How prove this stronger than Weitzenbock's inequality:$(ab+bc+ac)(a+b+c)^2\ge 12\sqrt{3}\cdot S\cdot(a^2+b^2+c^2)$ In $\Delta ABC$,$$AB=c,BC=a,AC=b,S_{ABC}=S$$
show that
$$(ab+bc+ac)(a+b+c)^2\ge 12\sqrt{3}\cdot S\cdot(a^2+b^2+c^2)$$
I know this  Weitzenböck's_inequality
$$a^2+b^2+c^2\ge 4\sqrt{3}S$$
But my inequality is stronger than this Weitzenbock's inequality.
my try:
let the semiperimeter, inradius, and circumradius be $s,r,R$ respectively
$$a+b+c=2s,ab+bc+ac=s^2+4Rr+r^2,S=rs$$
$$\Longleftrightarrow (s^2+4Rr+r^2)4s^2\ge 12\sqrt{3}\cdot rs[4s^2-2(s^2+4Rr+r^2)]$$
$$\Longleftrightarrow  s^3+4Rrs+r^2s\ge 6\sqrt{3}rs^2-24\sqrt{3}Rr^2-6\sqrt{3}r^3$$
and use this Gerretsen inequality:
$$r(16R-5r)\le s^2\le 4R^2+4Rr+3r^2$$
and Euler inequality
$$R\ge 2r$$
But seems is not usefull,
Thank you 

It follows immediately from known
$$2(ab+ac+bc)-a^2-b^2-c^2\geq4\sqrt3S$$
because it remains to prove that 
$$\frac{(ab+ac+bc)(a+b+c)^2}{3(a^2+b^2+c^2)}\geq\sum_{cyc}(2ab-a^2).$$ 
Let $a^2+b^2+c^2=k(ab+ac+bc)$.
Hence, $k\geq1$ and we need to prove that
$$\frac{k+2}{3k}\geq2-k$$ or
$$(k-1)(3k-2)\geq0,$$
which is obvious.
Done!
