In the 30th IMO training questions, there is a problem:
For positive numbers $x 、 y 、 z$ satisfying $x^{2}+y^{2}+z^{2}=1$, find the minimum value of $\frac{x}{1-x^{2}}+\frac{y}{1-y^{2}}+\frac{z}{1-z^{2}}$.
Given $a_{i} \in \mathbf{R}^{+}(i=1,2, \cdots, n, n \geqslant 3)$, $\sum_{i=1}^{n} a_{i}^{n-1}=1$. Then $\sum_{i=1}^{n} \frac{a_{i}^{n-2}}{1-a_{i}^{n-1}} \geqslant \frac{n}{n-1} \sqrt[n-1]{n}$.

Inspired by this, the author found that it can be further generalized to:

Given $a_{i} \in \mathbf{R}^{+}(i=1,2, \cdots, n, n \geqslant 3), \alpha_{1}$. $\alpha_{2}, k \in \mathbf{N}, c > a_{1}^{k a_{2}}$, and
$$
\begin{array}{l}
\sum_{i=1}^{n} a_{i}^{a_{1}+a_{2}}=n\left(\frac{c}{k+1}\right)^{\frac{a_{1}+a_{2}}{k a_{2}}} . \text { Then } \\
\sum_{i=1}^{n} \frac{a_{i}^{a_{1}}}{c-a_{i}^{k a_{2}}} \geqslant \frac{n}{k}\left(\frac{c}{k+1}\right)^{\frac{a_{1}-k a_{2}}{k a_{2}}} .
\end{array}
$$

Proof: Let $x_{1}=a_{i}^{a_{2}}\left(c-a_{i}^{k a_{2}}\right)$, then
$$
\begin{aligned}
x_{i}^{k} & =a_{i}^{k a_{2}}\left(c-a_{i}^{k a_{2}}\right)^{k} \\
& =\frac{1}{k}\left(k a_{i}^{k a_{2}}\right)\left(c-a_{i}^{k a_{2}}\right) \cdots\left(c-a_{i}^{k a_{2}}\right) \\
& \leqslant \frac{1}{k}\left[\frac{k a_{i}^{k a_{2}}+\left(c-a_{i}^{k a_{2}}\right)+\cdots+\left(c-a_{i}^{k a_{2}}\right)}{k+1}\right]^{k+1} \\
& =\frac{1}{k}\left(\frac{k c}{k+1}\right)^{k+1} .
\end{aligned}
$$

Then $x_{i} \leqslant\left[\frac{1}{k}\left(\frac{k c}{k+1}\right)^{k+1}\right]^{\frac{1}{k}}=k\left(\frac{c}{k+1}\right)^{\frac{k+1}{k}}$,
$$
\begin{array}{c}
\frac{a_{i}^{a_{1}}}{c-a_{i}^{k a_{2}}}=\frac{a_{i}^{a_{1}+a_{2}}}{a_{i}^{a_{2}}\left(c-a_{i}^{k a_{2}}\right)} \\
=\frac{a_{i}^{\alpha_{1}+a_{2}}}{x_{i}} \geqslant \frac{a_{i}^{a_{1}+a_{2}}}{k\left(\frac{c}{k+1}\right)^{\frac{k+1}{k}}} .
\end{array}
$$

Noting that $\sum_{i=1}^{n} a_{i}^{\alpha_{1}+a_{2}}=n\left(\frac{c}{k+1}\right)^{\frac{a_{1}+a_{2}}{k a_{2}}}$, we get the desired result.
Reference:
[1] An Zhenping. A Generalization of an Extremal Problem. Middle School Mathematics, 1996(4)