Prove that: $a^{12}+b^{12}+c^{12} \geq \frac{2049}{8}\cdot a^4b^4c^4 $ Let $a,b,c$ be real and $a+b+c=0.$
Prove that:$$a^{12}+b^{12}+c^{12} \geq \frac{2049}{8}\cdot a^4b^4c^4$$
I know we can make the following:
Let $bc\geq0$ thus we need to prove
$$b^{12}+c^{12}+(b+c)^{12}\geq \frac{2049}{8} b^4c^4(b+c)^4$$or
$$\frac{(2b+c)^2(b+2c)^2(b-c)^2(4b^6+12b^5c+99b^4c^2+178b^3c^3+99b^2c^4+12bc^5+4c^6)}{8}\geq 0\text{,}$$which is true.
But I'm trying to find another solution. Can be?

Alternative proof:
If $abc = 0$, the desired inequality is clearly true.
If $abc \ne 0$, WLOG, assume that $abc = -1$.
WLOG, assume further that $a, b > 0$ and $c < 0$.
We need to prove that $a^{12} + b^{12} + c^{12} \ge \frac{2049}{8}$.
Let $q = ab + bc + ca$. It is easy to obtain
$a^{12} + b^{12} + c^{12} = 2q^6-24q^3+3$. The details are given at the end.
Thus, it suffices to prove that
$$2q^6-24q^3+3 \ge \frac{2049}{8}$$
or (since $q < 0$)
$$q^3 \le - 27/4.$$
We have $a^3 + qa + 1 = a^3 + (ab + bc + ca)a + 1 = a^2(a + b + c) + abc + 1 = 0$.
Thus, we have
$q = - a^2 - \frac{1}{a} = - (a^2 + \frac{1}{2a} + \frac{1}{2a})
\le -3 \sqrt[3]{a^2 \cdot \frac{1}{2a} \cdot \frac{1}{2a}} = - 3\sqrt[3]{1/4}$.
Thus, we have $q^3 \le - 27/4$.
We are done.

The details of $a^{12} + b^{12} + c^{12} = 2q^6-24q^3+3$:
From $a + b + c = 0$ and $abc = -1$, it is easy to verify that
\begin{align*}
 a^3 + qa + 1 &= 0, \\
 b^3 + qb + 1 &= 0, \\
 c^3 + qc + 1 &= 0.
\end{align*}
We have
\begin{align*}
 &a^2 + b^2 + c^2 = (a + b + c)^2 - 2q = -2q, \\
 &a^3 + b^3 + c^3 = -q(a + b + c)  - 3 = - 3, \\
 &a^4 + b^4 + c^4 = -q(a^2 + b^2 + c^2) - (a + b + c)  = 2q^2.
\end{align*}
Then, we have
\begin{align*}
 &a^{12} + b^{12} + c^{12}\\
  =\,& (-qa - 1)^4 + (-qb - 1)^4 + (-qc - 1)^4\\
 =\, & q^4(a^4 + b^4 + c^4) + 4q^3(a^3 + b^3 + c^3) + 6q^2(a^2 + b^2 + c^2) + 4q(a + b + c) + 3\\
 =\, & 2q^6-24q^3+3.
\end{align*}
