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Let $a,b,c>0$ with $a^2+b^2+c^2=3$. Prove that $$\frac{a^5}{b^3+c}+\frac{b^5}{c^3+a}+\frac{c^5}{a^3+c} \geq \frac{3a^2b^2c^2}{2}.$$ I have tried Cauchy-Schwarz inequality in Engel/Titu Andreescu form (see this link), but in any attempt I cannot appropriately use the condition of the problem. Then I tried to homogenize the inequality by using the condition $a^2+b^2+c^2=3$ planning to then finalize with, for example, Muirhead's inequality - also unsuccessfully.

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    $\begingroup$ what have u done?? $\endgroup$ – MAN-MADE Jun 24 '17 at 11:36
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By C-S, Vasc's inequality and AM-GM we obtain: $$\sum_{cyc}\frac{a^5}{b^3+c}\geq\frac{(a^3+b^3+c^3)^2}{\sum\limits_{cyc}(b^3a+ac)}\geq\frac{(a^3+b^3+c^3)^2}{3+3}\geq\frac{9a^2b^2c^2}{6}=\frac{3a^2b^2c^2}{2}.$$ The Vasc's inequality is the following: $$(a^2+b^2+c^2)^2\geq3(b^3a+c^3b+a^3c).$$ Also I used $a^2+b^2+c^2\geq ab+ac+bc.$

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