10. Let $x_{1}, x_{2}, \cdots, x_{n}$ be $n$ non-negative real numbers $\left(n>2, n \in \mathbf{N}^{*}\right)$, and $\sum_{i=1}^{n} x_{i}=n, \sum_{i=1}^{n} i x_{i}=2 n-2$, find the maximum value of the expression $S=\sum_{i=1}^{n} i^{2} x_{i}$.
10. Solution: Let $y_{i}=\sum_{j=i}^{n} x_{j}, i=1,2, \cdots, n$ then $y_{1}=n$ and $\sum_{i=1}^{n} y_{i}=2 n-2$. $S=\sum_{k=1}^{n} k^{2} x_{k}=\sum_{k=1}^{n-1} k^{2}\left(y_{k} -y_{k-1}\right)+n^{2} y_{n}=\sum_{k=1}^{n-1} k^{2} y_{k}-\sum_{k=2}^{n}(k-1)^{2} y_{k}+n^{2} y_{n}=y_{1}+\sum_{k=2}^{n-1}(2 k-1) y_{k}-(n-1)^{2} y_{n}+n^{2} y_{n}=\sum_{k=1}^{n}(2 k -1) y_{k}$.

Notice that, $y_{1}=n$, and $y_{2} \geqslant y_{3} \geqslant \cdots \geqslant y_{n}$. If there exists $i \in\{2,3, \cdots, n-1\}$ such that $y_{i}>y_{n}$ (and assume $i$ is the largest index for which this holds), then by the expression of $S$, the coefficient of $y_{i}$ is less than the coefficients of $y_{i-1}, \cdots, y_{n}$. Therefore, under the condition of keeping $y_{i}+y_{i-1}+\cdots+y_{n}$ unchanged, replacing each number with their arithmetic mean increases $S$. Hence, when $y_{2}=y_{3}=\cdots=y_{n}$, $S$ is maximized, at which point $S_{\max }=n+\frac{n-2}{n-1} \sum_{k=2}^{n}(2 k-1)=n^{2}-2$.

Note: This problem can also be solved by letting $y_{k}=\sum_{i=1}^{k} x_{2}$, and directly using Abel's transformation, we get $S=\sum_{k=1}^{n}(2 k-1) y_{k}$, at this time $y_{n}=n, y_{n-1} \geqslant y_{n-2} \geqslant \cdots \geqslant y_{1}$.