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 (
ab 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
abc
1
(a b c).
2018a b c
2018a b c
2018a b c
1 1 1
9
suy ra:
x y z x yz
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
bc
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:
và
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 ab
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 yz
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
Mà
( 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
____________________THANKS FOR READING____________________