18 Real numbers $x, y, z$ not all positive satisfy $k\left(x^{2}-x+1\right)\left(y^{2}-y+\right.$ 1) $\left(z^{2}-z+1\right) \geqslant(x y z)^{2}-x y z+1$, find the minimum value of the real number $k$.
Let $x=0, y=z=\frac{1}{2}$, we get $k \geqslant \frac{16}{9}$. Now we prove that $k=\frac{16}{9}$ satisfies the condition, i.e., we need to prove:
$$
\begin{aligned}
& \frac{16}{9}\left(x^{2}-x+1\right)\left(y^{2}-y+1\right)\left(z^{2}-z+1\right) \\
\geqslant & (x y z)^{2}-x y z+1 .
\end{aligned}
$$

If one of $x, y, z$ is 0, without loss of generality, let $x=0$, then
$$
\begin{aligned}
& \frac{16}{9}\left(x^{2}-x+1\right)\left(y^{2}-y+1\right)\left(z^{2}-z+1\right) \\
= & \frac{16}{9}\left(y^{2}-y+1\right)\left(z^{2}-z+1\right) \\
= & \frac{16}{9}\left[\left(y-\frac{1}{2}\right)^{2}+\frac{3}{4}\right]\left[\left(z-\frac{1}{2}\right)^{2}+\frac{3}{4}\right] \\
\geqslant & \frac{16}{9} \cdot \frac{3}{4} \cdot \frac{3}{4}=1 \\
= & (x y z)^{2}-x y z+1 .
\end{aligned}
$$

In this case, the inequality (*) holds.
If $x, y, z$ are all non-zero, without loss of generality, let $x \leqslant y \leqslant z$.
(i) $x(x y z)^{2}-x y z+1 . }
\end{aligned}
$$

In this case, the inequality (*) holds.
(iii) $x \leqslant y \leqslant z(x y z)^{2}-x y z+1 . }
\end{aligned}
$$

In this case, the inequality (*) holds.
In conclusion, the minimum value of the real number $k$ is $\frac{16}{9}$.