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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.
Your first step leads to a wrong inequality because it does not save the case of the equality occurring: $a=b=c=1.$
After your first step it's enough to prove that: $$2(1-ab+a^2b^2)(1-c+c^2)\geq1+abc+a^2b^2c^2,$$ which is wrong for $a=b=c=1.$
The Vasc's solution.
Since $$2(a^2-a+1)(b^2-b+1)\geq a^2b^2+1,$$ it's enough to prove that: $$3(a^2b^2+1)(c^2-c+1)\geq2(a^2b^2c^2+abc+1),$$ which is a quadratic inequality of $c$.
Can you end it now?
Another way.
It's enough to prove our inequality for non-negatives $a$, $b$ and $c$.
Now, since $$3(a^2-a+1)^3-a^6-a^3-1=(a-1)^4(2a^2-a+2)\geq0,$$ by Holder we obtain: $$\prod_{cyc}(a^2-a+1)\geq\prod_{cyc}\sqrt[3]{\frac{a^6+a^3+1}{3}}\geq\frac{1}{3}(a^2b^2c^2+abc+1).$$
Now, let $a\leq0$, $b\geq0$ and $c\geq0.$
Thus, after replacing $a$ on $-a$ we need to prove that: $$3\sum_{cyc}(a^2+a+1)(b^2-b+1)(c^2-c+1)\geq a^2b^2c^2-abc+1,$$ which follows from the previous inequality: $$3\sum_{cyc}(a^2+a+1)(b^2-b+1)(c^2-c+1)\geq$$ $$\geq3\sum_{cyc}(a^2-a+1)(b^2-b+1)(c^2-c+1)\geq a^2b^2c^2+abc+1\geq a^2b^2c^2-abc+1.$$
Two SOS solutions with the help of computer
According to Vasc's solution in @Michael Rozenberg's answer, we have a simple SOS expression: \begin{align} &3(a^2-a+1)(b^2-b+1)(c^2-c+1) - (1 + abc + a^2b^2c^2)\\ =\ & \frac{1}{8}(abc-3c+2)^2 + \frac{3}{8}(abc-2ab+c)^2 + \frac{3}{8}(a-1)^2(b-1)^2(2c-1)^2\\ &\quad + \frac{9}{8}(a-1)^2(b-1)^2 + \frac{3}{8}(a-b)^2(2c-1)^2 + \frac{9}{8}(a-b)^2. \end{align}
Without using Vasc's solution, I can obtain a complicated SOS expression $$ 3(a^2-a+1)(b^2-b+1)(c^2-c+1) - (1 + abc + a^2b^2c^2) = \frac{1}{2}z^\mathsf{T}Qz$$ where $z = [1, a, b, c, ab, ca, bc, abc]^\mathsf{T}$ and $$Q = \left(\begin{array}{rrrrrrrr} 4 & -3 & -3 & -3 & 2 & 2 & 2 & -1\\ -3 & 6 & 1 & 1 & -3 & -3 & -1 & 2\\ -3 & 1 & 6 & 1 & -3 & -1 & -3 & 2\\ -3 & 1 & 1 & 6 & -1 & -3 & -3 & 2\\ 2 & -3 & -3 & -1 & 6 & 1 & 1 & -3\\ 2 & -3 & -1 & -3 & 1 & 6 & 1 & -3\\ 2 & -1 & -3 & -3 & 1 & 1 & 6 & -3\\ -1 & 2 & 2 & 2 & -3 & -3 & -3 & 4 \end{array}\right).$$ Remarks: $Q$ is positive semidefinite.
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.
