(3) Let $x, y, z \in [0,1]$, then the maximum value of $M=\sqrt{|x-y|}+\sqrt{|y-z|}+$ $\sqrt{|z-x|}$ is $\qquad$
(3) $\sqrt{2}+1$ Hint: Let $0 \leqslant x \leqslant y \leqslant z \leqslant 1$, then $M=\sqrt{y-x}+$ $\sqrt{z-y}+\sqrt{z-x}$.
Since
$$
\begin{aligned}
\sqrt{y-x}+\sqrt{z-y} & \leqslant \sqrt{2[(y-x)+(z-y)]} \\
& =\sqrt{2(z-x)},
\end{aligned}
$$

we have $M \leqslant \sqrt{2(z-x)}+\sqrt{z-x}=(\sqrt{2}+1) \sqrt{z-x} \leqslant \sqrt{2}+1$.
Equality holds if and only if $y-x=z-y, x=0, z=1$, i.e., $x=0, y=\frac{1}{2}$, $z=1$.
$$
\text { Hence } M_{\max }=\sqrt{2}+1 \text {. }
$$