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 AM-GM $$\frac{a_1+\cdots+a_n}{n}\geq\sqrt[n]{a_1\cdots a_n}$$ Since $a,~ b,~ c$ are positive real numbers $$\frac{a^3+b^3+c^3}{3}\geq\sqrt[3]{a^3b^3c^3}=abc$$ $$a^3+b^3+c^3\geq3abc$$ $$2(a^3+b^3+c^3)\geq6abc\tag{1}$$ Now we want to prove that $$2a^2(b + c) + 2b^2(c + a) + 2c^2(a + b) ≥ a^3 + b^3 + c^3 + 9abc\tag{2}$$ First let $$ \begin{cases} a = y + z \\ b = z + x \\ c = x + y \end{cases} $$ With $x,~y,~z\geq0$, then the left side of $(2)$ becomes $$4x^3 + 4y^3 + 4z^3 + 10x^2(y + z) + 10y^2(z + x) + 10z^2(x + y) + 24xyz$$ And the right side becomes $$2x^3 + 2y^3 + 2z^3 + 12x^2(y + z) + 12y^2(z + x) + 12z^2(x + y) + 18xyz$$ Further simplify we have $$x^3 + y^3 + z^3 + 3xyz ≥ x^2(y + z) + y^2(z +x) + z^2(x + y)$$ which is Schur's inequality, so we have proved that $(2)$ holds true.
From $(2)$ we have $$6(a^2(b + c) + b^2(c + a) + c^2(a + b)) ≥ 3(a^3 + b^3 + c^3) + 27abc\tag{3}$$ Add $(1)$ and $(3)$ $$6\left(a^2b+a^2c+b^2a+b^2c+c^2a+c^2b \right)+2(a^3+b^3+c^3)\ge 3(a^3+b^3+c^3)+33abc$$ $$ 6\left(a^2b+a^2c+b^2a+b^2c+c^2a+c^2b \right)-\left(a^3+b^3+c^3\right)-33abc\ge0 $$ and we're done.
\begin{align} 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}\label{1} \end{align}
As @DeepSea suggested, we can replace the expressions in terms of side lengths $a,b,c$ with equivalent in terms of semiperimeter $\rho=\tfrac12(a+b+c)$, inradius $r$ and circumradius $R$ of the triangle, knowing that
\begin{align} a+b+c&=2\rho \tag{2}\label{2} ,\\ a^2+b^2+c^2&=2(\rho^2-r^2-4rR) \tag{3}\label{3} ,\\ abc&=4\rho\,r\,R \tag{4}\label{4} , \end{align} so \eqref{1} is becomes
\begin{align} 7(2\rho)^3-9(2\rho)\cdot2(\rho^2-r^2-4rR)-108\cdot4\rho\,r\,R &\ge0 ,\\ 20\rho^3+36\rho\,r^2-288\rho\,r\,R &\ge0 ,\\ 5\rho^2+9 r^2-72 rR &\ge0 \tag{5}\label{5} ,\\ \end{align}
And ve can divide \eqref{5} by $R^2$ and consider new $\rho,r$ that correspond to a scaled triangle with $R=1$:
\begin{align} 5\rho^2+9 r^2-72 r &\ge0 \tag{6}\label{6} . \end{align}
Using the left part of Gerretsen's Inequality,
\begin{align} r\,(16\,R-5\,r)&\le\rho^2 , \end{align}
we can check if/when \begin{align} 5\,r\,(16-5\,r)+9 r^2-72 r &\ge0 \end{align} instead of \eqref{6}, which simplifies to \begin{align} 1-2\,r &\ge0 , \end{align}
which holds for $r\in[0,\tfrac12]$, that is, for all valid triangles.
Hence, \eqref{1}.
By your work $$7(a+b+c)^3-9(a+b+c)(a^2+b^2+c^2)-108abc=$$ $$=\sum_{cyc}(12a^2b+12a^2c-2a^3-22abc)=\sum_{cyc}(a^2b+a^2c-2a^3+11(a^2b+a^2c-2abc))=$$ $$=\sum_{cyc}a^2(c-a-(a-b))+11(b^2c+a^2c-2abc))=$$ $$=\sum_{cyc}((a-b)(b^2-a^2)+11c(a-b)^2))=$$ $$=\sum_{cyc}(a-b)^2(11c-a-b)\geq\sum_{cyc}(a-b)^2(3c-a-b).$$ 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: $$\sum_{cyc}(a-b)^2(3c-a-b)\geq(a-b)^2(3c-a-b)+(a-c)^2(3b-a-c)\geq$$ $$\geq(a-b)^2(3c-a-b)+(a-b)^2(3b-a-c)=2(a-b)^2(b+c-a)\geq0.$$
