Example 12 (2001 China Western Mathematical Olympiad) Let $x, y, z$ be positive real numbers, and $x+y+z \geqslant xyz$. Find the minimum value of $\frac{x^{2}+y^{2}+z^{2}}{xyz}$.
First, note the case when the equality holds in the condition.
That is, when $x=y=z=\sqrt{3}$, we have $x+y+z=x y z$, and $\frac{x^{2}+y^{2}+z^{2}}{x y z}=\sqrt{3}$.
Next, we prove that the minimum value of $\frac{x^{2}+y^{2}+z^{2}}{x y z}$ is $\sqrt{3}$.
In fact, we have
$$
x^{2}+y^{2}+z^{2} \geqslant \frac{1}{3}(x+y+z)^{2} \geqslant\left\{\begin{array}{c}
\frac{1}{3}(x y z)^{2} \geqslant \sqrt{3} x y z, \text { if } x y z \geqslant 3 \sqrt{3}, \\
3 \sqrt[3]{(x y z)^{2}} \geqslant \sqrt{3} x y z, \text { if } x y z<3 \sqrt{3} .
\end{array}\right.
$$

Therefore, the minimum value of $\frac{x^{2}+y^{2}+z^{2}}{x y z}$ is $\sqrt{3}$.