INEQUALITY'S BLOG

July 26, 2010

(Van Khea):Problem 013

Filed under: the new math — KKKVVV @ 8:47 pm

Let a, b, c, x, y, z>0; x^2+y^2+z^2=1. Prove that for \displaystyle k\geq \frac{a+b+c}{9} we have:

\displaystyle ax+by+cz+\frac{k}{xyz}\geq (3k+\frac{a+b+c}{3})\sqrt{3}

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