<s>[INST] 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 
 [/INST] 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.
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