<s>[INST] 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.
 [/INST] 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:
$$3\sum_{cyc}(a^2+a+1)(b^2-b+1)(c^2-c+1)\geq$$
$$\geq3\sum_{cyc}(a^2-a+1)(b^2-b+1)(c^2-c+1)\geq a^2b^2c^2+abc+1\geq a^2b^2c^2-abc+1.$$
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