# Một số bài tập về bất đẳng thức (Phạm Công Thành Quảng Ngãi)

TEST 1: BUNHIAKOPXKI AND AM-GM
“In mathematics the art of proposing a question must be held of higher value than solving it.”
Georg Cantor

PROBLEM 1: If a,b are positive numbers such that: a 2  b 2  4 then:
2018a  2017b  16140

PROBLEM 2: If a, b, c are positive numbers then:
a

3

 2018a  b  c  2020
PROBLEM 3: If a, b, c are positive numbers then:
a(2a  b)
1
ac  c) 2
(CMATH)

 (b 

PROBLEM 4: Let a, b, c are positive numbers such that abc=1. Prove that:

a

8

1
1

 3a  2ab  6 2
2

PROBLEM 5: If a, b, c are positive numbers then:

1
18 x  2 y  4 z
1
1
1

6( x  y ) (2 y  3z )( z  18 x) 12 x  2 y  z 6 x  4 y  z 6 x  2 y  2 z
PROBLEM 6: Let a, b, c > 0 such that ab  bc  ca 

3abc
. Prove that:
2018

1
2018a 2  4024ab  2015b 2

1
2018

PROLEM 7: If a, b, c are positives numbers and ab+bc+ca=abc then:

b2  2a 2
 3
ab

1 1
3
1
 ab  bc  ca  abc

a 2  b 2  ab

PROBLEM 8: Let a, b, c are positive numbers such that 
b 2  c 2  bc

c 2  a 2  ca

Find the minimum value of the empression:

a 4  b4
1
ab
 a 2  b2  ab  3 ( c )2

PROBLEM 9: If a, b, c are positive numbers then:

1
8

3
2
2
2
( 2018abc  1)
( a  4b  2)( 4b  (1009c) 2  2)( (1009c) 2  a 2  2)
3

 a1  a2  ...  a2018  1
.
2
2
2
a1  a2  ...  a2018  1

PROBLEM 10: Let a1 , a2 ,..., a2018 are real numbers satisfying: 
Find the minimum value and maximum value of a2018 .

__________THE END__________
Good luck!

RESOLUTION

Pham Cong Thanh

PROBLEM 1: If a,b are positive numbers such that: a 2  b 2  4 then:
2018a  2017b  16140

Dễ thấy a>b>0
Ta có: 4  a 2  b2  (a  b)(a  b)
1 4035
4035
1
 4035  [
(a  b)][ ( a  b)]
2 2
2
2

 4. .

Áp dụng BĐT ab  (

ab 2
) suy ra:
2

4035
1
4035  [
(a  b)][ (a  b)]  [
2
2

4035
1
(a  b)  (a  b)
2018a  2017b 2
2
2
]2  (
)
2
2

 (2018a  2017b)2  16140
 2018a  2017b  16140 (đpcm).

4036

 a  4035
Dấu “=” xảy ra khi 
b  4034

4035

PROBLEM 2: If a, b, c are positive numbers then:
a

3

 2018a  b  c  2020

Ta có: 3  2017.
Áp dụng BĐT

3  2017.

a
abc
1

 (a  b  c).
2018a  b  c
2018a  b  c
2018a  b  c

1 1 1
9
  
suy ra:
x y z x yz
a
9
9
 (a  b  c).

2018a  b  c
2020(a  b  c) 2020
a

3

 2018a  b  c  2020 . (đpcm)

Dấu “=”xảy ra khi a=b=c.

PROBLEM 3: If a, b, c are positive numbers then:
a(2a  b)
1
ac  c) 2

 (b 

Ta có: (b  ac  c) 2  ( b . b  a . c  a .

c 2
)
a

Áp dụng BĐT Bunhiakopxki, ta có: ( b. b  a . c  a .

=>

a (2a  b)

(b  ac  c) 2

a
bc

c2
a

a2
ab  ac  c 2

c 2
c2
)  (2a  b)(b  c  )
a
a

c(2c  a)
b(2b  c)
b2
c2

Tương tự:

2
2
(a  ab  c)2 ab  bc  a
(a  bc  b) 2 ac  bc  b
a(2a  b)
a2
b2
c2

=> 
2
2
2
(b  ac  c)2 ab  ac  c ab  bc  a ac  bc  b
Áp dụng BĐT Cauchy- Schward, ta có:
a(2a  b)
a2
b2
c2
(a  b  c ) 2
(a  b  c ) 2

1
2
2
2
a 2  b 2  c 2  2(ab  bc  ca ) (a  b  c )2
ac  c)2 ab  ac  c ab  bc  a ac  bc  b

 (b 

=>đpcm.
Dấu “=” xảy ra khi a=b=c.

PROBLEM 4: Let a, b, c are positive numbers such that abc=1. Prove that:

a

8

1
1

 3a  2ab  6 2
2

Ta có:

a8  3a 2  2ab  6 = (a8  1  1  1)  3a 2  2ab  3  AM GM (a 2  1)  2ab  2  AM GM 2( a  ab  1)
=>

1
1

a  3a  2ab  6 2(a  ab  1)
8

2

Tương tự:

=>

a

8

1
1
1
1

và 8
2
c  3c  2ca  6 2(c ca  1)
b  3b  2bc  6 2(b  bc  1)
8

2

1
1
1
 .
 3a  2ab  6 2
ab  a  1
2

Với abc=1 dễ chứng minh

=>

a

8

1
1

 3a  2ab  6 2
2

1

 ab  a  1  1

=>đpcm
Dấu “=” xảy ra khi a=b=c=1.

PROBLEM 5: If a, b, c are positive numbers then:

1
18 x  2 y  4 z
1
1
1

6( x  y ) (2 y  3z )( z  18 x) 12 x  2 y  z 6 x  4 y  z 6 x  2 y  2 z

BĐT 

1
(2 y  3z )  ( z  18 x)
1
1
1

6( x  y) (2 y  3z )( z  18 x) 12 x  2 y  z 6 x  4 y  z 6 x  2 y  2 z

1
1
1
1
1
1

6( x  y) 2 y  3z z  18 x 12 x  2 y  z 6 x  4 y  z 6 x  2 y  2 z

Đặt a=6x; b=2y; c=z
BĐT 

1

1

 a  3b   2a  b  c

Ta dễ dàng chứng minh BĐT trên.
Thật vậy: Áp dụng BĐT

1 1
4
 
ta có:
a b ab

1
1
4
2

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

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

c  3a a  2b  c 2(2a  b  c) 2a  b  c
Cộng vế tương ứng => đpcm.
Dấu “=” xảy ra khi a=b=c hay 6x=2y=z

PROBLEM 6: Let a, b, c > 0 such that ab  bc  ca 

Ta có:

1

2018a  4024ab  2015b

1



2018a 2  4024ab  2015b 2

Áp dụng BĐT

=>

1
2

2

3abc
. Prove that:
2018

1
2018

1
(2a  b) 2  2014(a  b) 2



1
2a  b

1 1 1
9
  
x y z x yz
1

9

1 3

3

3

1 1

1

1

 2a  b  9  a  a  b  9 ( a  b  c )  3 ( a  b  c )

Lại có ab  bc  ca 

3abc
2018

1 1 1
3
  
 a b c 2018

1

2018a  4024ab  2015b

 Đpcm
Dấu “=” xảy ra khi a  b  c  2018 .
2

2

1
2018

PROLEM 7: If a, b, c are positives numbers and ab+bc+ca=abc then:

b2  2a 2
 3
ab

Áp dụng BĐT Bunhiakopxki cho bộ số (1; 2) và (b; 2.a) , ta có:

(1  2)(b2  2a 2 )  Bunhiakopxki (b  2a)2

b 2  2a 2 

b  2a
3

b 2  2a 2 b  2a

ab
3ab
Tương tự:

c 2  2b 2 c  2b

;
bc
3bc

a 2  2c 2 a  2c

ca
3ca

b  2a c  2b a  2c 3(ab  bc  ca)
b 2  2a 2

 3 (đpcm)
ab
3ab
3bc
3ca
3abc

Dấu “=” xảy ra khi a=b=c=3.

1 1
3
1
 ab  bc  ca  abc

a 2  b 2  ab

PROBLEM 8: Let a, b, c are positive numbers such that 
b 2  c 2  bc

c 2  a 2  ca

Find the minimum value of the empression:

a 4  b4
1
ab
 a 2  b2  ab  3 ( c )2

Ta có: (a 2  b2 )(a  b)2  0  a 4  b 4  2a 3b  2ab3  2a 2b 2  0  a 4  b4  2ab(a 2  b 2  ab)

a 4  b4
 2ab

a 2  b 2  ab
b4  c 4
c4  a4

2
bc
 2ca
Tương tự: 2
;
b  c 2  bc
c 2  a 2  ca

a 4  b4
  2
  2ab (1)
a  b 2  ab
2

ab
a 2b 2 b 2 c 2 c 2 a 2
Lại có: ( )  2  2  2  2(a 2  b 2  c 2 )  AM GM 3 a 2
c
c
a
b
2

1
ab
( )   a 2 (2)

3
c
1
1 1
3
 

Mặt khác:
=> a  b  c  3
ab bc ca abc

a 4  b4
1
ab
 a2  b2  ab + 3 ( c )   a2   2ab  ( a)2  9 .
2

Từ (1) và (2) suy ra:
Vậy GTNN của

a 4  b4
1
ab
 a 2  b2  ab  3 ( c )2 là 9.

Dấu “=” xảy ra khi a=b=c=1.

PROBLEM 9: If a, b, c are positive numbers then:

1
8

3
2
2
2
( 2018abc  1)
( a  4b  2)( 4b  (1009c) 2  2)( (1009c) 2  a 2  2)
3

Đặt x=a; y=2b; z=1009c
BĐT 

1
8

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

Thật vậy, ta có:
(1  x)(1  y )(1  z )  xyz  ( xy  yz  xz )  ( x  y  z )  1  AM GM xyz  3 3 x 2 y 2 z 2  3 3 xyz  1  ( 3 xyz  1)3

1
1

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

Lại có

(1  x)(1  y )(1  z )  (1  x)(1  y ). (1  y )(1  z ). (1  x)(1  z ) 

( x  y  2)( y  z  2)( x  z  2)
8

( x  y  2)( y  z  2)( x  z  2) ( 2( x2  y 2 )  2)( 2(y2  z 2 )  2)( 2(z 2  x2 )  2) ( x2  y 2  2)( y2  z 2  2)( z 2  x 2  2)

8
8
8
 (1  x)(1  y )(1  z ) 

( x 2  y 2  2)( y2  z 2  2)( z 2  x 2  2)
8

1
8

(2)
2
2
2
(1  x)(1  y )(1  z ) ( x  y  2)( y  z 2  2)( z 2  x 2  2)
Từ (1) và (2) suy ra: đpcm.

 a1  a2  ...  a2018  1
.
2
2
2
a1  a2  ...  a2018  1

PROBLEM 10: Let a1 , a2 ,..., a2018 are real numbers satisfying: 

Find the minimum value and maximum value of a2018 .

 a1  a2  ...  a2018  1
 a1  a2  ...  a2017  1  a2018

 2
2
2
2
2
2
2
a1  a2  ...  a2018  1 a1  a2  ...  a2017  1  a2018

Ta có: 

Áp dụng BĐT Bunhiakopxki, ta có:

(1  1  ...  1)(a12  a2 2  ...  a2017 2 )  ( a1  a2  ...  a2017 ) 2
2
2
2
2
 2017(a1  a2  ...  a2017 )  ( a1  a2  ...  a2017 )
2
2
 2017(1  a2018 )  (1  a2018 )
2
 2018a2018  2a2018  2016  0

 1  a2018  

1008
.
1009

Vậy GTNN của a2018 

1008
và GTLN của a2018 là 1.
1009