INEQUALITY'S BLOG

July 17, 2011

Problem 239 (vankhea)

Filed under: Problem by Van Khea — KKKVVV @ 6:17 am

Prove that for a, b, c, p, q>0 we have: \displaystyle \frac{1}{a\sqrt[3]{pb+qc}}+\frac{1}{b\sqrt[3]{pc+qa}}+\frac{1}{c\sqrt[3]{pa+qb}}\geq \frac{3}{(abc)^{\frac{4}{9}}.\sqrt[3]{p+q}}

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