Olympiad Inequality Problem Consider three positive reals $x,y,z$ such that $xyz=1$.
How would one go about proving: 
$$\frac{x^5y^5}{x^2+y^2}+\frac{y^5z^5}{y^2+z^2}+\frac{x^5z^5}{x^2+z^2}\ge \frac{3}{2}$$ 
I really dont know even where to begin! It looks a BIT like Nesbitts? Maybe?

Edit. I posted a new "proof" (now deleted), before I realized I was adddressing the wrong question. I think the original proof is ok, I messed up something in trying to simplify the approach. 
Hint.
Since this is an olympiad problem, it is likely that there is a proof without using calculus. I am sketching one here, but I have a feeling it can improved vastly.
Step 1. Make the substitution $x = 1/a$, $y = 1/b$ and $z = 1/c$, so that $abc = 1$. We are now left with the expression
$$
\sum \frac{1}{a^3b^3 (a^2+b^2)} = \sum \frac{c^3}{a^2+b^2} = \frac{a^3}{b^2+c^2} + \frac{b^3}{c^2+a^2} + \frac{c^3}{a^2+b^2}.
$$
Step 2. Assume the order $a \leq b \leq c$ without loss of generality. Then show that 
$$
\frac{a^2}{b^2+c^2} \leq  \frac{b^2}{c^2+a^2} \leq \frac{c^2}{a^2+b^2}.
$$
Now, by the "Chebyshev sum inequality", we have:
$$
\frac{a^3}{b^2+c^2} + \frac{b^3}{c^2+a^2} + \frac{c^3}{a^2+b^2}
\geq \frac{1}{3} (a+b+c) \cdot \left( \frac{a^2}{b^2+c^2} + \frac{b^2}{c^2+a^2} + \frac{c^2}{a^2+b^2} \right).
$$
Step 3. For any $u,v,w > 0$, prove that
$$
\frac{u}{v+w} + \frac{v}{w+u} + \frac{w}{u+v} \geq \frac{3}{2}.
$$
Step 4. Conclude the inequality by plugging in the third inequality in the second. 
