15. (23rd All-Soviet Union Mathematical Olympiad, 1989) Given that $x, y, z$ are positive numbers, and satisfy the equation $x y z(x+y+z)=1$, find the minimum value of the expression $(x+y)(y+z)$.
15. From the known conditions and the arithmetic-geometric mean inequality, we have
$$
(x+y)(y+z)=(x+y+z) y+x z=\frac{1}{x y z} \cdot y+x z=\frac{1}{x z}+x z \geqslant 2 \text{. }
$$

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

Therefore, the minimum value of the expression is 2.