INEQUALITY'S BLOG

August 1, 2010

Problem 039

Filed under: Uncategorized — KKKVVV @ 6:52 am

Let f:R\longrightarrow R_{0}^{+} and f''>0. For a, b, c be positive real numbers and satisfying a\leq b\leq c; ab+bc+ca=1 ; Prove that:

a^2f(b)+b^2f(c)+c^2f(a)\geq f(3abc)

Proof

From the problem 038 we have

a^2f(b)+b^2f(c)+c^2f(a)\geq abf(c)+bcf(a)+caf(b)

From Jensen’s inequality for ab+bc+ca=1 we have

abf(c)+bcf(a)+caf(b)\geq f(abc+bca+cab)=f(3abc)

\Longrightarrowa^2f(b)+b^2f(c)+c^2f(a)\geq f(3abc)

Therefore the proof is completed.

 

Van Khea   

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