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 

Square both sides, put$S^2=\dfrac{(a+b+c)(a+b-c)(a+c-b)(b+c-a)}{16}$ in, we have:
$(ab+bc+ac)^2(a+b+c)^4\ge 27\cdot (a+b+c)(a+b-c)(a+c-b)(b+c-a)(a^2+b^2+c^2)^2$
with brutal force method(BW method), WOLG let $a=$Min{$a,b,c$},$b=a+u,c=a+v,u\ge0,v\ge0$ , put in the inequality and rearrange them, we have:
$ \iff k_6a^6+k_5a^5+k_4a^4+k_3a^3+k_2a^2+k_1a+k_1a+k_0 \ge0 \iff $
$k_6=+486v^2-486uv+486u^2 \\k_5=1404v^3-648uv^2-648u^2v+1404u^3\\k_4=1935v^4-360uv^3-945u^2v^2-360u^3v+1935u^4\\k_3=1572v^5-60uv^4-588u^2v^3-588u^3v^2-60u^4v+1572u^5\\k_2=760v^6+78uv^5-210u^2v^4-352u^3v^3-210u^4v^2+78u^5v+760u^6\\k_1=216v^7+4uv^6+32u^2v^5-140u^3v^4-140u^4v^3+32u^5v^2+4u^6v+216u^7\\k_0=27v^8+u^2v^6+4u^3v^5-48u^4v^4+4u^5v^3+u^6v^2+27u^8$
with  $u^n+v^n\ge u^{n-1}v+uv^{n-1} \ge u^{n-2}v^2+u^2v^{n-2} \ge ...$ , all $k_i \ge 0$ above , and it is trivial that when and only when $u=v=0 \implies k_i=0$
QED.
