<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] 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.
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