The ninth question: Given a positive integer $n$ greater than 1, let $x_{1}, x_{2}, \ldots, x_{n}$ be real numbers, and $\sum_{i=1}^{n} x_{i}^{2}=1$. Try to find the maximum value of $\sum_{\mathrm{k}=1}^{\mathrm{n}} \mathrm{kx}_{\mathrm{k}}{ }^{2}+\sum_{1 \leq \mathrm{i}<\mathrm{j} \leq \mathrm{n}}(\mathrm{i}+\mathrm{j}) \mathrm{x}_{\mathrm{i}} \mathrm{x}_{\mathrm{j}}$.
For any positive real number $\lambda$, we have
$$
\begin{array}{l}
\sum_{k=1}^{n} k x_{k}^{2}+\sum_{1 \leq i<j \leq n}(i+j) x_{i} x_{j}=\left(\sum_{i=1}^{n} x_{i}\right)\left(\sum_{i=1}^{n} i x_{i}\right)=\frac{1}{\lambda}\left(\sum_{i=1}^{n} \lambda x_{i}\right)\left(\sum_{i=1}^{n} i x_{i}\right) \\
\leq \frac{\left(\sum_{i=1}^{n}(\lambda+i) x_{i}\right)^{2}}{4 \lambda} \leq \frac{\left(\sum_{i=1}^{n}(\lambda+i)^{2}\right)\left(\sum_{i=1}^{n} x_{i}^{2}\right)}{4 \lambda}=\frac{\sum_{i=1}^{n}(\lambda+i)^{2}}{4 \lambda}=\frac{n \lambda^{2}+n(n+1) \lambda+\frac{n(n+1)(2 n+1)}{6}}{4 \lambda}
\end{array}
$$

We take $\lambda=\sqrt{\frac{(n+1)(2 n+1)}{6}}$, then according to (*), we have
$$
\sum_{k=1}^{n} k x_{k}^{2}+\sum_{1 \leq i<j \leq n}(i+j) x_{i} x_{j} \leq \frac{n(n+1) \sqrt{\frac{(n+1)(2 n+1)}{6}}+\frac{2 n(n+1)(2 n+1)}{6}}{\sqrt[4]{\frac{(n+1)(2 n+1)}{6}}}=\frac{n(n+1)}{4}+\frac{n}{2} \sqrt{\frac{(n+1)(2 n+1)}{6}}
$$

According to the equality condition of the Cauchy-Schwarz inequality, the equality holds if and only if $x_{i}=\left(i+\sqrt{\frac{(n+1)(2 n+1)}{6}}\right) \sqrt{\frac{1}{\sum_{i=1}^{n}\left(\sqrt{\frac{(n+1)(2 n+1)}{6}}+i\right)^{2}}}(1 \leq i \leq n)$.