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 
 A: Your inequality is equivalent to $12\sqrt{3}S \le f(a,b,c)$ where
$$f(a,b,c)=\frac{(a+b+c)^2(ab+bc+ac)}{a^2+b^2+c^2}.$$
One link of Hadwiger's inequality (credit to Macavity for the link) says that $12\sqrt{3}S \le g(a,b,c)$ where, defining as in the statement there $Q=(a-b)^2+(b-c)^2+(c-a)^2,$ $g$ is defined as
$$g(a,b,c)=(a+b+c)^2-2Q.$$
This latter inequality implies yours, as we will argue. It is also strictly stronger for any nonequilateral triangle.
Now since $f,g$ are each symmetric and homogeneous we may assume that $a=1,b=s,c=t$ and that $s,t\ge 1.$ (This amounts to taking $a$ as the shortest side.) The $g$ bound is better if we show $f-g \ge 0.$ We have after algebra (I used maple here) an expression whose numerator factors into two second degree terms in $s,t.$
$$f(1,s,t)-g(1,s,t)=\frac{AB}{s^2+t^2+1},$$
where $A,B$ are factors to be analyzed. It turns out that 
$$A=(s-t)^2+(s-1)(t-1), \\ B=3(s-t)^2+(2s-1)(2t-1)+2.$$
Since $s,t \ge 1$ the factor $B$ is positive in any case, while the factor $A$ is nonnegative. Note that the only way for $A$ to be zero is if $s=t$ because of the square term, and then $s=t=1$ since the term $(s-1)(t-1)=(s-1)^2$ occurs as a summand. This gives, assuming the two bounds $f,g$ agree on a given triangle, the case of an equilateral triangle.
Conclusion: Your inequality does indeed hold, but it seems more involved to state than the (known) Hadwiger inequality which is in fact stronger.
A: 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.
A: 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!
