Given $x 、 y 、 z \in[0,1]$. Find
$$
A=\sqrt{3|x-y|}-\sqrt{|y-z|}-\sqrt{3|z-x|}
$$

the maximum and minimum values.
Prove (1) If $0 \leqslant x \leqslant y \leqslant z \leqslant 1$, then
$$
A=\sqrt{3(y-x)}-\sqrt{z-y}-\sqrt{3(z-x)} .
$$

Notice that, $(y-x)+(z-y)=z-x$.
Let $a^{2}=y-x, b^{2}=z-y, c^{2}=z-x$. Then $a^{2}+b^{2}=c^{2}(a, b, c \in[0,1])$.
Let $a=c \cos \theta, b=c \sin \theta\left(\theta \in\left[0, \frac{\pi}{2}\right]\right)$. Then
$$
\begin{aligned}
A & =\sqrt{3} a-b-\sqrt{3} c \\
& =(\sqrt{3} \cos \theta-\sin \theta-\sqrt{3}) c \\
& =\left[2 \sin \left(\frac{\pi}{3}-\theta\right)-\sqrt{3}\right] c .
\end{aligned}
$$

Since $\theta \in\left[0, \frac{\pi}{2}\right], \frac{\pi}{3}-\theta \in\left[-\frac{\pi}{6}, \frac{\pi}{3}\right)$, so,
$$
-\sqrt{3}-1 \leqslant 2 \sin \left(\frac{\pi}{3}-\theta\right)-\sqrt{3} \leqslant 0 \text {. }
$$

Since $c \in[0,1]$, then
$$
-\sqrt{3}-1 \leqslant(-\sqrt{3}-1) c \leqslant A \leqslant 0 \text {. }
$$

When and only when $\theta=\frac{\pi}{2}, c=1$, i.e., $x=y=0, z=1$, $A_{\text {min }}=-\sqrt{3}-1$; when $y=z$, $A_{\text {max }}=0$.
(2) If $0 \leqslant x \leqslant z \leqslant y \leqslant 1$, then
$$
A=\sqrt{3(y-x)}-\sqrt{y-z}-\sqrt{3(z-x)} \text {. }
$$

Notice that, $(y-z)+(z-x)=y-x$.
Let $a^{2}=y-x, b^{2}=y-z, c^{2}=z-x$. Then $a^{2}=b^{2}+c^{2}(a, b, c \in[0,1])$.
Let $b=a \cos \alpha, c=a \sin \alpha\left(\alpha \in\left[0, \frac{\pi}{2}\right]\right)$. Then
$$
\begin{aligned}
A & =\sqrt{3} a-b-\sqrt{3} c \\
= & (\sqrt{3}-\cos \alpha-\sqrt{3} \sin \alpha) a \\
= & {\left[\sqrt{3}-2 \sin \left(\alpha+\frac{\pi}{6}\right)\right] a . }
\end{aligned}
$$

When and only when $\alpha=\frac{\pi}{3}, a=1$, i.e., $x=0, y=1, z=\frac{3}{4}$, $A_{\text {min }}=\sqrt{3}-2$; when and only when $\alpha=0, a=1$, i.e., $z=x=0, y=1$, $A_{\text {max }}=\sqrt{3}-1$.
(3) If $0 \leqslant y \leqslant x \leqslant z \leqslant 1$, then
$$
A=\sqrt{3(x-y)}-\sqrt{z-y}-\sqrt{3(z-x)} \text {. }
$$

Notice that, $(x-y)+(z-x)=z-y$.
Let $a^{2}=x-y, b^{2}=z-y, c^{2}=z-x$. Then $b^{2}=a^{2}+c^{2}(a, b, c \in[0,1])$.
Let $a=b \cos \beta, c=b \sin \beta\left(\beta \in\left[0, \frac{\pi}{2}\right]\right)$. Then
$$
\begin{aligned}
A & =\sqrt{3} a-b-\sqrt{3} c \\
& =(\sqrt{3} \cos \beta-1-\sqrt{3} \sin \beta) b \\
& =\left[\sqrt{6} \sin \left(\frac{\pi}{4}-\beta\right)-1\right] b .
\end{aligned}
$$

When and only when $\beta=\frac{\pi}{2}, b=1$, i.e., $x=y=0, z=1$, $A_{\text {min }}=-\sqrt{3}-1$; when and only when $\beta=0, b=1$, i.e., $x=z=1$, $y=0$, $A_{\text {max }}=\sqrt{3}-1$.
(4) If $0 \leqslant y \leqslant z \leqslant x \leqslant 1$, i.e.,
$A=\sqrt{3(x-y)}-\sqrt{z-y}-\sqrt{3(x-z)}$.
This case is the same as (2).
(5) If $0 \leqslant z \leqslant x \leqslant y \leqslant 1$, then
$$
A=\sqrt{3(y-x)}-\sqrt{y-z}-\sqrt{3(x-z)} \text {. }
$$

This case is the same as (3).
(6) If $0 \leqslant z \leqslant y \leqslant x \leqslant 1$, then
$$
A=\sqrt{3(x-y)}-\sqrt{y-z}-\sqrt{3(x-z)} \text {. }
$$

This case is the same as (1).
In summary, $A_{\text {max }}=\sqrt{3}-1, A_{\text {min }}=-\sqrt{3}-1$.
(Xu Shenglin, Huazhong Normal University, "Mathematical Communications" Editorial Department, 430079)