1. Given that $x$, $y$, $z$ are positive real numbers, and $x y z(x+y+z)=1$. Then the minimum value of $(x+y)(y+z)$ is $\qquad$
\begin{array}{l}=1.2. \\ (x+y)(y+z)=y^{2}+(x+z) y+x z \\ =y(x+y+z)+x z=y \cdot \frac{1}{x y z}+x z \\ =\frac{1}{x z}+x z \geqslant 2 .\end{array}