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?

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!

