Các câu hỏi Bất đẳng thức hay lớp 9 và lớp 10

BÊt ®¼ng thøc

bÊt ®¼ng thøc
Bµi 1: Chøng minh c¸c bÊt ®¼ng thøc sau b»ng ph¬ng ph¸p chuyÓn vÒ tæng d¹ng b×nh ph¬ng:
2
2
2
2
2
2
2
2
2
2
2
a. a + b ≥ 2ab ; ( a + b ) ≥ 4ab ; 2 a + b ≥ ( a + b ) ; a ± ab + b ≥ 0; 2a + b + c ≥ 2a ( b + c )

(

)

a b
1 1
4
1
4
a2
a 2 1  a  b 
; ≥
; ≥ 2a − b; 2 ≥ − ; 1 + 1 +  ≥ 4∀a; b > 0
b. + ≥ 2; + ≥

2
b a
a b a + b ab ( a + b ) b
b
b a  b  a 
2
2
c. 3( ab + bc + ca ) ≤ ( a + b + c ) ≤ 3( a 2 + b 2 + c 2 )
d. ( ab + bc + ca ) ≥ abc( a + b + c )
2
2
2
2
2
2
2
2
e. a + b + c ≥ a ( b + c )
f. a + b + c + d + e ≥ a ( b + c + d + e )
2
2
2
g. a + b + 4 ≥ ab + 2( a + b )
h. 3ab ≤ ( a + b − 2 ) + 6( a + b )
2
2
2
2
2
2

i. 4a + b + c ≥ 4a ( c − b ) − 2bc
k. a + 4b + 4c ≥ 4( ab − ac + 2bc )

Bµi 2: Chøng minh c¸c bÊt ®¼ng thøc b»ng ph¬ng ph¸p biÕn ®æi t¬ng ®¬ng:
3
3
3
2
2
3
3
3
2
2
a. a + b ≥ a b + b a ; 4 a + b ≥ ( a + b ) ; 3a ≥ ( 2a − b ) a + ab + b ∀a , b > 0

(

)

(

)

b. a + b ≥ ab + ba ; a + b ≥ ab + ba ; a + b ≥ a b + b a ; a
4

4

3

(

3

)

(

6

6

)(

5

5

8

8

)(

3

5

) (

5

3

) (

10

+ b10 ≥ a 3 b 7 + b 7 a 3

)(

)

(

c. ( a + b ) a 3 + b 3 ≤ 2 a 4 + b 4 ; a 2 + b 2 a 4 + b 4 ≥ a 3 + b 3 ; a 5 + b 5 a 3 + b 3 ≤ 2 a 8 + b 8

(

)(

)

d. ( ax + by ) ≤ a 2 + b 2 x 2 + y 2 ;
2

a 2 + x 2 + b2 + y2 ≥

2

( a + b)

2

+ ( x + y)

)

2

5a 3 − b 3
19a 3 − b 3
3a 3 + 7b 3
a 3 b3
2
2
≤
2
a
−
b
;
≤
2
a
−
b
;

≥
a
+
2
b
−
ab
;
+ 2 ≥ a 2 + b 2 ( a , b > 0)
2
2
2
3a + ab
5a + ab
2a + 3b
b
a
4
2
2
4
4
2
2 2
4
4
f. 8( a + b ) ≥ ( a + b ) ; ( a + b ) ≥ ab( a + b ) ; a + b + 2 ≥ 4ab; ( a + b + c + d ) ≥ ( a + c )( b + d )
e.

2

3

a 2b
2 1
2
9
 a 3 + b 3   a 2 + b 2  a 2 + 2ab
g. 
− 3
≤ ; 2 +
≥
∀a , b > 0
 ≥
; 2
2
2
3
2
2a + b
3 4a
( a + b ) 4( a + 2ab )
 2   2  2a + b
Bµi 3: Chøng minh c¸c bÊt ®¼ng thøc b»ng ph¬ng ph¸p ph©n tÝch b×nh ph¬ng ( ∀a , b, c > 0 ) :
2
a 2 b2 c2
b2 c2
a b c a
 a b c  bc ca ab
+

+
≥a+b+c
a. 2 + 2 + 2 + 3 ≥ 2 + + ; 2 + 2 + 2 ≥ 2 − + ;
b
c
a
c
a
b
c
b c a b
c a b a
3
b. ( a + b )( b + c )( c + a ) ≥ 8abc
c. ( a + b + c ) ≥ a 3 + b 3 + c 3 + 24abc
a
b
c
3 bc
ca
ab
a+b+c
1 1 1
+
+
≥ ;
+
+
≤
d. ( a + b + c )  + +  ≥ 9;

b+c c+a a+b 2 b+c c+a a+b
2
a b c
2
2
2
2
2
2
2
2
2
2
a
b
a +b
b +c
c +a
a + b + c2 a 2 + b2
2
2
e.
f.
+
+ 7( a + b ) ≥ 8 2( a + b )
+
+
≤3
b
c

a+b
b+c
c+a
a+b+c a+b
Bµi 4: Chøng minh r»ng ∀a , b, c, d > 0 ta cã:
1
1
2
( ab ≥ 1) ; 1 2 + 1 2 + 1 2 + 1 2 ≥ 4 ( abcd ≥ 1)
+
≥
a.
2
2
1+ a
1 + b 1 + ab
1+ a
1 + b 1 + c 1 + d 1 + abcd
1
1
1
3
( abc ≥ 1); 1 2 + 1 2 ≤ 2 ( ab ≤ 1)
+
+
≥
b.
2
2
2

1+ a
1 + b 1 + c 1 + abc
1+ a
1 + b 1 + ab
2+b+c 2+c+a 2+a +b
1
2
3
6
( a, b, c > 1)
+
+
≥ 6;
+
+
≥
c.
6
3
2
1+ a
1+ b
1+ c
1 + a 1 + b 1 + c 1 + abc
1
1
1
1
1
1

( a, b, c ≤ 1)
+
+
≤
+
+
d.
3
3
3
2
2
1 + a 1 + b 1 + c 1 + ab 1 + bc 1 + ca 2
1
1
1
1
1
1
1
4
( abcd = 1)
+
≥
;
+
+
+
≥
e.

2
2
2
2
2
2
(1 + a ) (1 + b ) 1 + ab (1 + a ) (1 + b ) (1 + c ) (1 + d ) 1 + abcd
Bµi 5: Chøng minh r»ng ∀a , b, c > 0 ta cã:
n +1
n +1
a. ∀a ∈ [ 0;1] : 27a 2 (1 − 2a ) ≤ 1; ( n + 1) a n (1 − na ) ≤ 1; 27a 2 (1 − a ) ≤ 4; ( n + 1) a n (1 − a ) ≤ n n
1


BÊt ®¼ng thøc

b. ∀a ∈ [ 0;1] :3 3a (1 − a 2 ) ≤ 2; 4 3 4 a (1 − a 3 ) ≤ 3; ( n + 1) n n + 1 a (1 − a n ) ≤ n

2a − 1; 3 3a − 2 ; n na − n + 1 ≤ a; ∀a.b.c = 1 : 3 3b − 2 5 5c − 4 7 7c − 6 ≤ 1

c.
d.

3

(1 + x )(1 + y )(1 + z ) ≥ 1 +

3

abc ; (1 + a 3 )(1 + b 3 )(1 + c 3 ) ≥ (1 + ab 2 )(1 + bc 2 )(1 + ca 2 )

(

)

x 1 x 2 ...x n + n y1 y 2 ...y n ≤ n ( x 1 + y1 )( x 2 + y 2 )...( x n + y n ) ∀x i ; y i > 0 i = 1; n
Bµi 6: Chøng minh r»ng ∀a , b, c > 0 ta cã:
a
b
c
3 3 2
a
b
1
2
2
a. 2
(
)
( 3a 2 + 3b 2 = 2)
+
+
≥
a
+
b
+
c
;
+

≥
2
2
2
2
2
2
2
b +c
c +a
a +b
2
1 + 3b 1 + 3a
3
2
2
2
2
2
3
a
b
c
4 4 3
a
9b
3
3
b.
(

)
+
+
≥
a
+
b
+
c
=
1
;
+
≥ 4 3 4 ( a 3 + 3b 3 = 1)
3
3
3
3
3
3
1− a 1− b 1− c
3
b
a + 2b
1 1 1 3 3 2
c. + + +
( a + b 2 + c 2 ) ≥ 3 3 ( 0 < a, b, c < 1); 13 + 13 + 13 + 768( a + b + c) ≥ 768
a b c
2
2

a
b
c
1
1
1
3
1

+
+
≥ 12  a 3 + b 3 + c 3 =
& 0 < a , b, c < 
d.
1 − 3a 1 − 3b 1 − 3c
64
3

Bµi 7: Chøng minh r»ng ∀a , b, c > 0 ta cã:
3
3
3
2
2
2
4
4
4
5
5

5
2 3
2 3
3 3
a. a + b + c ≥ ab + bc + ca ; a + b + c ≥ abc( a + b + c ) ; a + b + c ≥ a b + b c + c a
a 2 b 2 c 2 a c b a 3 b 3 c 3 ab bc ca a 5 b 5 c 5 a 2 b 2 b 2 c 2 c 2 a 2
b. 2 + 2 + 2 ≥ + + ;
+
+
≥
+
+ ;
+
+
≥
+
+
b
c
a
c b a bc ca ab c
a
b bc ca ab
c
a
b
3
3
3
2

2
2
5
5
5
a
b
c
a
b
c 1
1
1
1
1
1 a
b
c
c. 3 + 3 + 3 ≥
+
+ ; 3 + 3 + 3 ≥ 2 + 2 + 2 ; 2 + 2 + 2 ≥ a 3 + b3 + c3
b
c
a
bc ca ab a
b
c a b bc ca b
c
a
2

2
2
2
2
2
3
3
3
4
4
4
( b + c) + ( c + a ) ; a + b + c ; a + b + c ≥ a + b + c
a
b
c ( a + b)
d.
+
+ ;
+
b
c
a
4c
4a
4b
bc ca ab b 2 c c 2 a a 2 b
a2
b2
c2
a2

b2
c2
a + b + c a3
b3
c3
a 2 + b2 + c2
e.
+
+
;
+
+
≥
;
+
+
≥
b+c c+a a+b a+b b+c c+a
2
a + 2b b + 2c c + 2a
3
Bµi 8: Cho x , y, z > 0 vµ a , b, c > 0 . Chøng minh c¸c bÊt ®¼ng thøc sau:
e.

n

a.
Bµi 8: Cho x , y, z > 0 vµ a , b, c > 0 . Chøng minh c¸c bÊt ®¼ng thøc sau:

 1

 3
1 1
4 1 1 1
1
1 
9
3
3 
;

+ ≥
; + + ≥ 2
+
+
; 
+
+
x y x+y x y z
x
+
y
y
+
z
z
+
x
x
+
y

+
z
x
+
2
y
y
+
2
z
z
+
2
x




1
1
1
9
2
2
2
9
+ 2
+ 2
≥
; 2

+ 2
+ 2
≥ 2
b. 2
2
x + 2 yz y + 2zx z + 2xy ( x + y + z ) x + yz y + zx z + xy x + y 2 + z 2
9
1
1
1
11 1 1
≤
+
+
≤  + + 
c.
4( a + b + c ) a + 2b + c b + 2c + a c + 2a + b 4  a b c 
2a + b 2b + c 2c + a
a + 3b
b + 3c
c + 3a 12 a + 3b b + 3c c + 3a 12
+
+
≥ 3;
+
+
≥ ;
+
+
≥

d.
2c + b 2a + c 2b + a
2a + 3c 2b + 3a 2c + 3b 5 2b + c 2c + a 2a + b 3
Bµi 9: Cho x , y > 0 & x + y = 2 . Chøng minh r»ng:
2
2
3
3
4
4
n
n
2
2
n n
2
2
a. xy ≤ 1; x + y ≥ 2; x + y ≥ 2; x + y ≥ 2; x + y ≥ 2; xy( x + y ) ≤ 2; x y ( x + y ) ≤ 2
1 1
1
1
2 2
4 4
3
2
2
3
3
3
3

≥5
b. + ≥ 2; 2 + 2 ≥ 2; x + y + + ≥ 6; x + y + + ≥ 10; x y + y x +
x y
x
y
x y
x y
xy
 2 2  2 2 
 2 4  2 4 
2 
2
3
4

+ 2
≥5
c.  x +  y +  ≥ 9;  x +  y +  ≥ 9;  x + 2  y + 2  ≥ 25;
2
x 
y
y
x
y
x
xy
x
+
y










a.

2


BÊt ®¼ng thøc

Bµi 10: Cho x , y > 0 & x + y ≤ 1 . Chøng minh c¸c bÊt ®¼ng thøc sau:

1
1
2
3
m
n
+ 2
≥
6
;
+
≥
14

;
+
≥ 4 m + 2 n ( 2m ≥ n > 0 )
8 xy x + y 2
xy x 2 + y 2
xy x 2 + y 2
2
1
a+b
c
+ 2
+ 4xy ≥ 11;
+ 2
+ 16cxy ≥ 4a + 8b + 2c( a ≥ b > 0; c > 0 )
b.
2
xy x + y
xy
x + y2
 2 1  2 1 
1 
1
1 
1 


c. 1 − 2 1 − 2  ≥ 9;  4 + 2  4 + 2  ≥ 64;  4 x + 2  4 y + 2  ≥ 25
y 
x 
y 

y 
x 
 x 



(

a. xy x + y
2

2

) ≤ 1;


1 
1 
125x 4 + 2 125y 4 + 2  ≥ 125
y 
x 

Cho x , y, z > 0 . Chøng minh c¸c bÊt ®¼ng thøc sau:
a. x + y + z ≥ 3 3 xyz +

(

)

2

x − y ≥ 3 3 xyz
Bµi 3: Cho x , y, z > 0 & 2( x + y + z ) ≤ 3 . Chøng minh c¸c bÊt ®¼ng thøc sau:
1 1 1 15
1
1
1 99
3
3
3
a. x + y + z + + + ≥
b. x + y + z + 2 + 2 + 2 ≥
x y z 2
x
y
z
8
2

2

2

2


1 
1 
1
c.  2 x +  +  2 y +  +  2z +  ≥ 27

y 
z 
x

e. 3 + 2 x + 3 + 2 y + 3 + 2 x ≤ 6

2

2


1 
1 
1
d.  8x 3 +  +  8 y 3 +  +  8z 3 +  ≥ 27
y 
z 
x

f. 3 9 x + 7 y + 3 9 y + 7 z + 3 9z + 7 x ≤ 6

g. 4 13 + 3 2 x + 4 13 + 3 2 y + 4 13 + 3 2z ≤ 6 h. x +
2

1
1
1 3 17
+ y2 + 2 + z2 + 2 ≥
2
y

z
x
2

Bµi 4: Chøng minh r»ng ∀a , b, c > 0 ta cã:

ab
bc
ca
a 2 − c2 b2 − a 2 c2 − b2
+ 2
+ 2
≤1
+
+
≥0 2
c + 2ab a + 2bc b + 2ca
b+c
c+a
a+b
a
b
c
3 2
+
+
≥
2
ca + cb
ab + ac

bc + ba
Bµi 6: Cho a , b, c ∈ R & x , y, z > 0 . Chøng minh r»ng:
a 2 b2 ( a + b)
+
≥
a.
x
y
x+y
x
y
z
+
+
≥1
c.
x + 2 y y + 2z z + 2 x
1
4
9
36
+
+
≥
e.
x +1 y + 2 z + 3 x + y + z + 6
25
81
+
≥ 7(19x + 2 y = 14 )

g.
19x + 5 2 y + 9
Bµi 7: Cho a , b, c > 0 & a + b + c = 1 .

a 2 b 2 c 2 ( a + b + c)
+
+ ≥
b.
x
y
z
x+y+z
x
y
z
1
+
+
≥
d.
x + 2 y + 3z y + 2z + 3x z + 2 x + 3y 2
x2
y2
z2
+
+
≥1
f. 2
z + yz + x 2 x 2 + zx + y 2 y 2 + xy + z 2
x2

y2
z2
3
+
+
≥
h.
( x + y )( x + z ) ( y + z )( y + x ) ( z + x )( z + y ) 4

2

1. T×m gi¸ trÞ nhá nhÊt cña:
3
3
3
a. T1 = a + b + c
c. T3 =

8

2

b. T2 =
8

a + bc + b + ca + c + ab − ab − bc − ca

8

a

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

c. T4 =

2

2. T×m gi¸ trÞ lín nhÊt cña:
a. S1 = a + b + c

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

b. S1 = 12abc + a 2 + b 2 + c 2
3

BÊt ®¼ng thøc

c. S3 =

ab
1 + 3c 2

+

bc
1 + 3a 2

+

ca

d. S 4 =

ab
bc
ca
+
+
a + b + 2c b + c + 2a c + a + 2b

1 + 3b 2
Bµi 8: Cho a, b, c lµ c¸c sè thùc d¬ng sao cho abc = 1 . CMR:
a

b
c
3
1
1
1
+
+
≥
+
+
≤1
a.
b.
( a + 1)( b + 1) ( b + 1)( c + 1) ( c + 1)( a + 1) 4
a+2 b+2 c+2
Bµi 9: Cho x , y, z ∈ ( 0;1] . Chøng minh r»ng:
x
y
z
x
y
z
1
+
+
≤1
+
+
≤

a.
b. 2
1 + x + yz 1 + y + zx 1 + z + xy
y + z 2 + 4 z2 + x 2 + 4 x 2 + y2 + 4 2
1 1 1 4
+ + + ( a + b + c) ≥ 7
Bµi 10: Cho a , b, c > 0 vµ a 2 + b 2 + c 2 = 3 . Chøng minh r»ng:
a b c 3
1 
 1 1 1  225
Bµi 11: Cho a , b, c ∈  , 2 . CMR: ( a + b + c )  + +  ≤
2 
 a b c  16
Bµi 13: Cho a, b, c lµ 3 c¹nh tam gi¸c:
a. ( a + b − c )( b + c − a )( c + a − b ) ≤ abc b.

a +b−c +n b+c−a +n c+a −b ≤n a +n b +n c
1
1
1
1 1 1
a2
b2
c2
+
+
≥
+
+
c.

d.
+
+
≥a+b+c
a+b−c b+c−a c+a −b a b c b+c−a c+a −b a +b−c
Bµi 15: Cho a , b, c > 0 & a + b + c = 1 . CMR:
2
2
2
2
3
3
3
a. 4( a 3 + b 3 + c 3 − 3abc ) ≥ 3( a − b )
b. 5( a + b + c ) ≤ 6( a + b + c ) + 1
5
a 2 b2 c2
2
2
2
3
3
3
c.
d. 3( a + b + c ) + 2( a + b + c ) + 12abc ≥
+
+ ≥ 3( a 2 + b 2 + c 2 )
3
b
c

a
3
3
2
2
2( a + b ) ≥ ( a + b ) ( a + b ) ;
a 2 ( b + c − a ) + b 2 ( c + a − b ) + c 2 ( a + b − c ) ≤ 3abc ∀a; b; c ≥ 0
n

3
a6
b6
c6
a 5 b 5 c5
b3 c3 a 2 b 2 c2
3
3
3 a
+
+
≥ ab + bc + ca 2 + 2 + 2 ≥ a + b + c 2 + 2 + 2 ≥ +
+
b 2c2 c2a 2 a 2 b 2
b
c
a
b
c
a
b

c
a
2

a 5 b 5 c5
a 3 b3 c3
a b c
1 1 1
2
2
2
(
)
+
+
≥
a
b
+
b
c
+
c
a
+
+
≥
ab
+
bc

+
ca
+
+
≥
a
+
b
+
c


 + + 
b2 c2 a 2
b
c
a
b
c
a


a b c
1
1
1
27
1
1
1

9
3
+
+
≥
+
+
≥
2
2
b( a + b ) c( b + c ) a ( c + a ) 2( a + b + c )
ab bc ca ( a + b + c ) + 2
3
4
+ 2
+ 16xy ≥ 24;
xy x + y 2

4