The second question: Let $x_{i} \geqslant 0(i=1,2, \cdots, n)$, and $\sum_{i=1}^{n} x_{i}^{2}+2 \sum_{1 \leqslant k<j \leqslant n} \sqrt{\frac{k}{j}} x_{k} x_{j}=1$. Find the maximum and minimum values of $\sum_{i=1}^{n} x_{i}$.
Solution: The conclusion is: the minimum value of $\sum_{i=1}^{n} x_{i}$ is 1, and the maximum value is $\sqrt{\sum_{i=1}^{n}(\sqrt{i}-\sqrt{i-1})^{2}}$.
$$
\text { By } \begin{aligned}
1 & =\sum_{i=1}^{n} x_{i}^{2}+2 \sum_{1 \leqslant k<j<n} \sqrt{\frac{k}{j}} x_{k} x_{j} \\
& =\left(\sum_{i=1}^{n} x_{i}\right)^{2}-2 \sum_{1 \leqslant k<j \leqslant n}\left(1-\sqrt{\frac{k}{j}}\right) x_{k} x_{j} \\
& \leqslant\left(\sum_{i=1}^{n} x_{i}\right)^{2},
\end{aligned}
$$

Therefore, by $x_{i} \geqslant 0$ we know $\sum_{i=1}^{n} x_{i} \geqslant 1$, and the equality holds when all $x_{k} x_{j}(k<j)$ are zero, which means one of the $x_{i}$ is 1 and the rest are zero. Thus, the minimum value of $\sum_{i=1}^{n} x_{i}$ is 1.
$$
\begin{array}{l} 
\text { Also, } 1= \sum_{i=1}^{n} i\left(\frac{x_{i}}{\sqrt{i}}\right)^{2}+2 \sum_{1 \le k<j \leqslant n} k\left(\frac{x_{k}}{\sqrt{k}} \cdot \frac{x_{j}}{\sqrt{j}}\right) \\
= \sum_{i=1}^{n} \frac{x_{i}}{\sqrt{i}}\left(\sum_{j=1}^{i} j \frac{x_{j}}{\sqrt{j}}+i \sum_{j=i+1}^{n} \frac{x_{j}}{\sqrt{j}}\right) \\
=\sum_{i=1}^{n}\left(\sum_{j=i}^{n} \frac{x_{j}}{\sqrt{j}}\right)^{2} . \\
\text { Let } y_{i}=\sum_{j=i}^{n} \frac{x_{j}}{\sqrt{j}}, i=1,2, \cdots, n \text {, then all } y_{i} \geqslant
\end{array}
$$
0, and the condition becomes $\sum_{i=1}^{n} y_{i}^{2}=1$.
By $y_{i}=\sum_{j=i}^{n} \frac{x_{j}}{\sqrt{j}}, i=1,2, \cdots, n$, we have
$$
x_{n}=\sqrt{n} y_{n}, x_{i}=\sqrt{i}\left(y_{i}-y_{i+1}\right),
$$
$i=1,2, \cdots, n-1$. Let $y_{n+1}=0$, hence we have
$$
\begin{array}{l}
\sum_{i=1}^{n} x_{i}=\sum_{i=1}^{n} \sqrt{i}\left(y_{i}-y_{i+1}\right) \\
=\sum_{i=1}^{n} \sqrt{i y_{i}}-\sum_{i=1}^{n} \sqrt{i} y_{i+1} \\
=\sum_{i=1}^{n} \sqrt{i} y_{i}-\sum_{i=1}^{n} \sqrt{i-i} y_{i} \\
=\sum_{i=1}^{n}(\sqrt{i}-\sqrt{i-1}) y_{i} .
\end{array}
$$

By the Cauchy inequality, we have
$$
\begin{array}{l}
\left(\sum_{i=1}^{n} x_{i}\right)^{2} \leqslant \sum_{i=1}^{n}(\sqrt{i}-\sqrt{i-1})^{2} \cdot \sum_{i=1}^{n} y_{i}^{2} \\
=\sum_{i=1}^{n}(\sqrt{i}-\sqrt{i-1})^{2} . \\
\text { Thus, } \sum_{i=1}^{n} x_{i} \leqslant \sqrt{\sum_{i=1}^{n}(\sqrt{i}-\sqrt{i-1})^{2}},
\end{array}
$$

and the equality holds when there exists a constant $\lambda$ such that $y_{i} = \lambda(\sqrt{i}-\sqrt{i-1})$ for $i=1,2, \cdots, n$.
Substituting into $\sum_{i=1}^{n} y_{i}^{2}=1$ we can solve for
$$
\lambda=\frac{1}{\sqrt{\sum_{i=1}^{n}(\sqrt{i}-\sqrt{i-1})^{2}}} \text {. }
$$

Thus, $y_{1}=\frac{\sqrt{i}-\sqrt{i-1}}{\sqrt{\sum_{i=1}^{n}(\sqrt{i}-\sqrt{i-1})^{2}}}, i=1,2$,
$\cdots, n$. Therefore,
$$
\begin{array}{l}
x_{i}=\sqrt{i}\left(y_{i}-y_{i+1}\right) \\
=\frac{2 i-\sqrt{i^{2}+i}-\sqrt{i^{2}-i}}{\sqrt{\sum_{i=1}^{n}(\sqrt{i}-\sqrt{i-1})^{2}}} .
\end{array}
$$

Hence, when $x_{i}=\frac{2 i-\sqrt{i^{2}+i}-\sqrt{i^{2}-i}}{\sqrt{\sum_{i=1}^{n}(\sqrt{i}-\sqrt{i-1})^{2}}}$,
$\sum_{i=1}^{n} x_{i}$ achieves its maximum value $\sqrt{\sum_{i=1}^{n}(\sqrt{i}-\sqrt{i-1})^{2}}$.