The 17th question: Given an integer $n \geq 2$, it is known that real numbers $x_{1}, x_{2}, \ldots, x_{n}$ satisfy $\sum_{i=1}^{n} i x_{i}=1$,
(1) Find the minimum value of $\sum_{k=1}^{n} x_{k}^{2}+\sum_{1 \leq i<j \leq n} x_{i} x_{j}$;
(2) Find the minimum value of $\sum_{k=1}^{n} x_{k}^{2}+\sum_{1 \leq i<j \leq n}\left(x_{i}-x_{j}\right)^{2}$.
Question 17:
Solution: (1) Let $\mathrm{S}=\sum_{\mathrm{k}=1}^{\mathrm{n}} \mathrm{x}_{\mathrm{k}}^{2}+\sum_{1 \leq \mathrm{i}<\mathrm{j} \leq \mathrm{n}} \mathrm{x}_{\mathrm{i}} \mathrm{x}_{\mathrm{j}}$, then $2 \mathrm{~S}=\sum_{\mathrm{k}=1}^{\mathrm{n}} \mathrm{x}_{\mathrm{k}}^{2}+\left(\sum_{\mathrm{k}=1}^{\mathrm{n}} \mathrm{x}_{\mathrm{k}}\right)^{2}$. According to the condition, for any $\alpha \in \mathbb{R}$, we have
$$
\begin{array}{l}
1=\sum_{k=1}^{n}(k-\alpha) x_{k}+\alpha\left(\sum_{k=1}^{n} x_{k}\right) \\
\leq \sqrt{\left[\sum_{k=1}^{n}(k-\alpha)^{2}+\alpha^{2}\right]\left(\sum_{k=1}^{n} x_{k}^{2}+\left(\sum_{k=1}^{n} x_{k}\right)^{2}\right)} \\
=\sqrt{\left[(n+1) \alpha^{2}-n(n+1) \alpha+\frac{n(n+1)(2 n+1)}{6}\right] \cdot 2 S}
\end{array}
$$

Taking $\alpha=\frac{n}{2}$ in (1), we get $S \geq \frac{6}{n(n+1)(n+2)}$. According to the equality condition of the Cauchy-Schwarz inequality, when $\frac{x_{1}}{1-\frac{n}{2}}=\frac{x_{2}}{2-\frac{n}{2}}=\ldots=$ $\frac{x_{n}}{n-\frac{n}{2}}=\frac{\sum_{k=1}^{n} x_{k}}{\frac{1}{2}}$, i.e., $x_{k}=\frac{6(2 k-n)}{n(n+1)(n+2)}(1 \leq k \leq n)$, the equality holds.
(2) According to the condition, for any real number $\alpha$, we have
$$
\begin{array}{l}
1=\sum_{i=1}^{n} i x_{i}=\sum_{i=1}^{n} i x_{j}-\sum_{1 \leq i<j \leq n} \alpha(i-j)\left(x_{i}-x_{j}\right)+\sum_{1 \leq i<j \leq n} \alpha(i-j)\left(x_{i}-x_{j}\right) \\
=\sum_{i=1}^{n} i x_{1}-\sum_{i=1}^{n} \alpha x_{i}\left[\sum_{j=1}^{n}(i-j)\right]+\sum_{1 \leq i<j \leq n} \alpha(i-j)\left(x_{i}-x_{j}\right) \\
=\sum_{i=1}^{n} i x_{i}-\sum_{i=1}^{n} \alpha x_{i}\left[n i-\frac{n(n+1)}{2}\right]+\sum_{1 \leq i<j \leq n} \alpha(i-j)\left(x_{i}-x_{j}\right) \\
=\sum_{i=1}^{n}\left[i+\left(\frac{n(n+1)}{2}-n i\right) \alpha\right] x_{i}+\sum_{1 \leq i<j \leq n} \alpha(i-j)\left(x_{i}-x_{j}\right)
\end{array}
$$

According to the Cauchy-Schwarz inequality, we know:
$$
\begin{array}{l}
1=\left\{\sum_{i=1}^{n}\left[i+\left(\frac{n(n+1)}{2}-n i\right) \alpha\right] x_{i}+\sum_{1 \leq i<j \leq n} \alpha(i-j)\left(x_{i}-x_{j}\right)\right\}^{2} \\
\leq\left\{\sum_{i=1}^{n}\left[i+\left(\frac{n(n+1)}{2}-n i\right) \alpha\right]^{2}+\sum_{1 \leq i<j \leq n}[\alpha(i-j)]^{2}\right\}\left[\sum_{i=1}^{n} x_{i}^{2}+\sum_{1 \leq i<j \leq n}\left(x_{i}-x_{j}\right)^{2}\right]
\end{array}
$$

Taking $\alpha=\frac{1}{n+1}$ in (2), we get
$$
\begin{array}{l}
1 \leq \frac{n(n+2)(3 n+1)}{n+1}\left[\sum_{i=1}^{n} x_{i}^{2}+\sum_{1 \leq i<j \leq n}\left(x_{i}-x_{j}\right)^{2}\right] \\
\Rightarrow \sum_{i=1}^{n} x_{i}^{2}+\sum_{1 \leq i<j \leq n}\left(x_{i}-x_{j}\right)^{2} \geq \frac{12}{n(n+2)(3 n+1)}
\end{array}
$$

According to the equality condition of the Cauchy-Schwarz inequality, the equality holds if and only if $x_{i}=\left(i+\frac{n(n+1)}{2}\right) m(1 \leq i \leq n)$, where
$$
m=\frac{12}{n(n+1)(n+2)(3 n+1)} .
$$