Team Selection Test for IMO 2010

We have $\sqrt[4]{\frac{(a^2+b^2)(a^2-ab+b^2)}{2}}\le \frac{a^2+b^2}{a+b}$ for all nonnegative real numbers $a$, $b$ as this inequality is equivalent to $(a+b{)}^4(a^2-ab+b^2)\le 2(a^2+b^2{)}^3$ which is in turn equivalent to $(a-b{)}^4(a^2+ab+b^2)\ge 0$.

$$\frac{a^2+b^2}{a+b}+\frac{b^2+c^2}{b+c}+\frac{c^2+a^2}{c+a}\le \frac{2}{3}(a^2+b^2+c^2)\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)$$ Without loss of generality we may assume that $a\ge b\ge c$. Then we also have $a^2+b^2\ge a^2+c^2\ge b^2+c^2$ and $\frac{1}{a+b}\le \frac{1}{a+c}\le \frac{1}{b+c}$. The result follows by the Rearrangement Inequality.