Example 3.12 (2006 China National Team Training Problem) For $x, y, z$ not all positive, satisfying
$$k\left(x^{2}-x+1\right)\left(y^{2}-y+1\right)\left(z^{2}-z+1\right) \geqslant(x y z)^{2}-x y z+1$$

Find the minimum value of the real number $k$.
For a 3-variable 6th-degree symmetric inequality, the equality often holds when two numbers are equal or one of them is 0. After testing, it is found that when \( x = y = \frac{1}{2}, z = 0 \), \( k \) is minimized, at which point \( k = \frac{16}{9} \). We now prove that the inequality holds when \( k = \frac{16}{9} \).

Considering the direct proof is quite challenging, we attempt to construct a local inequality by examining the two-variable case.
We prove that when at least one of \( x, y \) is less than or equal to 0, the following holds:
\[
\begin{array}{l}
\frac{4}{3}\left(x^{2}-x+1\right)\left(y^{2}-y+1\right) \geqslant(x y)^{2}-x y+1 \Leftrightarrow \\
\left(x^{2} y^{2}-4 x^{2} y+4 x^{2}\right)-\left(4 x y^{2}-7 x y+4 x\right)+\left(4 y^{2}-4 y+1\right) \geqslant 0
\end{array}
\]
When at least one of \( x, y \) is less than or equal to 0, without loss of generality, assume \( x \leqslant 0, y > 0 \), then the original inequality is equivalent to
\[
x^{2}(y-2)^{2}-4 x(y-1)^{2}-x y+(2 y-1)^{2} \geqslant 0
\]

This is clearly true.
When \( x \leqslant 0, y \leqslant 0 \), it is equivalent to
\[
x^{2}(y-2)^{2}-4 x y^{2}+7 x y-4 x+(2 y-1)^{2} \geqslant 0
\]

This is also true. Therefore, we can deduce that
\[
\begin{array}{l}
\frac{16}{9}\left(x^{2}-x+1\right)\left(y^{2}-y+1\right)\left(z^{2}-z+1\right) \geqslant \\
\frac{4}{3}\left(z^{2}-z+1\right)\left((x y)^{2}-x y+1\right) \geqslant(x y z)^{2}-x y z+1
\end{array}
\]

Thus, the minimum value of \( k \) is \( \frac{16}{9} \).