Example 3 Given that $x, y, z$ are non-negative real numbers, not all zero. Find
$$
u=\frac{\sqrt{x^{2}+y^{2}+x y}+\sqrt{y^{2}+z^{2}+y z}+\sqrt{z^{2}+x^{2}+z x}}{x+y+z}
$$

the minimum value.
Let $z_{1}=x+\frac{y}{2}+\frac{\sqrt{3}}{2} y \mathrm{i}$,
$$
z_{2}=y+\frac{z}{2}+\frac{\sqrt{3}}{2} z \mathrm{i}, z_{3}=z+\frac{x}{2}+\frac{\sqrt{3}}{2} x \mathrm{i},
$$

where $x \geqslant 0, y \geqslant 0, z \geqslant 0$. Then
$$
\begin{array}{l}
\sqrt{x^{2}+y^{2}+x y}+\sqrt{y^{2}+z^{2}+y z}+\sqrt{z^{2}+x^{2}+z x} \\
=\left|z_{1}\right|+\left|z_{2}\right|+\left|z_{3}\right| \geqslant\left|z_{1}+z_{2}+z_{3}\right| \\
=\left|\frac{3}{2}(x+y+z)+\frac{\sqrt{3}}{2}(x+y+z) \mathrm{i}\right| \\
=\sqrt{3}|x+y+z| .
\end{array}
$$

Therefore, $u \geqslant \sqrt{3}$.
Equality holds if and only if $\arg z_{1}=\arg z_{2}=\arg z_{3}$, i.e., $x=y=z>0$. Thus, $u_{\text {min }}=\sqrt{3}$.

Note: The key to this solution is the clever elimination of $x+y+z$ by utilizing the cyclic symmetry of the given algebraic expressions in $x, y, z$.