inequality 9

Problem: let $a,b,c \in \mathbb{R^{+*}}$ , such as $a+b+c=3$ prove that:

$4(ab+bc+ca) +\frac{a^2b^2}{a+b}+\frac{a^2c^2}{a+c}+\frac{b^2c^2}{b+c} \leq \frac{27}{2}$

Solution: let put $a=3x$ and $b=3y$ and $c=3z$
then the inequality becomes;
$S=4\sum xy +3\sum \frac{x^2y^2}{x+y} \leq \frac{3}{2}$ with $x+y+z=1$ .
We have $\frac{x^2y^2}{x+y} \leq \frac{xy(x+y)}{4}$ because $(x+y)^2 \geq 4xy$
then $S\leq 4\sum xy +\frac{3}{4}\left(\sum xy(x+y)\right) = 4\sum+\frac{3}{4}\left(\sum xy -3xyz\right)$
$\Leftrightarrow S\leq \frac{19}{4}\left(\sum xy \right) - \frac{9xyz}{4}$
By Shur, $\sum xy \leq \frac{1+9xyz}{4} \Leftrightarrow \frac{19}{4}\left(\sum xy \right) \leq \frac{19+171xyz}{16}$
then $S\leq \frac{19+135xyz}{16}$ by AM-GM $xyz \leq \frac{1}{27}$ therefore: