, 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.