Prove that in every triangle the inequality $a^3r_a + b^3r_b + c^3r_c \ge 8S(2R-r)^2 $ takes place Prove that in every triangle the inequality $$a^3r_a + b^3r_b + c^3r_c \ge 8S(2R-r)^2 $$ takes place, with the usual notations ($a,b,c$ lengths of sides, $r_a, r_b, r_c$ radii of corresponding excircles, $R$ radius of circumscribed circle and $r$ radius of inscribed circle)
I've tried to solve it using Chebyshev (Rearrangement Inequality) or Jensen Inequality (for $f(x)=\frac{x^3}{p-x}$), however I always end up minoring the inequality too much, or obtaining an inequality which I am not able to prove.

Let $a=y+z$, $b=x+z$ and $c=x+y$.
Thus, $x>0$, $y>0$ and $z>0$ and in the standard notation we need to prove that
$$\sum_{cyc}\frac{a^3\cdot2S}{b+c-a}\geq8S\left(\frac{abc}{2S}-\frac{2S}{a+b+c}\right)^2$$ or
$$\sum_{cyc}\frac{a^3}{b+c-a}\geq\frac{(4abc(a+b+c)-16S^2)^2}{16S^2(a+b+c)^2}$$ or
$$\sum_{cyc}\frac{a^3}{b+c-a}\geq\frac{\left(4abc-\prod\limits_{cyc}a+b-c)\right))^2}{(a+b+c)\prod\limits_{cyc}(a+b-c)}$$ or
$$(x+y+z)\sum_{cyc}(x+y)^3xy\geq2\left(\sum_{cyc}(x^2y+x^2z)\right)^2$$ or
$$\sum_{cyc}(x+y)\sum_{cyc}(x+y)^3xy\geq4\left(\sum_{cyc}(x^2y+x^2z)\right)^2.$$
Now, by C-S
$$\sum_{cyc}(x+y)\sum_{cyc}(x+y)^3xy\geq\left(\sum_{cyc}(x+y)^2\sqrt{xy}\right)^2.$$
Id est, it's enough to prove that
$$\sum_{cyc}(x+y)^2\sqrt{xy}\geq2\sum_{cyc}(x^2y+x^2z)$$ or 
$$\sum_{cyc}(x+y)^2\sqrt{xy}\geq2\sum_{cyc}(x^2y+xy^2)$$ or
$$\sum_{cyc}(x+y)\sqrt{xy}(\sqrt{x}-\sqrt{y})^2\geq0.$$
Done!
