$\frac{ab}{a^{5}+b^{5}+ab}= \frac{1}{c(a^{5}+b^{5})+a^{2}b^{2}c^{2}}\leq \frac{1}{c(a^{2}b^{2}(a+b))+a^{2}b^{2}c^{2}}= \frac{c}{a+b+c}$ similarly for other fractions, adding these up you get the inequality

What about \[a = \frac{x}{y}\] etc. and then Muirhead? (with previous basic transormations to "clear" form) From what I calculated fast in my notebook seems to work, tho it's way more brutalforce and less elegant solution than proposed by Megus. (and takes 5 more minutes)

$a^{5}+b^{5}\ge a^{2}b^{2}(a+b)$ ? Prove it

Beat wrote: $a^{5}+b^{5}\ge a^{2}b^{2}(a+b)$ ? Prove it Since $a, b > 0$ we can use AM-GM, $a^{5}+a^{5}+a^{5}+b^{5}+b^{5}\ge 5a^{3}b^{2}\iff \frac{3}{5}a^{5}+\frac{2}{5}a^{5}\ge a^{3}b^{2}$ $a^{5}+a^{5}+b^{5}+b^{5}+b^{5}\ge 5a^{2}b^{3}\iff \frac{2}{5}a^{5}+\frac{3}{5}a^{5}\ge a^{2}b^{3}$ Now add the two inequalities to get $a^{5}+b^{5}\ge a^{3}b^{2}+a^{2}b^{3}= a^{2}b^{2}(a+b)$ Would you really need to write that in an olympiad solution?

Beat wrote: $a^{5}+b^{5}\ge a^{2}b^{2}(a+b)$ ? Prove it Monotonic crap and its obvious. btw. Xevarion your AM-GM proof is boring. Freaking boring. Erm, from what i calculated in my mind multiplying everything by everything and then using abc = 1 also works.

Also and perhaps simpler $a^{5}+b^{5}\geq a^{3}b^{2}+a^{2}b^{3}$ by rearrangement on $a^{3},b^{3}: a^{2},b^{2}$.

Use the identity $a^{5}+b^{5}=(a+b)(a^{4}-a^{3}b+a^{2}b^{2}-ab^{3}+b^{4})$ and upon simplifying you need to prove $a^{4}+b^{4}\geq a^{3}b+b^{4}a$,which,by putting all the terms in the left side,is equivalent to: \[(a^{3}-b^{3})(a-b)\geq0\] .I think it is clear how to finish this last one..

about $a^{5}+b^{5}\geq a^{2}b^{2}(a+b)$ we can say that it is true according to Muirhead: $(5,0)\succ(3,2)$

Megus wrote: $\frac{ab}{a^{5}+b^{5}+ab}= \frac{1}{c(a^{5}+b^{5})+a^{2}b^{2}c^{2}}\leq \frac{1}{c(a^{2}b^{2}(a+b))+a^{2}b^{2}c^{2}}= \frac{c}{a+b+c}$ similarly for other fractions, adding these up you get the inequality nice puzzle proof!