Example 1 Given that $x$, $y$, $z$ are positive integers, and $x y z(x+y+z)=1$. Find the minimum value of the expression $(x+y)(y+z)$.

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Explanation: This is a typical example of constructing a geometric figure to solve a problem. As shown in Figure 1, construct $\triangle ABC$, where the lengths of the three sides are
$$
\left\{\begin{array}{l}
a=x+y, \\
b=y+z, \\
c=z+x .
\end{array}\right.
$$

Then its area is
$$
\begin{aligned}
\triangle & =\sqrt{p(p-a)(p-b)(p-c)} \\
& =\sqrt{(x+y+z) x y z}=1 .
\end{aligned}
$$

On the other hand,
$$
(x+y)(y+z)=a b=\frac{2 \triangle}{\sin C} \geqslant 2 .
$$

Therefore, when and only when $\angle C=90^{\circ}(\sin C=1)$, $a b$ takes the minimum value of 2. At this time, by the Pythagorean theorem, we have
$$
(x+y)^{2}+(y+z)^{2}=(x+z)^{2} \text {, }
$$

which means $y(x+y+z)=x z$ when $(x+y)(y+z)$ takes the minimum value. For example, when $x=z=1, y=\sqrt{2}-1$, $(x+y)(y+z)=2$.

This construction is indeed quite ingenious and should be recognized for its training value, but for solving this problem, a direct algebraic method is much simpler.

The following two problems from the former Soviet Union's math competitions are also classic examples of constructing geometric figures to solve problems, and they can both find simpler algebraic solutions.
1. Let $a, b, c$ be positive numbers, and
$$
\left\{\begin{array}{l}
a^{2}+a b+\frac{b^{2}}{3}=25, \\
\frac{b^{2}}{3}+c^{2}=9, \\
a^{2}+a c+c^{2}=16 .
\end{array}\right.
$$

Find the value of $a b+2 b c+3 a c$.
2. Positive numbers $a, b, c, A, B, C$ satisfy the conditions $a+A=$ $b+B=c+C=k$. Prove that $a B+b C+c A<k^{2}$.