11. (Mathematical Bulletin, Issue 5, 2006) Given: $x, y, z \in \mathbf{R}_{+}$, and $x y z(x+y+z)=1$, find the minimum value of $(x+y)(y+z)(z+x)$.
11. From $x y z(x+y+z)=x^{2} y z+x y z(y+z)=1$

we get
$$
x^{2}+x(y+z)=\frac{1}{y z} .
$$

Therefore, $(x+y)(y+z)(z+x)=\left(x^{2}+x z+y z+x y\right)(y+z)$
$$
\begin{array}{l}
=\left[x^{2}+x(y+z)+y z\right](y+z)=\left(\frac{1}{y z}+y z\right)(y+z) \\
=\frac{1}{z}+\frac{1}{y}+y^{2} z+y z^{2}=\frac{1}{3 y}+\frac{1}{3 y}+\frac{1}{3 y}+\frac{1}{3 z}+\frac{1}{3 z}+\frac{1}{3 z}+y^{2} z+z^{2} \\
\geqslant 8 \sqrt[8]{\left(\frac{1}{3 y}\right)^{3}\left(\frac{1}{3 z}\right)^{3}\left(y^{2} z\right)\left(y z^{2}\right)}=8 \sqrt[8]{\left(\frac{1}{3}\right)^{6}}=\frac{8}{3} \sqrt[3]{3},
\end{array}
$$

Equality holds if and only if $y^{2} z=z^{2} y=\frac{1}{3 y}=\frac{1}{3 z}$. Therefore, $y=z=\sqrt[1]{\frac{1}{3}}$, and from the given condition, we have: $x=y=z=\sqrt[1]{\frac{1}{3}}$.