Let $a$, $b$ and $c$ be positive numbers such that $a^2+b^2+c^2+d^2=4$. Prove that: $$\frac{a^3}{b^2+c^2}+\frac{b^3}{c^2+d^2}+\frac{c^3}{d^2+a^2}+\frac{d^3}{a^2+b^2}\geq2$$ I tried C-S, but without success.
If $a^2+b^2+c^2+d^2=4$ so$\sum\limits_{cyc}\frac{a^3}{b^2+c^2}\geq2$
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inequality
contest-math
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0i think it is $d^2+a^2$. I tried to edit post but couldnt – 2017-02-02
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0@Markoff Chainz I fixed my post. Thank you! – 2017-02-02