4. Let $x$, $y$, $z$ be positive real numbers, and $x+y+z \geqslant xyz$. Find the minimum value of $\frac{x^{2}+y^{2}+z^{2}}{xyz}$.
(Feng Zhigang)
4. Notice that, when $x=y=z=\sqrt{3}$, $x+y+z=x y z$, and $\frac{x^{2}+y^{2}+z^{2}}{x y z}=\sqrt{3}$.
Below we prove that the minimum value of $\frac{x^{2}+y^{2}+z^{2}}{x y z}$ is $\sqrt{3}$.
In fact, we have
$x^{2}+y^{2}+z^{2} \geqslant \frac{1}{3}(x+y+z)^{2}$
$\geqslant\left\{\begin{array}{l}\frac{1}{3}(x y z)^{2} \geqslant \sqrt{3} x y z, \\ 3 \sqrt[3]{(x y z)^{2}} \geqslant \sqrt{3} x y z, \text { if } x y z \geqslant 3 \sqrt{3}, \\ x y z<3 \sqrt{3} .\end{array}\right.$
Therefore, the minimum value of $\frac{x^{2}+y^{2}+z^{2}}{x y z}$ is $\sqrt{3}$.