11. Let $x, y, z \in \mathbf{R}^{+}$, and satisfy $x y z(x+y+z)=1$, find the minimum value of $(x+y)(x+z)$.
11. Given $x^{2}+x(y+z)=\frac{1}{y z}$, thus
$$
(x+y)(x+z)=\left[x^{2}+x(y+z)\right]+y z=\frac{1}{y z}+y z \geqslant 2 \sqrt{\frac{1}{y z} \cdot y z}=
$$
2.
The equality holds if and only if $y z=1$.
Therefore, the minimum value of $(x+y)(x+z)$ is 2.