2. Let $n \geqslant 2$, find the maximum and minimum value of the product $x_{1} x_{2} \cdots x_{n}$ under the conditions: (1) $x_{i} \geqslant \frac{1}{n}, i=1,2, \cdots, n$; (2) $x_{1}^{2}+x_{2}^{2}+\cdots+$ $x_{n}^{2}=1$.
2. Solution (1) Find the minimum value of the product $x_{1} x_{2} \cdots x_{n}$.

Let $\left(x_{1}, x_{2}, \cdots, x_{n}\right)$ be any array that satisfies the conditions of the problem. Consider the new array $\left(x_{1}^{(1)}, x_{2}^{(1)}, \cdots, x_{n}^{(1)}\right)$, where
$$
x_{1}^{(1)}=x_{1}, x_{2}^{(1)}=x_{2}, \cdots, x_{n-2}^{(1)}=x_{n-2}, x_{n-1}^{(1)}=\sqrt{x_{n-1}^{2}+x_{n}^{2}-\frac{1}{n^{2}}}, x_{n}^{(1)}=\frac{1}{n} .
$$

This array clearly satisfies $x_{i}^{(1)} \geqslant \frac{1}{n}, i=1,2, \cdots, n, \sum_{i=1}^{n}\left(x_{i}^{(1)}\right)^{2}=1$.
Moreover, because $x_{n-1}^{2} x_{n}^{2}-\left(x_{n-1}^{(2)} x_{n}^{(1)}\right)^{2}=x_{n-1}^{2} x_{n}^{2}-\left(x_{n-1}^{2}+x_{n}^{2}-\frac{1}{n^{2}}\right) \frac{1}{n^{2}}=\left(x_{n-1}^{2}-\frac{1}{n^{2}}\right)\left(x_{n}^{2}-\frac{1}{n^{2}}\right) \geqslant 0$.
Thus, $x_{1} x_{2} \cdots x_{n} \geqslant x_{1}^{(1)} x_{2}^{(1)} \cdots x_{n}^{(1)}$.
Next, let $x_{1}^{(2)}=x_{1}^{(1)}, x_{2}^{(2)}=x_{2}^{(1)}, \cdots, x_{n-3}^{(2)}=x_{n-3}^{(1)}$,
$$
x_{n-2}^{(2)}=\sqrt{\left(x_{n-2}^{(1)}\right)^{2}+\left(x_{n-1}^{(1)}\right)^{2}-\frac{1}{n^{2}}}, x_{n-1}^{(2)}=x_{n}^{(2)}=\frac{1}{n} \text {. }
$$

Similarly, we get $x_{1}^{(1)} x_{2}^{(1)} \cdots x_{n}^{(1)} \geqslant x_{1}^{(2)} x_{2}^{(2)} \cdots x_{n}^{(2)}$.
Repeat this process $n-1$ times, and finally obtain the array $\left(x_{i}^{(n-1)}, \cdots, x_{n}^{(n-1)}\right)$, where
$$
x_{1}^{(n-1)}=\frac{\sqrt{n^{2}-n+1}}{n}, x_{2}^{(n-1)}=\cdots, x_{n}^{(n-1)}=\frac{1}{n} \text {. }
$$

And $x_{i}^{(n-1)} \geqslant \frac{1}{n}, i=1,2, \cdots, n \cdot \sum_{i=1}^{n}\left(x_{i}^{(n-1)}\right)^{2}=1$,
$$
x_{1} x_{2} \cdots x_{n} \geqslant x_{1}^{(n-1)} x_{2}^{(n-1)} \cdots x_{n}^{(n-1)} \geqslant \frac{\sqrt{n^{2}-n+1}}{n^{n}} .
$$

That is, for any array $\left(x_{1}, x_{2}, \cdots, x_{n}\right)$ that satisfies the conditions, we have $x_{1} x_{2} \cdots x_{n} \geqslant \frac{\sqrt{n^{2}-n+1}}{n^{n}}$.
And when $x_{1}=\frac{\sqrt{n^{2}-n+1}}{n}, x_{2}=\cdots=x_{n}=\frac{1}{n}$, the equality holds, so the minimum value is $\frac{\sqrt{n^{2}-n+1}}{n^{n}}$.
(2) By the AM-GM inequality, we have $x_{1}^{2} x_{2}^{2} \cdots x_{n}^{2} \leqslant\left(\frac{x_{1}^{2}+\cdots+x_{n}^{2}}{n}\right)^{n}=\frac{1}{n^{n}}$.
That is, when $x_{1}=x_{2}=\cdots=x_{n}=\frac{1}{\sqrt{n}}$, $\left(x_{1} x_{2} x_{3} \cdots x_{n}\right)_{\text {max }}=n^{-\frac{n}{2}}$.