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$a,b,c> 0$ prove $$\frac{a^3}{b+c}+\frac{b^3}{c+a}+\frac{c^3}{a+b}\ge\frac{a^2+b^2+c^2}{2}$$

I tried AM-GM but can't reached to solution please help me

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    $\begingroup$ Have you tried subtracting the right hand side from the left hand side and showing it's non-negative ? $\endgroup$ – user399481 Dec 15 '16 at 16:44
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Using the Angel form of the Cauchy-Schwarz inequality: $\displaystyle \sum_{cyclic}\dfrac{a^3}{b+c} = \displaystyle \sum_{cyclic}\dfrac{(a^2)^2}{ab+ac}\ge \dfrac{(a^2+b^2+c^2)^2}{2(ab+bc+ca)}\ge \dfrac{a^2+b^2+c^2}{2}$ since $a^2+b^2+c^2 \ge ab+bc+ca$ which is clear.

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Hint If you want to use AM-GM, $$\frac{a^3}{b+c}+\frac{a(b+c)}4\ge a^2$$

Add three such inequalities to sum and then use the rearrangement $ab+bc+ca\le a^2+b^2+c^2$.

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