OMCC 2019 Problem 5

Let $a,\ b$ and $c$ be positive real numbers so that $a+b+c=1$ . Show that $a\sqrt{a^2+6bc}+b\sqrt{b^2+6ac}+c\sqrt{c^2+6ab}\leq\frac{3\sqrt{2}}{4}$

Dataset text AI-MO AoPS. Quoted with source.

Solutions

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