BMO 2017

First we remark that $$a^5+b^5\ge ab(a^3+b^3).$$ Indeed $$\begin{array}{rl}{a}^5+b^5\ge ab(a^3+b^3)& \iff a^5-a^{4}b-a{b}^4+b^5\ge 0\\ & \iff (a-b)(a^4-b^4)\ge 0\\ & \iff (a-b{)}^2(a^2+b^2)(a+b)\ge 0.\end{array}$$ We rewrite the inequality as $$\frac{1}{a^5+b^5+ab{c}^3}+\frac{1}{b^5+c^5+bc{a}^3}+\frac{1}{c^5+a^5+ca{b}^3}\le 1$$ On the other hand the following inequality is true $$a^5+b^5+ab{c}^3\ge ab(a^3+b^3+c^3),$$ and similar for the other two. Finally, using AM-GM we get: $$\begin{array}{rl}& \frac{1}{a^5+b^5+c^2}+\frac{1}{b^5+c^5+a^2}+\frac{1}{c^5+a^5+b^2}\\ & \le \frac{1}{a^3+b^3+c^3}\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)=\frac{a+b+c}{a^3+b^3+c^3}\\ & \le \frac{a+b+c}{(a+b+c{)}^3}=\frac{9}{(a+b+c{)}^2}\le \frac{9}{(3\sqrt[3]{abc}{)}^2}=1.\end{array}$$