INEQUALITY'S BLOG

July 26, 2010

(Van Khea):Problem 012

Filed under: the new math — KKKVVV @ 7:28 pm

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

\displaystyle a+b+c+\frac{k}{abc}\geq (3k+1)\sqrt{3} ; \forall{k\geq \frac{1}{3}}

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