20. Let $n(n \geqslant 3)$ be a positive integer, and let positive numbers $\alpha$ and $\beta$ satisfy $\frac{\alpha}{\beta}=\frac{n-1}{n-2}$. Also, $x_{i} \in \mathbf{R}_{+}(i=1,2, \cdots, n)$, and $\sum_{i=1}^{n} x_{i}^{\alpha}=1$, find the minimum value of $\sum_{i=1}^{n} \frac{x_{i}^{\beta}}{1-x_{i}^{\alpha}}$.
20. Let $y_{i}=x_{i}^{\sigma-\beta}\left(1-x_{i}^{a}\right)$.

By the weighted arithmetic-geometric mean inequality, we have $y_{i}^{\alpha}=\left(\frac{\alpha-\beta}{\alpha}\right)^{\alpha-\beta}\left(\frac{\alpha}{\alpha-\beta} x_{i}^{\alpha}\right)^{\alpha \beta}\left(1-x_{i}^{\alpha}\right)^{\circ}$
$$
\leqslant\left(\frac{\alpha-\beta}{\alpha}\right)^{\alpha-\beta}\left[\frac{(\alpha-\beta) \cdot \frac{\alpha}{\alpha-\beta} x_{i}^{o}+\alpha\left(1-x_{i}^{q}\right)}{(\alpha-\beta)+\alpha}\right]^{2-\beta}=\left(\frac{\alpha-\beta}{\alpha}\right)^{\alpha-\beta}\left(\frac{\alpha}{2 \alpha-\beta}\right)^{2 \alpha-\beta} \text {. }
$$

Then $y_{i} \leqslant\left(\frac{\alpha-\beta}{\alpha}\right)^{\frac{2-\beta}{\alpha}}\left(\frac{\alpha}{2 \alpha-\beta}\right)^{\frac{2 \alpha-\beta}{\alpha}}=\left(\frac{1}{n-1}\right)^{\frac{1}{n-1}}\left(\frac{n-1}{n}\right)^{\frac{n}{n-1}}=(n-1) n^{-\frac{n}{n-1}}$, hence $\sum_{i=1}^{n} \frac{x_{i}^{\beta}}{1-x_{i}^{a}}=\sum_{i=1}^{n} \frac{x_{i}^{a}}{x_{i}^{-\beta}\left(1-x_{i}^{a}\right)} \geqslant(n-1)^{-1} \cdot n_{n-1}^{n-1} \cdot \sum_{i=1}^{n} x_{i}^{a}=\frac{n}{n-1} \cdot \sqrt[n-1]{n}$. The equality holds when $\frac{\alpha}{\alpha-\beta} x_{i}^{o}=1-x_{i}^{\alpha}$, i.e., $x_{i}^{\alpha}=\frac{\alpha-\beta}{2 \alpha-\beta}=\frac{1}{n}$, which means $x_{i}=\left(\frac{1}{n}\right)^{\frac{1}{\alpha}}$. Therefore, the minimum value of $\sum_{i=1}^{n} \frac{x_{i}^{\beta}}{1-x_{i}^{\beta}}$ is $\frac{n}{n-1} \cdot \sqrt[n-1]{n}$.