INEQUALITY'S BLOG

November 5, 2010

Inequality: problem 51

Filed under: the new math — KKKVVV @ 12:53 pm

(Van Khea): Let a, b, c\geq 0. Prove that:

\displaystyle \sqrt{ab+bc+ca}.\sqrt[3]{a^2+b^2+c^2}.\sqrt[6]{a+b+c}\geq \frac{\sqrt{2}}{2}\biggl(a^{\frac{4}{3}}\sqrt{b+c}+b^{\frac{4}{3}}\sqrt{c+a}+c^{\frac{4}{3}}\sqrt{a+b}\biggl)

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