Example 7 Let $x_{1}, x_{2}, \cdots, x_{n} \in \mathbf{R}(n \geqslant 2)$, and satisfy $\sum_{i=1}^{n} x_{i}^{2}+\sum_{i=1}^{n-1} x_{i} x_{i+1}=1$. Try to find $\max _{1 \leq k \leq n}^{\max }\left|x_{k}\right|$.
Solution: First, find the maximum value of $\left|x_{k}\right|$ for a given $k$.
In fact, the given equation is
$$
\begin{array}{l}
x_{1}^{2}+\left(x_{1}+x_{2}\right)^{2}+\left(x_{2}+x_{3}\right)^{2}+\cdots+ \\
\left(x_{n-1}+x_{n}\right)^{2}+x_{n}^{2}=2 .
\end{array}
$$

Notice the identity
$$
\begin{aligned}
\left|x_{k}\right|= & \mid x_{1}-\left(x_{1}+x_{2}\right)+\cdots+ \\
& (-1)^{k-1}\left(x_{k-1}+x_{k}\right) \mid, \\
\left|x_{k}\right|= & \mid x_{n}-\left(x_{n-1}+x_{n}\right)+\cdots+ \\
& (-1)^{n-k}\left(x_{k}+x_{k+1}\right) \mid .
\end{aligned}
$$

By the Cauchy-Schwarz inequality, we have
$$
x_{k}^{2} \leqslant k\left[x_{1}^{2}+\left(x_{1}+x_{2}\right)^{2}+\cdots+\left(x_{k-1}+x_{k}\right)^{2}\right] \text {, }
$$
$$
\begin{aligned}
x_{k}^{2} \leqslant & (n-k+1)\left[x_{n}^{2}+\left(x_{n-1}+x_{n}\right)^{2}+\cdots+\right. \\
& \left.\left(x_{k}+x_{k-1}\right)^{2}\right] .
\end{aligned}
$$

From equations (4) and (5), combined with equation (1), we get
$$
\frac{x_{k}^{2}}{k}+\frac{x_{k}^{2}}{n-k+1} \leqslant 2 \text {. }
$$

Solving this, we get $\left|x_{k}\right| \leqslant \sqrt{\frac{2 k(n+1-k)}{n+1}}$
$$
(k=1,2, \cdots, n) \text {. }
$$

By the condition for equality in the Cauchy-Schwarz inequality, we know that equation (6) holds with equality if and only if equations (4) and (5) hold with equality simultaneously, i.e.,
$$
\begin{array}{l}
x_{1}=-\left(x_{1}+x_{2}\right)=x_{2}+x_{3}=\cdots \\
=(-1)^{k-1}\left(x_{k-1}+x_{k}\right), \\
x_{k}+x_{k+1}=-\left(x_{k+1}+x_{k+2}\right)=\cdots \\
=(-1)^{n-k} x_{n} .
\end{array}
$$

From equation (7), for $1<i \leqslant k$, we have
$$
x_{1}=(-1)^{i-1}\left(x_{i-1}+x_{i}\right) \text {, }
$$

i.e.,
$$
x_{i}+x_{i-1}=(-1)^{i-1} x_{1} \text {. }
$$

Then $(-1)^{i} x_{i}-(-1)^{i-1} x_{i-1}=-x_{1}$.
Thus, $(-1)^{i} x_{i}=-x_{1}+\left(-x_{1}\right)(i-1)=-x_{1} i$.
Therefore, $x_{i}=(-1)^{i-1} i x_{1}(i=2,3, \cdots, k)$.
Let $i=k$, we get $x_{1}=\frac{x_{k}}{k}(-1)^{k-1}$. So,
$$
x_{i}=x_{k}(-1)^{i-k} \cdot \frac{i}{k}(i=1,2, \cdots, k-1) \text {. }
$$

Similarly, from equation (8), we can find
$$
x_{j}=x_{k}(-1)^{j-k} \cdot \frac{n+1-j}{n-k+1}(j=k+1, \cdots, n)
$$

The above shows that there exists a set of real numbers $x_{1}, x_{2}, \cdots, x_{n}$ that satisfy the given conditions and make equation (6) hold with equality.
Therefore, for a given $k$, we have
$\max \left|x_{k}\right|=\sqrt{\frac{2 k(n+1-k)}{n+1}}(1 \leqslant k \leqslant n)$.
As $k$ varies from 1 to $n$, it is not difficult to prove that
$k(n+1-k) \leqslant\left[\frac{n+1}{2}\right]\left[\frac{n+2}{2}\right]$.
Equality holds if and only if $k=\left[\frac{n+1}{2}\right]$.
$$
\begin{array}{l}
\text { Hence } \max _{1 \leqslant k<n} \max \left|x_{k}\right| \\
=\sqrt{\frac{2}{n+1}\left[\frac{n+1}{2}\right]\left[\frac{n+2}{2}\right]},
\end{array}
$$

where $[x]$ is the floor function.
Comment: The key is to pay attention to equations (2) and (3), then use the Cauchy-Schwarz inequality to derive equations (4) and (5). Thus, it connects the condition equation
$$
\sum_{i=1}^{n} x_{i}^{2}+\sum_{i=1}^{n-1} x_{i} x_{i+1}=1
$$