<s>[INST] Find the smallest $\alpha$ such that, for all $x,y,z$, $\alpha\,\left(x^2-x+1\right)\left(y^2-y+1\right)\left(z^2-z+1\right)\ge(xyz)^2+|xyz|+1$. 
Find the smallest $\alpha\in\mathbb{R}$ such that, for all $x,y,z\in\mathbb{R}$, the following inequality holds $$\alpha\,\left(x^2-x+1\right)\left(y^2-y+1\right)\left(z^2-z+1\right)\ge(xyz)^2+|xyz|+1\,.$$

This question is inspired by this thread. According to this related link, it suffices to assume that $x$, $y$, and $z$ are positive.  Mathematica indicates that $\alpha=3$ is the lowest possible value, where the sole equality case is $x=y=z=1$.  I am looking for a nice solution.  
 [/INST] If $x=y=z=1$ we get $\alpha\geq3$.
But for $\alpha=3$ it's enough to prove our inequality for non-negatives $x$, $y$ and $z$.
Since $3(x^2-x+1)^3-(x^6+x^3+1)=(x-1)^4(2x^2-x+2)\geq0$, 
our inequality follows from Holder:
$$3\prod_{cyc}(x^2-x+1)\geq\sqrt[3]{\prod\limits_{cyc}(x^6+x^3+1)}\geq x^2y^2z^2+xyz+1$$
Done!
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