11. (1) If real numbers $x, y, z$ satisfy $x^{2}+y^{2}+z^{2}=1$, prove: $|x-y|+|y-z|+|z-x| \leqslant 2 \sqrt{2}$;
(2) If 2023 real numbers $x_{1}, x_{2}, \cdots, x_{2023}$ satisfy $x_{1}^{2}+x_{2}^{2}+\cdots+x_{2023}^{2}=1$,
find the maximum value of $\left|x_{1}-x_{2}\right|+\left|x_{2}-x_{3}\right|+\cdots+\left|x_{2022}-x_{2023}\right|+\left|x_{2023}-x_{1}\right|$.
(1) By symmetry, without loss of generality, assume $x \geqslant y \geqslant z$, then $|x-y|+|y-z|+|z-x|=x-y+y-z+x-z = 2(x-z)$. By the Cauchy-Schwarz inequality, we have $(x-z)^{2} \leqslant 2\left(x^{2}+z^{2}\right) \leqslant 2\left(x^{2}+y^{2}+z^{2}\right)=2$, with equality when $x=\frac{\sqrt{2}}{2}, y=0, z=-\frac{\sqrt{2}}{2}$. Therefore, $|x-y|+|y-z|+|z-x| \leqslant 2 \sqrt{2}$.
(2) From $x_{1}^{2}+x_{2}^{2}+\cdots+x_{2023}^{2}=1$, we know that $x_{i}$ can take $a$ or $-a$. Among $x_{1}, x_{2}, x_{2023}$, there must be two numbers that are not of opposite signs, without loss of generality, assume $x_{2} \geqslant x_{1} \geqslant 0$.
Then $\left|x_{1}-x_{2}\right|+\left|x_{2}-x_{3}\right|+\cdots+\left|x_{2022}-x_{2023}\right|+\left|x_{2023}-x_{1}\right| \leqslant\left|x_{1}-x_{2}\right|+\left|x_{2}\right|+\left|x_{3}\right|+\cdots+\left|x_{2023}\right|+\left|x_{1}\right|=$ $\left|x_{1}-x_{2}\right|+\left|x_{1}\right|+\left|x_{2}\right|+2\left(\left|x_{3}\right|+\cdots+\left|x_{2023}\right|\right)$
$=2\left(x_{2}+\left|x_{3}\right|+\cdots+\left|x_{2023}\right|\right)$. By the Cauchy-Schwarz inequality, we have $\left(x_{2}+\left|x_{3}\right|+\cdots+\left|x_{2023}\right|\right)^{2}$
$\leqslant 2022\left(x_{2}^{2}+x_{3}^{2}+\cdots+x_{2023}^{2}\right) \leqslant 2022\left(x_{1}^{2}+x_{2}^{2}+\cdots+x_{2023}^{2}\right)=2022$, with equality when
we can take $x_{1}=0, x_{3}=\cdots=x_{2023}=-\frac{1}{\sqrt{2022}}, x_{2}=\cdots=x_{2022}=\frac{1}{\sqrt{2022}}$.
Therefore, the maximum value of $\left|x_{1}-x_{2}\right|+\left|x_{2}-x_{3}\right|+\cdots+\left|x_{2022}-x_{2023}\right|+\left|x_{2023}-x_{1}\right|$ is $2 \sqrt{2022}$.