INEQUALITY'S BLOG

August 1, 2010

Problem 035

Filed under: Uncategorized — KKKVVV @ 1:25 am

Let a\geq b\geq c>0; a+b+c=1. Prove that:

\displaystyle \frac{ab^2}{\sqrt{b+c}}+\frac{bc^2}{\sqrt{c+a}}+\frac{ca^2}{\sqrt{a+b}}\geq \frac{3\sqrt{3}abc}{\sqrt{2}}

 

Van Khea

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