Example 7 Given positive real numbers $x, y, z$ satisfy
$$
\left\{\begin{array}{l}
x^{2}+x y+y^{2}=9, \\
y^{2}+y z+z^{2}=16, \\
z^{2}+z x+x^{2}=25 .
\end{array}\right.
$$

Find the value of $x y+y z+z x$.
As shown in Figure 1, with $P$ as the endpoint,
draw rays $P A, P B, P C$, such that
$$
\begin{array}{l}
\angle A P B=\angle B P C \\
=\angle C P A=120^{\circ} .
\end{array}
$$

Let the lengths of segments $P A, P B, P C$ be $x, y, z$ respectively.
By the Law of Cosines, we have
$$
\begin{array}{l}
A B=\sqrt{x^{2}+x y+y^{2}}=3, \\
B C=\sqrt{y^{2}+y z+z^{2}}=4, \\
A C=\sqrt{z^{2}+z x+x^{2}}=5 .
\end{array}
$$

Thus, $A B^{2}+A C^{2}=B C^{2}$
$$
\begin{array}{l}
\Rightarrow \angle B A C=90^{\circ} \\
\Rightarrow S_{\triangle A B C}=6 .
\end{array}
$$
$$
\begin{array}{l}
\text { Also, } S_{\triangle A B C}=S_{\triangle P A B}+S_{\triangle P B C}+S_{\triangle P A C} \\
=\frac{1}{2} x y \sin 120^{\circ}+\frac{1}{2} y z \sin 120^{\circ}+\frac{1}{2} z x \sin 120^{\circ} \\
=\frac{\sqrt{3}}{4}(x y+y z+z x)=6,
\end{array}
$$

Therefore, $x y+y z+z x=8 \sqrt{3}$.