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Một số bài tập về bất đẳng thức (Phạm Công Thành Quảng Ngãi). Tài liệu tổng hợp 1 số bài tập về bất đẳng thức Bunhiakopxki và AmGm ( Cauchy)..............................................................................................................................
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 b then: 2018a 2017b 16140 PROBLEM 2: If a, b, c are positive numbers then: a 2018a b c 2020 PROBLEM 3: If a, b, c are positive numbers then: a(2a b) 1 ac c) (CMATH) (b PROBLEM 4: Let a, b, c are positive numbers such that abc=1 Prove that: a 1 3a 2ab 2 PROBLEM 5: If a, b, c are positive numbers then: 18 x y z 1 6( x y ) (2 y 3z )( z 18 x) 12 x y z x y z x y z PROBLEM 6: Let a, b, c > such that ab bc ca 3abc Prove that: 2018 2018a 4024ab 2015b 2018 PROLEM 7: If a, b, c are positives numbers and ab+bc+ca=abc then: b2 2a ab 1 1 ab bc ca abc a b ab PROBLEM 8: Let a, b, c are positive numbers such that b c bc c a ca Find the minimum value of the empression: a b4 ab a b2 ab ( c )2 PROBLEM 9: If a, b, c are positive numbers then: 2 ( 2018abc 1) ( a 4b 2)( 4b (1009c) 2)( (1009c) a 2) a1 a2 a2018 2 a1 a2 a2018 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 b then: 2018a 2017b 16140 Dễ thấy a>b>0 Ta có: a b2 (a b)(a b) 4035 4035 4035 [ (a b)][ ( a b)] 2 2 Áp dụng BĐT ab ( ab ) suy ra: 4035 4035 [ (a b)][ (a b)] [ 2 4035 (a b) (a b) 2018a 2017b 2 ]2 ( ) 2 (2018a 2017b)2 16140 2018a 2017b 16140 (đpcm) 4036 a 4035 Dấu “=” xảy b 4034 4035 PROBLEM 2: If a, b, c are positive numbers then: a 2018a b c 2020 Ta có: 2017. Áp dụng BĐT 2017. a abc (a b c). 2018a b c 2018a b c 2018a b c 1 suy ra: x y z x yz a 9 (a b c) 2018a b c 2020(a b c) 2020 a 2018a b c 2020 (đpcm) Dấu “=”xảy a=b=c PROBLEM 3: If a, b, c are positive numbers then: a(2a b) 1 ac c) (b Ta có: (b ac c) ( b b a c a c ) a Áp dụng BĐT Bunhiakopxki, ta có: ( b b a c a => a (2a b) (b ac c) a bc c2 a a2 ab ac c c c2 ) (2a b)(b c ) a a c(2c a) b(2b c) b2 c2 Tương tự: 2 (a ab c)2 ab bc a (a bc b) ac bc b a(2a b) a2 b2 c2 => 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 ) (a b c ) 1 2 a b c 2(ab bc ca ) (a b c )2 ac c)2 ab ac c ab bc a ac bc b (b =>đpcm Dấu “=” xảy a=b=c PROBLEM 4: Let a, b, c are positive numbers such that abc=1 Prove that: a 1 3a 2ab 2 Ta có: a8 3a 2ab = (a8 1) 3a 2ab AM GM (a 1) 2ab AM GM 2( a ab 1) => 1 a 3a 2ab 2(a ab 1) Tương tự: => a 1 1 c 3c 2ca 2(c ca 1) b 3b 2bc 2(b bc 1) 1 3a 2ab ab a Với abc=1 dễ chứng minh => a 1 3a 2ab 2 ab a =>đpcm Dấu “=” xảy a=b=c=1 PROBLEM 5: If a, b, c are positive numbers then: 18 x y z 1 6( x y ) (2 y 3z )( z 18 x) 12 x y z x y z x y z BĐT (2 y 3z ) ( z 18 x) 1 6( x y) (2 y 3z )( z 18 x) 12 x y z x y z x y z 1 1 1 6( x y) y 3z z 18 x 12 x y z x y z x y z Đặt a=6x; b=2y; c=z BĐT 1 a 3b 2a b c Ta dễ dàng chứng minh BĐT Thật vậy: Áp dụng BĐT 1 ta có: a b ab 1 a 3b a b 2c 2(a 2b c) a 2b c 1 b 3c 2a b c 2(a b 2c ) a b 2c 1 c 3a a 2b c 2(2a b c) 2a b c Cộng vế tương ứng => đpcm Dấu “=” xảy a=b=c hay 6x=2y=z PROBLEM 6: Let a, b, c > such that ab bc ca Ta có: 2018a 4024ab 2015b 2018a 4024ab 2015b Áp dụng BĐT => 2 3abc Prove that: 2018 2018 (2a b) 2014(a b) 2a b 1 x y z x yz 3 1 1 2a b a a b ( a b c ) ( a b c ) Lại có ab bc ca 3abc 2018 1 a b c 2018 2018a 4024ab 2015b Đpcm Dấu “=” xảy a b c 2018 2 2018 PROLEM 7: If a, b, c are positives numbers and ab+bc+ca=abc then: b2 2a ab Áp dụng BĐT Bunhiakopxki cho số (1; 2) (b; 2.a) , ta có: (1 2)(b2 2a ) Bunhiakopxki (b 2a)2 b 2a b 2a b 2a b 2a ab 3ab Tương tự: c 2b c 2b ; bc 3bc a 2c a 2c ca 3ca b 2a c 2b a 2c 3(ab bc ca) b 2a (đpcm) ab 3ab 3bc 3ca 3abc Dấu “=” xảy a=b=c=3 1 1 ab bc ca abc a b ab PROBLEM 8: Let a, b, c are positive numbers such that b c bc c a ca Find the minimum value of the empression: a b4 ab a b2 ab ( c )2 Ta có: (a b2 )(a b)2 a b 2a 3b 2ab3 2a 2b a b4 2ab(a b ab) a b4 2ab a b ab b4 c c4 a4 bc 2ca Tương tự: ; b c bc c a ca a b4 2ab (1) a b ab ab a 2b b c c a Lại có: ( ) 2(a b c ) AM GM 3 a c c a b ab ( ) a (2) c 1 Mặt khác: => a b c ab bc ca abc a b4 ab a2 b2 ab + ( c ) a2 2ab ( a)2 Từ (1) (2) suy ra: Vậy GTNN a b4 ab a b2 ab ( c )2 Dấu “=” xảy a=b=c=1 PROBLEM 9: If a, b, c are positive numbers then: 2 ( 2018abc 1) ( a 4b 2)( 4b (1009c) 2)( (1009c) a 2) Đặt x=a; y=2b; z=1009c BĐT ( xyz 1)3 ( x y 2)( y z 2)( z x 2) Thật vậy, ta có: (1 x)(1 y )(1 z ) xyz ( xy yz xz ) ( x y z ) AM GM xyz 3 x y z 3 xyz ( xyz 1)3 1 (1) ( 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) Mà ( x y 2)( y z 2)( x z 2) ( 2( x2 y ) 2)( 2(y2 z ) 2)( 2(z x2 ) 2) ( x2 y 2)( y2 z 2)( z x 2) 8 (1 x)(1 y )(1 z ) ( x y 2)( y2 z 2)( z x 2) 8 (2) 2 (1 x)(1 y )(1 z ) ( x y 2)( y z 2)( z x 2) Từ (1) (2) suy ra: đpcm a1 a2 a2018 2 a1 a2 a2018 PROBLEM 10: Let a1 , a2 , , a2018 are real numbers satisfying: Find the minimum value and maximum value of a2018 a1 a2 a2018 a1 a2 a2017 a2018 2 2 2 a1 a2 a2018 a1 a2 a2017 a2018 Ta có: Áp dụng BĐT Bunhiakopxki, ta có: (1 1)(a12 a2 a2017 ) ( a1 a2 a2017 ) 2 2 2017(a1 a2 a2017 ) ( a1 a2 a2017 ) 2 2017(1 a2018 ) (1 a2018 ) 2018a2018 2a2018 2016 a2018 1008 1009 Vậy GTNN a2018 1008 GTLN a2018 1009 THANKS FOR READING ... a, b, c are positives numbers and ab+bc+ca=abc then: b2 2a ab Áp dụng BĐT Bunhiakopxki cho số (1; 2) (b; 2.a) , ta có: (1 2)(b2 2a ) Bunhiakopxki (b 2a)2 b 2a b 2a b 2a