<s>[INST] 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.
 [/INST] \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}.
</s>