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?
 A: Sorry, I don't know latex. But, my solution is the following:


*

*Get rid of denominators.

*Multiply by 2.

*Expand de products. You'll get cyclic sums.

*Left side will be something like: cyc(7, 7, 0) + cyc(7, 5, 2) + cyc(5, 5, 4)

*Homogenize right side by multiplicating by cubic root of xyz to the 8th power. That way the sums of exponents of both sides of this inequality will be 14. There will be 24 terms on the left side and 24 terms on the right side.

*Finally, you have to prove:
2( cyc(7, 7, 0) + cyc(7, 5, 2) + cyc(5, 5, 4) ) ≥ 3 ( cyc(20/3, 14/3, 8/3) + 2*cyc(14/3, 14/3, 14/3)


*

*But, that happens to be Muirhead's Theorem. Done!

A: 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. 
A: Look at this pic. or solution. 
Hope you got a simple standard approach...
A: By Holder $$\sum_{cyc}\frac{x^5y^5}{x^2+y^2}\geq\frac{(xy+xz+yz)^5}{(1+1+1)^3\sum\limits_{cyc}(x^2+y^2)}=\frac{(xy+xz+yz)^5}{54(x^2+y^2+z^2)}.$$
Hence, it remains to prove that $$(xy+xz+yz)^5\geq81(x^2+y^2+z^2).$$
Let $x+y+z=3u$, $xy+xz+yz=3v^2$ and $xyz=w^3$.
Thus, since $xy+xz+yz$ and $x^2+y^2+z^2$ does not depend on $w^3$, 
it's enough to prove the last inequality for a maximal value of $w^3$, 
which happens for equality case of two variables.
Let $y=x$. Hence, $z=\frac{1}{x^2}$ and we need to prove that
$$(x^3+2)^5\geq81x(2x^6+1).$$
Let $x^3=t$.
Hence, we need to prove that $f(t)\geq0$, where
$$f(t)=5\ln(t+2)-\ln(2t^2+1)-\frac{1}{3}\ln{t}-4\ln3$$
But $$f'(t)=\frac{5}{t+2}-\frac{4t}{2t^2+1}-\frac{1}{3t}=\frac{(t-1)(2t-1)(4t-1)}{3t(t+2)(2t^2+1)}$$
and since $f\left(\frac{1}{4}\right)>0$, we are done!
