China Mathematical Competition

We may assume $0\le x\le y\le z\le 1$. Then $$M=\sqrt{y-x}+\sqrt{z-y}+\sqrt{z-x}.$$ Since $$\sqrt{y-x}+\sqrt{z-y}\le \sqrt{2[(y-x)+(z-y)]}=\sqrt{2(z-x)},$$ we have $$M\le \sqrt{2(z-x)}+\sqrt{z-x}=(\sqrt{2}+1)\sqrt{z-x}\le \sqrt{2}+1.$$ The equality holds if and only if $y-x=z-y$, $x=0$, $z=1$ (i.e. $x=0$, $y=\frac{1}{2}$, $z=1$). Therefore, the answer is $M_{\max }=\sqrt{2}+1$.