INEQUALITY'S BLOG

November 22, 2010

Problem 66: van khea

Filed under: Uncategorized — KKKVVV @ 1:07 am

Let a, b, c>0 ; abc=8. Prove that:
\displaystyle \frac{\sqrt[3]{a+2b}}{c^3\sqrt{2a+b}}+\frac{\sqrt[3]{b+2c}}{a^3\sqrt{2b+c}}+\frac{\sqrt[3]{c+2a}}{b^3\sqrt{2c+a}}\geq \frac{3}{8\sqrt[6]{6}}

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