<s>[INST] Prove that inequality $\sum _{cyc}\frac{a^2+b^2}{a+b}\le \frac{3\left(a^2+b^2+c^2\right)}{a+b+c}$ 
Let $a>0$,$b>0$ and $c>0$. Prove that:
   $$\dfrac{a^2+b^2}{a+b}+\dfrac{b^2+c^2}{b+c}+\dfrac{c^2+a^2}{c+a}\le \dfrac{3\left(a^2+b^2+c^2\right)}{a+b+c}.$$


$$\Leftrightarrow \frac{(a^2+b^2)(a+b+c)}{a+b}+\frac{(b^2+c^2)(a+b+c)}{b+c}+\frac{(c^2+a^2)(a+b+c)}{c+a}\leq 3(a^2+b^2+c^2)$$
$\Leftrightarrow 2(a^2+b^2+c^2)+\frac{c(a^2+b^2)}{a+b}+\frac{a(b^2+c^2)}{b+c}+\frac{b(a^2+c^2)}{a+c}\leq 3(a^2+b^2+c^2)$
$\Leftrightarrow \frac{c(a^2+b^2)}{a+b}+\frac{a(b^2+c^2)}{b+c}+\frac{b(a^2+c^2)}{a+c}\leq a^2+b^2+c^2$
$\Leftrightarrow \frac{c(a+b)^2-2abc}{a+b}+\frac{a(b+c)^2-2abc}{b+c}+\frac{b(a+c)^2-2abc}{a+c}\leq a^2+b^2+c^2$
$\Leftrightarrow 2(ab+bc+ac)\leq a^2+b^2+c^2+2abc\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{a+c}\right)$
I can't continue. Help me 
 [/INST] The following quantity is clearly positive
\begin{eqnarray*}
\sum_{perms} a^2 b (a-b)^2 \geq 0
\end{eqnarray*}
This can be rearranged to
\begin{eqnarray*}
3(a^2+b^2+c^2)(a+b)(b+c)(c+a) \geq (a+b+c) \left( \sum_{cyc} (a^2 +b^2)(c+a)(c+b) \right)
\end{eqnarray*}
Now divide this by $(a+b+c)(a+b)(b+c)(c+a)$ and we have 
\begin{eqnarray*}
\frac{3(a^2+b^2+c^2)}{(a+b+c)}  \geq \frac{a^2 +b^2}{a+b}+ \frac{b^2 +c^2}{b+c}+ \frac{c^2 +a^2}{c+a}. 
\end{eqnarray*}
</s>