Processing math: 100%

quyhktn-qa1's inequality

陈洪葛 posted @ 12 年前 in 不等式 , 945 阅读

Let a,b,c be nonnegative real numbers such that a+b+c=3.Prove that
113a2+(bc)2+113b2+(ca)2+113c2+(ab)2313.

Proof:(by scaleye,taiwan)

The inequality is equivalent to 113a2+(bc)2+113b2+(ca)2+113c2+(ab)22713(a+b+c)2.
By Cauchy Schwarz Inequality,
113a2+(bc)2=(a+13b+13c)2[13a2+(bc)2](a+13b+13c)2272(a+b+c)2[13a2+(bc)2](a+13b+13c)2.
It suffices to show that
1327(a+b+c)4[13a2+(bc)2](a+13b+13c)2.
After expanding and simplifying, it is equivalent to
65(a3b+ab3)122a2b2+8a2bc
or
61ab(ab)2+2(a3b+ab3+a3c+ac34a2bc)0.
Which is true by AM-GM.
Equality holds when a=b=c=1  or a=3, b=c=0 and any  permutations


登录 *


loading captcha image...
(输入验证码)
or Ctrl+Enter