11. (15 points) Let $x_{1}, x_{2}, \cdots, x_{n} \in \mathbf{R}^{+}$, define $S_{n} = \sum_{i=1}^{n}\left(x_{i}+\frac{n-1}{n^{2}} \frac{1}{x_{i}}\right)^{2}$,
(1) Find the minimum value of $S_{n}$;
(2) Under the condition $x_{1}^{2}+x_{2}^{2}+\cdots+x_{n}^{2}=1$, find the minimum value of $S_{n}$;
(3) Under the condition $x_{1}+x_{2}+\cdots+x_{n}=1$, find the minimum value of $S_{n}$, and provide a proof.
11. Solution: (1) $S_{n} \geqslant \sum_{i=1}^{n}\left(2 \sqrt{\frac{n-1}{n^{2}}}\right)^{2}=4 \sum_{i=1}^{n} \frac{n-1}{n^{2}}=4 \frac{n-1}{n}$.
(When $x_{i}=\frac{\sqrt{n-1}}{n}$, the minimum value is achieved)
(2) $S_{n}=\sum_{i=1}^{n}\left(x_{i}^{2}+2 \frac{n-1}{n^{2}}+\frac{(n-1)^{2}}{n^{4}} \frac{1}{x_{i}^{2}}\right)$
$=1+2 \frac{n-1}{n}+\frac{(n-1)^{2}}{n^{4}} \sum_{i=1}^{n} \frac{1}{x_{i}^{2}} \geqslant 1+2 \frac{n-1}{n}+$ $\frac{(n-1)^{2}}{n^{2}}=\left(1+\frac{n-1}{n}\right)^{2}$.
(When $x_{1}=x_{2}=\cdots=x_{n}=\frac{1}{\sqrt{n}}$, the minimum value $\left(1+\frac{n-1}{n}\right)^{2}$ is achieved)
(3) Since
$$
\begin{array}{l}
{\left[\sum_{i=1}^{n} 1 \cdot\left(x_{i}+\frac{n-1}{n^{2}} \frac{1}{x_{i}}\right)\right]^{2} \leqslant\left(\sum_{i=1}^{n} 1^{2}\right) \cdot \sum_{i=1}^{n}} \\
\left(x_{i}+\frac{n-1}{n^{2}} \frac{1}{x_{i}}\right)^{2} . \\
\text { Therefore, } S_{n}=\sum_{i=1}^{n}\left(x_{i}+\frac{n-1}{n^{2}} \frac{1}{x_{i}}\right)^{2} \geqslant \frac{1}{n} \\
{\left[\sum_{i=1}^{n}\left(x_{i}+\frac{n-1}{n^{2}} \frac{1}{x_{i}}\right)\right]^{2} \geqslant \frac{1}{n}\left[1+\frac{n-1}{n^{2}} \cdot n^{2}\right]^{2}=n .}
\end{array}
$$

Therefore, $S_{n}=\sum_{i=1}^{n}\left(x_{i}+\frac{n-1}{n^{2}} \frac{1}{x_{i}}\right)^{2} \geqslant \frac{1}{n}$
(When $x_{1}=x_{2}=\cdots=x_{n}=\frac{1}{n}$, the minimum value $n$ is achieved)