Let $a$, $b$ and $c$ be positive real numbers, prove that \[\frac{2a^2 b^2}{a^5+b^5}+\frac{2b^2 c^2}{b^5+c^5}+\frac{2c^2 a^2}{c^5+a^5}\le\frac{a+b}{2ab}+\frac{b+c}{2bc}+\frac{c+a}{2ca}\]
Let $a$, $b$ and $c$ be positive real numbers. Prove that$$\frac{2a^2b^2}{b^5+c^5}+\frac{2b^2c^2}{c^5+a^5}+\frac{2c^2a^2}{a^5+b^5} \le\frac{a+b}{2bc}+\frac{b+c}{2ca}+\frac{c+a}{2ab}$$
Let's prove first that $\frac{2a^{2}b^{2}}{a^{5} + b^{5}} \le \frac{a+b}{2ab}$.
Assume that $a \ge b$
$\frac{2a^{2}b^{2}}{a^{5} + b^{5}} \le \frac{a+b}{2ab}$. Implies that:
$4a^{3}b^{3} \le a^{6} + b^{5}a + a^{5}b + b^{6} $ $\implies$ $0 \le (a^{3} - b^{3})^{2} + ab(a^{2} - b^{2})^{2} $
Now we can do the same for the other two inequalities and the result will follow.