TITLE: Prove that $\frac{ab}{a^5+b^5+ab}+\frac{bc}{b^5+c^5+bc}+\frac{ca}{c^5+a^5+ca} \leq 1.$
QUESTION [6 upvotes]: The following problem was on the IMO 1996 shortlist :

Let $a,b,c$ be positive real numbers such that $abc = 1$. Prove that
$$\dfrac{ab}{a^5+b^5+ab}+\dfrac{bc}{b^5+c^5+bc}+\dfrac{ca}{c^5+a^5+ca} \leq 1.$$

I tried factoring out things but that didn't seem to work. I don't see how to factor the denominator so I get stuck.

REPLY [6 votes]: since
$$a^5+b^5\ge a^2b^3+a^3b^2$$
so
$$\sum\dfrac{ab}{a^5+b^5+ab}\le\sum\dfrac{1}{ab^2+a^2b+1}=\sum\dfrac{abc}{ab^2+a^2b+abc}=\sum\dfrac{c}{a+b+c}=1$$