<s>[INST] Find the smallest number $\alpha$, such that for all $x,y,z$ $\alpha(x^2-x+1)(y^2-y+1)(z^2-z+1)\ge(xyz)^2-xyz+1$ 
Find a smallest number $\alpha$, such that for all $x,y,z$ (not all of which are positive) inequality
$$\alpha(x^2-x+1)(y^2-y+1)(z^2-z+1)\ge(xyz)^2-xyz+1$$

My work so far:
Let $f(t)=t^2-t+1$. Then $f(t) \ge \frac34$.
If $x=0, y=z=\frac12$, then $$\alpha\ge \frac{16}9$$
 [/INST] From your
$a(x^2-x+1)^3\ge x^6-x^3+1
$,
since
$(x+1)(x^2-x+1)
=x^3+1
$
and
$(x^3+1)(x^6-x^3+1)
=x^8+1
$,
$a(\frac{x^3+1}{x+1})^3\ge \frac{x^9+1}{x^3+1}
$
or
$a\ge \frac{(x^9+1)(x+1)^3}{(x^3+1)^4}
$.
According to Wolfy,
this has a maximum of
$2.1547$
at $x \approx 0.435421
$
and
$2.29663
$.
These are the roots of
$0
= x^6-2 x^5-x^4+x^2+2 x-1
$.
The exact roots are
$-1, 1,
\frac12 (1\pm\sqrt{2} 3^{1/4}+\sqrt{3}),
\frac12 (1\pm i\sqrt{2} 3^{1/4}-\sqrt{3})
$.
So $a$ must be at least
$2.1547$.
</s>