INEQUALITY'S BLOG

July 27, 2010

(Van Khea):Problem 016

Filed under: the new math — KKKVVV @ 5:19 pm

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

\displaystyle \frac{a^{3}}{bc}+\frac{b^{3}}{ca}+\frac{c^{3}}{ab}+k.\frac{a^{5}+b^{5}+c^{5}}{abc(a+b+c)}\geq k+3 ; \forall{k\geq 3}

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