Example 11 Let $x_{i} \geqslant 0, i=1,2, \cdots, n$, satisfy
$$\sum_{i=1}^{n} x_{i}^{2}+2 \sum_{1 \leq j<k \leqslant n} \sqrt{\frac{j}{k}} x_{j} x_{k}=1 .$$

Find the minimum and maximum values of $x_{1}+x_{2}+\cdots+x_{n}$.
Solve: From
$$\left(x_{1}+x_{2}+\cdots+x_{n}\right)^{2}=\sum_{i=1}^{n} x_{i}^{2}+2 \sum_{1 \leqslant j<k \leqslant n} x_{j} x_{k} \geqslant 1$$

Take $x_{1}=1, x_{2}=\cdots=x_{n}=0$, then the minimum value of $x_{1}+x_{2}+\cdots+x_{n}$ is 1. Let $y_{i}=$ $\frac{x_{i}}{\sqrt{i}}$, then the condition becomes
$$\sum_{i=1}^{n} i y_{i}^{2}+2 \sum_{1 \leqslant j<k \leqslant n} j y_{j} y_{k}=1$$

which is equivalent to
$$\sum_{i=1}^{n}\left(y_{i}+y_{i+1}+\cdots+y_{n}\right)^{2}=1$$

Let $t_{i}=y_{i}+y_{i+1}+\cdots+y_{n}$, then $y_{i}=t_{i}-t_{i+1}, x_{i} \geqslant 0$, which implies $y_{i} \geqslant 0, t_{i}$ is non-increasing, so $x_{i}=\sqrt{i} y_{i}=\sqrt{i}\left(t_{i}-t_{i+1}\right)$. Let $t_{i+1}=0$, then
$$x_{1}+x_{2}+\cdots+x_{n}=\sum_{i=1}^{n} \sqrt{i}\left(t_{i}-t_{i+1}\right)=\sum_{i=1}^{n} t_{i}(\sqrt{i}-\sqrt{i-1})$$

Therefore, by $\sum_{i=1}^{n} t_{i}^{2}=1$ and the Cauchy-Schwarz inequality, we get
$$\begin{aligned}
\left(x_{1}+x_{2}+\cdots+x_{n}\right)^{2} & =\left[\sum_{i=1}^{n} t_{i}(\sqrt{i}-\sqrt{i-1})\right]^{2} \\
& \leqslant \sum_{i=1}^{n}(\sqrt{i}-\sqrt{i-1})^{2}\left(\sum_{i=1}^{n} t_{i}^{2}\right) \\
& =\sum_{i=1}^{n}(\sqrt{i}-\sqrt{i-1})^{2}
\end{aligned}$$

Equality holds if and only if $\frac{t_{1}}{1}=\frac{t_{2}}{\sqrt{2}-1}=\cdots=\frac{t_{n}}{\sqrt{n}-\sqrt{n-1}}$ and $x_{1}^{2}+x_{2}^{2}+\cdots+$ $x_{n}^{2}=1$, which uniquely determines $\left(t_{1}, t_{2}, \cdots, t_{n}\right)$ and thus uniquely determines $\left(x_{1}, x_{2}, \cdots, x_{n}\right)$. Therefore, the maximum value of $x_{1}+$ $x_{2}+\cdots+x_{n}$ is $\sqrt{\sum_{i=1}^{n}(\sqrt{i}-\sqrt{i-1})^{2}}$.