Example 9 If $x, y, z \in \mathbf{R}_{+}$, find
$$
\frac{\sqrt{x^{2}+y^{2}+x y}+\sqrt{y^{2}+z^{2}+y z}+\sqrt{z^{2}+x^{2}+z x}}{x+y+z}
$$

the minimum value.
Explanation: Through the formula we get
$$
\begin{array}{l}
x^{2}+y^{2}+x y=\left(x+\frac{1}{2} y\right)^{2}+\left(\frac{\sqrt{3}}{2} y\right)^{2}, \\
y^{2}+z^{2}+y z=\left(y+\frac{1}{2} z\right)^{2}+\left(\frac{\sqrt{3}}{2} z\right)^{2}, \\
z^{2}+x^{2}+z x=\left(z+\frac{1}{2} x\right)^{2}+\left(\frac{\sqrt{3}}{2} x\right)^{2} .
\end{array}
$$

Construct complex numbers for substitution.
$$
\begin{array}{l}
\text { Let } z_{1}=\left(x+\frac{1}{2} y\right)+\frac{\sqrt{3}}{2} y \mathrm{i}, \\
z_{2}=\left(y+\frac{1}{2} z\right)+\frac{\sqrt{3}}{2} z \mathrm{i}, \\
z_{3}=\left(z+\frac{1}{2} x\right)+\frac{\sqrt{3}}{2} x \mathrm{i} .
\end{array}
$$

By $\left|z_{1}\right|+\left|z_{2}\right|+\left|z_{3}\right| \geqslant\left|z_{1}+z_{2}+z_{3}\right|$ $=\left|\frac{3}{2}(x+y+z)+\frac{\sqrt{3}}{2}(x+y+z) \mathrm{i}\right|$, then $\frac{\sqrt{x^{2}+y^{2}+x y}+\sqrt{y^{2}+z^{2}+y z}+\sqrt{z^{2}+x^{2}+z}}{x+y+z}$ $\geqslant \frac{\left|\frac{3}{2}(x+y+z)+\frac{\sqrt{3}}{2}(x+y+z) \mathrm{i}\right|}{x+y+z}$ $=\sqrt{3}$.
The equality holds if and only if $x=y=z$.
Therefore, when $x=y=z$, the minimum value sought is $\sqrt{3}$.