1 1 9 9 9 31 1 1 (1 ) (1 ) (1 ) 3 ( ) 6 2

x y z x y z x y z

Bi gii

Bi 2: Cho x, y, z 0 tho mn x y z 3 . Chng minh rng:

2 2 2

x y z 3 1 1 12 1 x 1 y 1 z1 x 1 y 1 z

Bi 3: Cho a, b, c > 0 sao cho abc = 1. Tm GTNN ca 2 2 2 2 2 2

bc ca abPa b a c b a b c c a c b

Bi 1: Chng minh rng vi a,b,c 0 ta c

3 3 3 3 3 3

1 1 1 1abca b abc b c abc c a abc

(1)

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Bin i 2 2 2

bc ca abPa (b c) b (a c) c (a b)

=2 2 2

1 1 1a b c

b c a c a bbc ac ab

= 2 2 2

1 1 1a b c

1 1 1 1 1 1b c c a a b

n y th mi th tr ln nh nhng hn nhiu ri

Ta ch cn t 1xa

, 1yb

, 1zc

, khi xyz = 1 v 2 2 2x y zP

y z z x x y

p dng BT Csi ta c: 2x y z x

y z 4

, 2y z x y

z x 4

, 2z x y z

x y 4

x y zP x y z2

3x y z 3 3P xyz

2 2 2

. Vy min

3P2

khi x = y = z = 1

Bi gii

Ta c (1) tng ng

3 3 3 3 3 3 3 3 3 2 2 2 3 2 3 2 2 3a b c a b b c c a ab bc ca a b c ab c a bc .

p dng BT csi cho ba s dng ta c:

3 3 3 2a b b 3ab

3 3 3 2b c c 3bc

3 3 3 2c a a 3ca

T suy ra 3 3 3 2 2 23(a b c ) 3(ab bc ca ) 3 3 3 2 2 2a b c ab bc ca (2)

Tng t ta c 3 3 3 3 3 3 3 2 2 3 3 2a b b c c a ab c a bc a b c (3)

T (2) v (3) suy ra: 3 3 3 3 3 3 3 3 3 2 2 2 3 2 3 2 2 3a b c a b b c c a ab bc ca a b c ab c a bc

Bi 4: Cho a, b, c > 0. Chng minh rng:

3 3 3 2 2 21 a 1 b 1 c 1 ab 1 bc 1 ca (1)

Bi 5: Cho a, b, c > 0: 1 1 1 3a b c . Chng minh rng 1 a 1 b 1 c 8 (1)

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Bi gii

T 1 1 1 3a b c suy ra 3 2 2 23abc ab bc ca 3 a b c abc 1

Bin i (1) 1 1 1 81 1 1 0a b c abc

(2)

Ta c VT(2) = 1 1 1 1 1 1 71a b c ab bc ca abc

= 1 1 1 74ab bc ca abc

3 2 2 2

1 74 3abca b c

3 74

abc abc (do abc 1 )

44 0abc

(do abc 1 ) (pcm).

Bi gii

Ta c: 3 2 3 2 3 2 3 2 3 2 3 2

2 y 2 y2 x 2 z 2 x 2 zx y y z z x 2 x y 2 y z 2 z x

=

= 1 1 1 1 1 1x y y z z x

2 2 2 2 2 2

1 1 1 1 1 1 1 1 12 2 2x y y z z x

= 2 2 2

1 1 1x y z

Bi gii

Theo BT Csi ta c 1 1x y 4x y

, x,y 0 v

2

1 4xy (x y)

, x,y 0

Do 2 2 2 2

1 1 1 1 1ab 2ab 2aba b a b

2 2 2

4 4 2 4 62(a b) 2ab a b

.

Bi 6: Cho x, y, z > 0. Chng minh rng:

3 2 3 2 3 2 2 2 2

2 y2 x 2 z 1 1 1x y y z z x x y z

Bi 7: Cho a, b > 0 tho mn a + b = 1. Chng minh rng

2 2

1 1 6ab a b

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Bi gii

Gp nhng bi ton dng ny ta thng s dng BT sau (BT Mincpxki)

22 2 2 2 2( ) x a y b x y a b (1) vi mi a,b,x,y

22 2 2 2 2 2 2( ) x a y b z c x y z a b c (2) vi mi a,b,c,x,y,z

Chng minh BT (1) tht n gin, c th a ra 2 cch nh sau:

Cch 1: Bin i tng ng (1) bng cch bnh phng hai v v a v BT ng 2( ) 0 bx ay

Cch 2: S dng BT vc t

Vi mi vc t ,

u v ta c

u v u v (*)

(v 22 2 22 2

2 . 2 .

u v u v u v u v u v u v )

t ( ; )

u x a , ( ; )

v y b . p dng (*) suy ra ngay BT (1)

p dng hai ln BT (1) ta suy ra BT (2), v cng suy ra cc BT tng qut cho b 2n

s

Tr li bi ton: p dng BT (2) ta c:

2

2 2 2 22 2 2

1 1 1 1 1 1P x y z (x y z)x y zx y z

22

3 313 xyz 3

xyz

99tt

Vi 23 ( )t xyz2

x y z 10 t3 9

.

n y th tng dng hm s l kh r rng ri

Xt hm s 9f (t ) 9tt

vi 1t 0,9

Bi 8: (Khi A- 2003). Cho x, y, z l ba s dng tho mn x y z 1 . Chng minh rng:

2 2 22 2 2

1 1 1x y z 82x y z

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Ta c /2

9 1f (t ) 9 0, t 0,9t

f (t ) gim trn 10,

9

1f (t ) f 829

.

Vy P f(t ) 82 .

Bi gii

T gi thit v BT cn chng minh cho php ta ngh ngay ti BT c bn sau:

* Vi a,b dng ta c 1 1 4a b a b

, hay 1 1 1 1

a b 4 a b

.

Du = xy ra khi v ch khi a = b

* p dng kho lo BT trn ta c

1 1 1 1 1 1 1 1 1 1 1 1 12x y z 4 2x y z 4 2x 4 y z 8 x 2y 2z

(1)

1 1 1 1 1 1 1 1 1 1 1 1 1x 2y z 4 2y x z 4 2y 4 x z 8 y 2x 2z

(2)

1 1 1 1 1 1 1 1 1 1 1 1 1x y 2z 4 2z y x 4 2z 4 y x 8 z 2y 2x

(3)

Cng theo tng v cc BT (1), (2) v (3) ta c

1 1 1 1 1 1 1 12x y z x 2y z x y 2z 4 x y z

Bi gii

* Mu cht ca bi ton l phi bit phn tch cc c s v tri

212 15. 35 4

; 212 20. 45 3

; 215 20. 54 3

* n y p dng BT Csi cho hai s dng ta c

Bi 9: (Khi A 2005). Cho x, y, z l cc s thc dng tho mn iu kin 1 1 1 4x y z .

Chng minh rng: 1 1 1 12x y z x 2y z x y 2z

Bi 10: (Khi B 2005). Chng minh rng vi mi x R , ta c x x x

x x x12 15 20 3 4 55 4 3

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12 15 12 152 . 2.35 4 5 4

x x x xx (1)

12 20 12 20. 2.45 3 5 3

x x x xx (2)

15 20 15 202 . 2.54 3 4 3

x x x xx (3)

Cng cc BT (1), (2), (3) ta c x x x

x x x12 15 20 3 4 55 4 3

(pcm).

Du = xy ra khi v ch khi x=0.

Bi gii

* p dng BT C si cho ba s dng ta c

3 3 3 331 x y 3 1.x .y 3xy 3 31 x y 3

xy xy

(1)

* Cng hai BT tng t ta c

3 3 3 3 3 31 x y 1 y z 1 z x 3 3 3xy yz zx xy yz zx

* Mt khc: 33 3 3 3 3 33 3 3xy yz zx xy yz zx

3 3 3 3 3 31 x y 1 y z 1 z x 3 3

xy yz zx

Bi gii

Bi 11: (Khi D 2005). Cho cc s thc dng x, y, z tho mn xyz = 1.

Chng minh rng: 3 3 3 3 3 31 x y 1 y z 1 z x 3 3

xy yz zx

Bi 12: (D b D 2005). Cho x, y, z l ba s thc tho mn x y z 0 . Chng minh rng

x y z3 4 3 4 3 4 6

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* Bi ton ny c cht gn gi vi bi 11, nu tinh ta thy rng

x y z x y z 04 .4 .4 4 4 1

* Ta c : 4x x x3 4 1 1 1 4 4 4 4 8x x x3 4 2 4 2 4

* Cng hai BT tng t ta c

38 8 8x y z x y z x y z83 4 3 4 3 4 2 4 4 4 6 4 .4 .4 6

Du = xy ra khi v ch khi x=y=z=0.

Bi gii

* Ta c 3

43

x x x x1 x 1 43 3 3 3

Tng t : 3

43 3

y y y y y1 1 4x 3x 3x 3x 3 .x

V

3

4 3

9 3 3 3 31 1 4y y y y y

26

43

9 31 16yy

Do 2

3 3 6

43 3 3 3

y 9 x y 3(1 x) 1 1 256 256x 3 3 x yy

Bi gii

Cch 1

Ta c

3 3

a 3b 1 1 1a 3b (a 3b)1.1 (a 3b 2)3 3

Bi 13: (D b A 2005). Chng minh rng vi mi x,y 0 , ta c

2y 9(1 x) 1 1 256x y

Bi 14: (D b B 2005). Cho a, b, c l ba s dng tho mn 3a b c4

.

Chng minh rng

3 3 3a 3b b 3c c 3a 3

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3 3

b 3c 1 1 1b 3c (b 3c)1.1 (b 3c 2)3 3

3 3

c 3a 1 1 1c 3a (c 3a)1.1 (c 3a 2)3 3

3 3 3 1 1 3a 3b b 3c c 3a 4(a b c) 6 4 6 3

3 3 4

Du = xy ra

3a b c 1a b c44a 3b b 3c c 3a

Cch 2

t

3x a 3b 3x a 3b

3y b 3c 3y b 3c

3z c 3a 3z c 3a

3 3 33x y z 4(a b c) 4 34

. Ta cn chng minh x y z 3

BT ny l qu n gin ri! ( cc em t chng minh xem nh).

Bi gii

T 0 x 1 suy ra 2x x

p dng BT C si cho ba s dng ta c

2 21 1 1y x yx 2 yx x y4 4 4

1x y y x4

(pcm).

Du bng xy ra

2

2

0 x y 1

x x1yx4

x 11y4

Bi 15: (D b 2 B 2005). Chng minh rng nu 0 x y 1 ta c

1x y y x4

.

Bi 16: (D b 2 D 2005). Cho x, y, z dng tho mn x.y.z 1 . Chng minh rng

2 2 2x y z 31 y 1 z 1 x 2

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Bi gii

Ta c

2 2x 1 y x 1 y2 x1 y 4 1 y 4

Cng hai BT tng t ta c

2 2 2x 1 y y 1 z z 1 x x y z

1 y 4 1 z 4 1 x 4

2 2 23x y z 3(x y z) 3 3 3 9 3 33. xyz

1 y 1 z 1 x 4 4 4 4 4 4 2 (pcm).

Bi gii

t xa 3 , yb 3 , zc 3 (a, b, c 0)

Ta c : x y z3 3 3 1 1 1 1 1a b c

ab bc ca abc (*)

BT cn chng minh tr thnh:

2 2 2a b c a b ca bc b ca c ab 4

3 3 3

2 2 2

a b c a b c4a abc b abc c abc

(2)

Ta c: 2 2a abc a ab bc ca (a b)(a c) (do ab bc ca abc )

Tng t 2b abc (b a)(b c) v 2c abc (c a)(c b)

Do (2)

3 3 3a b c a b c(a b)(a c) (b a)(b c) (c a)(c b) 4

(3)

p dng BT C si cho ba s dng ta c

3 33

a a b a c a 33. a(a b)(a c) 8 8 64 4

Tng t

3 33

b b c b a b 33. b(b c)(b a) 8 8 64 4

3 33

c c a c b c 33. c(c a)(c b) 8 8 64 4

Bi 17: (D b A 2006). Cho x,y,z dng tho mn x y z3 3 3 1 . Chng minh rng:

x y z x y z

x y z y z x z x y

9 9 9 3 3 343 3 3 3 3 3

(1)

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39

Cng tng v ca ba BT ta c:

3 3 3a b c 1 3(a b c)(a b c)(a b)(a c) (b a)(b c) (c a)(c b) 2 4

3 3 3a b c a b c(a b)(a c) (b a)(b c) (c a)(c b) 4

, suy ra (3) c chng minh

Bi ton c chng minh hon ton.

Bi gii

* Gi thit xyz=1 thng c rt nhiu k thut x l tinh t, vic dng n th no

cn ph thuc vo tng bi ton. Trong bi ton ny c mt lin h nh gia t v mu, c

th ta thc hin li gii nh sau:

Ta c 2 2 2 4x (y z) x y x z 2 x yz 2x x (do xyz=1)

Tng t 2y (z x) 2y y v 2z (x y) 2z z

2y y2x x 2z zPy y 2z z z z 2x x x x 2y y

* n y t v mu qu gn gi ri. bi ton n gin hn ta thc hin

php i bin khng lm thay i gi thit sau: a x x;b y y;c z z , khi abc=1.

Lc 2a 2b 2cPb 2c c 2a a 2b

* Cc bn hy chng minh BT sau nh (rt quan trng y, qu nhiu p dng hay)

a b c 3mb nc mc na ma nb m n

(vi mi a,b,c dng; m,n l hng s dng cho trc)

* p dng BT ny ta c ngay P 2 (cng chng cn n abc=1 na)

Du bng xy ra khi x=y=z=1. Vy GTNN ca P bng 2.

Bi 18: (Khi A 2007). Cho x, y, z l cc s thc dng tho mn xyz=1.

Tm gi tr nh nht ca 2 2 2( ) ( ) ( )

2 2 2

x y z y z x z x yPy y z z z z x x x x y y

Bi 19: (Khi B 2007). Cho x, y, z l cc s dng thay i. Tm gi tr nh nht ca

biu thc 1 1 12 2 2

x y zP x y zyz zx xy

.

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Bi gii

Ta c 2 2 2 2 2 2x y z x y zP

2 2 2 xyz

M 2 2 2 2 2 2

2 2 2 x y y z z xx y z xy yz zx2 2 2

2 2 2x y z xy yz zx 1 1 1

xyz xyz x y z

2 2 2 2 2 2x y z 1 1 1 x 1 y 1 z 1P2 2 2 x y z 2 x 2 y 2 z

tng dng hm s li n rt t nhin

Xt hm s 2t 1f (t )

2 t vi t >0 / 2

1( ) 0 1f t t tt

Lp bng bin thin ta c 3f (t )2

, t 0 . Suy ra 9P2

.

Du bng xy ra khi v ch khi x y z 1 . Vy GTNN ca P bng 92

.

Bi gii

Bt ng thc cho tng ng vi

a b

b aa b ln 1 4 ln 1 41 4 1 4a b

Xt hm s xln 1 4

f (x)x

vi x>0. Ta c: x x x x

2 x

4 ln 4 (1 4 ) ln 1 4f '(x) 0

x (1 4 )

Suy ra f(x) nghch bin trn (0; ) .

Bi 20: (Khi D- 2007). Cho a b 0. Chng minh rng b a

a ba b

1 12 22 2

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Do f(x) nghch bin trn (0; ) v a b 0 nn f (a) f (b) v ta c iu phi chng

minh.

Bi gii

Ta c 2 2

2 2 2 22(x 6xy) 2(x 6xy)P

1 2xy 2y x y 2xy 2y

* Nu y=0 th 2x 1 P 2 .

* Nu y 0 th

2

2

2 2

x x2 6y y 2t 12tP

t 2t 3x x2 3y y

n y c th dng iu kin c nghim ca phng trnh bc hai hoc dng o

hm bng cch kho st hm s 2

22t 12tf (t)t 2t 3

, vi t .

Ta tm ra max3 1P 3 x , y10 10

hoc 3 1x , y10 10

min3 2P 6 x , y13 13

hoc 3 2x , y13 13

Bi gii

Ta s p dng BT sau: 2(a b) 4ab

Ta c 22 2

(x y)(1 xy) (x y)(1 xy) 1P(1 x) (1 y) 4(x y) (1 xy)

. Suy ra 1 1P

4 4

Bi 21: (Khi B- 2008). Cho hai s thc x,y thay i tho mn h thc 2 2x y 1 .

Tm GTLN, GTNN ca 2

22(x 6xy)P

1 2xy 2y

.

Bi 22: Khi D- 2008). Cho hai s thc khng m x,y thay i . Tm GTLN, GTNN ca

2 2(x y)(1 xy)P(1 x) (1 y)

.

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Khi x=0, y=1 th 1P4

. Khi x=1, y=0 th 1P4

.

Vy GTNN ca P bng 14

, GTLN ca P bng 14

.

Bi gii

Ta c 3 3 34(a b ) (a b) a,b 0 nn 3 33 4(a b ) a b

p dng BT ny ta c

2 2 2 2 2 2

x y z x y zP (x y) (y z) (x z) 2 2 x y zy z x y z x

n y ta dng BT csi cho 6 s ta c

62 2 2

x y zP 2.6 x.y.z. . . 12y z x

ng thc xy ra khi x=y=z=1. Vy GTNN ca P bng 12.

Bi gii

Ta c 3 33x 4(y z ) x y z , suy ra 3 33

y z y zx y zx 4(y z )

Cng hai BT tng t ta c pcm.

Bi 23 (D b A- 2007): Cho x,y,z dng. Tm gi tr nh nht ca biu thc

3 3 3 3 3 33 3 32 2 2

x y zP 4(x y ) 4(y z ) 4(z x ) 2y z x

Bi 24: Cho x,y,z dng. Chng minh rng

3 3 3 3 3 33 3 3

y z x z x yP 2x 4(y z ) y 4(x z ) z 4(x y )

Bi 25: Cho a,b,c dng tho mn iu kin 3 1a b c 12 abc

. Tm gi tr nh nht ca

3 3 3

2 2 2

a b b c c aP1 ab 1 bc 1 ca

.

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Bi gii

Bin i 2 2 2 2a b c (a b c)P 1 1 1 a b cb c a (a b c)

ab bc ca abc

= 2(a b c)

1(a b c) 1abc

Theo gi thit ta c 1 21 (a b c)abc 3

, suy ra 2(a b c) 3P 2 2(a b c). (a b c)

3

Du bng xy ra khi

4 3 3 2a b c

2a a 1 0 (a 1)(2a a a 1) 0 a 11 21 (a b c)abc 3

Vy GTNN ca P bng 32

khi a b c 1 .

Bi gii

Ci hay v kh ca bi ton l gi thit (x + y)3 + 4xy 2 (1)

Biu thc A ch cha cc s hng x2 + y2 v x2y2. iu ny cho ta ngh ti vic nh gi

tng x+y v qua nh gi c tng bnh phng x2 + y2 .

C th ta bin i gi thit (1) nh sau:

Ta c 3 2 3 3 2(x y) (x y) (x y) 4xy 2 (x y) (x y) 2 0 x y 1 .

Suy ra 2

2 2 (x y) 1x y2 2

. Ta bin i A nh sau:

2 2 22 2 2 2 2 2 2 2 2 2 2 2

2 2 2 2 2

(x y )A 3 (x y ) x y 2(x y ) 1 3 (x y ) 2(x y ) 14

9 (x y ) 2(x y ) 14

t t = x2 + y2 th t 12

. Xt hm 29 1f (t) t 2t 1, t4 2

ta tm ra 9min f (t)16

khi 1t2

.

Bi 26 (H KB-2009). Cho cc s thc x, y thay i v tho mn (x + y)3 + 4xy 2. Tm gi tr nh nht ca biu thc :

A = 3(x4 + y4 + x2y2) 2(x2 + y2) + 1.

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Vy min9 1A khi x y

16 2 .

Bi gii

Trc ht ta bin i P theo tch xy. Ta c

P= (4x2 + 3y)(4y2 + 3x) + 25xy = 16x2y2 + 12(x3 + y3) + 34xy

= 16x2y2 + 12[(x + y)3 3xy(x + y)] + 34xy = 16x2y2 + 12(1 3xy) + 34xy = 16x2y2 2xy + 12

t t = x.y, v x, y 0 v x + y = 1 nn 104

t . Khi S = 16t2 2t + 12

Xt hm f(t)=16t2 2t + 12 vi 1t 0;4

, ta tm ra min, max ca hm f(t)

p s ca bi ton l: Max P = 252

khi x = y = 12

Min P = 19116

khi 2 3x

42 3y

4

hay 2 3x

42 3y

4

Bi gii

T gi thit ta c (x+y)(x+z)=4yz.

Do BT tng ng vi 3 3 3(x y) (x z) 12yz(y z) 5(y z) (1)

Mt khc

22 2 2

3 3

3(y z)x (y z)x 3yz 4x 4(y z)x 3(y z) 04

(2x y z) 2x 3(y z) 0 2x y z 8x (y z)

p dng BT c bn: 3 3 3(x y) 4(x y ) ta c 3 3 3 3 3 3 3VT(1) 4(x y ) 4(x z ) 12yz(y z) 8x 4(y z) 5(y z) (pcm)

Du ng thc xy ra khi v ch khi x=y=z.

Bi 27 (H KD-2009). Cho cc s thc khng m x, y thay i v tha mn x + y = 1. Tm GTLN, GTNN ca P = (4x2 + 3y)(4y2 + 3x) + 25xy.

Bi 28 (H KA-2009): Chng minh rng vi mi s thc dng x, y, z tha mn x(x + y + z) = 3yz, ta c

3 3 3x y x z 3 x y x z y z 5 y z .

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Bi gii Ta bin i P theo mt hm s ca bin t=ab+bc+ca. Ta c BT c bn 2 2 2 2 2 2 2 23( ) ( )a b b c c a ab bc ca t Suy ra 2 3 2 1 2P t t t .

Mt khc 2( ) 3( )a b c ab bc ca suy ra 103

t .

Xt hm 2( ) 3 2 1 2f t t t t vi 103

t .

Ta c 2 (2 3) 1 2 2'( ) 2 3

1 2 1 2t tf t t

t t

; '( ) 0 (2 3) 1 2 2f t t t (1)

t 1 2u t th 1 13

u , pt (1) tr thnh: 3 4 2 0u u (2)

D chng minh c 3 4 2 0u u vi mi 1 13

u , do (2) v nghim, tc l (1) v

nghim. T suy ra ngay '( ) 0f t vi mi 103

t .

Do ( ) (0) 2f t f . Vy minP=2 khi chng hn a=b=0 v c=1.

Bi gii Tp xc nh ca hm s 2;5D .

Ta c 2 2

2 4 2 3'( )2 4 21 2 3 10

x xf xx x x x

, vi ( 2;5)x .

2 2'( ) 0 ( 2 4) 3 10 ( 2 3) 4 21f x x x x x x x Suy ra 2 2 2 2( 2 4) ( 3 10) ( 2 3) ( 4 21)x x x x x x (1)

Khai trin ta c 2815151 104 29 013

xx x

x

.

Th li ch c 13

x l nghim ca (1).

Lp bng bin thin ca hm s f(x) suy ra 1 200 98minf(x)3 3

f

.

Cc thi nm 2011, 2012, 2013 cc bn c th tham kho -p n ca b gio dc ti

http://www.thituyensinh.vn

Bi 29 (H KB-2010). Cho a,b,c khng m tha mn a+b+c=1. Tm GTNN ca: 2 2 2 2 2 2 2 2 23( ) 3( ) 2P a b b c c a ab bc ca a b c

Bi 30 (H KD-2010). Tm gi tr nh nht ca hm s 2 2( ) 4 21 3 10f x x x x x

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Trn y ch l mt cch tip cn vi phng php v cch hc v BT m ti mun

gii thiu vi cc bn, c bit l cc em hc sinh lp 12 chun b thi H- C. Cn mt s

chuyn su hn ti s tip tc gi ti cc bn sau. Hy vng cht t kin thc ny s gip

cc em hc sinh 12 t c kt qu cao nht trong k thi H sp ti. Chc cc em thnh

cng. Mi thc mc hay chia s hy lin h vi ti theo a ch

Email: [email protected]; ST: 0912960417

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MT S BI TP CHO CC BN LUYN TP

1. Cho a,b,c dng tha mn a+b+c=1. Tm GTNN ca 3 3 31P a b c4

2. Cho a,b,c khng m tha mn: a+b+c=1. Chng minh rng:

2ab bc ca 2abc27

3. Cho x y z 0 . Chng minh rng: x z y x y zz y x y z x

4. Cho a>b>c>0. Chng minh rng: 3 2 3 2 3 2 2 3 2 3 2 3a b b c c a a b b c c a

5. Cho x,y,z dng. Chng minh rng: 4 4 4 2 2 2 2 2 2(x y z ) xyz(x y z) xy(x y ) yz(y z ) xz(x z )

6. Cho x,y,z khng m tha mn 2 2 2x y z 1 . Chng minh rng:

6(y z x) 27xyz 10

7. Cho a,b,c dng v k>0 cho trc. Tm GTNN ca

a b b c a c kabcPc a b (a b)(b c)(a c)

8. Cho x,y,z dng tha mn 3(x y z) 32xyz . Tm GTLN, GTNN ca 4 4 4

4

x y zP(x y z)

.

9. Cho dy 4 3 2nu n 20n 0,5n 13n n 1 . Tm s hng ln nht ca dy v tnh s

hng y.

10. Cho x,y l cc s t nhin. Tm GTLN ca hm s

1 1f (x, y)x y 1 (x 1)(y 1)

11. Cho a, b, c l ba s thc dng. Chng minh bt ng thc: 2a b c

b c c a a b

.

12. Cho ba s dng a, b, c tha a + b + c 2. Chng minh :

2 2 21 1 1 1

a bc b ca c ab abc

13. Cho ba s dng a, b, c tha: 1 1 1 2a b c . Chng minh bt ng thc:

1 1 1 13 3 3a b b c c a

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14. Xt cc s thc dng x, y, z tha mn iu kin x + y + z = 1.

Tm gi tr nh nht ca biu thc: 2 2 2x (y z) y (z x) z (x y)P

yz zx xz

15. Tm gi tr ln nht v gi tr nh nht ca hm s 4 2 2

2 2

2 1 1 11 1 2x x xy

x x

16. Tm gi tr ln nht v gi tr nh nht ca hm s 21 1y x x

17. Cho x, y, z l cc s thc thuc on [0; 1] v tha mn x + y + z = 1. Tm gi tr ln nht ca

biu thc P = x2 + y2 + z2.

18. Cho x, y, z l cc s thc tha mn x2 + y2 + z2 = 1. Tm gi tr ln nht, nh nht ca biu thc:

P = x3 + y3 + z3 3xyz.

19. Cho hai s thc ,x y thay i v tho mn iu kin 2 2 11x y . Hy tm gi tr ln nht v gi

tr nh nht ca biu thc 2P x xy .

20. Cho hai s thc x, y thay i v tho mn 2 2 8x y . Hy tm gi tr ln nht v gi tr nh

nht ca biu thc 3 3 3P x y xy .

21. Cho x, y l hai s thc thay i v tho mn iu kin: 2 2x y x y . Hy tm gi tr ln nht

v gi tr nh nht ca biu thc 3 3A x y .

22. Cho ,x y l hai s thc thay i v tho mn iu kin: 2 2 2 2 2x y x y . Hy tm gi tr

ln nht v gi tr nh nht ca biu thc 2 2A x y .

23. Cho ba s thc dng x, y, z. Tm gi tr nh nht ca biu thc:

x y z y z z x x yPy z z x x y x y z

.

24. Cho cc s thc x, y, z thuc khong (0; 1) v tha mn: xy + yz + zx = 1. Tm gi tr nh nht

ca biu thc: 2 2 2(1 )(1 )(1 )xyz

x y z

T

25. Cho cc s thc dng x, y tha mn: 1 1 2x y . Tm gi tr nh nht ca biu thc:

3 3 2 2( 6) ( 6)A x y x y y x

26. Cho 3 s thc dng x, y, z tha mn: x + y + z 1. Tm gi tr ln nht ca biu thc:

P = xy + yz + zx 2xyz

27. Cho cc s thc dng x, y, z. Chng minh: 3 3 3 3 3 3

2 2 2

5 5 53 3 3

x y y z z x x y zxy x yz y zx z

.

28. Cho ba s thc dng a, b, c tha mn: a.b.c = 1. Tm gi tr ln nht ca biu thc:

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ab bc caTa b ab b c bc c a ca

29. Cho hai s thc x, y khc khng, tha mn: 4 2x yy x y x . Tm gi tr ln nht v gi tr nh

nht ca biu thc: 2 2 3T x y x y

30. Cho ba s thc dng x, y, z tha mn 3x y z . Tm gi tr ln nht ca biu thc:

2 2 21 1 1T x x y y z z

31. Cho cc s thc khng m zyx ,, tho mn 3222 zyx . Tm gi tr ln nht ca biu

thc

zyxzxyzxyA

5 .

32. Cho x,y R v x, y > 1. Tm gi tr nh nht ca 3 3 2 2

( 1)( 1)x y x y

Px y

33. Cho x, y, z l cc s thc dng ln hn 1 v tho mn iu kin xy + yz + zx 2xyz

Tm gi tr ln nht ca biu thc A = (x - 1)(y - 1)(z - 1).

34. Chng minh: 1 1 1 12x y zx y z

vi mi s thc x , y , z thuc on 1;3 .

35. Cho ba s thc dng a, b, c tha mn 3 3 3

2 2 2 2 2 2 1a b c

a ab b b bc c c ca a

Tm gi tr ln nht ca biu thc S = a + b + c

36. Cho x, y l hai s dng thay i tho iu kin 4(x + y) 5 = 0. Tm gi tr nh nht ca biu

thc S = 4 14

x y

.

37. Chng minh rng: 2

cos 2 , .2

xxe x x x R

38. Cho a, b, c l 3 cnh ca tam gic c chu vi bng 3. Tm gi tr nh nht ca biu thc:

3 3 3( ) ( ) ( )

3 3 3

a b c b c a c a bP

c a b.

39. Cho ba s thc a, b, c tho mn abc= 22 . Chng minh rng:

4224466

2244

66

2244

66

acacac

cbcbcb

bababa

40. Cho a,b,c l ba s thc dng ty tho mn a+b+c = 2 .Tm gi tr ln nht ca biu thc :

2 2 2ab bc caPc ab a bc b ca

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41. Cho x, y, z l cc s dng tha mn x y z xyz . Tm gi tr nh nht ca biu thc:

2 2 2

1 2 5Px y z

42. Tm gi tr ln nht, gi tr nh nht ca hm s

2 4

22

3 8cos x 12cos xy1 2cos x

43. Tm GTLN ca 4 4

4 4

2( )1

x xf xx x

44. Cho cc s thc dng a, b, c. Tm gi tr ln nht ca biu thc

2 2 2

1 21 1 1

Pa b ca b c

45. Cho a,b,c dng tho mn 2 2 2 1a b c . CMR: 3 3 3

2 2 2

1 1 1 2( )3 a b ca b c abc

.

46. Cho x,y,z dng tho mn 3x y z . Tm GTNN ca biu thc

3 3 3

4 4 4

2 1 8 4 2 2 1 8 4 2 2 1 8 4 2

x y zPy y x z z y x x z

.

47. Cho a,b,c dng tho mn 2 2 2 2 2 2 2 2 23a b b c c a a b c . Tm GTLN ca biu thc

2 22009 2011 2007( ) 2009 2011bc a c a b c bc a bPa bc

.

48. Cho x,y,z khng m. Tm GTLN ca 1 1

1 (1 )(1 )(1 )P

x y z x y z

49. Cho , ,x y z l cc s thc dng, tho mn 3x y z . Tm gi tr nh nht ca biu thc 3 3 3

(2 ) (2 ) (2 )x y zP

y z x z x y x y z

.

50. Cho ba s dng a,b,c c tng bng 3 . Chng minh rng

3 3 3 3 3 3

2 2 2 2 2 2

3

3 4 11 3 4 11 3 4 11 5

a b b c c aa ab b b bc c c ca a

51. Cho a, b, c l ba s dng tha mn 12

a b c . Tm gi tr ln nht ca biu thc:

a b b c b c a c a c a bP

a b b c a c b c a c a b a c a b b c

52. Cho x > 0, y > 0 tha mn 2 2 3x y xy x y xy . Tm gi tr nh nht ca biu thc

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51

22 2 (1 2 ) 3

2xyP x yxy

.

53. Cho x, y, z 0 tho mn 0x y z . Tm gi tr nh nht ca

3 3 3

316x y zP

x y z

54. Chng minh 2 2 2 1

2a b c ab bc ca a b c

a b b c c a

vi mi s dng ; ;a b c .

55. Cho a,b, c dng v a2+b2+c2=3. Tm gi tr nh nht ca biu thc 3 3 3

2 2 23 3 3a b cP

b c a

56. Cho x, y, z l cc s thc dng ln hn 1 v tho mn iu kin 1 1 1 2x y z

Tm gi tr ln nht ca biu thc A = (x - 1)(y - 1)(z - 1).

57. Cho cba ,, > 0. Chng minh rng:

3333 )111)((3cba

cbaac

cb

ba

.

58. Tm GTLN ca hm s 2 2 22 4 2( ) sin os 2sin

1 4 3(1 4 ) 3(1 4 )x x xf x cx x x

.

59. Cho x, y, z, t, s l cc s thc thay i tha mn 0 x y z t s v 1x y z t s .

Tm gi tr ln nht ca biu thc s ( )( y )zt ts ztP x yy s xz .

60. Cho , , 0x y z v 34

x y z . T m gi tr ln nht ca biu thc:

33 33 3 3E x y y z z x

61. Cho ba s thc , ,a b c thuc khong 0;2 v tho mn: 4ab bc ca abc

Chng minh rng: 2 2 24 4 4 3 3a b c

62. Cho 2 s thc x, y tha mn : 2 2 1 1x y x y .

Tm GTLN, GTNN ca F = 2(1 )( ) ( )2 2

xy x yx yx y y xx y

.

63. Cho 2 s thc x, y tha mn : 2 24 1x y . Tm GTLN, GTNN ca 2( 1) 4 ( 1)

2( 1)x y x yP

x y

.

64. Ty theo tham s m, tm gi tr nh nht ca biu thc P= 2 2( 2 1) (2 3)x y x my . Vi

,x y

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65. Cho 2 s thc khng m x,y tha mn x2 + y2 + xy = 3. Tm gi tr ln nht , nh nht ca biu

thc P = x3 + y3 ( x2 + y2).

66. Cho a, b, c, d l cc s thc khng m, khc nhau tng i mt, tha mn iu kin

ab + bc + ca = 4. Chng minh rng 1)(

1)(

1)(

1222

accbba

.

67. Vi a, b, c l nhng s thc dng, chng minh rng

cbaabbac

caacb

bccba

1

535353 222222.

68. Vi x, y, z l nhng s thc dng tha mn x + y + z = xyz, chng minh rng

49

11

11

12

222

zyx.

69. Cho cc s thc khng m a, b.Chng minh rng:

( a2 + b + 43 ) ( b2 + a +

43 ) ( 2a +

21 ) ( 2b +

21 ).

70. Cho cc s thc dng x,y,z . Chng minh rng: 0222

xzzxz

zyyzy

yxxyx

71. Cho cc s dng x, y, z tha mn xyz + x + y z = 0. Tm gi tr ln nht ca biu thc:

P = 1

21

31

2222

zyx

72. Xt cc tam thc bc hai f(x) = a x2 + bx + c, trong a < b v f(x) 0 vi mi x R.

Hy tm gi tr nh nht ca biu thc M = ab

cba .

73. Cho ba s dng a, b, c tha mn iu kin : ab + bc + ca = 2abc.

Chng minh rng: 222 )12(1

)12(1

)12(1

ccbbaa 21

74. Bit rng 3

21 yxyx . Tm gi tr nh nht ca biu thc M = 1

12

1

yx

.

75. Cc s dng a,b,c tha mn abc = 1.Chng minh rng: )(23111 222 cbacba .

76. Cho ba s khng m a, b, c tha mn: a + b + c = 1. Chng minh rng: 2a + b2 + c2 + abc12 1

77. Cho x, y l cc s dng tha mn 3111 yxxy

. Tm gi tr ln nht ca biu thc:

M = 22111

)1(3

)1(3

yxyxxyx

yxy

.

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