6.75 Given that $x, y, z$ are positive numbers, and satisfy the equation
$$
x y z(x+y+z)=1 \text{, }
$$

find the minimum value of the expression $(x+y)(y+z)$.
[Solution] From the given conditions and the arithmetic-geometric mean inequality, we have
$$
\begin{aligned}
(x+y)(y+z) & =(x+y+z) y+x z \\
& =\frac{1}{x y z} \cdot y+x z \\
& =\frac{1}{x z}+x z \\
& \geqslant 2 .
\end{aligned}
$$

On the other hand, when $x=1, y=\sqrt{2}-1, z=1$, the given equation is satisfied, and
$$
(x+y)(y+z)=2 \text{.}
$$

Therefore, the minimum value of the expression is 2.