Problem 22. Find the minimum value of the expression $(x+y) \cdot(x+z),$ if $x, y, z-$ are positive numbers and $x \cdot y \cdot z \cdot(x+y+z)=1$.
Solution. Let $a=y+z ; b=x+z ; c=x+y$. Consider a triangle with sides $a, b, c$ (it is not difficult to verify that such a triangle exists for any $x, y, z$). Then:

$$
\begin{gathered}
(x+y) \cdot(x+z)=b \cdot c \\
x y z(x+y+z)=(p-a)(p-b)(p-c) p=S^{2}=1
\end{gathered}
$$

But $b \cdot c=\frac{2 S}{\sin A}=\frac{2}{\sin A}$, and $|\sin A| \leq 1$. In this case, it is clear that the minimum value of the product $b \cdot c$ is 2. Therefore, the minimum value of the expression $(x+y) \cdot(x+z)$ under the given conditions is 2.