Question 4 Given that $x, y, z$ are all positive numbers, and
$$
x y z=\frac{1}{2}, x^{2}+y^{2}+z^{2} \leqslant 2 \text{. }
$$

Then the maximum value of $x^{4}+y^{4}+z^{4}$ is
【Analysis】This problem discusses the maximum and minimum values of a ternary function. The idea is to eliminate variables and transform it into a unary function to solve.
Solution Without loss of generality, assume $x^{2} \leqslant y^{2} \leqslant z^{2}$. Then $x^{2} \leqslant \frac{2}{3}$.
Since $x^{2}+y^{2}+z^{2} \leqslant 2$, we have,
$$
\begin{array}{l}
x^{4}+y^{4}+z^{4}=x^{4}+\left(y^{2}+z^{2}\right)^{2}-2 y^{2} z^{2} \\
\leqslant x^{4}+\left(2-x^{2}\right)^{2}-\frac{1}{2 x^{2}}=2 x^{4}-4 x^{2}-\frac{1}{2 x^{2}}+4 .
\end{array}
$$

Let $t=x^{2}$. Then $t \in\left(0, \frac{2}{3}\right]$.
$$
\begin{array}{l}
\text { By } f(t)=2 t^{2}-4 t-\frac{1}{2 t}+4 \\
=\frac{3}{2}-\frac{(1-t)(2 t-1)^{2}}{2 t} \leqslant \frac{3}{2},
\end{array}
$$

The equality holds if and only if $t=\frac{1}{2}$.
Therefore, when and only when $x=\frac{1}{\sqrt{2}}, y=\frac{1}{\sqrt{2}}, z=1$, $x^{4}+y^{4}+z^{4}$ achieves its maximum value $\frac{3}{2}$.