INEQUALITY'S BLOG

July 27, 2010

(van Khea):Problem 014

Filed under: the new math — KKKVVV @ 4:59 pm

Let a, b, c>0 ; a^2+b^2+c^2=1. Prove that:

\displaystyle ab+bc+ca+\frac{k}{abc}\geq 3k\sqrt{3}+1\displaystyle \forall{k\geq \frac{1}{3\sqrt{3}}}

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