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 $$\alpha\ge \frac{16}9$$

In the starting formulation the answer is $\frac{16}{9}$.
If $x$, $y$ and $z$ are non-positives so after replacing $x\rightarrow-x$...
we need to prove that 
$$\frac{16}{9}(x^2+x+1)(y^2+y+1)(z^2+z+1)\geq x^2y^2z^2-xyz+1$$
which is true for all non-negatives $x$, $y$ and $z$ because we'll prove now that even
$$\frac{16}{9}(x^2+x+1)(y^2-y+1)(z^2-z+1)\geq x^2y^2z^2+xyz+1$$.
If $x\leq0$, $y\leq0$ and $z\geq0$ we need to prove that
$$\frac{16}{9}(x^2+x+1)(y^2+y+1)(z^2-z+1)\geq x^2y^2z^2-xyz+1$$ 
for non-negatives $x$, $y$ and $z$, which follows from
$$\frac{16}{9}(x^2+x+1)(y^2-y+1)(z^2-z+1)\geq x^2y^2z^2+xyz+1$$
again.
Now we'll prove it:
$$\begin{align}
&16(x^2+x+1)(y^2-y+1)(z^2-z+1)-9(x^2y^2z^2+xyz+1)
\\&\phantom{aaa}=(16(y^2-y+1)(z^2-z+1)-9y^2z^2)x^2+(16(y^2-y+1)(z^2-z+1)-9yz)x
\\
&\phantom{aaaaa}+16(y^2-y+1)(z^2-z+1)-9
\\&\phantom{aaa}
\geq\left(16\left(\frac{3}{4}y^2+\left(\frac{y}{2}-1\right)^2\right)\left(\frac{3}{4}z^2+\left(\frac{z}{2}-1\right)^2\right)-9y^2z^2\right)x^2+
+(3y\cdot3z-9yz)x
\\&\phantom{aaaaa}+
16\left(\frac{3}{4}+\left(\frac{1}{2}-y\right)^2\right)\left(\frac{3}{4}+\left(\frac{1}{2}-z\right)^2\right)-9
\\&\phantom{aaa}\geq0
\end{align}$$
