Example 4 Given that $x, y, z$ are all positive numbers, and $x y z \cdot (x+y+z)=1$. Then, the minimum value of $(x+y)(y+z)$ is . $\qquad$
Analysis: To find the minimum value of the original expression, that is, to find the minimum value of $x y+y^{2}+$ $x z+y z$, we can achieve this by appropriately combining terms so that the sum becomes the sum of two terms, and the product of these two terms should be a constant.
$$
\text { Solution: } \begin{array}{l} 
\because(x+y)(y+z)=\left(x y+y^{2}+y z\right) \\
+x z \\
= y(x+y+z)+x z, \\
\text { and } x z>0, y(x+y+z)>0, \\
\therefore y(x+y+z)+x z \\
\geqslant 2 \sqrt{y(x+y+z) \cdot x z}=2, \\
\therefore(x+y)(y+z) \text { has a minimum value of } 2 .
\end{array}
$$