Example 2-1 Given that $x, y, z$ are positive numbers, and $xyz(x+y+z)=1$.
Find the minimum value of $(x+y)(y+z)$.
(1989, All-Soviet Union Mathematical Competition)
Solution 1: $(x+y)(y+z)$
$$
\begin{array}{l}
=x z+y(x+y+z) \\
\geqslant 2 \sqrt{x y z(x+y+z)}=2 .
\end{array}
$$

When $x=z=1, y=\sqrt{2}-1$, $y(x+y+z) = xz$, $(x+y)(y+z)$ takes the minimum value 2.
Solution 2: As shown in Figure 1, construct
$\triangle ABC$, with side lengths
$$
\left\{\begin{array}{l}
a=x+y, \\
b=y+z, \\
c=z+x .
\end{array}\right.
$$

Then the area of the triangle is
$$
\begin{array}{l}
S=\sqrt{p(p-a)(p-b)(p-c)} \\
=\sqrt{x y z(x+y+z)}=1 \text { (constant). } \\
\text { Also, }(x+y)(y+z)=a b=\frac{2 S}{\sin C} \geqslant 2, \text { when and }
\end{array}
$$

only when $\sin C=1\left(\angle C=90^{\circ}\right)$, $(x+y)(y+z)$ takes the minimum value 2.

This problem's integration of algebra and geometry has sparked our interest in exploration, providing some insights.
(1) Figure 1 and the corresponding transformation (1) serve as a bridge between the positive numbers $x, y, z$ and $\triangle ABC$, representing a stable approach to integrating algebra and geometry;
(2) Solution 2 tells us: Among triangles with a constant area, the product of two sides is minimized when their included angle is a right angle (right-angled triangle);
(3) Solution 1 tells us: $(x+y)(y+z)$ takes the minimum value under the condition
$$
\begin{array}{l}
x z=y(x+y+z) \\
\Leftrightarrow(z+x)^{2}=(x+y)^{2}+(y+z)^{2} .
\end{array}
$$

Combining this with the converse of the Pythagorean theorem, we see that $\triangle ABC$ is a right-angled triangle, which is consistent with the geometric observation. However, it also provides new insights, namely that from the constant area
$$
\begin{array}{c}
\sqrt{x y z(x+y+z)}=S \text { and equation (2), we get } \\
x z=y(x+y+z)=S .
\end{array}
$$

In this right-angled triangle, $y$ equals the radius of the inscribed circle, so $y(x+y+z)=S$ means that "the area of a right-angled triangle is equal to the product of the semiperimeter and the radius of the inscribed circle."
And $x z=S$ is the following conclusion.
Conclusion: The area of a right-angled triangle is equal to the product of the segments into which the hypotenuse is divided by the points of tangency of the inscribed circle.