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PPHHNN II:: LL TTHHUUYYTT CCNN BBNN
1. TNH CHT BT NG THC
TT iu kin Tnh cht c im1. a b
a cb c
> >> Bc cu
2. a b a c b c< + < + Cng 2 v BT vi mt s bt k3. c > 0 a b ac bc> > 4. c < 0 a b ac bc> <
Nhn hay chia 2 v BT vi mt s
5. a ba c b d
c d
> + > +
> Cng theo v 2 BT cng chiu
6. 0
0b
d
>>
a b ac bd c d
> >
> Nhn theo v 2 BT cng chiu
7.n nguyn dng
2 1 2 1
2 20
n n
n n
a b a b
a b a b
+ + < a a x a
12.0a >
x ax a
x a
13. a b a b a b + +
BT tr tuyt i
14. N> M A N < Suy t 2. v 4.15. 0
0N> >
>
A A
N< Suy t 9. v 3.
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2. CC BT NG THC THNG DNG BT NG THC CSI:
*Du hiu s dng: S dng khi cc s khng mL: Cho n s khng m a1 , a2 , . . . , an Ta c:
1 2
1 2 1 2 1 2
1 2 1 21 2 1 2
1 2
...
... ... ...
... ...... ...
ng thc xy ra ...
n
nn
n n n
n
n nnn n
n
a a a
a a a n a a a hay a a an
a a a a a aa a a hay a a a
n n
a a a
+ + +
+ + +
+ + + + + +
= = =
HQ: Cho n s dng a1 , a2 , . . . , an Ta c:
+ + + + + +
+ + + + + + = = =
2
2
1 2 1 2 1 2 1 2
1 2
1 1 1 1 1 1 1 1... ...
... ...ng thc xy ra ...
n n n n
n
nhay
a a a a a a a a a n a a aa a a
BT NG THC BUNHIAKPSKI:
*Du hiu s dng: Tng ca cc bnh phngL: Cho n cp s a1 , a2 , . . . , an v b1 , b2 , . . . , bn . Ta c:
( ) ( ) ( )
2 2 2 2 2 21 1 2 2 1 2 1 2
2 2 2 2 2 2 21 1 2 2 1 2 1 2
... ... ...
... ... ...
n n n n
n n n n
a b a b a b a a a b b b
hay a b a b a b a a a b b b
+ + + + + + + + +
+ + + + + + + + +
ng thc xy ra 1 21 2
... n
n
a a a
b b b = = =
*Ch : A & ( i k>0)A th A A k k A k V
HQ: Trong khng gian n chiu cho 2 vect 1 2 1 2( ; ;...; ) , ( ; ;...; )n nu a a a v b b b= =
: . .Tac u v u v
ng thc xy ra Hai vect trn cng phng
Gii thch: ( ): . . .cos ,Theonh ngha uv u v u v =
BT NG THC VECT ( Phng php hnh hc )
*Du hiu s dng: ng ca cc bnh phngT
1/ . ng thc xy ra Hai vect cng hnga b a b+ +
M rng: a b c a b c + + + +
2/ . . . ng thc xy ra Hai vect cng phngu v u v
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PPHHNN IIII:: LLUUYYNN TTPP CCNN BBNN
II.. CCHHNNGG MMIINNHH BBTT DDAA VVOO NNHH NNGGHHAA VV TTNNHH CCHHTT CCBBNN
1. Cho a, b > 0 chng minh: + +
33 3
a b a b2 2 (*)
Gii:
(*) + +
33 3a b a b0
2 2 ( )( )+
23a b a b 0
8. PCM.
2. Chng minh:+ +
2 2a b a b
2 2()
Gii: a + b 0 , () lun ng.
a + b > 0 , () + + +
2 2 2 2a b 2ab a b
04 2
( )
2
a b0
4, ng.
Vy:+ +
2 2a b a b
2 2.
3. Cho a + b 0 chng minh:+ +
3 3
3a b a b
2 2
Gii:+ +
3 3
3a b a b
2 2
( )+ +
3 3 3a b a b
8 2
( )( ) 2 23 b a a b 0 ( ) ( ) + 23 b a a b 0 , PCM.
4. Cho a, b > 0 . Chng minh: + +a b
a bb a
()
Gii:
() + +a a b b a b b a ( ) ( ) a b a a b b 0 ( )( ) a b a b 0 ( ) ( ) +
2a b a b 0 , PCM.
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5. Chng minh: Vi a b 1: + ++ +2 2
1 1 2
1 ab1 a 1 b
Gii:
+ ++ +2 2
1 1 2
1 ab1 a 1 b +
+ ++ +2 21 1 1 1
01 ab 1 ab1 a 1 b
( ) ( ) ( ) ( ) + + + + +
2 2
2 2ab a ab b 01 a 1 ab 1 b 1 ab ( )( )( ) ( )( )( ) + + + + +2 2
a b a b a b 01 a 1 ab 1 b 1 ab
+ + + 2 2b a a b
01 ab 1 a 1 b
( )( )
+ + + +
2 2
2 2
b a a ab b ba0
1 ab 1 a 1 b
( ) ( )
( )( ) ( )
+ + +
2
2 2
b a ab 10
1 ab 1 a 1 b, PCM.
V : a b 1 ab 1 ab 1 0.
6. Chng minh: ( )+ + + + +2 2 2a b c 3 2 a b c ; a , b , c RGii:
( )+ + + + +2 2 2a b c 3 2 a b c ( ) ( ) ( ) + + 2 2 2
a 1 b 1 c 1 0 . PCM.
7. Chng minh: ( )+ + + + + + +2 2 2 2 2a b c d e a b c d e Gii:
( )+ + + + + + +2 2 2 2 2a b c d e a b c d e
+ + + + + + + 2 2 2 22 2 2 2a a a aab b ac c ad d ae e 04 4 4 4
+ + +
2 2 2 2a a a a
b c d e 02 2 2 2
. PCM
8. Chng minh: + + + +2 2 2x y z xy yz zx Gii:
+ + + +2 2 2x y z xy yz zx + + 2 2 22x 2y 2z 2xy 2yz 2zx 0
( ) ( ) ( ) + + 2 22x y x z y z 0
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9. a. Chng minh:+ + + +
a b c ab bc ca
; a,b,c 03 3
b. Chng minh:+ + + +
22 2 2a b c a b c
3 3
Gii:
a. Chng minh: + + + + a b c ab bc ca ; a,b,c 03 3
+ + + +2 2 2a b c ab bc ca
+ + + + + + + + +
=
2 2 2 2a b c a b c 2ab 2bc 2ca ab bc ca
3 9 3
+ + + +
a b c ab bc ca
3 3
b. Chng minh:
+ + + +
22 2 2a b c a b c
3 3
( ) ( )+ + = + + + + +2 2 2 2 2 2 2 2 23 a b c a b c 2 a b c
( ) ( ) + + + + + = + + 22 2 2a b c 2 ab bc ca a b c
+ + + +
22 2 2a b c a b c
3 3
10. Chng minh: + + +2
2 2a b c ab ac 2bc4
Gii:
+ + +2
2 2a b c ab ac 2bc4
( ) + + 2
2 2a a b c b c 2bc 04
( )
2a
b c 02
.
11. Chng minh: + + + +2 2a b 1 ab a b Gii:
+ + + +2 2a b 1 ab a b + + 2 22a 2b 2 2ab 2a 2b 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 .
12. Chng minh: + + +2 2 2x y z 2xy 2xz 2yz Gii:
+ + +2 2 2x y z 2xy 2xz 2yz + + + 2 2 2x y z 2xy 2xz 2yz 0 (x y + z)2 0.
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13. Chng minh: + + + + +4 4 2 2x y z 1 2xy(xy x z 1) Gii:
+ + + + +4 4 2 2x y z 1 2x(xy x z 1) + + + + 4 4 2 2 2 2x y z 1 2x y 2x 2xz 2x 0
( ) ( ) ( ) + + 2 2 22 2x y x z x 1 0 .
14. Chng minh: Nu a + b 1 th: + 3 3 1a b
4
Gii: a + b 1 b 1 a b3 = (1 a)3 = 1 a + a2 a3
a3 + b3 = +
21 1 1
3 a2 4 4
.
15. Cho a, b, c l s o di 3 cnh ca 1 tam gic. Chng minh:a. ab + bc + ca a2 + b2 + c2 < 2(ab + bc + ca).b. abc (a + b c)(a + c b)(b + c a)c. 2a2b2 + 2b2c2 + 2c2a2 a4 b4 c4 > 0
Gii:a. ab + bc + ca a2 + b2 + c2 < 2(ab + bc + ca). ab + bc + ca a2 + b2 + c2 (a b)2 + (a c)2 + (b c)2 > > > a b c , b a c , c a b
> +2 2 2a b 2bc c , > +2 2 2b a 2ac c , > +2 2 2c a 2ab b a2 + b2 + c2 < 2(ab + bc + ca).
b. abc (a + b c)(a + c b)(b + c a)
( )> 22 2a a b c ( )( )> + + 2a a c b a b c
( )> 22 2b b a c ( )( )> + + 2b b c a a b c
( )> 22 2c c a b ( )( )> + + 2c b c a a c b
( ) ( ) ( )> + + + 2 2 22 2 2a b c a b c a c b b c a
( )( )( )> + + + abc a b c a c b b c a
c. 2a2b2 + 2b2c2 + 2c2a2 a4 b4 c4 > 0 4a2b2 + 2c2(b2 + a2) a4 b4 2a2b2 c4 > 0 4a2b2 + 2c2(b2 + a2) (a2 + b2)2 c4 > 0 (2ab)2 [(a2 + b2) c2]2 > 0 [c2 (a b)2][(a + b)2 c2] > 0 (c a + b)(c + a b)(a + b c)(a + b + c) > 0 . ng
V a , b , c l ba cnh ca tam gic c a + b > 0 , c + a b > 0 , a + b c > 0 , a + b + c > 0.
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IIII.. CCHHNNGG MMIINNHH BBTT DDAA VVOO BBTT CCSSII
1. Chng minh: + + + (a b)(b c)(c a) 8abc ; a,b,c 0 Gii: p dng bt ng thc Csi cho hai s khng m:
+ a b 2 ab , + b c 2 bc , + a c 2 ac
( )( )( )+ + + =2 2 2a b b c a c 8 a b c 8abc .
2. Chng minh: + + + + 2 2 2(a b c)(a b c ) 9abc ; a,b,c 0 Gii: p dng bt ng thc Csi cho ba s khng m:
+ + 3a b c 3 abc , + + 32 2 2 2 2 2a b c 3 a b c
( )( )+ + + + =32 2 2 3 3 3a b c a b c 9 a b c 9abc .
3. Chng minh: ( )( )( ) ( )+ + + +331 a 1 b 1 c 1 abc vi a , b , c 0
Gii:
( )( )( )+ + + = + + + + + + +1 a 1 b 1 c 1 a b c ab ac bc abc.
+ + 3a b c 3 abc , + + 3 2 2 2ab ac bc 3 a b c
( )( )( ) ( )+ + + + + + = +33 2 2 23 31 a 1 b 1 c 1 3 abc 3 a b c abc 1 abc
4. Cho a, b > 0. Chng minh: +
+ + +
m mm 1a b1 1 2
b a, vi m Z+
Gii:
+ + + + + + = + + =
m m m m mm m 1a b a b b a1 1 2 1 . 1 2 2 2 4 2
b a b a a b
5. Chng minh: + + + + bc ca ab
a b c ; a,b,c 0a b c
Gii: p dng BT Csi cho hai s khng m:
+ =2bc ca abc
2 2ca b ab
, + =2bc ba b ac
2 2ba c ac
, + =2ca ab a bc
2 2ab c bc
+ + + +bc ca ab
a b ca b c
.
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6. Chng minh:+
6 9
2 3x y 3x y 16 ; x,y 04
Gii:+
6 9
2 3x y 3x y 16 ; x,y 04
+ + 6 9 2 3x y 64 12x y ( ) ( )+ + 3 32 3 3 2 3x y 4 12x y
p dng BT Csi cho ba s khng m:
( ) ( )+ + =3 32 3 3 2 3 2 3x y 4 3x y 4 12x y .
7. Chng minh: + +
4 22
12a 3a 1
1 a.
Gii:
+ +
4 22
12a 3a 1
1 a + + + +
+4 4 2 2
2
1a a a 1 4a
1 a.
p dng BT Csi cho 4 s khng m: +
+
4 4 22
1a , a , a 1,
1 a
( )+ + + + + =+ +
4 4 2 4 4 2 242 2
1 1a a a 1 4 a a a 1 4a
1 a 1 a
8. Chng minh: ( )> 1995a 1995 a 1 , a > 0Gii:
( )> 1995a 1995 a 1
> + >1995 1995a 1995a 1995 a 1995 1995a
+ > + = + + + + =
19951995 1995 1995 1995
1994 soa 1995 a 1994 a 1 1 ... 1 1995 a 1995a
9. Chng minh: ( ) ( ) ( )+ + + + + 2 2 2 2 2 2a 1 b b 1 c c 1 a 6abc .Gii:
( ) ( ) ( )+ + + + + = + + + + +2 2 2 2 2 2 2 2 2 2 2 2 2 2 2a 1 b b 1 c c 1 a a a b b b c c c a p dng bt ng thc Csi cho 6 s khng m:
+ + + + + =62 2 2 2 2 2 2 2 2 6 6 6a a b b b c c c a 6 a b c 6abc
10. Cho a , b > 0. Chng minh: + + + + + + +2 2 2 2 2 2
a b c 1 1 1 1
2 a b ca b b c a c
Gii:
=+2 2a a 1
2ab 2ba b, =
+2 2b b 1
2bc 2cb c, =
+2 2c c 1
2ac 2aa c
Vy:
+ + + + + + +2 2 2 2 2 2
a b c 1 1 1 1
2 a b ca b b c a c
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11. Cho a , b 1 , chng minh: + ab a b 1 b a 1.Gii:
( ) ( )= + = + a a 1 1 2 a 1, b b 1 1 2 b 1
ab 2b a 1 , ab 2a b 1
+ ab a b 1 b a 1
12. Cho x, y, z > 1 v x + y + z = 4. Chng minh: xyz 64(x 1)(y 1)(z 1)Gii:
( ) ( )= + = + + + x x 1 1 x 1 x y z 3
( ) ( ) ( ) ( ) ( ) ( )( )= + + + 24x 1 x 1 y 1 z 1 4 x 1 y 1 z 1
Tng t: ( )( ) ( ) 24y 4 x 1 y 1 z 1 ; ( )( )( )
24z 4 x 1 y 1 z 1
xyz 64(x 1)(y 1)(z 1).
13. Cho a > b > c, Chng minh: ( )( ) 3
a 3 a b b c c .Gii:
( ) ( ) ( )( )= + + 3a a b b c c 3 a b b c c
14. Cho: a , b , c > 0 v a + b + c = 1. Chng minh:a) b + c 16abc.b) (1 a)(1 b)(1 c) 8abc
c)
+ + +
1 1 11 1 1 64
a b c
Gii:
a) b + c 16abc.
+
2b c
bc2
( )+
= =
2 22b c 1 a
16abc 16a 16a 4a 1 a2 2
( ) ( )( ) ( ) ( ) = = = + 2 224a 1 a 1 a 4a 4a 1 a 1 1 2a 1 a b c
b) (1 a)(1 b)(1 c) 8abc
(1 a)(1 b)(1 c) = (b + c)(a + c)(a + b) =2 bc.2 ac.2 ab 8abc
c)
+ + +
1 1 11 1 1 64
a b c
+ + +
+ =
4 21 a a b c 4 a bc1
a a a
+ 4 21 4 ab c
1b b
+ 4 21 4 abc
1c c
+ + +
1 1 11 1 1 64
a b c
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15. Cho x > y > 0 . Chng minh:( )
+
1x 3
x y y
Gii:
( )( )
( )
( )
= + + =
3
x y y1VT x y y 3 3
x y y x y y
16. Chng minh:
a)+
+
2
2
x 22
x 1,x R b)
+
x 86
x 1, x > 1 c)
+
+
2
2
a 54
a 1
Gii:
a)+
+
2
2
x 22
x 1 + +2 2x 2 2 x 1 + + +2 2x 1 1 2 x 1
b) +
x 8x 1
= + = + = x 1 9 9 9x 1 2 x 1 6
x 1 x 1 x 1
c. ( ) ( )+ + + = +2 2 2a 1 4 2 4 a 1 4 a 1+
+
2
2
a 54
a 1
17. Chng minh:+ +
+ + >+ + +
ab bc ca a b c; a, b, c 0
a b b c c a 2
Gii: V : + a b 2 ab
=+
ab ab ab
a b 22 ab, =
+bc bc bc
b c 22 bc, =+
ac ac ac
a c 22 ac
+ + + +a b c ab bc ca , da vo: + + + +2 2 2a b c ab bc ca .
+ + + +
+ + + + +
ab bc ca ab bc ac a b c
a b b c c a 2 2
18. Chng minh: + + +
2 2
4 4
x y 1
41 16x 1 16y, x , y R
Gii:
( )
= =+ +
2 2 2
4 2 2
x x x 1
81 16x 2.4x1 4x
( )
= =+ +
2 2 2
4 2 2
y y y 1
81 16y 2.4y1 4y
+ + +
2 2
4 4
x y 1
41 16x 1 16y
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19. Chng minh: + + + + +a b c 3
b c a c a b 2; a , b , c > 0
Gii: t X = b + c , Y = c + a , Z = a + b.
a + b + c = 12
(X + Y + Z)
+ + +
= = =Y Z X Z X Y X Y Z
a , b , c2 2 2
+ + = + + + + + + + +
a b c 1 Y X Z X Z Y3
b c a c a b 2 X Y X Z Y Z [ ] + + =
1 32 2 2 3
2 2.
Cch khc:
+ + = + + + + + + + + + + +
a b c a b c1 1 1 3
b c a c a b b c a c a b
( ) ( ) ( )[ ] = + + + + + + + + + +
1 1 1 1a b b c c a 3
2 b c a c a b
p dng bt ng thc Csi cho ba s khng m:
( ) ( ) ( )[ ]
+ + + + + + + = + + +
1 1 1 1 9 3a b b c c a 32 b c a c a b 2 2
20. Cho a , b , c > 0. C/m: + + + + + + + +3 3 3 3 3 3
1 1 1 1
abca b abc b c abc c a abc
Gii:
( )( ) ( )+ = + + +3 3 2 2a b a b a ab a a b ab
( ) ( )+ + + + = + +3 3a b abc a b ab abc ab a b c , tng t
( ) ( )+ + + + = + +3 3b c abc b c bc abc bc a b c
( ) ( )+ + + + = + +3 3c a abc c a ca abc ca a b c
( ) ( ) ( ) + + + + = + + + + + + + + 1 1 1 1 a b cVT
ab a b c bc a b c ca a b c a b c abc
21. p dng BT Csi cho hai s chng minh:
a. + + + 4a b c d 4 abcd vi a , b , c , d 0 (Csi 4 s)
b. + + 3a b c 3 abc vi a , b , c 0 , (Csi 3 s )Gii:
a. + + + 4a b c d 4 abcd vi a , b , c , d 0 (Csi 4 s)
+ + a b 2 ab , c d 2 cd
( ) ( )+ + + 4a b cd 2 ab cd 2 2 ab. cd 4 abcd b. + + 3a b c 3 abc vi a , b , c 0 , (Csi 3 s )
+ + + +
+ + + 4a b c a b c
a b c 4. abc3 3
+ + + +
4a b c a b c
abc3 3
+ + + +
4a b c a b c
abc3 3
+ +
3a b c
abc3
+ + 3a b c 3 abc .
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22. Chng minh: + + + +3 3 3 2 2 2a b c a bc b ac c ab ; a , b , c > 0Gii:
+ 3 2a abc 2a bc , + 3 2b abc 2b ac , + 3 2c abc 2c ab
( )+ + + + +3 3 3 2 2 2a b c 3abc 2 a bc b ac c ab
( ) ( )+ + + +3 3 3 2 2 22 a b c 2 a bc b ac c ab , v : + + 3 3 3a b c 3abc Vy: + + + +3 3 3 2 2 2a b c a bc b ac c ab
23. Chng minh: + + 3 942 a 3 b 4 c 9 abc Gii: p dng bt ng thc Csi cho 9 s khng m:
= + + + + + + + + 3 3 3 94 4 4 4VT a a b b b c c c c 9 abc
T BI 24 N BI 38 C TH DNG PHNG PHP O HM
24. Cho = +x 18
y2 x
, x > 0. nh x y t GTNN.
Gii:
p dng BT Csi cho hai s khng m: = + =x 18 x 18
y 2 . 62 x 2 x
Du = xy ra = = = 2x 18
x 36 x 62 x
, chn x = 6.
Vy: Khi x = 6 th y t GTNN bng 6
25. Cho = + >
x 2y ,x 1
2 x 1. nh x y t GTNN.
Gii:
= + +
x 1 2 1y
2 x 1 2
p dng bt ng thc Csi cho hai s khng m
x 1 2
,2 x 1
:
= + + + =
x 1 2 1 x 1 2 1 5y 2 .
2 x 1 2 2 x 1 2 2
Du = xy ra ( )=
= = =
2 x 3x 1 2x 1 4
x 1(loai)2 x 1
Vy: Khi x = 3 th y t GTNN bng 52
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26. Cho = + > +
3x 1y , x 1
2 x 1. nh x y t GTNN.
Gii:
+
= + +
3(x 1) 1 3y
2 x 1 2
p dng bt ng thc Csi cho hai s khng m( )+
+3 x 1 1
,2 x 1 :
( ) ( )+ += + =
+ +3 x 1 1 3 3 x 1 1 3 3
y 2 . 62 x 1 2 2 x 1 2 2
Du = xy ra ( )
( )
= + = + =
+=
2
6x 1
3 x 1 1 2 3x 12 x 1 3 6
x 1(loai)3
Vy: Khi = 6x 1
3
th y t GTNN bng 3
6
2
27. Cho = + >
x 5 1y ,x
3 2x 1 2. nh x y t GTNN.
Gii:
= + +
2x 1 5 1y
6 2x 1 3
p dng bt ng thc Csi cho hai s khng m
2x 1 5
,6 2x 1
:
+= + + + =
2x 1 5 1 2x 1 5 1 30 1
y 2 .6 2x 1 3 6 2x 1 3 3
Du = xy ra ( )
+= = =
+=
2
30 1x
2x 1 5 22x 1 306 2x 1 30 1
x (loai)2
Vy: Khi+
=30 1
x2
th y t GTNN bng+30 1
3
28. Cho = +x 5
y1 x x
, 0 < x < 1 . nh x y t GTNN.
Gii:
( ) +
= + = + + + = + x 5 1 x 5x x x 1 x 1 x
f(x) 5 5 2 5 5 2 5 51 x x 1 x x 1 x x
Du = xy ra
= = =
2x 1 x x 5 5
5 5 x1 x x 1 x 4
(0 < x < 1)
Vy: GTNN ca y l +2 5 5 khi
=5 5
x4
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29. Cho+
=3
2
x 1y
x, x > 0 . nh x y t GTNN.
Gii:
+
= + = + + =3
32 2 2 2 3
x 1 1 x x 1 x x 1 3x 3
2 2 2 2 4x x x x
Du = xy ra = = 2x x 12 2 x
= 3x 2 .
Vy: GTNN ca y l3
3
4khi = 3x 2
30. Tm GTNN ca+ +
=2x 4x 4
f(x)x
, x > 0.
Gii:
+ +
= + + + =
2x 4x 4 4 4x 4 2 x. 4 8x x x
Du = xy ra =4
xx
x = 2 (x > 0).
Vy: GTNN ca y l 8 khi x = 2.
31. Tm GTNN ca = +2 32
f(x) xx
, x > 0.
Gii:
+ = + + + + =
322 2 2 22 5
3 3 3 3 52 x x x 1 1 x 1 5x 5
3 3 3 3 27x x x x
Du = xy ra = =2
53
x 1x 3
3 x x = 2 (x > 0).
Vy: GTNN ca y l5
5
2 7khi = 5x 3 .
32. Tm GTLN ca f(x) = (2x 1)(3 5x)
Gii: f(x) = 10x2 + 11x 3 =
= +
22 11x 11 1 110 x 3 10 x
10 20 40 40
Du = xy ra =1 1
x2 0
Vy: Khi =1 1
x2 0
th y t GTLN bng 14 0
.
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33. Cho y = x(6 x) , 0 x 6 . nh x y t GTLN.Gii: p dng BT Csi cho 2 s khng m x v 6 x (v 0 x 6):
( ) ( )= + 6 x 6 x 2 x 6 x x(6 x) 9 Du = xy ra x = 6 x x = 3 Vy: Khi x = 3 th y t GTLN bng 9.
34. Cho y = (x + 3)(5 2x) , 3 x 52
. nh x y t GTLN
Gii:
y = (x + 3)(5 2x) = 12
(2x + 6)(5 2x)
p dng BT Csi cho 2 s khng m 2x + 6 v 5 2x ,
53 x
2:
( ) ( ) ( )( )= + + + 11 2x 6 5 2x 2 2x 6 5 2x 12
(2x + 6)(5 2x) 1218
Du = xy ra 2x + 6 = 5 2x = 1x 4
Vy: Khi = 1
x4
th y t GTLN bng121
8.
35. Cho y = (2x + 5)(5 x) , 5
x 52
. nh x y t GTLN
Gii:
y = (2x + 5)(5 x) = 12
(2x + 5)(10 2x)
p dng BT Csi cho 2 s khng m 2x + 5 , 10 2x , 5 x 52 :
( ) ( ) ( )( )+ + + 2x 5 10 2x 2 2x 5 10 2x 12
(2x + 5)(10 2x) 6 2 5
8
Du = xy ra 2x + 5 = 10 2x = 5x4
Vy: Khi =5
x4
th y t GTLN bng625
8
36. Cho y = (6x + 3)(5 2x) , 1
2
x 5
2
. nh x y t GTLN
Gii: y = 3(2x + 1)(5 2x)
p dng BT Csi cho 2 s khng m 2x + 1 , 5 2x ,
1 5x
2 2:
( ) ( ) ( )( )+ + + 2x 1 5 2x 2 2x 1 5 2x (2x + 1)(5 2x) 9 Du = xy ra 2x + 1 = 5 2x x = 1 Vy: Khi x = 1 th y t GTLN bng 9.
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37. Cho =+2x
yx 2
. nh x y t GTLN
Gii:
+ =2 22 x 2 2x 2x 2 + 2
1 x
2 2 2 x
1y
2 2
Du = xy ra = 2
x 2 & x > 0 x= 2 Vy: Khi =x 2 th y t GTLN bng
1
2 2.
38. Cho( )
=+
2
32
xy
x 2. nh x y t GTLN
Gii:
+ = + + 32 2 2x 2 x 1 1 3 x .1.1 ( )
( )+
+
232 232
x 1x 2 27x
27x 2
Du = xy ra = =
2
x 1 x 1 Vy: Khi = x 1 th y t GTLN bng
1
27.
IIIIII.. CCHHNNGG MMIINNHH BBTT DDAA VVOO BBTT BBUUNNHHIIAACCPPXXKKII
1. Chng minh: (ab + cd)2 (a2 + c2)(b2 + d2) BT BunhiacopxkiGii :
(ab + cd)2 (a2 + c2)(b2 + d2)
+ + + + +2 2 2 2 2 2 2 2 2 2 2 2a b 2abcd c d a b a d c b c d
+ 2 2 2 2a d c b 2abcd 0 ( ) 2
ad cb 0 .
2. Chng minh: + sin x cos x 2 Gii : p dng BT Bunhiacopski cho 4 s 1 , sinx , 1 , cosx :
+ =sinx cosx ( ) ( )+ + + =2 2 2 21. sinx 1. cosx 1 1 sin x cos x 2
3. Cho 3a 4b = 7. Chng minh: 3a2 + 4b2 7.Gii :
p dng BT Bunhiacopski cho 4 s 3 , 3 a , 4 , 4 b :
( )( )+ = + + +2 23a 4b 3. 3a 4. 4b 3 4 3a 4b 3a2 + 4b2 7.
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4. Cho 2a 3b = 7. Chng minh: 3a2 + 5b2725
47.
Gii :
= 2 3
2a 3b 3a 5b3 5
p dng BT Bunhiacopski cho 4 s
2 3
, 3 a , , 5b3 5 :
( )
+ +
2 22 3 4 93 a 5b 3a 5b3 53 5
3a2 + 5b27 3 5
4 7.
5. Cho 3a 5b = 8. Chng minh: 7a2 + 11b22464
13 7.
Gii :
= 3 5
3a 5b 7 a 11b7 11
p dng BT Bunhiacopski cho 4 s 3 5, 7 a , , 11b7 11
:
( )
+ +
2 23 5 9 257 a 11b 7a 11b7 117 11
7a2 + 11b22 4 6 4
13 7.
6. Cho a + b = 2. Chng minh: a4 + b4 2.Gii : p dng BT Bunhiacopski:
( )( )
= + + +
2 2
2 a b 1 1 a b a2
+ b
2
2 ( ) ( )( ) + + +2 2 4 42 a b 1 1 a b a4 + b4 2
7. Cho a + b 1 Chng minh: + 2 21
a b2
Gii :
( ) ( ) + + + + 2 2 2 2 2 21
1 a b 1 1 a b a b2
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PPHHNN IIIIII.. TTHHII II HHCCNM 1999
1. (H Y Dc TP HCM 1999)
Cho 3 s a, b, c khc 0. Chng minh: + + + +2 2 2
2 2 2
a b c a b c
b c ab c a
Gii:p dng BT Csi ta c:
* + + =2 2 2 2 2 2
32 2 2 2 2 2
a b c a b c3 . . 3
b c a b c a(1)
* + 2
2
a a1 2
bb; +
2
2
b b1 2
cc; +
2
2
c c1 2
aa
+ + + +
2 2 2
2 2 2
a b c a b c2 3
b c ab c a(2)
Kt hp (1) v (2) ta c: + + + +
2 2 2
2 2 2
a b c a b c2 2
b c ab c a
+ + + +2 2 2
2 2 2
a b c a b c
b c ab c a
2. (H Hng hi 1999)Cho x, y, z 0 v x + y + z 3. Chng minh rng:
+ + + ++ + ++ + +2 2 2
x y z 3 1 1 1
2 1 x 1 y 1 z1 x 1 y 1 z
Gii:
Do (x 1)2 0 nn x2 + 1 2x + 22x
1 x 1
Tng t ta cng c:+ 22y
1 y 1;
+ 22 z
1 z 1
Do :+ 22 x
1 x+
+ 22 y
1 y+
+ 22 z
1 z 3
Hay: + + + + +2 2 2x y z 3
21 x 1 y 1 z(1)
p dng BT Csi cho 3 s khng m ta c:
+ ++ + +
=+ + + + + +3 3
1 1 1
1 11 x 1 y 1 z
3 (1 x)(1 y)(1 z) (1 x)(1 y)(1 z)
+ + ++ +
+ + +
33 (1 x)(1 y)(1 z)1 1 1
1 x 1 y 1 z
+ + + + +(1 x) (1 y) (1 z)
3 2
+ ++ + +3 1 1 1
2 1 x 1 y 1 z(2)
Kt hp (1) v (2) ta c BT cn chng minh.
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3. (H An ninh HN khi D 1999)Cho 3 s x, y, z thay i, nhn gi tr thuc on [0;1]. Chng minh rng:
2(x3 + y3 + z3) (x2y + y2z + z2x) 3 (*)
Gii:V 0 x, y, z 1 nn x2 x3; y2 y3; z2 z3.Suy ra: 2(x3 + y3 + z3) (x2y + y2z + z2x) 2(x2 + y2 + z2) (x2y + y2z + z2x)Do nu ta chng minh c:
2(x2 + y2 + z2) (x2y + y2z + z2x) 3 (1)th (*) ng.Ta c: (1 y)(1 + y x2) 0 x2 + y2 x2y 1 0 (2)
Du = (2) xy ra
=
= =
y 1
x 1
y 0
Tng t ta cng c: x2 + z2 z2x 1 0 (3)y2 + z2 y2z 1 0 (4)
Cng (2), (3), (4) v theo v ta c:2(x2 + y2 + z2) (x2y + y2z + z2x) 3
Vy (1) ng (*) ng
Nhn xt: Du = (*) xy ra (x; y; z) { }(1;1;1),(1;1;0),(1;0;1),(0;1;1)
NM 2000
4. (HQG HN khi A 2000)Vi a, b, c l 3 s thc bt k tho iu kin a + b + c = 0.Chng minh rng: 8a + 8b + 8c 2a + 2b + 2c
Gii:t x = 2a, y = 2b, z = 2c th x, y, z > 0..kin a + b + c = 0 xyz = 2a+b+c = 1, do theo BT Csi: x + y + z 3Mt khc: x3 + 1 + 1 3x x3 3x 2Tng t: y3 3y 2; z3 3z 2 x3 + y3 + z3 3(x + y + z) 6 = (x + y + z) + 2(x + y + z 3) x + y + z
8a
+ 8b
+ 8c
2a
+ 2b
+ 2c
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5. (HQG HN khi D 2000)Vi a, b, c l 3 s thc dng tho iu kin: ab + bc + ca = abc. Chng minh rng:
+ + ++ +
2 2 2 2 2 2b 2a c 2b a 2c3
ab bc ca
Gii:
Ta c: + += = +2 2 2 2
2 2 2 2b 2a b 2a 1 12.ab a b a b
t x = 1a
; y = 1b
; z = 1c
th gi thit>
+ + =
a,b,c 0
ab bc ca abc
>
+ + =
x,y,z 0
x y z 1
v pcm + + + + + 2 2 2 2 2 2x 2y y 2z z 2x 3 Theo BT Bunhiacopxki ta c:
3(x2 + 2y2) = 3(x2 + y2 + y2) (x + y + y)2 + +2 2 1x 2y (x 2y)
3
Vit 2 BT tng t, ri cng li, ta c:
+ + + + + + + =2 2 2 2 2 2 1x 2y y 2z z 2x (3x 3y 3z) 33
ng thc xy ra x = y = z = 13
a = b = c = 3
6. (H Bch khoa HN khi A 2000)
Cho 2 s a, b tho iu kin a + b 0. Ch. minh rng:+ +
33 3a b a b
2 2
Gii:
Ta c:+ +
33 3a b a b
2 2 4(a3 + b3) (a + b)3
(a + b) [4(a2 + b2 ab) (a2 + b2 + 2ab)] 0 (a + b)(3a2 + 3b2 6ab) 0 (a + b)(a b)2 0BT cui cng ny ng, nn BT cn chng minh l ng.ng thc xy ra a = b.
7. (HSP TP HCM khi DE 2000)Cho 3 s a, b, c bt k. Chng minh cc BT:
a) a2
+ b2
+ c2
ab + bc + ca b) (ab + bc + ca)2
3abc(a + b + c)Gii:a) a2 + b2 2ab; b2 + c2 2bc; c2 + a2 2ca
a2 + b2 + c2 ab + bc + ca.ng thc xy ra a = b = cb) (ab + bc + ca)2 = (ab)2 + (bc)2 + (ca)2 + 2(abbc + bcca + caab)
abbc + bcca + caab + 2abc(a + b + c) = 3abc(a + b + c)
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8. (H Nng nghip I khi A 2000)Cho 3 s dng a, b, c tho iu kin abc = 1.
Tm gi tr nh nht ca biu thc: P= + ++ + +2 2 2 2 2 2bc ca ab
a b a c b c b a c a c b
Gii: Ta c: = = = + + ++
2
2 2 22
1bc bc 1 a
1 11 1a b a c a (b c) ab cb c
t x = 1a
; y = 1b
; z = 1c
th gi thit
a, b, c > 0
abc = 1
>
x, y, z 0
xyz=1v P = + +
+ + +
2 2 2x y z
y z z x x y
Theo BT Bunhiacopxki ta c:
(y + z + z + x + x + y).P
+ + + + +
+ + +
2x y z
y z. z x. x y.
y z z x x y
2(x + y + z).P (x + y + z)2 P 12
(x + y + z) =31 1
.3 xyz .32 2
P 32
Nu P = 32
th x = y = z = 1 a = b = c = 1. o li, nu a = b = c = 1 th P = 32
.
Vy minP = 32
9. (H Thu li II 2000)Chng minh rng vi mi s dng a, b, c ta u c:
(a + 1).(b + 1).(c + 1) ( )+331 abc
Gii:(a + 1).(b + 1).(c + 1) = 1 + a + b + c + ab + bc + ca + abc
1 + 3 + 3 2 2 23 abc 3 a b c + abc = ( )+331 abc
10. (H Y HN 2000)
Gi s x, y l hai s dng tho iu kin + =2 3
6x y
.
Tm gi tr nh nht ca tng x + y.
Gii: ( )
+ = + + +
22 2 3 2 3
2 3 . x . y (x y)x y x y = 6(x + y) x + y
( )+2
2 3
6
Gi tr( )+
22 3
6t c
( )
=
+ + =
2
2 3: x : y
x y
2 3x y
6
+=
+ =
2( 2 3)x
6
3( 2 3)y
6
Vy min(x + y) =+5 2 6
6
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11. (H An Giang khi D 2000)Cho cc s a, b, c 0. Chng minh: ac + 1 + bc + 1 ab(ac 1 + bc 1)
Gii:Gi s a b 0 ac(a b) bc(a b) ac + 1 + bc + 1 ab(ac 1 + bc 1)
12. (H Ty Nguyn khi AB 2000)
CMR vi mi x, y, z dng v x + y + z = 1 th xy + yz + zx > +18xyz
2 xyz
Gii:p dng BT Csi cho 6 s dng ta c:
2 = x + y + z + x + y + z 6 3 xyz (1) v xy + yz + zx 32 2 23 x y z (2)
Nhn cc BT (1) v (2) v theo v ta c:2(xy + yz + zx) 18xyz (3)Mt khc ta c: xyz(xy + yz + zx) > 0 (4)Cng cc BT (3) v (4) v theo v ta c:
(xy + yz + zx)(2 + xyz) > 18xyz xy + yz + zx >+
18xyz
2 xyz(v 2 +xyz > 0)
13. (H An Ninh khi A 2000)Chng minh rng vi mi s nguyn n 3 ta u c: nn + 1 > (n + 1)n
Gii:
Ta c: 34 = 81, 43 = 64 34 > 43 BT cn chng minh ng vi n = 3.
Vi n > 3, pcm n >+
nn 1
n
+
n1
1n
< n (1)
Ta c: +
n1
1n
==n
kn k
k 0
1C
n= 1 +
++ + +
2 n
n n(n 1) 1 n(n 1)...(n n 1) 1. ... .
n 2! n!n n
= 1 + 1 +
+ +
1 1 1 1 2 n 11 ... 1 1 ... 1
2! n n! n n n 0 th:
+ +
1 1 4x y x y
(1). Du = xy ra x = y.
p dng (1) ta c: + = +
1 1 4 4
p a p b p a p b c
+ = + 1 1 4 4
p b p c p b p c a, + =
+
1 1 4 4
p c p a p c p a b
Cng 3 BT trn v theo v, ta c:
+ + + +
1 1 1 1 1 12 4
p a p b p c a b c pcm
Du = xy ra a = b = c.
20. (H Nng nghip I HN khi A 2001)Cho 3 s x, y, z > 0. Chng minh rng:
+ + + ++ + +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
Gii:p dng BT Csi cho 2 s dng x3, y2 ta c:
x3 + y2 2 =3 2x y 2xy x =+3 2
2 x 2 x 1xy2xy xx y
p dng BT Csi cho 2 s dng2 2
1 1,
x yta c:
+
2 2
1 1 1 1
xy 2 x y
+ +
3 2 2 2
2 x 1 1 1
2x y x y
Tng t ta cng c:
+
+ 3 2 2 2
2 y 1 1 1
2y z y z
;
+
+ 3 2 2 2
2 z 1 1 1
2z x z x
Suy ra: + + + ++ + +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
Du = xy ra = = =
= = =
3 2 3 2 3 2x y y z z xva va
x y y z z x x = y = z = 1
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21. (H PCCC khi A 2001)Ch. minh rng vi a 2, b 2, c 2 th: + + ++ + >b c c a a blog a log b log c 1
Gii:Trc ht ch rng nu a > 1, x > 1 th hm s y = alo g x l ng bin v dng.
Do hm s y = logxa =a
1
log xl nghch bin.
V vai tr ca a, b, c l nh nhau, nn ta c th gi thit a b c. Ta c:VT= + + + + + + ++ + + + =b c c a a b a b a b a b a blog a log b log c log a log b log c log abc
V a, b, c 2 nn abc 2ab = ab + ab > a + bDo VT loga+babc > loga+b(a + b) = 1.
22. (H Quc gia HN khi D 2001)Ch. minh rng vi mi x 0 v vi mi > 1 ta lun c: x + 1 x.T chng minh rng vi 3 s dng a, b, c bt k th:
+ + + +3 3 3
3 3 3
a b c a b c
b c ab c a Gii:
Xt f(x) = x x + 1 (x 0)f(x) = (x 1 1); f(x) = 0 x = 1
Vy vi x 0 v > 1 th f(x) 0 hay x + 1 x.
BT cn chng minh: + + + +
3 3 32 2 2a b c a b c
b c a b c a
p dng BT chng minh vi = 32
, ta c:
+
3
2a 1 3 a.
b 2 2 b;
+
3
2b 1 3 b.
c 2 2 c;
+
3
2c 1 3 c.
a 2 2 a
Mt khc, theo BT Csi ta c:
+ +
3 3 3
2 2 21 a b c 3
2 b c a 2
Cng 4 BT trn, v theo v, ta c:
+ + + + + +
3 3 3
2 2 23 a b c 3 3 a b c 3
2 b c a 2 2 b c a 2
Suy ra: + + + +
3 3 3
2 2 2a b c a b c
b c a b c a
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23. (H Thi Nguyn khi D 2001)
Cho a 1, b 1. Chng minh rng: + a b 1 b a 1 ab (*)Gii:
BT (*)
+ a b 1 b a 1
1ab ab
+
1 1 1 11 1 1
b b a a(1)
Theo BT Csi ta c:
+ =
1 111 1 1b b
1b b 2 2
+ =
1 11
1 1 1a a1
a a 2 2
Cng 2 BT li ta c BT cn chng minh.
Du = xy ra
= =
= =
1 1 11
b b 2
1 1 11a a 2
a = b = 2.
24. (H Vinh khi A, B 2001)Chng minh rng nu a, b, c l di ba cnh ca mt tam gic c chu vi bng 3 th:
3a2 + 3b2 + 3c2 + 4abc 13Gii:
Ta c: 3 2a = a + b + c 2a = b + c a > 0.Do theo BT Csi ta c:
(3 2a)(3 2b)(3 2c)
+ +
33 2a 3 2b 3 2c
3 = 1 27 9(2a + 2b + 2c) + 3(4ab + 4bc + 4ca) 8abc 1 27 54 + 12(ab + bc + ca) 8abc 1 4abc 6(ab + bc + ca) 14 3(a2 + b2 + c2) + 4abc 3(a2 + b2 + c2) + 6(ab + bc + ca) 14 = 3(a + b +c)2 14 =13ng thc xy ra 3 2a = 3 2b = 3 2c a = b = c = 1.
25. (H Y Thi Bnh khi A 2001)
Cho a, b, c l nhng s dng v a + b = c. Ch. minh rng: + >2 2 2
3 3 3a b c Gii:
T gi thit ta c: +a b
c c= 1 0 +
2 2
3 3a b a b
c c c c= 1
T suy ra: + >2 2 2
3 3 3a b c
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NM 200226. (i hc 2002 d b 1)
Gi x, y, z l khong cch t im M thuc min trong ca ABC c 3 gc nhn ncc cnh BC, CA, AB. Chng minh rng:
+ ++ +
2 2 2a b cx y z
2R(a, b, c l cc cnh ca ABC, R l bn knh
ng trn ngoi tip). Du = xy ra khi no?
Gii: + + = + +1 1 1
x y z . ax . by . cza b c
+ +
1 1 1(ax+by+cz)
a b c
+ +
1 1 1.2S
a b c=
+ +
1 1 1 abc
a b c 2R=
+ +ab bc ca2R
+ +2 2 2a b c
2R
Du = xy ra = =
= =
a b c
x y z
ABC eu
M trung vi trong tam G cua ABC
27. (i hc 2002 d b 3)
Gi s x, y l hai s dng thay i tho mn iu kin x + y = 54 . Tm gi tr nh nht
ca biu thc: S = +4 1
x 4y
Gii:
Cch 1: S = + + + + 5
1 1 1 1 1 5
x x x x 4y x.x.x.x.4y
+ + + +5.5
x x x x 4y= 5
minS = 5
=
=
+ =
1 1
x 4y
x 4y
5x y4
=
=
x 1
1y
4
Cch 2: S = +
4 1
x 5 4x= f(x), 0 < x < 5
4
f(x) = +2 2
4 4
x (5 4x); f(x) = 0
=
<
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28. (i hc 2002 d b 5)Gi s a, b, c, d l 4 s nguyn thay i tho mn 1 a < b < c < d 50.
Chng minh bt ng thc:+ +
+ 2a c b b 50
b d 50b
v tm gi tr nh nht ca biu thc: S = +a c
b d.
Gii:V a 1, d 50 v c > b (c, b N) nn c b + 1 thnh th:
S = +a c
b d
++
1 b 1
b 50=
+ +2b b 5050b
Vy BT ca ra c chng minh.
Du = xy ra
=
= = +
a 1
d 50
c b 1
tm minS, ta t+ +2b b 5050b = + +
b 1 1
50 b 50 v xt hm s c bin s lin tc x:
f(x) = + +x 1 1
50 x 50(2 x 48)
f(x) =
=2
2 2
1 1 x 50
50 x 50x; f(x) = 0
=
2x 50
2 x 48 =x 5 2
Bng bin thin:
2
Chuyn v biu thc f(b) =+ +2b b 5050b
(2 b 48, b N)
T BBT suy ra khi b bin thin t 2 n 7, f(b) gim ri chuyn sang tng khi b binthin t 8 n 48. Suy ra minf(b) = min[f(7); f(8)].
Ta c f(7) =+
=49 57 53
350 175; f(8) =
+= >
64 58 61 53
400 200 175
Vy minS = 5 31 7 5 khi
= = = =
a 1
b 7
c 8
d 5 0
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29. (i hc 2002 d b 6)
Cho tam gic ABC c din tch bng 32
. Gi a, b, c ln lt l di cc cnh BC, CA,
AB v ha, hb, hc tng ng l di cc ng cao k t cc nh A, B, C. Chngminh rng:
+ + + +
a b c
1 1 1 1 1 13
a b c h h h
Gii:
Ta c din tch tam gic: S = = =a b c1 1 1
ah bh ch2 2 2
ha =2S
a; hb =
2S
b; hc =
2S
c + + = + +
a b c
1 1 1 1(a b c)
h h h 2S
+ + + + = + + + +
a b c
1 1 1 1 1 1 1 1 1 1(a b c)
a b c h h h 2S a b c
p dng BT Csi ta c: (a + b + c)
+ +
1 1 1
a b c 9
v v S =3
2, nn ta c:
+ + + + =
a b c
1 1 1 1 1 1 93
a b c h h h 3
NM 2003
30. (CGT II 2003 d b)
Cho 3 s bt k x, y, z. CMR: + + + + +2 2 2 2 2 2x xy y x xz+z y yz+z
Gii:Trong mt phng to Oxy, xt cc im:
A
+
y 3x ; z
2 2 , B
+
3 30; y z
2 2, C
y z;0
2 2
Ta c: AB =
+ + = + +
222 2y 3x y x xy y
2 2
AC = + + = + +
222 2z 3x z x xz z
2 2
BC =
+ + = +
222 2y z 3 (y z) y yz+z
2 2 2
Vi 3 im A, B, C ta lun c: AB + AC BC
+ + + + +2 2 2 2 2 2x xy y x xz+z y yz+z
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31. (i hc khi A 2003)Cho x, y, z l 3 s dng v x + y + z 1. Chng minh rng:
+ + + + + 2 2 22 2 2
1 1 1x y z 82
x y z
Gii:
Vi mi
u,v ta c: + +
u v u v (*)
t = = =
1 1 1a x; ; b y; ; c z;
x y z
p dng bt ng thc (*), ta c: + + + + + +
a b c a b c a b c
Vy P = + + + + +2 2 2
2 2 2
1 1 1x y z
x y z
+ + + + +
22 1 1 1(x y z)
x y z
Cch 1:
Ta c: P
+ + + + +
22 1 1 1
(x y z) x y z ( )
+
22
33
1
3 xyz 3 xyz = +9
9t t
vi t = 23( xyz) 0 < t + +
2x y z 1
3 9
t Q(t) = 9t +9
tQ(t) = 9
2
9
t< 0, t
10;
9Q(t) gim trn
10;
9
Q(t) Q
1
9= 82. Vy P Q(t) 82
Du "=" xy ra x = y = z = 13
.
Cch 2: Ta c:
(x + y + z)2 +
+ +
21 1 1
x y z= 81(x + y + z)2 +
+ +
21 1 1
x y z 80(x + y + z)2
18(x + y + z).
+ +
1 1 1
x y z 80(x + y + z)2 162 80 = 82
Vy P 82
Du "=" xy ra x = y = z = 13
.
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32. (i hc khi A 2003 d b 1)
Tm gi tr ln nht v gi tr nh nht ca hm s:y = sin5x + 3 cosxGii:
Tm max: y = sin5x + 3 cosx sin4x + 3 cosx (1)
Ta chng minh: sin4x + 3 cosx 3 , x R (2)
3 (1 cosx) sin4x 0 3 (1 cosx) (1 cos2x)2 0 (1 cosx).[ 3 (1 cosx)(1 + cosx)2] 0 (3)Theo BT Csi ta c:
(1 cosx)(1 + cosx)(1 + cosx) = 12
(2 2cosx)(1 + cosx)(1 + cosx)
=
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NM 2005
34. (i hc khi A 2005)
Cho x, y, z l cc s dng tho mn : + + =1 1 1
4x y z
.
Chng minh rng:+ +
+ + + +
1 1 11
2x+y+z x 2y z x y 2z Gii:
Vi a, b > 0 ta c:
4ab (a + b)2+
+1 a b
a b 4ab
+ +
1 1 1 1
a b 4 a b
Du "=" xy ra khi v ch khi a = b.p dng kt qu trn ta c:
+ +
1 1 1 1
2x+y+z 4 2x y z
+ +
1 1 1 1 1
4 2x 4 y z=
+ +
1 1 1 1
8 x 2y 2z(1)
Tng t:
+ + + +
1 1 1 1
x 2y z 4 2y x z
+ +
1 1 1 1 1
4 2y 4 x z=
+ +
1 1 1 1
8 y 2z 2x(2)
+ + + +
1 1 1 1
x y 2z 4 2z x y
+ +
1 1 1 1 1
4 2z 4 x y=
+ +
1 1 1 1
8 z 2x 2y(3)
Vy:
+ + + + + + + +
1 1 1 1 1 11
2x+y+z x 2y z x y 2z 4 x yz= 1
Ta thy trong cc bt ng thc (1), (2), (3) th du "=" xy ra khi v ch khi
x = y = z. Vy ng thc xy ra khi v ch khi x = y = z = 34
.
35. (i hc khi B 2005)Chng minh rng vi mi x R, ta c:
+ + + +
x x xx x x12 15 20 3 4 5
5 4 3
Khi no ng thc xy ra?Gii:
p dng bt ng thc Csi cho 2 s dng ta c:
+
x x x x12 15 12 15
2 .
5 4 5 4
+
x x12 15
5 4
2.3x (1)
Tng t ta c:
+
x x12 20
5 3 2.4x (2)
+
x x15 20
4 3 2.5x (3)
Cng cc bt ng thc (1), (2), (3), chia 2 v ca bt ng thc nhn c cho 2 tac pcm.ng thc xy ra (1), (2), (3) l cc ng thc x = 0.
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36. (i hc khi D 2005)Cho cc s dng x, y, z tho mn xyz = 1. Chng minh rng:
+ + + + + ++ +
3 3 3 3 3 31 x y 1 y z 1 z x3 3
xy yz zx
Khi no ng thc xy ra?Gii:
p dng bt ng thc Csi cho 3 s dng ta c:
1 + x3 + y3 3 3 33 1.x .y = 3xy + +
3 31 x y 3
xy xy(1)
Tng t:+ +
3 31 y z 3
yz yz(2);
+ +
3 31 z x 3
zx zx(3)
Mt khc + + 33 3 3 3 3 3
3xy yz zx xy yz zx
+ + 3 3 3 3 3xy yz zx
(4)
Cng cc bt ng thc (1), (2), (3), (4) ta c pcm.ng thc xy ra (1), (2), (3), (4) l cc ng thc x = y = z = 1.
37. (i hc khi A 2005 d b 1)
Cho 3 s x, y, z tho x + y + z = 0. CMR: + + + + +x y z3 4 3 4 3 4 6Gii:
Ta c: 3 + 4x = 1 + 1 + 1 + 4x 4 4 x4 + = 84x x x3 4 2 4 2 4
Tng t: + 8y y3 4 2 4 ; + 8z z3 4 2 4
Vy + + + + +x y z3 4 3 4 3 4 2 + + 8 8 8x y z4 4 4
3 8 x y z6 4 .4 .4 6 + +24 x y z4 = 6
38. (i hc khi A 2005 d b 2)
Chng minh rng vi mi x, y > 0 ta c: ( )
+ + +
2y 9
1 x 1 1x y
256
ng thc xy ra khi no?Gii:
Ta c: 1 + x = 1 + + + 3
43
x x x x43 3 3 3
v 1 +yx
= 1 + + + 34
3 3y y y y4
3x 3x 3x 3 x
1 +9
y= 1 + + +
3
43
3 3 3 34
y y y y
+
2 6
43
9 31 16
y y
Vy: ( )
+ + +
2y 9
1 x 1 1x y
2563 3 6
43 3 3 3
x y 3. .
3 3 x y= 256
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39. (i hc khi B 2005 d b 1)
Cho 3 s dng a, b, c tho mn: a + b + c = 34
. Chng minh rng:
+ + + + + 3 3 3a 3b b 3c c 3a 3 Khi no ng thc xy ra?
Gii: Cch 1:
Ta c: + + ++ = + +3 a 3b 1 1 1(a 3b).1.1 (a 3b 2)3 3
+ + ++ = + +3
b 3c 1 1 1(b 3c).1.1 (b 3c 2)
3 3;
+ + ++ = + +3
c 3a 1 1 1(c 3a).1.1 (c 3a 2)
3 3
Suy ra: [ ]+ + + + + + + +3 3 31
a 3b b 3c c 3a 4(a b c) 63
+
1 34. 6
3 4= 3
Du "=" xy ra
+ + = + = + = +
3a b c
4a 3b b 3c c 3a=1
a = b = c = 14
Cch 2:
t x = +3 a 3b x3 = a + 3b; y = +3 b 3c y3 = b + 3c;
z = +3 c 3a z3 = c + 3a
x3 + y3 + z3 = 4(a + b + c) = 4. 34
= 3. BT cn ch. minh x + y + z 3
Ta c: x3 + 1 + 1 3 3 3x .1.1 = 3x ; y3 + 1 + 1 3 33 y .1.1 = 3y ;
z3 + 1 + 1 3 3 3z .1.1 = 3z 9 3(x + y + z) (v x3 + y3 + z3 = 3)Vy x + y + z 3
Du "=" xy ra
= = =
+ + =
3 3 3x y z 1
3a b c
4
+ = + = +
a 3b b 3c c 3a=1
3a+b+c=
4 a = b = c = 1
4
40. (i hc khi B 2005 d b 2)
Chng minh rng nu 0 y x 1 th 1
x y y x4
.
ng thc xy ra khi no?Gii:
Ta c: 0 x 1 x x2
1
x y y x4
+1
x y y x4 (1)
Theo BT Csi ta c: + + =2 21 1 1
y x yx 2 yx . x y4 4 4
1
x y y x4
Du "=" xy ra
=
=
= =
2
2
0 y x 1 x 1x x 1
y1 4yx4
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41. (i hc khi D 2005 d b 2)
Cho x, y, z l 3 s dng v xyz = 1. CMR: + + + + +
2 2 2x y z 3
1 y 1 z 1 x 2
Gii:
Ta c:+ +
+ =
+ +
2 2x 1 y x 1 y2 . x
1 y 4 1 y 4
+ ++ =
+ +
2 2y 1 z y 1 z2 . y
1 z 4 1 z 4
+ ++ =
+ +
2 2z 1 x z 1 x2 . z
1 x 4 1 x 4
Cng 3 bt ng thc trn, v theo v, ta c: + + +
+ + + + + + + + + +
2 2 2x 1 y y 1 z z 1 xx y z
1 y 4 1 z 4 1 x 4
+ +
+ + + + ++ + +
2 2 2x y z 3 x y z
x y z1 y 1 z 1 x 4 4
+ +
3(x y z) 3
4 4
= =3 3 9 3 3
.34 4 4 4 2
(v x + y + z 3 3 xyz = 3)
Vy: + + + + +
2 2 2x y z 3
1 y 1 z 1 x 2.
NM 2006
42. (CBC Hoa Sen khi A 2006)Cho x, y, z > 0 v xyz = 1. Chng minh rng: x3 + y3 + z3 x + y + z.
Gii:
x3 + y3 + z3 3 3 3 33 x y z 2(x3 + y3 + z3) 6
x3 + 1 + 1 3 3 3x x3 + 2 3x (1)
Tng t: y3 + 1 + 1 3 33 y y3 + 2 3y (2)
z3 + 1 + 1 3 3 3z z3 + 2 3z (3)Cng (1), (2), (3) v theo v suy ra bt ng thc cn chng minh.
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43. (CKTKT Cn Th khi A 2006)Cho 3 s dng x, y, z tho x + y + z 1.
Tm gi tr nh nht ca biu thc: A = x + y + z + + +1 1 1
x y z
Gii: Cch 1:
Theo BT Csi: 1 x + y + z 3 3 xyz > 0 , + + 31 1 1 3
x y z xyz
T : A 3 3 xyz +3
3
xyz
t: t = 3 xyz , iu kin: 0 < t 13
Xt hm s f(t) = 3t + 3
tvi 0 < t 1
3, f(t) = 3
2
3
t=
2
2
3(t 1)
t< 0, t
10;
3
Bng bin thin:1
3
T bng bin thin ta suy ra: A 10. Du "=" xy ra khi x = y = z = 13
Vy Amin = 10 t c khi x = y = z =1
3
.
Cch 2:
Theo BT Csi: 1 x + y + z 3 3 xyz > 0 3
1
xyz 3
x + 1 2
9x 3, y +
1 2
9y 3, z +
1 2
9z 3
T : A= + + + + + + + +
1 1 1 8 1 1 1x y z
9x 9y 9z 9 x y z 2 +
3
8 3
9 xyz 10
Du "=" xy ra khi x = y = z = 13
.Vy Amin = 10 t c khi x = y = z =1
3
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44. (CSPHCM khi ABTDM 2006)
Cho x, y l hai s thc dng v tho x + y = 54
.
Tm gi tr nh nht ca biu thc: A = +4 1
x 4y.
Gii:
Ta c: x + y =5
4 4x + 4y 5 = 0
A = +4 1
x 4y= + +
4 14x+ 4y 5
x 4y A 2
4.4x
x+ 2
1.4y
4y 5 A 5
Du "=" xy ra
=
= + = >
44x
x1
4y4y
5x y
4x,y 0
=
=
x 1
1y
4
. Vy Amin = 5.
45. (CKTKT Cn Th khi B 2006)Cho 4 s dng a, b, c, d. Chng minh bt ng thc:
+ + ++ + + + + + + +
a b c d
a b c b c d c d a d a b< 2
Gii:V a, b, c, d > 0 nn ta lun c:
+ < + =+ + + + + +
a c a c
1a b c c d a a c a c v + < + =+ + + + + +
b d b d
1b c d d a b b d b d Cng v theo v cc BT trn ta c pcm.
46. (CKT Cao Thng khi A 2006)
Chng minh rng nu x > 0 th (x + 1)2
+ + 2
1 21
xx 16.
Gii:
Ta c: (x + 1)2
+ +
2
1 21
xx 16 (1) (x + 1)2
+
21
1
x 16
(x + 1)
+
11
x 4 (do x > 0) (x + 1)2 4x (x 1)2 0 (2)
(2) lun ng nn (1) c chng minh.
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47. (CKTKTCN1 khi A 2006)
Cho 3 s dng a, b, c. Ch. minh rng:+ + + + + +
+ + a b c a b c a b c
9a b c
Gii:
Xt v tri ca BT cho: VT =+ + + + + + + +
b c a c a b1 1 1
a a b b c c
= 3 +
+ + + + +
b a c a c b
a b a c b c
Do a, b, c > 0 nn theo BT Csi ta c:
+ =b a b a
2 . 2a b a b
; + =b c b c
2 . 2c b c b
; + =c a c a
2 . 2a c a c
Khi : VT 3 + 2 + 2 + 2 = 9 (pcm).
48. (CKTYT1 2006)Cho cc s thc x, y thay i tho mn iu kin: y 0; x2 + x = y + 12.Tm gi tr ln nht, nh nht ca biu thc: A = xy + x + 2y + 17
Gii:y 0, x2 + x = y + 12 x2 + x 12 0 4 x 3y = x2 + x 12 A = x3 + 3x2 9x 7t f(x) = A = x3 + 3x2 9x 7 vi 4 x 3f(x) = 3x2 + 6x 9 ; f(x) = 0 x = 1 hoc x = 3
f(4) = 13, f(3) = 20, f(1) = 12, f(3) = 20Vy maxA = 20 (x = 3, y = 0), minA = 12 (x = 1, y = 10).
49. (CBC Hoa Sen khi D 2006)Cho x, y, z > 0; x + y + z = xyz. Tm gi tr nh nht ca biu thc A = xyz.
Gii:
Ta c: x + y + z 3 3 xyz xyz 3 3 xyz (xyz)2 27 xyz 3 3
Du "=" xy ra x = y = z = 3 .Vy minA = 3 3 .
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50. (i hc khi A 2006)Cho 2 s thc x 0, y 0 thay i v tho mn iu kin:
(x + y)xy = x2 + y2 xy.
Tm gi tr ln nht ca biu thc: A = +3 3
1 1
x y.
Gii:
Cch 1:T gi thit suy ra: + = + 2 2
1 1 1 1 1
x y xyx y.
t 1x
= a, 1y
= b, ta c: a + b = a2 + b2 ab (1)
A = a3 + b3 = (a + b)(a2 ab + b2) = (a + b)2T (1) suy ra: a + b = (a + b)2 3ab.
V ab +
2a b
2nn a + b (a + b)2 + 2
3(a b)
4
(a + b)2 4(a + b) 0 0 a + b 4 Suy ra: A = (a + b)2 16
Vi x = y = 12
th A = 16. Vy gi tr ln nht ca A l 16.
Cch 2:t S = x + y, P = xy vi S2 4P 0. T gi thit S, P 0.
Ta c: SP = S2 3P P =+
2S
S 3
A = +3 3
1 1
x y=
+3 3
3 3
x y
x y=
+ + 2 2
3 3
(x y)(x y xy)
x y=
+ 2
3 3
(x y) xy
x y=
+ 2
2 2
(x y)
x y
A =+
=
2
2
S S 3
SP
k: S2 4P 0 S2 +
24S
S 3 0 S2
+
S 1
S 3 0 +
S 1
S 3 0 (v S0)
<
S 3
S 1(*)
t h = f(S) =+S 3S
h = 2
3
S< 0, S tho (*)
T bng bin thin, ta c: 0 < h 4 v h 1, S tho (*).
M A = h MaxA = 16 khi x = y = 12
(S = 1, P = 14
).
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Cch 3:
(x + y)xy =
+
2 2y 3yx
2 4> 0
++ =
1 1 x y
x y xy> 0
A = +3 3
1 1
x y=
+3 3
3 3
x y
x y=
+
21 1
x y = +
1 1A
x y
D chng minh c: + +
3 3 3a b a b2 2
(vi a + b > 0)
du "=" xy ra khi a = b.
p dng vi a =1
x, b =
1
y, ta c:
++
333 1 11 1x yx y
2 2
3A A
2 2 A 16.
Du "=" xy ra khi = =1 1
2x y
. Vy Max A = 16.
Cch 4:A =
2
2
S
P, suy ra = =
2S 3S
AP S SP
S2 4P 0 S2 4 2S SP
3 0
P1
S1 43
0 P 1
S 4(chia cho S2)
Nn: A =2
2
S
P 16. Vy Max A = 16 (khi x = y = 1
2).
51. (i hc khi B 2006)Cho x, y l cc s thc thay i. Tm gi tr nh nht ca biu thc:
A = ( ) ( ) + + + + + 2 22 2x 1 y x 1 y y 2 Gii:Trong mpOxy, xt M(x 1; y), N(x + 1; y).
Do OM + ON MN nn:
( ) ( ) + + + + + = +2 22 2 2 2x 1 y x 1 y 4 4y 2 1 y
Do : A 2 + + 21 y y 2 = f(y)
Vi y 2 f(y) = 2 + 21 y + 2 y f(y) =+2
2y
y 1 1
f(y) = 0 2y = + 21 y = +
2 2y 04y 1 y
y = 13
Do ta c bng bin thin nh trn
Vi y 2 f(y) 2 + 21 y 2 5 > 2 + 3 .Vy A 2 + 3 vi mi s thc x, y.
Khi x = 0 v y = 13
th A = 2 + 3
Nn gi tr nh nht ca A l 2 + 3 .
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NM 2007
52. (i hc khi A 2007)
Cho x, y, z l cc s thc dng thay i v tha mn iu kin xyz =1.
Tm gi tr nh nht ca biu thc:2 2 2( ) ( ) ( )
2 2 2
x y z y z x z x yP
y z z z z x x x x y y
+ + += + +
+ + +
Gii:
Ta c: 2 2 2( ) 2 . : ( ) 2 , ( ) 2y z x x y z x y y z x y z z + + + Tngt
22 2
2 2 2
y y x x z z P
y y z z z z x x x x y y + +
+ + +
t: 2 , 2 , 2a x x y y b y y z z c z z x x= + = + = +
Suy ra4 2 4 2 4 2
, ,9 9 9
c a b a b c b c a x x y y z z
+ + + = = =
Do : 2 4 2 4 2 4 2 2 4 69 9 9 9 9
c a b a b c b c a c a b a b cPb c a b c a
+ + + + + = + + + + +
2(4.3 3 6) 2
9 + =
1 1 2 2 1 4 1 3c a b c a b a b
Dob c a b c a b a
+ + = + + + + =
Du "=" xy ra x = y = z = 1. Vy gi tr nh nht ca P l 2.
53. (i hc khi B 2007)Cho x, y, z l ba s thc dng thay i. Tm gi tr nh nht ca biu thc:
1 1 12 2 2
x y z P x y z
yz zx xy
= + + + + +
Gii:
Ta c :2 2 2 2 2 2
2 2 2
x y z x y z P
xyz
+ += + + + .
Do2 2 2 2 2 2
2 2 2
2 2 2
x y y z z x x y z xy yz zx
+ + ++ + = + + + +
2 2 2
1 1 12 2 2 x y z P
x y z + + + + +
Xt hm s :2 1
( ) , 02
t f t t
t= + >
Lp bng bin thin ca hm s ny ta c3
( ) , 0.2
f t t >
Suy ra :9
2P . Du bng xy ra x = y = z = 1. Vy gi tr nh nht ca P l 9 .
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54. (i hc khi D 2007)
Cho1 1
0. : 2 22 2
b a
a b
a ba b CMR
> + +
Gii:
( ) ( ) ( ) ( )ln 1 4 ln 1 41 12 2 1 4 1 42 2
a bb a
b aa b a ba b a b
+ + + + + +
Xt hm s :( ) ( ) ( )
( )/
2
ln 1 4 4 ln 4 1 4 ln 1 4( ) , 0. ( ) 0
1 4
x x x x x
x f x x f x
x x
+ + += > = 0 nn f(a) f(b)Vy ta c iu phi chng minh.
NM 2009
55. (i hc khi A 2009)Chng minh rng vi mi s thc dng x,y,z tha mn x( x + y + z )= 3xy, ta c:
3 3 3( ) ( ) 3( )( )( ) 5( )x y x z x y x z y z y z + + + + + + + +
Gii:t a = x + y , b = z + c , c = x + y , ,
2 2 2
b c a c a b a b c x y z
+ + + = = =
T iu kin x(x + y + z) = 3yz 2 2 2 2 2 24 ( ) 3( ) (*)a b c b c a b bc c = + + = + Ta c bt ng thc cho tng ng:
3 3 35 3a b c abc + + 25 ( ) 3 (**)a a b c bc + +
T (*)
2
22 2 2 2 2 ( )2 2( ) 2 2
2
a bc
b ca b c bc b c a b c
+
= + + +
2
2
2 ( )3 3
a a b c
a bc
+
(**) ng pcm
Du = xy ra khi a= b= c tc l x = y = z.
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56. (i hc khi B 2009)
Cho cc s thcx, y thay i v tho mn ( ) 4 2 x y xy+
+ +
Tm gi tr nh nht ca biu thc ( ) ( )4 4 2 2 2 23 2 1A x y x y x y= + + + +
Gii: Kt hp ( ) ( ) ( ) ( )3 2 3 2
4 2 & 4 2 1 y xy x y xy suy ra x y x y x y+ + + + + + +
( ) ( ) ( ) ( ) ( )
( ) ( ) ( ) ( ) ( )
24 4 2 2 2 2 2 2 4 4 2 2
2 2 22 2 2 2 2 2 2 2 2 2
3 33 2 1 2 12 23 3 9
2 1 2 12 2 4
A x y x y x y x y x y x y
x y x y x y A x y x y
= + + + + = + + + + +
+ + + + + + + +
t 2 2t x y= + , ta c( )
2
2 2 1 1
2 2 2
x y x y t
++
Do :29 2 1
4t t +
Xt 2 /
1;2
9 9 1 1 9( ) 2 1 ; ( ) 2 0, min ( )
4 2 2 2 16
t t t t f t t t f t f +
= + = > = =
9
16A . ng thc xy ra
1
2x y = = . Vy gi tr nh nht ca A l
9
16
57. (i hc khi D 2009)Cho cc s thc khng mx; y thay i v tho mn 1x y+ =
Tm gi tr ln nht v gi tr nh nht ca biu thc ( ) ( )2 24 3 4 3 25S x y y x xy= + + +
Gii: Do x + y = 1 nn ( )2 2 3 316 12 9 25S x y x y xy xy= + + + +
( ) ( )32 2 2 216 12 3 34 16 2 12 x y x y xy x y xy x y xy = + + + + = +
t t = xy ta c2
2 1 116 2 12; 0 0;2 4 4
x yS t t xy t
+ = + =
Xt hm s 21
( ) 16 2 12 0;4
f t t t = +
lien tuc tren oan
/ / 1 1 191 1 25( ) 32 2 ; ( ) 0 ; (0) 12 , ,16 16 16 4 2
f t t f t t f f f = = = = = =
110;0;
44
1 25 1 191max ( ) ; min ( )
4 2 16 16 f t f f t f
= = = =
Gi tr ln nht ca S bng ( )125 1 1
; ;42 2 2
x ykhi x y
xy
+ = = =
Gi tr nh nht ca S bng191
16
( ) ( )1
2 3 2 3 2 3 2 3; ; ; ;1
4 4 4 416
x y
khi x y hay x yxy
+ = + + = = =
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58. (Cao ng ABD 2009)Cho a v b l hai s thc tha mn 0 1a b< < < .Chng minh rng: 2 2ln ln ln lna b b a a b >
Gii:2 2
2 2
ln lnln ln ln ln
1 1
a ba b b a a b
a b >
+ +
Do hm s f(t) ng bin trn khong (0;1)
M 0
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NM 2010
60. (Cao ng ABD 2010)Cho hai s thc dng thay ix, ytha mn iu kin 3 1y+
Tm gi tr nh nht ca biu thc1 1
Ax xy
= +
Gii:
1 1 1 2 1 2 4 8 82 8
2 ( ) 32 ( )A
x x x y x x y x x y x y xy x x y= + + = =
+ + + + ++
ng thc xy ra1
4x y = = . Vy gi tr nh nht ca A bng 8.
61. (i hc B 2010)Cho cc s thc khng m a, b, c tha mn: a + b + c = 1.Tm gi tr nh nht ca biu thc
M = 3(a2b2+b2c2+c2a2) + 3(ab + bc + ca) + 2 2 22 a b c+ + .Gii:
t t = ab + bc + ca, ta c: a2 + b2 + c2 ab + bc + ca 1 = (a + b + c)2 = a2 + b2 + c2 + 2(ab + bc + ca) 3(ab + bc + ca)
a2 + b2 + c2 = 1 2t v1
03
t
Theo B.C.S ta c : t2 = (ab + bc + ca)2 3(a2b2 + b2c2 + c2a2)
M 2 3 2 1 2 ( )t t t f t + + =
f(t) =2
2 31 2
tt
+
f (t) =3
22
(1 2 )t
< 0, t
10,
3
f(t) l hm gim
/ / 1 11( ) 2 33 3
f t f =
> 0 f tng f(t) f(0) = 2, t 1
0,3
M 2, a, b, c khng m tha a + b + c = 1Khi a = b = 0 v c = 1 th M = 2. Vy min M = 2.
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