Proving $3(1−a+a^2)(1−b+b^2)(1−c+c^2)≥1+abc+a^2b^2c^2$ My task was to prove the question above over real variables.
I thought that this minor inequality should help-
$$ 3(1 − a + a^2)(1 − b + b^2) ≥ 2(1 − ab + a^2 b^2). $$
which is true.
By this inequality, the original inequality is converted to-
$$ (1 - ab)^2 (1 - c)^2 + (ab-c)^2 + abc \geq 0 $$
This proves the inequality for $abc\geq 0$.
I want to prove this Inequality for $abc\lt0$. But I couldn't find a solution for $abc\lt0$.
Any extensions for $abc\lt0$ are thankfully accepted.

Here is a straightforward solution based on quadratic inequalities.
For simplicity, denote $A=1-a+a^2$ and $B=1-b+b^2$. We need to show that
$$
3AB(1-c+c^2) \geq 1+abc +a^2b^2c^2.
$$
This is equivalent to showing that
$$
(3AB-a^2b^2)c^2 - (3AB+ab)c + 3(AB-1) \geq 0.
$$
If we regard the left-hand side above as a quadratic function of $c$,
$$
f_{A,B,a,b}(c)= (3AB-a^2b^2)c^2 - (3AB+ab)c + 3(AB-1),
$$
it suffices to show $f_{A,B,a,b}(c)\geq 0$ for any real $a,b,c$. From the fact
$A=1-a+a^2 \geq \frac 3 4 a^2$ and $B\geq \frac 3 4 b^2$, we know the leading coefficient of $f_{A,B,a,b}$ is strictly positive, i.e., $3AB - a^2b^2 >0$. Now it remains to show the discriminant of $f_{A,B,a,b}$ is non-positive. Namely,
$$
(3AB +ab)^2 -4(3AB-a^2b^2)(3AB-1) \leq 0,
$$
or equivalently,
$$
4AB + 2ABab + 4ABa^2b^2 \leq a^2b^2 +9A^2B^2.
$$
By AM-GM inequality, we have
$$
2ABab \leq a^2b^2 + A^2B^2.
$$
Therefore, it suffices to show
$$
4AB + 4ABa^2b^2 \leq 8A^2B^2,
$$
or equivalently,
\begin{equation}\begin{split}
1+a^2b^2 \leq& 2AB \\
=& 2(1-a+a^2)(1-b+b^2), \\
\end{split}\end{equation}
which is also used in Vasc's solution provided by Michael Rozenberg.
