8. Find the least value of the expression $(x+y)(y+z)$, under the condition that $x, y, z$ are positive numbers satisfying the equation
$$
x y z(x+y+z)=1 \text {. }
$$
Solution. For arbitrary positive $x, y, z$ satisfying the equation $x y z(x+$ $y+z)=1$, we have
$$
(x+y)(y+z)=(x+y+z) y+x z=\frac{1}{x z}+x z \geq 2 .
$$

This inequality becomes an equality, for example, when $x=z=1$ and $y=\sqrt{2}-1$.