24. Given that $x, y, z$ are positive numbers, and satisfy $x y z(x+y+z)=1$, find the minimum value of the expression $(x+y)(x+z)$. (1989 All-Soviet Union Mathematical Olympiad Problem)
24. Since $(x+y)(x+z)=y z+x(x+y+z)$, by the AM-GM inequality we have
$$\begin{array}{l}
(x+y)(x+z)=y z+x(x+y+z) \geqslant 2 \sqrt{y z \cdot x(x+y+z)}= \\
2 \sqrt{x y z \cdot(x+y+z)}=2
\end{array}$$

Furthermore, when $x=\sqrt{2}-1, y=z=1$, the above expression achieves equality, hence the minimum value of $(x+y)(x+z)$ is 2.