INEQUALITY'S BLOG

November 5, 2010

Inequality: Problem 55

Filed under: the new math — KKKVVV @ 4:57 pm

(Van Khea): Let a, b, c>0 ; abc=1. Prove that:

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

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