INEQUALITY'S BLOG

November 22, 2010

Problem 68: van khea

Filed under: Uncategorized — KKKVVV @ 1:10 am

Let a, b, c>0; abc=8. Prove that:
\displaystyle \frac{a^3\sqrt{3a+4b+3c}}{4a+3b+4c}+\frac{b^3\sqrt{3a+3b+4c}}{4a+4b+3c}+\frac{c^3\sqrt{4a+3b+3c}}{3a+4b+4c}\geq \frac{24\sqrt{5}}{11}

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