For example, $5 x, y, z$ are real numbers, and satisfy $x+y+z=0, xyz=2$. Find the minimum value of $|x|+|y|+|z|$.
(1990, Beijing Junior High School Mathematics Competition).
Solution: From the problem, we know that among $x, y, z$, there are 2 negative numbers and 1 positive number. Without loss of generality, let $x>0, y<0, z<0$. When $x>0, -y>0, -z>0$, we have
$(-y)(-z) \leqslant \left[\frac{(-y)+(-z)}{2}\right]^{2}=\frac{x^{2}}{4}$,
which means $2 x=\frac{4}{(-y)(-z)} \geqslant \frac{4}{\frac{(-y)+(-z)}{2}}$
$$
=\frac{16}{x^{2}}.
$$

Thus, $x^{3} \geqslant 8$, solving which gives $x \geqslant 2$.
Therefore, $|x|+|y|+|z|=2 x \geqslant 4$,
which means the minimum value of $|x|+|y|+|z|$ is 4.