Prove that $\frac{6(a^2 + b^2 + c^2)}{a + b + c} \geq \frac{(a + b)^2}{b + c} + \frac{(b + c)^2}{c + a} + \frac{(a + c)^2}{a + b}$ Prove that if $a,b,c$ are the lengths of the edges of a given triangle, then the following inequality holds:
$\frac{6(a^2 + b^2 + c^2)}{a + b + c} \geq \frac{(a + b)^2}{b + c} + \frac{(b + c)^2}{c + a} + \frac{(a + c)^2}{a + b}$.
I have tried so far using the Ravi substitution, but the expression got even uglier. I have also tried making the substitutions: $x = a + b$, $y = b + c$, $z = a + c$, and after using $3(a^2 + b^2 + c^2) \geq (a + b + c)^2$, this resulted in showing that: $x + y + z \geq \frac{x^2}{y} + \frac{y^2}{z} + \frac{z^2}{x}$, where we also know (from triangle inequality) that: $z \geq \frac{x+y}{3}$ and the other 2 analogues, but at this point I got stuck. 

Let $c=\max\{a,b,c\}$, $a=x+u$, $b=x+v$ and $c=x+u+v$, where $x>0$, $u\geq0$ and $v\geq0.$
Thus, we need to prove that
$$8(u^2-uv+v^2)x^3+6(3u^3-2u^2v+uv^2+3v^3)x^2+$$
$$+(11u^4+2u^3v-15u^2v^2+26uv^3+11v^4)x+2(u^5+u^4v-3u^3v^2+u^2v^2+5uv^4+v^5)\geq0,$$
which is obvious.
Also, the $uvw$'s technique helps, but it's very complicated here.
