For $a,b,c \geqslant 0.$ Then $$9 ( a+b+c ) ^{2} ( ab+ac+bc ) ^{2}+108a^2b^2c^2-31abc ( a+b+c ) ^{3} \geqslant 0.$$
I use computer and found that the following stronger inequality holds for all reals of $a,b,c.$
$$\sum (a^2 -bc) \Big[9\, \left( a+b+c \right) ^{2} \left( ab+ac+bc \right) ^{2}+108\,{a}^{2 }{b}^{2}{c}^{2}-31\,abc \left( a+b+c \right) ^{3}\Big] \geqslant \frac{81}{4} \sum ab \prod \left( a-b \right) ^{2}$$
And also:
$$\sum (a^2 -bc) \Big[9\, \left( a+b+c \right) ^{2} \left( ab+ac+bc \right) ^{2}+108\,{a}^{2 }{b}^{2}{c}^{2}-31\,abc \left( a+b+c \right) ^{3}\Big] \geqslant {\frac {27}{4}}\, \left( a+b+c \right) ^{2} \prod \left( a-b \right) ^{2}$$
Any another inequality$?$
Best Answer
There is the following stronger version.
The equality occurs for $a=b=c$ and for $(a,b,c)=t(6+4\sqrt3,1,1)$, where $t\geq0$ and for any cyclic permutations of the last.
Even if we'll replace $4(3\sqrt3-4)$ on $4$, the BW does not help here!
By the way, this inequality we can prove by $uvw$ immediately:
it's equivalent to $f(v^2)\geq0,$ where $f$ increases.
Indeed, let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Thus, we need to prove that: $$729u^2v^4+108w^6-31\cdot27u^3w^3\geq4(3\sqrt3-4)(27u^3-27uv^2)w^3$$ or $f(v^2)\geq0,$ where $$f(v^2)=27u^2v^4+4w^6-31u^3w^3-4(3\sqrt3-4)(u^3-uv^2)w^3.$$ But $$f'(v^2)=54u^2v^2+4(3\sqrt3-4)uw^3\geq0,$$ which says that $f$ increases and it's enough to prove our inequality for the minimal value of $v^2$, which by $uvw$ happens for equality case of two variables.
Since our inequality is homogeneous and for $w^3=0$ it's obvious, it's e enough to assume $b=c=1$, which gives: $$9(a+2)^2(2a+1)^2+108a^2-31(a+2)^3a\geq4(3\sqrt3-4)(a^3-3a+2)a$$ or $$(a-1)^2(a-6-4\sqrt3)^2\geq0$$ and we are done.
It seems that this inequality is true for any reals $a$, $b$ and $c$,
but it's another problem (the previous reasoning does not help because $v^2$ can be negative).