<s>[INST] 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?
 [/INST] 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!
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