$9 \cdot 55$ Let $n \geqslant 2$, find the maximum and minimum value of the product $x_{1} x_{2} \cdots x_{n}$ under the conditions $x_{i} \geqslant \frac{1}{n}, i=1,2, \cdots$, $n$, and $x_{1}^{2}+x_{2}^{2}+\cdots+x_{n}^{2}=1$.
[Solution] First, find the maximum value. By the AM-GM inequality, we have
$$\sqrt[n]{x_{1}^{2} \cdot x_{2}^{2} \cdot \cdots \cdot x_{n}^{2}} \leqslant \frac{x_{1}^{2}+x_{2}^{2}+\cdots+x_{n}^{2}}{n}=\frac{1}{n}$$

and the equality holds when $x_{1}=x_{2}=\cdots=x_{n}=\frac{1}{\sqrt{n}}>\frac{1}{n}(n \geqslant 2)$. Therefore, the maximum value sought is $n^{-\frac{n}{2}}$.

Next, find the minimum value. Take any $x_{1}, x_{2}, \cdots, x_{n}$ that satisfy the conditions of the problem. Let
$$y_{1}=x_{1}, \cdots, y_{n-2}=x_{n-2}, y_{n-1}=\sqrt{x_{n-1}^{2}+x_{n}^{2}-\frac{1}{n^{2}}}, y_{n}=\frac{1}{n},$$

then $y_{i} \geqslant \frac{1}{n}, i=1,2, \cdots, n$, and $y_{1}^{2}+\cdots+y_{n}^{2}=x_{1}^{2}+\cdots+x_{n}^{2}=1$, i.e., $y_{1}$, $y_{2}, \cdots, y_{n}$ satisfy the conditions of the problem. Furthermore, since
$$\begin{aligned}
y_{n-1}^{2} y_{n}^{2}-x_{n-1}^{2} x_{n}^{2} & =\frac{1}{n^{2}}\left(x_{n-1}^{2}+x_{n}^{2}-\frac{1}{n^{2}}\right)-x_{n-1}^{2} x_{n}^{2} \\
& =-\left(x_{n-1}^{2}-\frac{1}{n^{2}}\right)\left(x_{n}^{2}-\frac{1}{n^{2}}\right) \leqslant 0
\end{aligned}$$

it follows that $y_{1} y_{2} \cdots y_{n-2} y_{n-1} y_{n} \leqslant x_{1} x_{2} \cdots x_{n-2} x_{n-1} x_{n}$. Repeating this process $n-1$ times, we get
$$\begin{aligned}
x_{1} x_{2} \cdots x_{n} & \geqslant\left(\frac{1}{n}\right)^{n-1} \sqrt{1-\frac{n-1}{n^{2}}} \\
& =\frac{\sqrt{n^{2}-n+1}}{n^{n}}
\end{aligned}$$

Clearly, when $x_{1}=x_{2}=\cdots=x_{n-1}=\frac{1}{n}, x_{n}=\frac{\sqrt{n^{2}-n+1}}{n}$, the equality in the above inequality holds. Therefore, under the conditions of the problem, the minimum value of the product $x_{1} x_{2} \cdots x_{n}$ is
$$\frac{\sqrt{n^{2}-n+1}}{n^{n}}$$