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.
 A: 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.
A: \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}.
A: 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:
$$\sum_{cyc}(a-b)^2(3c-a-b)\geq(a-b)^2(3c-a-b)+(a-c)^2(3b-a-c)\geq$$
$$\geq(a-b)^2(3c-a-b)+(a-b)^2(3b-a-c)=2(a-b)^2(b+c-a)\geq0.$$
