a+b/2ab >=2a²b²/( a^5+b^5) using MA>=MG and just do the same for the others

$$(a+b)(a^5+b^5)\geq 4a^3b^3\iff \frac{a+b}{2ab}\geq \frac{2a^2 b^2}{a^5+b^5}$$

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.