Example 11 (13th National High School Mathematics Winter Camp Problem) Let $n \geqslant 2, x_{1}, x_{2}, \cdots, x_{n}$ be real numbers, and $\sum_{i=1}^{n} x_{i}^{2}+$ $\sum_{i=1}^{n-1} x_{i} x_{i+1}=1$. For each fixed $k(k \in \mathbf{Z}, 1 \leqslant k \leqslant n)$, find the maximum value of $\left|x_{k}\right|$.
From the given conditions, we have $2 \sum_{i=1}^{n} x_{i}^{2}+2 \sum_{i=1}^{n-1} x_{i} x_{i+1}=2$, which means
$$
x_{1}^{2}+\left(x_{1}+x_{2}\right)^{2}+\left(x_{2}+x_{3}\right)^{2}+\cdots+\left(x_{n-2}+x_{n-1}\right)^{2}+\left(x_{n-1}+x_{n}\right)^{2}+x_{n}^{2}=2.
$$

By the Cauchy-Schwarz inequality, we get $\sqrt{\frac{x_{1}^{2}+\left(x_{1}+x_{2}\right)^{2}+\cdots+\left(x_{k-1}+x_{k}\right)^{2}}{k}}$
$$
\begin{array}{l}
\geqslant \frac{\left|x_{1}\right|+\left|x_{1}+x_{2}\right|+\cdots+\left|x_{k-1}+x_{k}\right|}{k} \\
\geqslant \frac{\left|x_{1}-\left(x_{1}+x_{2}\right)+\cdots+(-1)^{k-1}\left(x_{k-1}+x_{k}\right)\right|}{k}=\frac{\left|x_{k}\right|}{k},
\end{array}
$$

which implies
$$
x_{1}^{2}+\left(x_{1}+x_{2}\right)^{2}+\cdots+\left(x_{k-1}+x_{k}\right)^{2} \geqslant \frac{x_{k}^{2}}{k}.
$$

Similarly, we have $\sqrt{\frac{\left(x_{k}+x_{k+1}\right)^{2}+\cdots+\left(x_{n-1}+x_{n}\right)^{2}+x_{n}^{2}}{n-k+1}}$
$$
\begin{array}{l}
\geqslant \frac{\left|x_{k}+x_{k-1}\right|+\cdots+\left|x_{n-1}+x_{n}\right|+\left|x_{n}\right|}{n-k+1} \\
\geqslant \frac{\left|x_{n}-\left(x_{n-1}+x_{n}\right)+\cdots+(-1)^{n-k}\left(x_{k}+x_{k+1}\right)\right|}{n-k+1} \geqslant \frac{\left|x_{k}\right|}{n-k+1},
\end{array}
$$

which implies $\left(x_{k}+x_{k+1}\right)^{2}+\cdots+\left(x_{n-1}+x_{n}\right)^{2}+x_{n}^{2} \geqslant \frac{x_{k}^{2}}{n-k+1}$.
From (2) + (3) and (1), we get $2 \geqslant\left(\frac{1}{k}+\frac{1}{n-k+1}\right) x_{k}^{2}$,
which means
$$
\left|x_{k}\right| \leqslant \sqrt{\frac{2 k(n+1-k)}{n+1}} \quad (k=1,2, \cdots, n).
$$

From the above, we know that the equality in (4) holds if and only if $x_{1}=-\left(x_{1}+x_{2}\right)=x_{2}+x_{3}=\cdots=(-1)^{k-1}\left(x_{k-1}+x_{k}\right)$ and $x_{k}+x_{k+1}=$ $-\left(x_{k+1}+x_{k+2}\right)=\cdots=(-1)^{n-k} x_{n}$, which means $\left|x_{k}\right|=$ $\sqrt{\frac{2 k(n+1-k)}{n+1}}$ when $x_{i}=x_{k} \cdot(-1)^{i-k} \frac{-}{k}$ $(i=1,2, \cdots, k-1)$ and $x_{j}=x_{k} \cdot(-1)^{j-k} \cdot \frac{n+1-j}{n-k+1}(j=k+1, k+2, \cdots, n)$.
In summary, we have $\left|x_{k}\right|_{\max }=\sqrt{\frac{2 k(n+1-k)}{n+1}}$.