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B�T ��NG TH�C VÀ GIÁ TR� L�N NH�T, GIÁ TR� NH� NH�T�A BI�U TH�C ��I S . A. B�T ��NG TH�C.

I/ PH��NG PHÁP BI�N ��I T��NG ��NG. Thí d�: Cho a, b là hai s� th�c b�t kì. Ch�ng minh r�ng: 2 2a b ab a b 1 0+ − − − + ≥ . Gi�i: 2 2a b ab a b 1 0+ − − − + ≥ � 2 22a 2b 2ab 2a 2b 2 0+ − − − + ≥ �

2 2 2 2(a 2ab b ) (a 2a 1) (b 2b 1) 0− + + − + + − + ≥ � 2 2 2(a b) (a 1) (b 1) 0− + − + − ≥ (�úng) (D�u “=” x�y ra � a b 1= = )

II/ PH��NG PHÁP TAM TH�C B�C HAI. Thí d�: Cho a, b là hai s� th�c b�t kì. Ch�ng minh r�ng: 2 2a b ab a b 1 0+ − − − + ≥ . Gi�i: 2 2a b ab a b 1 0+ − − − + ≥ � 2 2a (b 1)a b b 1 0− + + − + ≥ .

Xét tam th�c bc hai: 2 2f (a) a (b 1)a b b 1= − + + − + . 2 2 2(1 b) 4(b b 1) 3(b 1) 0, b R∆ = + − − + = − − ≤ ∀ ∈

� f (a) 0, a, b R≥ ∀ ∈ (D�u “=” x�y ra � a b 1= = )

III/ PH��NG PHÁP QUY N�P. Thí d�: Cho a, b là hai s� th�c không âm và n là s� nguyên d�ng.

Ch�ng minh r�ng: n n na b a b

2 2

+ +� �≤� �

� � (*).

Gi�i: • V�i n = 1 thì (*) �úng.

• Gi� s (*) �úng v�i n k 1= ≥ , ngh�a là ta có: k k ka b a b

2 2

+ +� �≤� �

� �

�

k k ka b a b a b a b

2 2 2 2

� �+ + + +� � � � � �≤ � ��

� � � � � ��

k 1 k k k 1 k 1a b a b a b a b

2 2 2 2

+ + +� �+ + + +� � � �≤ ≤� ��

� � � ��

Trong �ó: k k k 1 k 1a b a b a b

2 2 2

+ +� �+ + +� �≤� ��

� �� ( )( ) ( )k k k 1 k 1a b a b 2 a b+ +

+ + ≤ +

� ( )k k(a b) a b 0− − ≥ � ( )2 k 1 k 2 k 1(a b) a a b ... b 0− − −− + + + ≥ (�úng).

Vy (*) �úng v�i m�i *n N∈ .

IV/ PH��NG PHÁP ÁP D�NG B�T ��NG TH�C CÔSI (C.S). B�t ��ng th�c Côsi:

• a 0, b 0∀ ≥ ≥ , ta ��: a b

ab.2

+≥ ��ng th�c ��y ra � a b= .

• a 0, b 0, c 0∀ ≥ ≥ ≥ , ta ��: 3a b cabc.

3

+ +≥ ��ng th�c ��y ra � a = b = c.

Thí d�: Cho a, b, c, u, v là n�m s� th�c b�t kì. Ch�ng minh r�ng: 2 2 2 2 2a b c u v a(b c u v)+ + + + ≥ + + + .

Gi�i: 2 2

2 2a ab 2 b ab ab

4 4+ ≥ = ≥ . T�ng t� �

22a

c ac4

+ ≥ ; … ��pcm.

(D�u “=” x�y ra �a 2b 2c 2u 2v= = = = )

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V/ PH��NG PHÁP ÁP D�NG B�T ��NG TH�C BUNHIAKÔPXKI CÔSI (B.C).

B�t ��ng th�c Bunhiac�pxki: • V�i hai c�p s��th�c b�t ��(a; b), (x; y), ta ��:

2 2 2 2 2(ax by) (a b )(x y )+ ≤ + + �2 2 2 2ax by (a b )(x y )+ ≤ + +

��ng th�c ��y ra � a b

0x y

= � ay − bx = 0 � a b

(x 0, y 0)x y

= ≠ ≠ .

• V�i hai c�p s��th�c b�t ��(a; b; c), (x; y; z), ta ��:

2 2 2 2 2 2 2(ax by cz) (a b c )(x y z )+ + ≤ + + + + �2 2 2 2 2 2ax by cz (a b c )(x y z )+ + ≤ + + + +

��ng th�c ��y ra � a b c

(x 0, y 0, z 0)x y z

= = ≠ ≠ ≠ .

Thí d�: Ch�ng minh r�ng: s inx cos x 2 , x R+ ≤ ∀ ∈ .

Gi�i: 2 2 2 21.s inx 1.cos x (1 1 )(s in x cos x)+ ≤ + + ��pcm.

(D�u “=” x�y ra � s inx cos x

1 1= � tanx 1= � x k , k Z

4

π= + π ∈ )

VI/ PH��NG PHÁP T�A �� VÉCT�.

• a b a b+ ≤ +� � � �

(còn g�i là b�t ��ng th�c tam giác)

• a.b a . b≤� � � �

v�i 1 2a(a ; a )�

, 1 2b(b ; b )�

� 2 2 2 21 1 2 2 1 2 1 2a b a b (a a )(b b )+ ≤ + + c�ng là B�T B.C

(D�u “=” x�y ra � ( )cos a, b 1= ±� �

� 1 2a(a ; a )�

và 1 2b(b ; b )�

cùng ph�ng)

Thí d�: Cho x, y, z là ba s� th�c b�t kì.

Ch�ng minh r�ng: 2 2 2 2 2 2x xy y x xz z y yz z (*)+ + + + + ≥ + + .

Gi�i: (*) �2 2

2 2 2 21 3 1 3x y y x z z y yz z

2 4 2 4� � � �

+ + + + + ≥ + +� � � ��

Ch�n 1 3

a x y; y2 2

� �+� ��

� �

�

; 1 3

b x z; z2 2

� �− −� ��

�

�1 1 3 3

a b y z; y z2 2 2 2

� �+ = − +� ��

� �

� �

Áp d�ng: a b a b+ ≥ +� � � �

��pcm.

VII/ PH��NG PHÁP V�N D�NG S� BI�N THIÊN CA HÀM S .

Thí d�: Ch�ng minh r�ng v�i m�i s� th�c x d�ng, ta có: 2x

x ln(1 x)2

− < + .

Gi�i: 2x

f (x) ln(1 x) x2

= + + − v�i x [0; )∈ + ∞ .

21 xf '(x) x 1 0, x (0; )

1 x 1 x= + − = > ∀ ∈ + ∞

+ + và f '(x) 0= � x 0= .

�Hàm f ��ng bi�n trên [0; )+ ∞ � v�i x > 0 thì f(x) > f(0) ��pcm.

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BÀI T�P. 1) Cho ∆ABC có ba góc nh�n.

a) Ch�ng minh r�ng: tan A tan B tan C tan A.tan B.tan C+ + = . b) Tìm giá tr� nh� nh�t c�a bi u th�c: P tan A.tan B.tan C= .

H�ng d!n: a)A B C

2 2 2

π +� �= − � �

� � �

A B Ctan cot

2 2 2� �

= +� ��

�

B C1 tan tanA 2 2tan

B C2 tan tan2 2

−

=

+

��pcm.

A (B C)= π − + � [ ]t anA tan (B C) tan(B C)= π − + = − + ��pcm.

b) P tan A tan B tan C tan A.tan B.tan C= + + = 3P tan A tan B tan C 3 tan A.tan B.tan C= + + ≥ � 3P 3 P≥ �P 3 3≥

�max P 3 3= khi A B C3

π= = = .

2) Cho A, B, C là ba góc c�a m"t tam giác.

a) Ch�ng minh r�ng:A B B C C A

tan .tan tan .tan tan .tan 12 2 2 2 2 2

+ + = .

b) Tìm giá tr� l�n nh�t c�a bi u th�c: A B C

P tan .tan . tan2 2 2

= .

H�ng d!n: a)A B C

2 2 2

π +� �= − � �

� � �

A B Ctan cot

2 2 2� �

= +� ��

�

B C1 tan tanA 2 2tan

B C2 tan tan2 2

−

=

+

��pcm.

b) Áp d�ng B�T C.S cho ba s� # v� trái câu a)

2 2 23A B C

VT 3 tan tan tan2 2 2

≥ �2

A B C1 27 tan tan tan

2 2 2� �

≥ � ��

� 21 27P≥ �1

P3 3

≤

�1

max P3 3

= khi A B C3

π= = = .

3) Cho tam giác có �" dài ba c$nh là a, b, c và p là n a chu vi. Ch�ng minh r�ng: a) 2 2 2a b c 2(ab bc ca)+ + < + + .

b) abc

(p a)(p b)(p c)8

− − − ≤ .

c) 1 1 1 1 1 1

2p a p b p c a b c

� �+ + ≥ + +� �

− − − � �.

d) a b c

3b c a c a b a b c

+ + ≥+ − + − + −

.

e) p a p b p c 3p− + − + − ≤ .

H�ng d!n: a) Ta có: b c a b c− < < + �2 2b c a− < � 2 2 2b c 2bc a+ − < . T�ng t� … ��pcm.

b) (p a) (p b) c

(p a)(p b)2 2

− + −− − ≤ = . T�ng t� … ��pcm.

c) Áp d�ng h% qu� c�a B�T C.S: 1 1 4

a b a b+ ≥

+�

1 1 4 4

p a p b 2p a b c+ ≥ =

− − − −. T�ng t� … ��pcm.

d) 3a b c abc

3 3.1 3b c a c a b a b c (b c a)(c a b)(a b c)

+ + ≥ ≥ =+ − + − + − + − + − + −

(vì abc (b c a)(c a b)(a b c)≥ + − + − + − , ch�ng minh t�ng t� câu a)

e) 1. p a 1. p b 1. p c (1 1 1)(3p a b c) 3p− + − + − ≤ + + − − − = (D�u “=” x�y ra � a b c= = )

4) Ch�ng minh r�ng trong m�i tam giác có chu vi không �&i thì tam giác �'u có di%n tích l�n nh�t.

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H�ng d!n: Gi� s ABC∆ có �" dài ba c$nh là a, b, c và a b c

p2

+ += (không �&i).

Di%n tích ABC∆ : S p(p a)(p b)(p c)= − − − .

3(p a) (p b) (p c) p

(p a)(p b)(p c)3 3

− + − + −− − − ≤ = �

3p(p a)(p b)(p c)

27− − − ≤

�

3 2p pS p(p a)(p b)(p c) p

27 3 3= − − − ≤ = �

2pmaxS

3 3= khi A B C

3

π= = = .

5) Cho tam giác có �" dài ba c$nh là a, b, c và di%n tích S. Ch�ng minh r�ng: 2 2 2a b c 4S 3+ + ≥ .

H�ng d!n: Bài tp 4 �2p

S3 3

≤ �2

a b c 1S

2 3 3

+ +� �≤ � ��

� 2 2 212 3 S a b c 2ab 2bc 2ca≤ + + + + +

�2 2 2 2 2 2 2 2 212 3 S a b c (a b ) (b c ) (c a )≤ + + + + + + + + � 2 2 24 3 S a b c≤ + + .

6) ABC∆ có �" dài ba c$nh là a, b, c; �" dài ba �(ng cao t�ng �ng là ha, hb, hc và di%n tích ABC∆

b�ng 1,5. Ch�ng minh r�ng:a b c

1 1 1 1 1 13

a b c h h h

� �� + + + + ≥� ��

� �� .

H�ng d!n: a

1S ah

2= �

a

1 a

h 2S= ; …

33

a b c

1 1 1 1 1 1 1 1 1 1 1 1(a b c) 3 .3 abc 3

a b c h h h 2S a b c 2(3 / 2) abc

� �� + + + + ≥ + + + + ≥ =� ��

� � � �� .

7) Cho x, y là hai s� th�c tùy ý. Ch�ng minh r�ng: 2 2x 5y 4xy 2x 6y 3 0+ − + − + > .

H�ng d!n: 2 2f (x) x 2(1 2y)x 5y 6y 3= + − + − + . 2' (y 1) 1 0, y R� �∆ = − − + < ∀ ∈ ��pcm.

8) Cho x, y, z là ba s� th�c tùy ý. Ch�ng minh r�ng: 2 2 2x 19y 6z 8xy 4xz 12yz 0+ + − − + ≥ .

H�ng d!n: 2 2 2f (x) x 2(4y 2z)x 19y 6z 12yz= − − + + + . 2 2

x' 3y 4zy 2z f (y)∆ = − + − = � 2y' 2z 0, z R∆ = − ≤ ∀ ∈ � x' f (y) 0, y, z R∆ = ≤ ∀ ∈ ��pcm.

9) Ch�ng minh r�ng v�i m�i s� th�c x, ta có: x x x

x x x12 15 203 4 5

5 4 3� � � � � �

+ + ≥ + +� � � � � ��

. Khi nào ��ng th�c

x�y ra?

H�ng d!n: x x x x

x12 15 12 152 . 2.3

5 4 5 4� � � � � � � �

+ ≥ =� � � � � � � ��

; t�ng t� … ��pcm.

��ng th�c x�y ra �x x x

12 15 20

5 4 3� � � � � �

= =� � � � � ��

� x 0= .

10) Cho x, y , z là các s� th�c d�ng. Ch�ng minh r�ng: 3x 2y 4z xy 3 yz 5 zx+ + ≥ + + .

H�ng d!n: x y

xy2

+≥ ;

3(y z)3 yz

2

+≥ ;

5(z x)zx

2

+≥ ��pcm.

11) Cho x, y, z là ba s� th�c d�ng th�a: x z> và y z> . Ch�ng minh r�ng:

z(x z) z(y z) xy− + − ≤ .

H�ng d!n: z(x z) z(y z) xy− + − ≤ �z(x z) z(y z)

1xy xy

− −+ ≤

z(x z) z(y z) 1 z x z 1 z y z 1 x y1

xy xy 2 y x 2 x y 2 x y

� � � � � �− − − −+ ≤ + + + = + =� � � � � �

� � � � � �.

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12) Cho ba s� d�ng x, y, z th�a: xy yz zx xyz+ + = . Ch�ng minh r�ng: 1 1 1 1

x(x 1) y(y 1) z(z 1) 2+ + ≥

− − −.

H�ng d!n: Gi� thi�t: xy yz zx xyz+ + = �1 1 1

1x y z

+ + = .

��t: 1

ax

= ; 1

by

= ; 1

cz

= . Khi �ó: a, b, c là ba s� d�ng th�a: a b c 1+ + = .

2

2 2

1 1 1 a1x(x 1) b c1 1x 1 xx y z

= = =− +� � � �

− +� � � ��

; 21 b

y(y 1) c a=

− +;

21 c

z(z 1) a b=

− +.

B�T Cô si: 2a b c

ab c 4

++ ≥

+;

2b c ab

c a 4

++ ≥

+;

2c a bc

a b 4

++ ≥

+ ��pcm.

(��ng th�c x�y ra khi và ch) khi x y z 3= = = )

13) Cho x, y, z là ba s� th�c d�ng. Ch�ng minh r�ng: 3 2 3 2 3 2 2 2 2

2 y2 x 2 z 1 1 1

x y y z z x x y z+ + ≤ + +

+ + +.

H�ng d!n: ��t: a x= , b y= , c z= .

3 2 3 2 3 2 6 4 6 4 6 4 3 2 3 2 3 2 2 2 2 2 2 2

2 y2 x 2 z 2a 2b 2c 2a 2b 2c 1 1 1

x y y z z x a b b c c a 2a b 2b c 2c a a b b c c a+ + ≤ + + ≤ + + = + +

+ + + + + +

2 2 2 4 4 4 2 2 2 2 2 2

1 1 1 1 1 1 1 1 1

x y z a b c a b b c c a+ + = + + ≥ + + ��pcm (D�u “=” x�y ra � x y z 1= = = )

14) Cho x, y, z là các s� d�ng th�a: 1 1 1

4x y z

+ + = . Ch�ng minh r�ng:

1 1 11

2x y z x 2y z x y 2z+ + ≤

+ + + + + +.

H�ng d!n: 1 1 1 1 1 1 1 1 1

2x y z 4 2x y z 4 2x 4 y z

� �� ≤ + ≤ + +� ��

+ + +� � � � �

1 1 1 1 1

2x y z 8 x 2y 2z

� �≤ + +� �

+ + � �

T�ng t�: 1 1 1 1 1

x 2y z 8 2x y 2z

� �≤ + +� �

+ + � �;

1 1 1 1 1

x y 2z 8 2x 2y z

� �≤ + +� �

+ + � � ��pcm.

15) Cho x, y, z là ba s� th�c d�ng th�a: xyz 1= . Ch�ng minh r�ng:1 1 1

1x y 1 y z 1 z x 1

+ + ≤+ + + + + +

.

H�ng d!n: ��t: 3x a= , 3y b= , 3z c= (abc = 1).

Ta có: 3 3 2 2a b (a b)(a ab b ) (a b)ab+ = + − + ≥ + (vì a b 0= > và 2 2a b ab ab+ − ≥ )

�3 3a b 1 (a b)ab abc ab(a b c) 0+ + ≥ + + = + + > �

3 3

1 1

a b 1 ab(a b c)≤

+ + + +

T�ng t�: 3 3

1 1

b c 1 bc(a b c)≤

+ + + +,

3 3

1 1

c a 1 ca(a b c)≤

+ + + +

�3 3 3 3 3 3

1 1 1 1 1 1 11

a b 1 b c 1 c a 1 (a b c) ab bc ca� �

+ + ≤ + + =� �+ + + + + + + + � �

�1 1 1

1x y 1 y z 1 z x 1

+ + ≤+ + + + + +

. ��ng th�c x�y ra khi và ch) khi x y z 1= = = .

16) Cho x, y, z là ba s� th�c d�ng th�a: xyz 1= . Ch�ng minh r�ng: 3 3 3 3 3 31 x y 1 y z 1 z x

3 3xy yz zx

+ + + + + ++ + ≥ .

H�ng d!n: Yêu c*u BT � 3 3 3 3 3 3z 1 x y x 1 y z y 1 z x 3 3 xyz+ + + + + + + + ≥

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3 3 3 331 x y 3 x y 3xy+ + ≥ = �3 31 x y 3 xy+ + ≥ � 3 3z 1 x y 3 z xy+ + ≥

T�ng t� � ( ) 2 2 23VT 3 z xy x yz y xz 3.3 xyz x y z 3 3≥ + + ≥ = .

17) Ch�ng minh r�ng v�i m�i s� t� nhiên m (m 2)≥ , ta có: 2ln m ln(m 1).ln(m 1)> − + .

H�ng d!n: 2ln m ln(m 1).ln(m 1)> − + (*). V�i m = 2 thì B�T (*) �úng.

Khi m > 2: (*) �ln m ln(m 1)

ln(m 1) ln m

+>

−

Xét hàm s� ln x

f (x)ln(x 1)

=−

v�i x > 2. Hàm f ngh�ch bi�n nên v�i m > 2 thì f (m) f (m 1)> + ��pcm.

18) Cho x, y, z là ba s� th�c d�ng th�a: xy yz zx 3+ + = . Ch�ng minh r�ng:

2 2 2

1 1 1 1

1 x (y z) 1 y (z x) 1 z (x y) xyz+ + ≤

+ + + + + +.

H�ng d!n: 23xy yz zx 3 (xyz)+ + ≥ � xyz 1≤

�2 21 x (y z) xyz x (y z) x(xy yz zx) 3x+ + ≥ + + = + + = �

2

1 1

1 x (y z) 3x≤

+ +

T�ng t� �2 2 2

1 1 1 1 1 1 1 xy yz zx 1

1 x (y z) 1 y (z x) 1 z (x y) 3 x y z 3xyz xyz

� � + ++ + ≤ + + = =� �

+ + + + + + � �

D�u “=” x�y ra �xyz 1

xy yz zx 3

= �

+ + =� � x y z 1= = = .

19) Cho x, y, z là ba s� th�c d�ng th�a: x 3y 5z 3+ + ≤ . Ch�ng minh r�ng: 4 4 43xy 625z 4 15yz x 4 5zx 81y 4 45 5 xyz+ + + + + ≥ .

H�ng d!n: 4 4 43xy 625z 4 15yz x 4 5zx 81y 4 45 5 xyz+ + + + + ≥

�2 2 2

2 2 2

4 4 4x 9y 25z 45

x 9y 25z+ + + + + ≥

V�i 2

a x;x

� ��

�

, 2

b 3y;3y

� ��

�

, 2

c 5z;5z

� ��

�

, 2 2 2

u a b c x 3y 5z;x 3y 5z

� �= + + = + + + +� �

� �

� � � �

Ta có: a b c a b c+ + ≥ + +� � � � � �

�

2

2 2 2 22 2 2

4 4 4 2 2 2x 9y 25z (x 3y 5z)

x 9y 25z x 3y 5z

� �+ + + + + ≥ + + + + +� �

� �

� ( )( )

22 2 2 3

22 2 23

4 4 4 36x 9y 25z 9 x3y5z

x 9y 25z x3y5z+ + + + + ≥ +

��t: ( )2

3t x3y5z= (x, y, z là ba s� th�c d�ng th�a: x 3y 5z 3+ + ≤ � 0 t 1< ≤ ).

( )( )

23

23

36 36 369 x3y5z 9t 36t 27t 72 27 45

t tx3y5z+ = + = + − ≥ − =

(��ng th�c x�y ra � t 1= � x 3y 5z 1= = = )

20) Cho x, y, z là ba s� th�c th�a: x y z 0+ + = . Ch�ng minh r�ng: x y z x y z8 8 8 2 2 2+ + ≥ + + .

H�ng d!n: 3 3 3x y z x y z a b c 08 8 8 3 8 .8 .8 3 8 3 8 3+ ++ + ≥ = = = � ( )x y z2 8 8 8 6 (1)+ + ≥

3x x x8 1 1 3 8 .1.1 3.2+ + ≥ = ; t�ng t� � x y z x y z8 8 8 6 3(2 2 2 ) (2)+ + + ≥ + +

(1) + (2) � ( ) ( )x y z x y z3 8 8 8 3 2 2 2+ + ≥ + + ��pcm.

21) Cho x, y là các s� th�c tùy ý. Ch�ng minh r�ng: 2 2 2 2x 4 x 2x y 1 y 6y 10 5+ + − + + + − + ≥ .

� ��


t

3t2 -8t + 4

-� +�

0

2/3

0

2

H�ng d!n: Ch�n ( )a x; 2�

; ( )b x 1; y− +�

; ( )c 1; 3 y− −�

. Áp d�ng: a b c a b c+ + ≥ + +� � � � � �

��pcm.

22) Cho các s� th�c d�ng x, y, z th�a mãn: x + y + z = 1. Ch�ng minh r�ng: 2 2 2 2 2 22x xy 2y 2y yz 2z 2z zx 2x 5.+ + + + + + + + ≥

H�ng d!n: Cách 1: Ch�n 1 7

u x y; x; y2 2

� �+� ��

� �

�

, 1 7

v y z; y; z2 2

� �+� ��

� �

�

, 1 7

u z x; z; x2 2

� �+� ��

� �

�

.

u v w u v w+ + ≥ + +� � ��

� �pcm.

Cách 2: ( ) ( ) ( )2 2 22 24(2x xy 2y ) 5 x y 3 x y 5 x y+ + = + + − ≥ +

Vì x, y > 0 � ( )2 2 52x xy 2y x y .

2+ + ≥ +

T�ng t�: ( )2 2 52y yz 2z y z

2+ + ≥ + ; ( )2 2 5

2z zx 2x z x2

+ + ≥ + ��pcm.

23) Cho x, y là hai s� th�c tùy ý. Ch�ng minh r�ng: 2 2

1 (x y)(1 xy) 1

2 (1 x )(1 y ) 2

+ −− ≤ ≤

+ +.

H�ng d!n: 2 2

1 (x y)(1 xy) 1

2 (1 x )(1 y ) 2

+ −− ≤ ≤

+ + �

2 2

2 2

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

2 (1 x )(1 y ) 2

− + −− ≤ ≤

+ + �

2 2

2 2

2x(1 y ) 2y(1 x )1

(1 x )(1 y )

− + −≤

+ +

Ch�n ( )2a 2x; 1 x−�

, ( )2b 1 y ; 2y−�

. Áp d�ng: a.b a . b≤� � � �

��pcm.

24) Cho x, y là hai s� th�c khác 0. Ch�ng minh r�ng: 2 2

2 2

x y x y3 8 10 0 (*)

y x y x

� � � �+ − + + ≥� � � �

� �� .

H�ng d!n: ��t x y

ty x

= + ( t 2)≤ �2 2

22 2

x yt 2

y x+ = − .

Khi �ó (*) tr# thành: 23t 8t 4 0− + ≥ �

23t 8t 4 0− + ≥ �úng v�i m�i t ( ; 2] [2; )∈ −∞ − ∪ + ∞ ��pcm.

� ��


B. GIÁ TR� L�N NH�T, GIÁ TR� NH� NH�T A BI�U TH�C ��I S .

� �+A V,�-.NG /0M C1C /23�45A 67M S8�M9T BI:N (n�u �;c). � PH+<NG =6>P TAM TH?C B@C HAI. � PH+<NG =6>P /0M T@P AB>�/23�45A 67M S8. � SC�-DNG /ENH CHFT 45A 4>C 67M S8��G�BI:T. � >P -DNG BFT �HNG TH?C.

BÀI T�P. 1) Cho a, b, c IJ ba s��KL�a �Mc �i'u ki%n: 0 a 1≤ ≤ , 0 b 1≤ ≤ , 0 c 1≤ ≤ . /�m NOM�KP�� l�n nh�t QJ�NOM�KP��RL��nh�t ��a bi u th�c M = a + b + c − ab − bc − ca.

H�ng d!n: M = (a − ab) + (b − bc) + (c − ca) = a(1 − b) + b(1 − c) + c(1 − a) ≥ 0. Khi a = b = c = 0 � M = 0 � min M = 0. (1 − a) ≥ 0, (1 −b) ≥ 0, (1 − c) ≥ 0 � (1 − a)(1 − b)(1 − c) ≥ 0 � a + b + c − ab − bc − ca ≤ 1 − abc � a + b + c − ab − bc − ca ≤ 1. Khi a 1, b 0, c [0, 1]= = ∈ � M = 1 � max M = 1.

2) /�m NOM�KP��RL��nh�t ��a bi u th�c M = cot4a + cot4b + 2tan2a. tan2b + 2. H�ng d!n: M = (cot2a − cot2b)2 + 2(cota.cotb − tana. tanb)2 + 6 ≥ 6.

Khi a b4

π= = �KL��M = 6 � min M = 6.

3) Cho �Mc s� th�c x, y, z KL�a �i'u ki%n: x y z 5+ + = . /�m NOM� KP�� RL�� nh�t ��a bi u th�c

A x 1 y 1 z 1 .= − + − + −

H�ng d!n: Ta �� a b a b , a, b R.− ≥ − ∀ ∈

x 1 x 1− ≥ − , y 1 y 1− ≥ − , z 1 z 1− ≥ − � x 1 y 1 z 1 x y z 3 2− + − + − ≥ + + − = .

� min A = 2, �$t �;c khi x y z 5 / 3= = = .

4) /�m NOM�KP��l�n nh�t QJ�NOM�KP��RL��nh�t ��a LJm s�� 4y sin x cos x= − .

H�ng d!n: �i'u ki%n: sinx ≥ 0 QJ�cosx ≥ 0.

4 4y sin x cos x sin x 1= − ≤ ≤ . Khi x2

π= �KL��y = 1 � max y = 1.

4y sin x cos x cos x 1= − ≥ − ≥ − . Khi x 0= �KL��y = −1 � min y = −1.

5) /�m NOM�KP��l�n nh�t ��a bi u th�c y 9x 4

Ax y

−−= + v�i x ≥ 4 QJ�y ≥ 9.

H�ng d!n: 4 x 4 x

4(x 4)2 2

+ −− ≤ = �

xx 4

4− ≤ �

x 4 1

x 4

−≤ . T�ng t��

y 9 1

y 6

−≤

� y 9x 4 1

Ax y 12

−−= + ≤ � max A =

1

12, �$t �;c khi QJ��L)�khi �

x 4 4

y 9 9

− = �

− =� �

x 8

y 18

= �

=�

6) Cho hai s�� th�c x, y KL�a 2 236x 16y 9 0+ − = . /�m NOM� KP�� l�n nh�t, NOM� KP�� RL��nh�t ��a bi u th�c A y 2x 5.= − +

H�ng d!n: Cách 1: A y 2x 5= − + � y 2x A 5= + − � 2 2100x 64(A 5)x 16(A 5) 9 0 (*)+ − + − − =

(*) ��nghi%m � 2' 900 576(A 5) 0∆ = − − ≥ � 5

A 54

− ≤ � 15 25

A4 4

≤ ≤

� max A = 25

4, �$t �;c khi

2x

55 9

y 2x4 20

= −��

�� = + =��

; min A = 15

4, �$t �;c khi

2x

55 9

y 2x4 20

=��

�� = − = −��

� ��


+∞ +∞

4

-2x

A

A'

-∞ +∞0

0

2

0

CMch 2: ( )2

2 2 2 21 1 1 1 25A 5 (2x y) 6x 4y (36x 16y )

3 4 9 16 16� � � �

− = − = − ≤ + + =� � � ��

� ...

CMch 2’: ��t 1 1

u ;3 4� �

−� ��

�

, ( )v 6x; 4y�

� u.v 2x y= −� �

, 5

u12

=�

, 2 2v 36x 16y 3= + =�

1 1 5A 5 2x y 6x 4y u.v u . v

3 4 4− = − = − = ≤ =

� � � �

(b�t ��ng th�c Bunhiac�pxki) � ...

7) /�m NOM�KP��l�n nh�t, NOM�KP��RL��nh�t ��a bi u th�c 2 2

x 2y 1A

x y 7

+ +=

+ +.

H�ng d!n: T�ng t��Mch 1 SJi 6)

� max 1

A2

= khi x = 1 QJ�y = 2; min 5

A14

= − khi x = −7/5 QJ�y = −3.

8) Cho x, y IJ�hai s��d�ng KL�a �i'u ki%n x + y = 5

4. /�m NOM�KP��RL��nh�t ��a bi u th�c

4 1A

x 4y= + .

H�ng d!n: CMch 1: 4 1 4 1 5

A 0 xx 4y x 5 4x 4

� �= + = + < <� �

− � � �

2 2

4 1A '(x)

x (5 4x)= − +

−

� min A = 5 khi x = 4y = 1.

CMch 2: 5 54

4 1 1 1 1 1 1 1 1 5.5A 5 5 5

x 4y x x x x 4y x 4y x.x.x.x.4y x x x x 4y= + = + + + + ≥ = ≥ =

+ + + +

CMch 3: 5 2 1 4 1 5 4 1

x y (x y)2 x 4y 4 x 4yx 2 y

� � � �= + ≤ + + = +� � � �

� � � � �

25 5A

4 4≤

9) Cho x, y IJ�hai s��d�ng. /�m NOM�KP��RL��nh�t ��a bi u th�c

2

y 9A (1 x) 1 1

x y

� �� = + + +� ��

.

H�ng d!n: T�ng t��Mch 2 SJi 8): 3

43

x x x x1 x 1 4

3 3 3 3+ = + + + ≥ ;

3

43 3

y y y y y1 1 4

x 3x 3x 3x 3 x+ = + + + ≥

3

43

9 3 3 3 31 1 4

y y y y ( y)+ = + + + ≥ �

26

43

9 31 16

yy

� �+ ≥� ��

� � � A 256≥

� min A = 256 khi x = 3�QJ�y = 9. 10) Cho x, y, z IJ� ba s�� d�ng KL�a x + y + z = 0. T�m NOM� KP�� RL�� nh�t ��a bi u th�c

x y zA 3 4 3 4 3 4= + + + + + .

H�ng d!n: T�ng t��Mch 2 SJi 8): 4x x x3 4 1 1 1 4 4 4+ = + + + ≥ � 4x x3 4 2 2+ ≥ . . .

� ( ) 34 4 4 4 12x y z x y z 0A 2 2 2 2 2.3 2 6 2 6+ +≥ + + ≥ = = � min A = 6 khi x = y = z = 0.

11) Cho �Mc s��th�c x, y KL�a �i'u ki%n: x + y = 2. /�m NOM�KP��RL��nh�t ��a bi u th�c A = 2x + 2y. H�ng d!n: CMch 1: x + y = 2 � y = 2 − x � A = x 2 x2 2 −

+ .

��t t = 2x > 0 �2t 4 4

A tt t

+= = + (t > 0).

� 2

2 2

4 t 4A '(t) 1

t t

−= − = . A’(t) = 0 � t = 2 (t = −2�IT$i).

� min A = 4, �$t �;c khi QJ��L)�khi t = 2 � x y 1= = .

CMch 2: x y x y 22 2 2 2 .2 2 2 4+ ≥ = = � min A = 4, �$t �;c khi QJ��L)�khi � x y 1= = .

� ��


+∞ +∞

9

-1x

A

A'

-∞ +∞1

0

3

0

12) /�m NOM�KP��nL� nh�t ��a bi u th�c: A = x + y v�i 1 4

1x y

+ = �QJ�x > 0, y > 0.

H�ng d!n: CMch 1: 1 4

1x y

+ = � 4x

yx 1

=−

. V�i y > 0 � 1

1x

< � x > 1 � 4x

A xx 1

= +−

(v�i x > 1)

� 2

2 2

4 (x 1) 4A '(x) 1

(x 1) (x 1)

− −= − =

− −

A’(x) = 0 � x = 3 (x = −1 IT$i).

� min A = 9, �$t �;c khi QJ��L)�khi � x 3

y 6

= �

=�

AO�i �Mch 2: 4x 4 4 4(x 1)

A x y x x 4 (x 1) 5 2 5 9x 1 x 1 x 1 x 1

−= + = + = + + = − + + ≥ + ≥

− − − −

� min A = 9, �$t �;c khi QJ��L)�khi � x 3

y 6

= �

=�

13) Cho x, y IJ hai s��không âm KL�a �i'u ki%n x + y = 1. /�m NOM�KP��l�n nh�t QJ�NOM�KP��RL��nh�t ��a bi u th�c A = 3x + 3y.

H�ng d!n: T�ng t��SJi 6) � max A = 4 khi x = 0 ho�c x = 1; min A = 2 3 khi 1

x2

= .

14) Cho x, y IJ�hai s��không âm KL�a �i'u ki%n x + y = 1. /�m NOM�KP��l�n nh�t QJ�NOM�KP��RL��nh�t ��a bi u th�c A = 3x + 9y.

H�ng d!n: T�ng t��SJi 7) � max A = 10 khi x = 0 ho�c x = 1; min A = 39

34

khi x 33 18= .

15) Cho x, y IJ�hai s��không âm KL�a �i'u ki%n x + y = 1. /�m NOM�KP��l�n nh�t QJ�NOM�KP��RL��nh�t ��a bi u

th�c x y

Ay 1 x 1

= ++ +

.

H�ng d!n: T�ng t�� max A = 1 khi (x = 0; y =1) ho�c (x = 1; y = 0); min A = 2

3 khi

1x y

2= = .

16) Cho x, y IJ�hai s��d�ng KL�a �i'u ki%n x + y = 1. /�m NOM�KP��RL��nh�t ��a bi u th�c 1

A xyxy

= + .

H�ng d!n: x y

xy2

+≥ �

2x y 1

xy2 4

+� �≤ =� ��

. ��ng th�c ��y ra � 1

x y2

= =

��t t = xy 1

0 t4

� �< ≤� �

� �. UVt LJm s��

1A f (t) t

t= = + v�i

1t 0;

4� �

∈� ��

� min A = 17

4 khi

1t

4= �

1x y

2= = .

17) /�m NOM�KP��RL��nh�t ��a LJm s��f(x) = 29

4x sin xx

π+ + trên �LT�ng (0; +∞).

H�ng d!n: 2 29 9

4x 2 4x 12x x

π π+ ≥ = π �QJ�sinx ≥ −1, ∀x∈R � f (x) 12 1, x R≥ π − ∀ ∈ .

3f 12 1

2

π� �= π −� �

� � �

x (0, )min f (x) 12 1

∈ +∞

= π − .

18) Cho ba s��d�ng x, y, z KL�a �i'u ki%n: x + y + z = xyz. /�m NOM�KP��RL��nh�t ��a bi u th�c A xyz.=

H�ng d!n: 33A x y z 3 xyz 3 A= + + ≥ = � A3 ≥ 27A � A2 ≥ 27 (Q��A > 0) � A 3 3≥ .

� min A = 3 3 , �$t �;c khi QJ��L)�khi � x y z 3= = = .

19) Cho x + y = 1 QJ�y ≥ 0. /�m NOM�KP��l�n nh�t ��a bi u th�c 2A xy .=

H�ng d!n: * N�u x ≤ 0 KL��A = xy2 ≤ 0.

� ��


* N�u x > 0 KL�: 1 = x + y = 2

3y y xy

x 32 2 4

+ + ≥ � 2 4xy

27≤

� max A = 4

27, �$t �;c khi QJ��L)�khi

x 1/ 3

y 2 / 3

= �

=�.

20) Cho ba s��d�ng x, y, z KL�a: x + y + z = 1. /�m NOM�KP��RL��nh�t ��a bi u th�c x y

Axyz

+= .

H�ng d!n: 1 = x + y + z = (x y) z 2 (x y)z+ + ≥ + � 1 4(x y)z≥ + � 2 2(x y) 4(x y) z+ ≥ + (1).

M�t �LMc: 2(x y) 4xy+ ≥ nên tW�(1) suy ra x y 16xyz+ ≥ � x y

16xyz

+≥

� min A = 16, �$t �;c khi 1

x y4

= = �QJ�1

z2

= .

21) Cho ba s��d�ng x, y, z KL�a: x + y + z = 1. /�m NOM�KP��l�n nh�t ��a bi u th�c A xy yz zx.= + +

H�ng d!n: 2 2x y 2xy+ ≥ , 2 2y z 2yz+ ≥ , 2 2z x 2zx+ ≥ � 2 2 2x y z xy yz zx+ + ≥ + + (1).

M�t �LMc: 1 = (x + y + z)2 = 2 2 2x y z 2(xy yz zx)+ + + + + (2)

(1) QJ�(2) � 1 3(xy yz zx)≥ + + � 1

xy yz zx3

+ + ≤ � max A = 1

3, �$t �;c khi

1x y z

3= = = .

22) Cho ba s�� d�ng x, y, z KL�a: x + y + z = 1. /�m NOM� KP�� l�n nh�t ��a bi u th�c A xyz(x y)(y z)(z x).= + + +

H�ng d!n: 3x y z 3 xyz+ + ≥ � 31 3 xyz≥ (1)

3(x y) (y z) (z x) 3 (x y)(y z)(z x)+ + + + + ≥ + + + � 32 3 (x y)(y z)(z x)≥ + + + (2)

(1) QJ�(2) � 332 9 xyz(x y)(y z)(z x) 9 A≥ + + + = � 8

A729

≤

� max A 8

729= , �$t �;c khi QJ��L)�khi

1x y z

3= = = .

23) Cho ba s�� không âm x, y, z KL�a �i'u ki%n: x + y + z ≤ 3. /�m NOM� KP�� l�n nh�t ��a bi u th�c

2 2 2

x y zA

1 x 1 y 1 z= + +

+ + +.

H�ng d!n: 2 2

2 2 2

x 1 2x 1 x (x 1)0

1 x 2 2(1 x ) 2(1 x )

− − − −− = = ≤

+ + + �

2

x 1

1 x 2≤

+ �

2 2 2

x y z 3A

1 x 1 y 1 z 2= + + ≤

+ + +

� max A = 3

2, �$t �;c khi QJ��L)�khi x y z 1= = = .

24) Cho ba s�� không âm x, y, z KL�a �i'u ki%n: x + y + z ≤ 3. /�m NOM� KP�� RL�� nh�t ��a bi u th�c 1 1 1

A1 x 1 y 1 z

= + ++ + +

.

H�ng d!n: 3

1 1 1 1 1 3A 3 3

(1 x) (1 y) (1 z)1 x 1 y 1 z 2(1 x)(1 y)(1 z)3

= + + ≥ ≥ ≥+ + + + ++ + + + + +

� min A = 3

2, �$t �;c khi QJ��L)�khi x y z 1= = = .

25) Cho �Mc s��th�c x, y KL�a x + y = 1, x ≥ 0, y ≥ 0. /�m NOM�KP��l�n nh�t, NOM�KP��RL��nh�t ��a bi u th�c 2 2A x y .= +

H�ng d!n: Theo gt � 0 x 1, 0 y 1≤ ≤ ≤ ≤ � 2 2x x, y y≤ ≤ � 2 2x y 1+ ≤

� max A = 1, �$t �;c khi QJ��L)�khi x 1

y 0

= �

=� ho�c

x 0

y 1

= �

=�.

1 = (x + y)2 = (1.x + 1.y)2 ≤ (12 + 12)(x2 + y2) � 1

A2

≥

� ��


+∞

0

x

A

A'

-∞ +∞

1/8

1/8

+∞

� min A = 1, �$t �;c khi QJ��L)�khi 1

x y2

= = .

26) Cho �Mc s� th�c x, y, z KL�a �i'u ki%n: xy + yz + zx = 1. /�m NOM� KP�� RL�� nh�t ��a bi u th�c 4 4 4A x y z .= + +

H�ng d!n: 33A x y z 3 xyz 3 A= + + ≥ = � A3 ≥ 27A � A2 ≥ 27 (Q��A > 0) � A 3 3≥ 2 2 2 2 2 2 2 2 2 2 21 (xy yz zx) (x y z )(x y z ) (x y z )= + + ≤ + + + + = + + � 2 2 21 x y z≤ + + (1).

2 2 2 2 2 2 2 4 4 4(x y z ) (1 1 1 )(x y z )+ + ≤ + + + + (2).

(1) QJ�(2) � 4 4 41 3(x y z )≤ + + � 4 4 4 1x y z

3+ + ≥ � min A =

1

3 khi �

3x y z

3= = = ± .

27) /�m NOM�KP��RL��nh�t ��a bi u th�c A = 4x2 + 9y2, bi�t r�ng 4x − 6y = 1 (x, y ∈R).

H�ng d!n: CMch 1: 1 4x 6y= − � 4x 1

y6

−= �

22 4x 1

A 4x 96

−� �= + � �

� �

� 4x 1 2

A '(x) 8x 9.2 16x 26 3

−� �= + ⋅ = −� �

� �. A’(x) = 0 � x =

1

8

� min A = 9, �$t �;c khi QJ��L)�khi � x 1/ 8

y 1/12

= �

= −�

CMch 2: 2 2 2 21 4x 6y 2(2x) ( 2)(3y) (2 2 )(4x 9y )= − = + − ≤ + +

� 2 2 1A 4x 9y

8= + ≥

� 1

min A8

= , �$t �;c khi QJ��L)�khi � 4x 6y 1

2x 3y

2 2

− = ��

=�� −

� 4x 6y 1

2x 3y 0

− = �

− =� �

x 1/ 8

y 1/12

= �

= −�

28) Cho x, y là hai s� th�c th�a: ( )2 22 x y xy 1+ = + . Tìm giá tr� l�n nh�t và giá tr� nh� nh�t c�a bi u th�c: 4 4x y

P2xy 1

+=

+.

H�ng d!n: Ta có 2xy 1 2 (x y) 2xy 4xy� �+ = + − ≥ − �1

xy5

≥ − .

2xy 1 2 (x y) 2xy 4xy� �+ = − + ≥ �1

xy3

≤ �1 1

t5 3

− ≤ ≤ .

��t t xy= �( )

22 2 2 2 2x y 2x y 7t 2t 1P

2xy 1 4(2t 1)

+ − − + += =

+ + v�i

1 1t ;

5 3� �

∈ −� � �

2

2

79( t t)P '

2(2t 1)

− −=

+

P ' 0= � t 0= ( t 1= − lo$i) � 1

max P4

= khi t xy 0= = . 2

min P15

= khi 1

t xy3

= = ho�c 1

t xy5

= = −

29) /�m NOM�KP��l�n nh�t, NOM�KP��RL��nh�t ��a bi u th�c A = 2x + 3y, bi�t r�ng 2x2 + 3y2 ≤ 5.

H�ng d!n: ( )2

2 2 2(2x 3y) 2. 2 x 3. 3 y (2 3)(2x 3y ) 25+ = + ≤ + + ≤ � 5 2x 3y 5− ≤ + ≤ .

� max A = 5, �$t �;c khi QJ��L)�khi x = y = 1; min A = −5, �$t �;c khi QJ��L)�khi x = y = −1. 30) Cho x, y z, là các s� th�c th�a: x 1, y 2, z 3≥ ≥ ≥ . Tìm giá tr� l�n nh�t c�a bi u th�c:

x y 2 z 3 y z 3 x 1 z x 1 y 2P

xyz

− − + − − + − −= .

H�ng d!n: y 2 z 3 x 1 y 2z 3 x 1

Pyz zx xy

− − − −− −= + +

1. x 1 1 (x 1) 10

x 2x 2

− + −≤ ≤ = . T�ng t�:

2. y 2 10

y 2 2

−≤ ≤ ;

3. z 3 10

z 2 3

−≤ ≤

� ��


�1 1 1 1 1 1 1 1 1 1

P2 2 42 2 2 3 2 3 2 2 6 3 2

� �≤ ⋅ + ⋅ + ⋅ = + +� �

� �.

Khi

x 1 1

y 2 2

z 3 3

− =��

− =��

− =��

�

x 2

y 4

z 6

= �

=�� =�

thì 1 1 1 1

P4 6 3 2

� �= + +� �

� � �

1 1 1 1min P

4 6 3 2

� �= + +� �

� �.

31) Cho ba s� th�c x, y, z l�n h�n m"t và xyz 8= . Tìm giá tr� nh� nh�t c�a bi u th�c: 1 1 1

P1 x 1 y 1 z

= + ++ + +

.

H�ng d!n: 1 1 2

1 x 1 y 1 xy+ ≥

+ + +

(vì 1 1 1 1

01 x 1 y1 xy 1 xy

− + − ≥+ ++ +

�( ) ( )

( )

2

y x xy 10

(1 x)(1 y) 1 xy

− −≥

+ + + �úng)

��ng th�c x�y ra khi và ch) khi x y= .

4 4 4 43 36 12

1 1 1 1 2 2 2 4

1 x 1 y 1 z 1 xyz 1 xy 1 xyz1 xyz 1 x y z+ + + ≥ + ≥ =

+ + + + + ++ +

3

3P 1

1 xyz≥ ≥

+ � min P 1= khi x y z 2= = = .

32) Cho x, y, z là ba s� th�c không âm th�a: 2 2 2x y z 3+ + = . Tìm giá tr� nh� nh�t c�a bi u th�c: 3 3 3

2 2 2

x y zP

1 x 1 y 1 z= + +

+ + +.

H�ng d!n: 3 3 2 6 2

32 2 3

x x 1 x x 3x3

4 2 16 22 1 x 2 1 x 2 2 2

++ + ≥ =

+ +;

T�ng t�: 3 3 2 2

2 2 2

y y 1 y 3y

4 22 1 y 2 1 y 2 2 2

++ + ≥

+ +; …

�

2 2 2 2 2 2

6

3(x y z ) 3(x y z )P

4 2 2 8

+ + + ++ ≥ �

6 3

3 9P

2 2 2 2+ ≥ �

9 3 3P

2 2 2 2 2≥ − =

�3

min P2

= khi x y z 1= = = .

33) Cho x, y, z là ba s� th�c d�ng th�a: 3 3 3

2 2 2 2 2 2

x y z1

x xy y y yz z z zx x+ + =

+ + + + + +. Tìm giá tr� l�n

nh�t c�a bi u th�c: P x y z= + + .

H�ng d!n: 3

2 2

x 2x y

x xy y 3

−≥

+ +(vì 3 2 23x (2x y)(x xy y )≥ − + +

�3 3 2 2x y x y xy 0+ − − ≥ � 2(x y)(x y) 0+ − ≥ ).

T�ng t�: 3

2 2

y 2y z

y yz z 3

−≥

+ +;

3

2 2

z 2z x

z zx x 3

−≥

+ +

�

3 3 3

2 2 2 2 2 2

x y z x y z

x xy y y yz z z zx x 3

+ ++ + ≥

+ + + + + + �P 3≤ �max P 3= khi x y z 1= = = .

34) Cho x, y, z là ba s� th�c không âm th�a: 2011 2011 2011x y z 3+ + = . Tìm giá tr� l�n nh�t c�a bi u th�c: 5 5 5P x y z= + + .

H�ng d!n: Áp d�ng B�T Cô si cho 2006 s� 1 và 5 s� x2011:

� ��


( )52001 2001 200120111 1 ... 1 x ... x 2011 x+ + + + + + ≥ � 2011 52006 5x 2011x+ ≥ .

T�ng t� � ( ) ( )2011 2011 2011 5 5 53.2006 5 x +y +z 2011 x y z+ ≥ + +

�6018 5.3 2011.P+ ≥ �P 3≤ . Khi x y z 1= = = thì P 3= �max P 3= . 35) Cho x, y, z là ba s� th�c d�ng th�a: xyz 1= . Tìm giá tr� nh� nh�t c�a bi u th�c:

y z xP

1 xy 1 yz 1 zx= + +

+ + +.

H�ng d!n: Vì: xyz 1= nên t�n t$i các s� d�ng a, b, c th�a: a b c

x ; y ; zb c a

= = = .

� a b c

Pb c c a a b

= + ++ + +

��t: X a b= + ; Y b c= + ; Z c a= + �1

a b c (X Y Z)2

+ + = + +

�Y Z X

a2

+ −= ;

X Z Yb

2

+ −= ;

X Y Zc

2

+ −=

�a b c Y Z X X Z Y X Y Z

Pb c c a a b 2X 2Y 2Z

+ − + − + −= + + = + +

+ + +

�1 X Y X Z Y Z 3

P 32 Y X Z X Z Y 2

� �� = + + + + + − ≥� � � � � ��

� � � � � � �

3min P

2= khi x y z 1= = = .

36) Cho x, y, z là ba s� th�c d�ng th�a: 2 2 2x y z 1+ + = . Tìm giá tr� nh� nh�t c�a bi u th�c:

2 2 2 2 2 2

x y zP

y x x z x y= + +

+ + +.

H�ng d!n: 2 2 2

2 2 232x (1 x ) (1 x )2x (1 x )

3

+ − + −≥ − � 2 2 23 2

2x (1 x )3

− ≤ � 2 2x(1 x )

3 3− ≤

�2

2

x 3 3x

1 x 2≥

− � 2

2 2

x 3 3x

y z 2≥

+. T�ng t�: 2

2 2

y 3 3y

z x 2≥

+; 2

2 2

z 3 3z

x y 2≥

+.

�2 2 2 2 2 2

x y z 3 3P

y x x z x y 2= + + ≥

+ + + �

3 3min P

2= khi

3x y z

3= = = .

37) Cho x, y, z là ba s� th�c không âm th�a: x y z 1+ + = . Tìm giá tr� l�n nh�t c�a bi u th�c: P xy yz zx 2xyz= + + − .

H�ng d!n: P xy yz zx 2xyz x(y z) (1 2x)yz x(1 x) (1 2x)yz= + + − = + + − = − + − . ��t: t yz= . 2 2(y z) (1 x)

0 t yz4 4

+ −≤ = ≤ = . Xét hàm s� f (t) x(1 x) (1 2x)t= − + − trên �o$n

2(1 x)0;

4

� �−� �

.

2(x 1 x) 1 7f (0) x(1 x)

4 4 27

+ −= − ≤ = < ;

22(1 x) 7 1 1 1 7f 2a a , a [0;1]

4 27 4 3 3 27

� �− � �� = − + − ≤ ∀ ∈� � � ��

� ��

�7

P xy yz zx 2xyz27

= + + − ≤ �7

max P27

= khi 1

x y z3

= = = .

38) Cho x, y , z là ba s� th�c không âm th�a: x y z 0+ + > . Tìm giá tr� nh� nh�t c�a bi u th�c: 3 3 3

3

x y 16zP

(x y z)

+ +=

+ +.

H�ng d!n: Ta có: 3

3 3 (x y)x y

4

++ ≥ (vì

33 3 (x y)

x y4

++ ≥ � 2(x y) (x y) 0− + ≥ ).

��t: a x y z= + + �3 3 3 3

3 33 3

(x y) 64z (a z) 64z4P (1 t) 64t

a a

+ + − +≥ = = − + (v�i

z0 t 1

a≤ = ≤ ).

Xét hàm s� 3 3f (t) (1 t) 64t= − + trên �o$n [0; 1].

� ��


2 2f '(t) 3 64t (1 t)� �= − − ; f '(t) 0= �1

t [0;1]9

= ∈ �t [0;1]

64min P min f (t)

81∈

= = khi x y 4z 0= = > .

39) Cho x, y, z là ba s� th�c l�n h�n 1 và th�a: 1 1 1

2x y z

+ + ≥ . Tìm giá tr� l�n nh�t c�a bi u th�c:

P (x 1)(y 1)(z 1)= − − − .

H�ng d!n: 1 1 1

2x y z

+ + ≥ �1 1 1 y 1 z 1 (y 1)(z 1)

1 1 2x y z y z yz

− − − −≥ − + − = + ≥ .

T�ng t� �1 (x 1)(z 1)

2y xz

− −≥ ;

1 (x 1)(y 1)2

z xy

− −≥

�1 (x 1)(y 1)(z 1)

8xyz xyz

− − −≥ �

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

8= − − − ≤ �

1maxP

8= khi

3x y z

2= = = .

40) Cho x, y, z, t là b�n s� d�ng th�a: x y z t 4+ + + = . Ch�ng minh r�ng:

2 2 2 2

x y z t2

1 y z 1 z t 1 t x 1 x y+ + + ≥

+ + + +.

H�ng d!n: 2 2

2 2

x xy z xy z xy z xy(1 z) xy xyzx x x x x

1 y z 1 y z 2 4 4 42y z

+= − ≥ − = − ≥ − = − −

+ +

(D�u “=” x�y ra � y z 1= = )

T�ng t�: 2

y yz yzty

1 z t 4 4≥ − −

+;

2

z zt ztxz

1 t x 4 4≥ − −

+;

2

t tx txyt

1 x y 4 4≥ − −

+

�2 2 2 2

x y z t xy yz zt tx xyz yzt ztx txy4

1 y z 1 z t 1 t x 1 x y 4 4

+ + + + + ++ + + ≥ − −

+ + + +

Ta có: 2

x y z txy yz zt tx (x z)(y t) 4

2

+ + +� �+ + + = + + ≤ =� �

� � (D�u “=” x�y ra � x z y t+ = + )

2 2x y z t

xyz yzt ztx txy xy(z t) zt(x y) (z t) (x y)2 2

+ +� � � �+ + + = + + + ≤ + + +� � � �

� � � �

�x y z t

xyz yzt ztx txy (x y)(z t) (x y)(z t)4 4

+ +� �+ + + ≤ + + + = + +� �

� �

�

2x y z t

xyz yzt ztx txy 42

+ + +� �+ + + ≤ =� �

� � (D�u “=” x�y ra � x y z t 1= = = = )

�2 2 2 2

x y z t 4 44 2

1 y z 1 z t 1 t x 1 x y 4 4+ + + ≥ − − =

+ + + +. (D�u “=” x�y ra � x y z t 1= = = = )

41) Cho x, y , z là ba s� th�c d�ng th�a: x y z xyz+ + = . Tìm giá tr� nh� nh�t c�a bi u th�c: xy yz zx

Pz(1 xy) x(1 yz) y(1 zx)

= + ++ + +

.

H�ng d!n: Cách 1: xy yz zx 1 1 1 1 1 1

Pz(1 xy) x(1 yz) y(1 zx) x y x x xyz y xyz z xyz

� �= + + = + + − + +� �

+ + + + + +� �

1 1 1 1 1 1P

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

� �= + + − + +� �

+ + + + + +� �

Áp d�ng: 1 1 1 1

a b 4 a b� �

≤ +� �+ � �

(D�u “=” x�y ra � a = b).

�1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

Px y x 4 2x 2y 2z y z z x x y x y x 4 x y x

� � � �≥ + + − + + + + + ≥ + + − + +� � � �

+ + +� � � �

1 1 1 1 1 1 1 3 1 1 1P

x y x 4 x y x 4 x y x

� � � �≥ + + − + + = + +� � � �

� � � �

� ��


Áp d�ng: 2(a b c) 3(ab bc ca), a, b, c.+ + ≥ + + ∀ (D�u “=” x�y ra � a = b = c).

�

21 1 1 1 1 1 x y z

3 3 3x y x xy yz zx xyz

� � � � � �+ ++ + ≥ + + = =� � � � � �

� � � � � � �

1 1 13

x y x+ + ≥ �

3 3P

4≥

�3 3

P4

= khi x y z 3= = = .

Cách 2: 1 1 1 1 1 1

Px y x 2x y z 2y x z 2z x y

� �= + + − + +� �

+ + + + + +� �

2 2 24 4 4

1 1 1 1 1 1P

x y x 4 x yz 4 xy z 4 xyz

� �� ≥ + + − + +� ��

1 1 1 1 2 1 1 1 2 1 1 1 2P

x y x 16 x y z x y z x y z

� �≥ + + − + + + + + + + +� �

� ��

3 1 1 1P

4 x y x

� �≥ + +� �

� �

42) /�m NOM�KP��l�n nh�t, NOM�KP��RL��nh�t ��a LJm s��y = f(x) = 2sin x 2 sin x+ − .

H�ng d!n: TX�: D = R. y = 2sin x 2 sin x 1 2 1 0, x R+ − ≥ − + − = ∀ ∈ .

min y = 0, �$t �;c khi QJ��L)�khi sinx = −1 � x k22

π= − + π .

M�t �LMc: 2 2 2sin x 2 sin x 1 1. sin x 2 sin x 2+ − ≤ + + − ≤

max y = 2, �$t �;c khi QJ��L)�khi 2sin x 2 sin x 1= − = � x k22

π= + π .

43) /�m NOM�KP��l�n nh�t, NOM�KP��RL��nh�t ��a LJm s��y = asinx + bcosx (v�i a.b ≠ 0).

H�ng d!n: TX�: D = R. 2 2 2 2y a sin x bcos x (a b )(sin x cos x)= + ≤ + +

� 2 2y a b≤ + � 2 2 2 2a b y a b− + ≤ ≤ +

� max y = 2 2a b+ , �$t �;c khi QJ��L)�khi

2 2a sin x bcos x a b

sin x cos x

a b

+ = +��

=��

(��nghi%m x)

min y = 2 2a b− + , �$t �;c khi QJ��L)�khi

2 2a sin x b cos x a b

atan x

b

+ = − +��

=��

(��nghi%m x)

44) /�m NOM�KP��l�n nh�t, NOM�KP��RL��nh�t ��a LJm s��y = 3sinx − 4cosx + 1. H�ng d!n: CMch 1: TX�: D = R. y 1 3sin x 4cos x 5− = − ≤ � 5 y 1 5− ≤ − ≤ � 4 y 6− ≤ ≤

� max y = 6, �$t �;c khi QJ��L)�khi 3sin x 4cos x 5

tan x 3/ 4

− = �

= −��

sin x 3/ 5

tan x 3/ 4

= �

= −�(��nghi%m x).

min y = −4, �$t �;c khi QJ��L)�khi 3sin x 4cos x 5

tan x 3/ 4

− = − �

= −��

sin x 3/ 5

tan x 3/ 4

= − �

= −�(��nghi%m x).

CMch 2: * N�u x (2k 1)= + π � y = 5.

* N�u x (2k 1)≠ + π � 2

2 2

2t 1 tsin x , cos x

1 t 1 t

−= =

+ +

xt tan

2� �

=� ��

� 2

2

3t 6t 5y

t 1

− + +=

+

45) /�m NOM�KP��l�n nh�t, NOM�KP��RL��nh�t ��a LJm s��sin x 2cos x 1

y3sin x 4cos x 5

+ +=

+ +

H�ng d!n: TX�: D = R. sin x 2cos x 1

y3sin x 4cos x 5

+ +=

+ + � y(3sin x 4cos x 5) sin x 2cos x 1+ + = + +

� 2 21 5y (3y 1)sin x (4y 2)cos x (3y 1) (4y 2)− = − + − ≤ − + − � y 1/ 3≤

� ��


� max y = 1


y 1/ 3

3y 1tan x

4y 2

= �

−�=� +�

� tanx = 0 � x k= π

6Jm s��không ��NOM�KP��RL��nh�t.

46) /�m NOM�KP��l�n nh�t, NOM�KP��RL��nh�t ��a LJm s�� y (3sin x 4cos x)( 3 sin x cos x)= + − .

H�ng d!n: TX�: D = R. ( )2 2y 3 3 sin x 3 4 3 sin x cos x 4cos x= − − −

( )1 cos 2x sin 2x 1 cos 2x

y 3 3 3 4 3 42 2 2

− +� � � �= − − −� � � �

� � � �

3 3 4 3 3 4 3 3

y 2 sin 2x cos 2x2 2 2

� � � �− ++ − = −� � � ��

� � � � �

3 3y 2 5

2+ − ≤ �

3 3 3 37 y 3

2 2− + ≤ ≤ +

� max y = 3 3

32

+ , min y = 3 3

72

− + , �$t �;c khi � 4 3 3

tan x4 3 3

−=

+ (��nghi%m x).

47) Cho �Mc s��th�c x, y, z, t KL�a �i'u ki%n: x2 + y2 = 1 QJ�z2 + t2 = 1. T�m NOM�KP��l�n nh�t QJ�NOM�KP��RL��nh�t ��a bi u th�c A xz yt.= +

H�ng d!n: 2 2 2 2 2(xz yt) (x y )(z t ) 1+ ≤ + + = � 1 xz yt 1− ≤ + ≤

� max A = 1, �$t �;c khi QJ��L)�khi � 2

x y z t2

= = = = ±

min A = −1, �$t �;c khi QJ��L)�khi � 2

x y z t2

= = − = − = ±

48) Trong nhXng nghi%m th�c��a h%�

2 2

2 2

x y 9

z t 16

xt yz 12

+ =�

+ =�� + ≥�

. 6Yy K�m nghi%m � �A = x + z �$t NOM�KP��l�n nh�t.

H�ng d!n: 2 2 2 2 2(xt yz) (x y )(z t ) 9.16 144+ ≤ + + = = � xt yz 12+ ≤ .

xt yz 12+ = � x y

kt z

= = � xt yz 12+ = � 2 2 2 2

22 2 2 2

x y x y 9k

t z z t 16

+= = = =

+ �

3k

4= ±

� 2 2 2 2 2 2 2 2 2 2x y z t x 2xz z y 2xz t (x z) (y t) 9 16 25+ + + = + = + − + = + + − = + = (Q��xz = yt).

� 2(x z) 25+ ≤ �Q�� 2(y t) 0− ≥

� max A = 5, �$t �;c khi QJ��L)�khi � y t

3x t

4

= ��

=��

� 9 16 12

x ; y ; z t .5 5 5

= = = =

49) Cho x, y, z IJ�ba N�c KZy [�KL�a��Mc �i'u ki%n: 0 x2

π< < , 0 y

2

π< < , 0 z

2

π< < , x y z+ + = π . /�m NOM�

KP��RL��nh�t ��a bi u th�c A = tanx. tany. tanz. H�ng d!n: GT � tanx > 0, tany > 0, tanz > 0 QJ�tan(x + y) = tan( π − z)

� tan x tan y

tan z1 tan x.tan y

−= −

− � tan x tan y tan z tan x.tan y.tan z+ + = .

� 33A tan x tan y tan z 3 tan x.tan y. tan z 3 A= + + ≥ = � 3A 3 A≥ � A 3 3≥

� min A = 3 3 , �$t �;c khi QJ��L)�khi � x y z2

π= = = .

50) Cho x, y, z IJ�ba N�c KZy [�KL�a �Mc �i'u ki%n: x 0≥ , y 0≥ , z 0≥ , x y z2

π+ + = . /�m NOM�KP��l�n nh�t

��a bi u th�c A 1 tan x.tan y 1 tan y.tan z 1 tan z. tan x= + + + + + .

� ��


H�ng d!n: AO��s �0 z2

π< < � tan x 0≥ , tan y 0≥ , tan z 0> QJ� x y

2

π+ ≠ �

tan(x y) tan z2

π� �+ = −� �

� �

� tan x.tan y tan y.tan z tan z.tan x 1+ + =

� 2 2 2 2A (1 1 1 )(1 tan x.tan y 1 tan y.tan z 1 tan z. tan x)≤ + + + + + + + � A2 ≤ 3.4 � A 2 3≤

Khi z = 0 KL��A 2 2 2 3= + < .

� max A = 2 3 , �$t �;c khi QJ��L)�khi � x y z6

π= = = .

51) Cho A, B, C là ba góc c�a m"t tam giác. Tìm giá tr� nh� nh�t c�a: A B C

P tan tan tan2 2 2

= + + .

H�ng d!n: 2 2A B A Btan tan 2 tan . tan

2 2 2 2+ ≥ ; …

�2 2 2A B C A B B C C A

tan tan tan tan . tan tan . tan tan . tan2 2 2 2 2 2 2 2 2

+ + ≥ + +

�2 2 2A B C

tan tan tan 12 2 2

+ + ≥ (vì A B B C C A

tan .tan tan .tan tan .tan 12 2 2 2 2 2

+ + = )

22 2 2 2 2 2A B C A B C A B B C C A

tan tan tan tan tan tan 2 tan tan tan tan tan tan2 2 2 2 2 2 2 2 2 2 2 2

� � � �+ + = + + + + +� � � �

� � � �

22 2 2A B C

tan tan tan 32 2 2

� �+ + ≥� �

� � � 2 2 2A B C

tan tan tan 32 2 2

+ + ≥

�min P 3= khi A B C3

π= = = .

52) /�m NOM�KP��l�n nh�t ��a bi u th�c A = cosx + cosy − cos(x − y).

H�ng d!n: CMch 1: 2x y x y x yA 2cos cos 2cos 1

2 2 2

+ − +� �= ⋅ − −� �

� �

� 2 x y x y x y2cos 2cos cos A 1 0

2 2 2

+ + −− ⋅ + − = . �ây IJ�ph�ng KP�nh bc hai theo 2 x y

cos2

+, ��

nghi%m � 2 x y' cos 2(A 1) 0

2

+∆ = − − ≥ � 2 x y

2A 2 cos 32

+≤ + ≤ �

3A

2≤

� max A = 3

2, �$t �;c khi x y z

3

π= = = .

CMch 2: A ≤ 3

2, ∀x, y � 2x y x y x y 3

2cos cos 2cos 12 2 2 2

+ − +� �⋅ − − ≤� �

� �

� 2 x y x y x y 12cos 2cos cos 0

2 2 2 2

+ + −− ⋅ + ≤ � 2 x y x y x y

4cos 4cos cos 1 02 2 2

+ + −− ⋅ + ≤

� 2

2 2x y x y x ycos cos sin 0

2 2 2

+ + −� �− + ≥� �

� �� max A =

3

2, �$t �;c khi x y z

3

π= = = .

53) Cho ∆ABC �� Mc N�c KL�a C ≤ B ≤ A ≤ 900. /�m NOM� KP�� RL�� nh�t ��a bi u th�c A B A B

M cos sin sin2 2 2

−= .

H�ng d!n: [ ]1

M 1 cos(A B) (cos A cos B)4

= + − − +

Ta c�: ( )1

sin C.cos(A b) sin(A B).cos(A B) sin 2B sin 2A sin A cos A sin Bcos B2

− = + − = + = + .

� sin A sin B

cos(A B) cos A cos Bsin C sin C

− = + .

� ��


GT � sin A sin B sin C 0≥ ≥ > �QJ� cos A 0, cosB 0≥ > � sin A sin B

1sin C sin C

≥ ≥

� sin A sin B

cos(A B) cos A cos B cos A cos Bsin C sin C

− = + ≥ + � cos(A B) cos A cos B 0− − − ≥

� 1

M4

≥ . Khi ∆ABC �'u (C ≤ B ≤ A ≤ 900) KL��1

M4

= � min M =1

4

(1

M4

= � ∆ABC �'u ho�c ∆ABC vuông K$i A)

54) Cho A, B, C IJ� ba N�c trong ��a m"t tam NOMc. /�m NOM� KP�� RL�� nh�t ��a bi u th�c 1 1 1

M2 cos 2A 2 cos 2B 2 cos 2C

= + ++ + −

.

H�ng d!n: 3

1 1 1 1M 3

2 cos 2A 2 cos 2B 2 cos 2C (2 cos 2A)(2 cos 2B)(2 cos 2C)= + + ≥

+ + − + + −

� 22

9 9M

6 cos 2A cos 2B cos 2C 1 17 cos (A B) 2 cos C cos(A B)

2 2

≥ =+ + − � �

+ − − + −� �

� 9 6

M1 572

≥ =

+

� min M = 6

5, �$t �;c khi QJ��L)�khi A = B = 300�QJ�C = 1200.

55) /�m NOM�KP��RL��nh�t ��a bi u th�c 22 2

1A log cos xy

cos xy

� �= +� �

� �.

H�ng d!n: �i'u ki%n cos2xy ≠ 0 � 2 22 2

1 1cos xy 2 cos xy 2

cos xy cos xy+ ≥ ⋅ = � 2A log 2 1≥ =

Khi xy = π � A = 1 � min A = 1. 56) /�m NOM�KP��l�n nh�t, NOM�KP��RL��nh�t ��a bi u th�c A = 2x − y − 2, bi�t r�ng (x; y) IJ�K�a �"��i m M

thu"c elip 2 2x y

14 9

+ = .

H�ng d!n: 2 2 2

2 2 2x y x y(2x y) 4 3 (4 3 ) 25

2 3 4 9

� �� − = ⋅ − ⋅ ≤ + + =� ��

� � � �� 5 2x y 5− ≤ − ≤ � 7 A 3− ≤ ≤ .

� max A = 3, �$t �;c khi QJ��L)�khi x 8 / 5

y 9 / 5

= �

= −�; min A = −7, �$t �;c khi QJ��L)�khi

x 8 / 5

y 9 / 5

= − �

=�

57) A�i (x; y) IJ� nghi%m ��a h%� ph�ng KP�nh: x my 2 4m

mx y 3m 1

− = − �

+ = +�. /�m NOM� KP�� l�n nh�t ��a bi u th�c

2 2A x y 2x= + − .

H�ng d!n: 2

2

3m 3m 2x

m 1

− +=

+,

2

2

4m 4m 1y

m 1

+ +=

+ �

2

2

19m 4m 1A f (m)

m 1

− += =

+

� max A = 10 85+

58) UMc ��nh m � �h%�ph�ng KP�nh: 2 2x y 9

(2m 1)x my m 1 0

+ =�

+ + + − =�� hai nghi%m th�c 1 1(x ; y ) , 2 2(x ; y )

sao cho bi u th�c 2 21 2 1 2A (x x ) (y y )= − + − �$t NOM�KP��l�n nh�t.

H�ng d!n: Bi u th�c A IJ�S�nh ph�ng �LT�ng �Mch giXa hai giao �i m ��a �(ng KP\n QJ��(ng th�ng.

A �$t NOM�KP��l�n nh�t � A = (2R)2 ��(ng th�ng �i qua tâm ��a �(ng KP\n � m =1. 59) Cho L�nh �L�p S.ABC ��SA = x, BC = y, �Mc �$nh �\n I$i �'u b�ng 1.

� ��


//

//

//

//

y/2

y/2

x/2

x/2

N

M

C

B

A

S

a) /]nh th �K]ch V ��a L�nh �L�p theo x, y. b) /�m x, y � �th �K]ch V IJ�l�n nh�t.

H�ng d!n: a) A�i M, N l*n l;t IJ�trung �i m ��a SA, BC. ∆SAB, ∆SAC cân �SA BM⊥ , SA CM⊥ �SA (BCM)⊥ .

� BCM BCM

1 xV SA.S S

3 3= = .

2xBM CM 1

4= = −

�∆BCM cân K$i M QJ�MN BC⊥

�

2 2 22 y x y

MN MC 12 4

+� �= − = −� �

� �

� 2 2

BCM

1 x yS y 1

2 4

+= − �

2 2xy x yV 1

6 4

+= −

b) 2 2x y 2xy xy

4 4 2

+≥ = � 2 2xy xy 1 2 xy 1 xy xy

V 1 x y 2 (2 xy)6 2 6 2 6 2 2

−� �≤ − = = −� �

� �

�

3xy xy

2 xy1 1 16 2 32 2V 26 2 6 27 27

� �+ + −� �

≤ ≤ =� ��

� max V = 2 3


2 2x y 2xy

xy2 xy

2

+ =��

= −��

� 2

x y3

= =

60) Trong không gian Oxyz cho �i m M(1; 2; 3). A�i (P) IJ�m�t ph�ng �i qua M QJ�c^t Ox, Oy, Oz l*n l;t K$i A(a; 0; 0), B(0; b; 0), C(0; 0; c) (v�i a > 0, b > 0, c > 0). /�m a, b, c � � th � K]ch kh�i t��di%n OABC IJ�RL��nh�t.

H�ng d!n: OABC

1 1V OA.OB.OC abc

6 6= = .

x y z(P) : 1

a b c+ + = . (P) �i qua M �

1 2 31

a b c+ + =

� 31 2 3 6

1 3a b c abc

= + + ≥ � 6

1 27abc

≥ ⋅ � abc 27.6≥ � OABC

1V abc 27

6= ≥

� min OABCV 27= , �$t �;c khi QJ��L)�khi 1 2 3 1

a b c 3= = = �

a 3

b 6

c 9

= �

=�� =�

61) Cho a, b, c , d IJ�b�n s��nguyên KL�a 1 a b c d 50≤ < < < ≤ . Ch�ng minh 2a c b b 50

b d 50b

+ ++ ≥ �QJ�K�m

NOM�KP��RL��nh�t ��a bi u th�c a c

Ab d

= + .

H�ng d!n: a 1, d 50, c b≥ ≤ > � c b 1≥ + (Q��b, c N∈ ) � 2a c 1 b 1 b b 50

b d b 50 50b

+ + ++ ≥ + =

��ng th�c ��y ra � a = 1, d = 50, c = b + 1. 2a c b b 50 b 1 1

Ab d 50b 50 b 50

+ += + ≥ = + + . UVt LJm s��

x 1 1f (x)

50 x 50= + + trên �T$n [2; 48].

f ’(x) = 0 � x 5 2= ∉N ( x 5 2= − �IT$i) � min f(x) = f(5 2 )

� min f(b) = min{f(7), f(8)} = f(7) =53

175 khi a = 1, b = 7, c = 8, d = 50.

62) /�m �Mc �i m 2x 2

M (C) : yx 1

+∈ =

− sao cho t&ng �LT�ng �Mch tW�M ��n hai �(ng ti%m cn ��a (C) IJ�

RL��nh�t. �Mp s�: M1(3; 4), M2(−1; 0).

� ��


f(y)

1/ 3

2+ 3

2-�

f '(y)

y

0

63) �/�m hai �i m M, N l*n l;t thu"c hai RLMnh ��a 2x x 1

(C) : yx 1

+ +=

+ sao cho �T$n MN RL��nh�t.

H�ng d!n: 1

(C) : y xx 1

= ++

��TC� x = −1. AO��s �1

M 1 a; 1 aa

� �− + − + +� ��

, 1

N 1 b; 1 bb

� �− − − − −� ��

(a 0, b 0)> > �IJ�hai �i m thu"c hai RLMnh ��a (C). � 2

2 2 1 1MN (a b) a b

a b� �

= + + + + +� ��

� 2

2 2 22 2 2 2

1 2 1 2 1MN (a b) 1 1 (a b) 2 4ab 2

ab ab a b ab a b

� �� = + + + = + + + ≥ + +� ��

� � � � � ��

( )2 1 1MN 4 2ab 2 4 2 2ab 2 4 2 2 2

ab ab

� �� ≥ + + ≥ ⋅ + = +� ��

� min MN = 2 2(1 2)+ , �$t �;c khi

QJ��L)�khi 4

1a b

2= = . Khi �� 4

4 4

1 1M 1 ; 1 2

2 2

� �− + − + +� ��

, 4

4 4

1 1N 1 ; 1 2

2 2

� �− − − − −� ��

.

64) Cho hai s��th�c x 0, y 0≠ ≠ thay �&i QJ�KL�a _Yn �i'u ki%n: 2 2(x y)xy x y xy.+ = + − �/�m NOM�KP��l�n

nh�t ��a bi u th�c: 3 3

1 1A

x y= + . (�'�thi �$i�L�c kh�i A n�m 2006)

H�ng d!n: 2 2(x y)xy x y xy+ = + − �2 2

1 1 1 1 1

x y x y xy+ = + − . ��t

1 1a, b

x y= = � 2 2a b a b ab+ = + − (1)

�3 3 2 2 2A a b (a b)(a b ab) (a b)= + = + + − = + .

(1) � 2a b (a b) 3ab+ = + − . Vì 2

a bab

2

+� �≤ � ��

� 2 23a b (a b) (a b)

4+ ≥ + − + � 2(a b) 4(a b) 0+ − + ≤

�0 (a b) 4≤ + ≤ � 2A (a b) 16= + ≤ . V�i 1

x y2

= = thì A 16= . Vy max A 16= .

65) Cho x, y IJ� �Mc s�� th�c thay �&i. /�m NOM� KP�� RL�� nh�t ��a bi u th�c: 2 2 2 2A (x 1) y (x 1) y y 2= − + + + + + − . (�'�thi �$i L�c kh�i B n�m 2006)

H�ng d!n: Ch�n a(1 x; y)−�

, b(1 x; y)+�

� 2A a b y 2 a b y 2 2 1 y y 2= + + − ≥ + + − = + + −� � � �

(D�u “=” x�y ra � x = 0). Xét hàm s� 2f (y) 2 1 y y 2= + + −

• y 2≥ : Hàm f luôn t�ng.

• y < 2: 2

2yf '(y) 1

1 y= −

+. f '(y) 0= �

1y

3=

�1

min A f 2 33

� �= = +� �

� �.

bdt gtln gtnn

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