INEQUALITY'S BLOG

November 5, 2010

Inequality: Problem 56

Filed under: the new math — KKKVVV @ 5:05 pm

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

\displaystyle \frac{a}{b}\sqrt{b+c}+\frac{b}{c}\sqrt{c+a}+\frac{c}{a}\sqrt{a+b}\leq \frac{\sqrt{2}}{\sqrt{3}}\biggl(\frac{b}{a}+\frac{c}{b}+\frac{a}{c}\biggl)^2

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