Three. (20 points)
Given $x_{i} \in R^{+} \cup\{0\} . i=1,2, \cdots, n, x_{1}^{2}+x_{2}^{2}+\cdots+x_{n}^{2}=1$
Find the maximum and minimum values of $\left[n x_{1}+(n-1) x_{2}+\cdots+x_{n}\right]\left[x_{1}+2 x_{2}+\cdots+n x_{n}\right]$.
Three, (20 points)
$$
\begin{aligned}
{\left[n x_{1}+\right.} & \left.(n-1) x_{2}+\cdots+x_{n}\right]\left[x_{1}+2 x_{2}+\cdots+n x_{n}\right] \\
& \leqslant\left\{\frac{\left\lfloor n x+(n-1) x_{2}+\cdots+x_{n}\right]+\left[x_{n}+2 x_{2}+\cdots+n x_{n}\right]}{2}\right\}^{2} \\
& =\left(\frac{n+1}{2}\right)^{2} \cdot\left(x_{1}+x_{2}+\cdots+x_{n}\right)^{2} \\
& \leqslant\left(\frac{n+1}{2}\right)^{2} \cdot n \cdot\left(x_{1}^{2}+x_{2}^{2}+\cdots+x_{n}^{2}\right) \\
& =\frac{(n+1)^{2} n}{4}
\end{aligned}
$$

When $x_{1}=x_{2}=\cdots=x_{n}=\sqrt{\frac{1}{n}}$, the maximum value is achieved.
Also, $2 \cdot(n-1)>n \quad 3 \cdot(n-2)>n \cdots k(n-k+1)>n$
So $\left[n x_{1}+(n-1) x_{2}+\cdots+x_{n}\right]\left[x_{1}+2 T_{2}+\cdots+n T_{n}\right]$
$$
\begin{array}{l}
\geqslant_{n}\left(x_{1}^{2}+x_{2}^{2}+\cdots+x_{n}^{2}\right) \\
=n
\end{array}
$$

When $x_{1}=1, x_{2}=\cdots=x_{n}=0$ or $x_{n}=1, x_{1}=x_{2}=\cdots=x_{n-1}=0$, the minimum value is achieved.