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+(b−c)2+113b2+(c−a)2+113c2+(a−b)2≥313.
Proof:(by scaleye,taiwan)
The inequality is equivalent to 113a2+(b−c)2+113b2+(c−a)2+113c2+(a−b)2≥2713(a+b+c)2.
By Cauchy Schwarz Inequality,
∑113a2+(b−c)2=∑(a+13b+13c)2[13a2+(b−c)2](a+13b+13c)2≥272(a+b+c)2∑[13a2+(b−c)2](a+13b+13c)2.
It suffices to show that
13⋅27(a+b+c)4≥∑[13a2+(b−c)2](a+13b+13c)2.
After expanding and simplifying, it is equivalent to
65∑(a3b+ab3)≥122∑a2b2+8∑a2bc
or
61∑ab(a−b)2+2∑(a3b+ab3+a3c+ac3−4a2bc)≥0.
Which is true by AM-GM.
Equality holds when a=b=c=1 or a=3, b=c=0 and any permutations