1. Given $x, y, z \in \mathbf{R}^{+}$ and satisfy $x y z(x+y+z)=1$. Then the minimum value of $(x+y)(y+z)$ is ( ).
(A) 2
(B) 3
(C) $\frac{4 \sqrt{3}}{3}$
(D) 1
\begin{array}{l}-1 \cdot(\mathrm{A}) . \\ (x+y)(y+z)=x y+x z+y^{2}+y z \\ =y(x+y+z)+x z=y \cdot \frac{1}{x y z}+x z \\ =\frac{1}{x z}+x z \geqslant 2 \sqrt{\frac{1}{x z} \cdot x z} \\ =2\left(x, y, z \in \mathbf{R}^{+}\right) .\end{array}