Cc ng dng ca bt ng thc Csi.

Mt S NG DNG CA BT NG THC C SING DNG 1: Chng minh bt ng thcBi ton s 1. Cho a, b, c > 0. Chng minh rng

*Phn tch:

V tri cha a, b, c > 0 v cc nghch o ca chng. V vy ta ngh n vic dng bt ng thc Csi.Li gii:

Cch 1: p dng bt ng thc Csi cho cc b s a, b, c v ta c:

Nhn tng v ca hai bt ng thc trn ta c:

(pcm).Cch 2:

Du "=" xy ra

Bi ton s 1.1 Chng minh cc bt ng thc:

a. (a, b, c > 0)

b.

Bi ton s 1.2 Chng minh rng:

a.

p dng BT Csi cho 2 s x2 +1 v 1.

b. > 1.p dng BT Csi cho 2 s x - 1 v 9.

c.

p dng BT Csi ta c

Nhn tng v ca 2 BT trn ta suy c pcm.Bi ton s 1.3 Chng minh rng:

a.

EMBED Equation.DSMT4

b.

p dng BT Csi cho 6 s .Bi ton s 1.4

a. n s dng a1, a2, ..., an. Chng minh rng:

b.Nu a1, a2,...., an dng v a1a2...an = 1 th a1+ a2 +...+ an

p dng BT Csi cho n s dng trn)Bi ton s 2. Chng minh bt ng Netbit

> 0.Gii.

t x= b + c, y = a + c, z = a +bKhi x, y, z > 0 v

Ta c:

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

Cch khc:

Khai thc bi ton:

Bng cch tng t, ta c th chng minh c cc bt ng thc sau: vi a, b, c dng ta c:

Bi ton s 2.2. Cho x, y > 0. Chng minh rng

EMBED Equation.3 (1)Phn tch:

Do x, y > 0 nn BT (1) c th suy ra t BT Csi hoc xt hiu.Gii

Cch 1: S dng BT Csic cho 2 s dng x, y:

Cch 2. Xt hiu ca 2 v:

(2)Do x > 0, y > 0 nn BT (2) lun ng.

Vy (1) lun ng. (pcm)

Khai thc bi ton:

Ta thy BT trn c lin quan n vic cng mu nn c th s dng chng minh BT sau:

Cho a, b, c l di 3 cnh ca mt tam gic, chng minh rng:

trong

Bi tp tng t:

Bi 1. Chng minh rng:

Bi 2. Cho a, b, c, d l cc s dng. Chng minh rng:

Bi 3. Cho . Chng minh rng:

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

Bi 5. Cho x, y, z > 0. Chng minh rng:

Bi 6. Cho a, b > 0. Chng minh rng:

Bi 7. Cho x, y > 0. Chng minh rng:

Bi 8. Cho x, y 0. Chng minh rng:

Bi 9. Cho a, b > 0. Chng minh rng:

p dng bt ng thc Csi chng minh BT trong tam gicBi ton s 3. Cho a, b, c l di cnh ca mt tam gic.

Chng minh rng:

Gii:

Cch 1.

t x = b + c a; y = a + c - b; z = a + b c.Khi x, y, z > 0 v

V tri:

Du bng xy ra

Cch 2. Nhn xt: Do a, b, c, l di 3 cnh ca tam gic nn ta c:

a + b - c > 0; a + c b > 0; b + c - a > 0

p dng BT Csi cho cc cp s dng:

Nhn thy cc v ca BT trn l cc s dng v 3 BT ny cng chiu, nhn tng v ca chng ta c:

Ta c:

EMBED Equation.3 Bi tp 3.1. Cho a, b, c l di 3 cnh ca mt tam gic ABC,

Chng minh rng: (*)

GiiV

chng minh (*) ta cn chng minh: (1)

Tht vy:

Ta c:

(pcm)

Bi tp 3.2. Chng minh rng

(*)Trong a, b, c l di 3 cnh ca mt tam gic.

GiiTa c

Tht vy:

Lun ng suy ra (1) ng

Tng t:

Do :

M:

(4)

Do:

T (3) v (4) suy ra iu phi chng minh.Cc bi tp khc:

Bi tp 3.3 Cho a, b, c l di 3 cnh ca 1 tam gic v c chu vi l 2. Chng minh rng: a2 + b2 + c2 + 2abc < 2.

Bi tp 3.4 Cho a, b, c l 3 cnh ca 1 tam gic. Chng minh rng:

Bi tp 3.5 Gi s a, b, c l di 3 cnh ca 1 tam gic.

Chng minh rng:

Bi tp 3.6 Gi s a, b, c l di 3 cnh ca 1 tam gic.

Chng minh rng

Bi tp 3.7. Cho a, b, c, d > 0 v a + b + c + d = 1

Chng minh rng:

NG DNG 2: ng dng bt ng thc Csi tm cc tr* Vi a 0, b 0 ta c , du = xy ra a = b

* Vi n s khng m: a1 , a2 , , an ta c:

Du = xy ra a1 = = an * T BT trn ta suy ra:

+ Nu a.b = k (const) th min(a + b) = 2 a = b

+ Nu a + b = k (const) th max(a.b) = a = b

* M rng i vi n s khng m:

+ Nu a1.a2an = k (const) th min(a1 + a2 + + an) = n

a1 = a2 = = an + Nu a1 + a2 + + an = k (const) th max(a1.a2an) =

a1 = a2 = = an V d: Cho x > 0, y > 0 tho mn:

Tm GTNN ca A =

Bi lm:

V x > 0, y > 0 nn > 0, > 0, > 0, > 0 . Ta c:

Vy min A = 4 x = y = 4

Nhn xt: Trong v d trn ta s dng BT Csi theo 2 chiu ngc nhau:

+ Dng dng iu kin tng t c

+ Dng lm gim tng dng kt qu

Khng phi lc no ta cng c th dng trc tip BT Csi i vi cc s trong bi. Ta c mt s bin php bin i mt biu thc c th vn dng BT Csi ri tm cc tr ca n:

* Cch 1: tm cc tr ca mt biu thc ta tm cc tr ca bnh phng biu thc .V d: Tm GTNN ca A =

Bi gii

iu kin:

Ta c: A2 = ( 3x 5 ) + ( 7 3x ) + 2

A2 ( 3x 5 + 7 3x ) + 2 = 4

Du = xy ra 3x 5 = 7 3x x = 2

Vy max A2 = 4 max A = 2 x = 2

Ta thy A c cho di dng tng ca 2 cn thc. Hai biu thc ly cn c tng khng i (bng 2). V vy, nu bnh phng A s xut hin hng t l 2 ln tch ca 2 cn thc. n y c th vn dng BT Csi

* Cch 2: Nhn v chia biu thc vi cng mt s khc 0V d: Tm GTLN ca A =

Bi gii:

iu kin: x 9. Ta c:

Du = xy ra

Vy max A =

Trong cch gii trn, x 9 c biu din thnh khi vn dng BT Csi tch ny tr thnh na tng: c dng kx c th rt gn cho x mu. ( s 3 c tm bng cch ly , s 9 c trong bi)

* Cch 3: Bin i biu thc cho thnh tng ca cc biu thc sao cho tch ca chng l mt hng s.

V d 1: ( Tch mt hng t thnh tng ca nhiu hng t bng nhau)

Cho x > 0, tm GTNN ca A =

Bi gii

A = =

A 4.2 = 8 ( du = xy ra )

Vy min A = 8 khi x = 2

V d 2: (Tch mt hng t cha bin thnh tng ca mt hng s vi mt hng t cha bin sao cho hng t ny l nghch o ca mt hng t khc c trong biu thc cho)

Cho 0 < x < 2, tm GTNN ca A =

Bi gii

Du = xy ra

Vy min A = 7

Trong cch gii trn ta tch thnh tng . Hng t nghch o vi nn khi vn dng BT Csi ta c tch ca chng l mt hng s.

* Cch 4: Thm mt hng t vo biu thc choV d: Cho x, y, z > 0 tho mn: x + y + z = 2

Tm GTNN ca P =

Bi gii

V x, y, z > 0 ta c:

p dng BT Csi i vi 2 s dng v ta c:

(1) . Tng t ta c:

Cng (1) + (2) + (3) ta c:

Du = xy ra

Vy min P = 1

Nhn xt: Ta thm vo hng t th nht c trong bi, khi vn dng BT Csi c th kh c (y + z). Cng nh vy i vi 2 hng t cn li ca bi. Du ng thc xy ra ng thi trong (1), (2), (3)

Nu ta ln lt thm (y + z), (x + z), (x + y) vo th ta cng kh c (y + z), (x + z), (x + y) nhng iu quan trng l khng tm c cc gi tr ca x, y, z du ca cc ng thc ng thi xy ra, do khng tm c GTNN ca P.

p dng cc cch trn cng vi vic s dng BT Csi ta c cc v d khc nh sau:

VD 1: Cho a, b, c > 0 tho mn: a + b + c = 1

Tm GTLN ca P =

Phn tch: a, b, c > 0

Do c th khai trin P ri c lng theo BT Csi

Bi gii

Cch 1:

p dng BT Csi cho 3 s dng ta c:

(1)

Mt khc:

(2)

(1) + (2) ta c: . Vy min P = 64

Cch 2:

Tng qut: cho S = a + b + c

tm GTLN ca P =

VD 2: Tm GTLN ca B =

Bi gii

max B =

VD 3: Cho 2 s dng x, y c x + y = 1

Tm GTNN ca B =

Bi gii

Ta c: B = = 1 +

Vy min B = 9

VD 4: Cho x, y, z > 0 tho mn:

Tm GTNN ca P = xyz

Bi gii

Ta c:

Tng t:

Vy max P =

VD 5: Cho M = 3x2 2x + 3y2 2y + 6 |x| + 1

Tnh gi tr ca M bit x, y l 2 s tho mn x.y = 1 v biu thc |x + y| t GTNN.

Bi gii:

Ta c:

Min |x + y| = 2 khi x = y, khi

Khi x = y = 1 hoc x = y = - 1

+ Khi x = y = 1 th M = 9

+ Khi x = y = - 1 th M = 17

VD 6:

Cho cc s thc khng m a1, , a5 tho mn: a1 + + a5 =1

Tm GTLN ca A = a1a2 + a2a3 + a3a4 + a4a5Bi gii

Ta c: A = a1a2 + a2a3 + a3a4 + a4a5 (a1 + a3 + a5)(a2 + a4)

Vy max A =

VD 7: Cho a, b > 0. Tm GTNN ca A = ( x > 0)

Bi gii

.

Du = xy ra

VD 8: Tm GTNN ca hm y = vi 0 < x < 1

Bi gii

Ta c: y = ( 0 < x < 1)

=

Du = xy ra

VD 9: Cho a, b > 0 cho trc.

Cc s x, y > 0 thay i sao cho

Tm x, y S = x + y t GTNN. Tm min S theo a, b.

Bi gii

Ta c:

M

VD 10: Tm GTNN ca P =

Bi gii

Ta c: P =

=

Suy ra min P = 64 x = 1 hoc x = - 3

Bi tp tng t

BT 1: Cho x, y > 0 tho mn x. y = 1. Tm GTLN ca A =

BT 2: Tm GTLN ca cc biu thc sau:

BT 3: Cho a, b, c > 0 tho mn . Tm GTLN ca biu thc Q = abc.

BT 4: Cho x, y > 0 tho mn x + y = 1. Tm GTNN ca biu thc

P =

BT 5: Tm GTNN ca cc biu thc sau:

BT 6: Cho x, y > 0 tho mn . Tm GTNN ca biu thc

E =

BT 7: Tm GTLN v GTNN ca A =

BT 8: Tm GTLN ca A = bit

BT 9: Cho a, b > 0 tho mn a. b = 216

Tm GTNN ca S = 6a + 4b

BT 10: Cho a, b > 0 tho mn .

Tm GTNN ca A =

BT 11: Cho a, b > 0 tho mn .

Tm GTNN ca S =

BT 12: Cho x, y, z 0 tho mn xy + yz + zx = 100.

Tm GTNN ca A = xyz

BT 13: Vi gi tr no ca a th tch xy nhn GTLN nu x, y, a l cc s thc tho mn

BT 14: Tm GTNN ca A = bit a > 0, x > 0

BT 15: Vi gi tr no ca s dng a th biu thc D t GTNN ?

A =

BT 16: Tm GTNN ca C =

BT 17: Tm GTLN ca E =

BT 18: Tm GTLN ca tch

Bit v

BT 19: Tm GTLN ca B =

BT 20: Tm GTNN ca N = bit rng x, y > 0

BT 21: Tm GTLN ca H = vi

BT 22: Tm GTLN ca biu thc:

P =

Vi mi x, y, z bin i nhng lun tho mn

BT 23: Tm GTNN ca ;

BT 24: Tm GTLN ca

BT 25: Tm GTLN ca vi x > 1

PAGE 16

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