For $a$, $b$, $c$ the sides of a triangle, show $ 7(a+b+c)^3-9(a+b+c)\left(a^2+b^2+c^2\right)-108abc\ge0$ 
If $a$, $b$, and $c$ are the three sidelengths of an arbitrary triangle, prove that the following inequality is true, with equality for equilateral triangles.
$$ 7\left(a+b+c\right)^3-9\left(a+b+c\right)\left(a^2+b^2+c^2\right)-108abc\ge0 \tag{1}$$

In expanded form:
$$ 6\left(a^2b+a^2c+b^2a+b^2c+c^2a+c^2b \right)-\left(a^3+b^3+c^3\right)-33abc\ge0 \tag{2}$$
This a part of an ongoing research in triangle geometry and related to solving a cubic equation.

By your work $$7(a+b+c)^2-9(a+b+c)(a^2+b^2+c^2)-108abc=$$
$$=\sum_{cyc}(12a^2b+12a^2c-2a^3-22abc)=\sum_{cyc}(a^2b+a^2c-2a^3+11(a^2b+a^2c-2abc))=$$
$$=\sum_{cyc}a^2(c-a-(a-b))+11(b^2c+a^2c-2abc))=$$
$$=\sum_{cyc}((a-b)(b^2-a^2)+11c(a-b)^2))=$$
$$=\sum_{cyc}(a-b)^2(11c-a-b)\geq\sum_{cyc}(a-b)^2(3c-a-b).$$
Now, let $a\geq b\geq c$.
Thus, since $3a-b-c>0$, $3b-a-c\geq b+c-a>0$ and $a-c\geq a-b$, we obtain:
$$\sum_{cyc}(a-b)^2(3c-a-b)\geq(a-b)^2(3c-a-b)+(a-c)^2(3b-a-c)\geq$$
$$\geq(a-b)^2(3c-a-b)+(a-b)^2(3b-a-c)=2(a-b)^2(b+c-a)\geq0.$$
