Think about Question 1: Let real numbers $x, y$ satisfy $-1<x<1, -1<y<1$, try to compare $\frac{1}{1-x^{2}}+\frac{1}{1-y^{2}}$ with $\frac{2}{1-xy}$.
Analyzing the structure of the given two expressions, by analogy with the formula for the sum of an infinite geometric series $S=\frac{a_{1}}{1-q}=a_{1}+a_{1} q+\cdots+a_{1} q^{n-1}+\cdots(|q|<1)$, a quick and concise solution can be obtained.
Solution Since $|x|<1,|y|<1$, it follows that $x^{2}<1, y^{2}<1$, hence $\frac{1}{1-x^{2}}=1+x^{2}+x^{1}+\cdots$,
$$
\begin{aligned}
\frac{1}{1-y^{2}}=1+y^{2} & +y^{1}+\cdots, \\
\frac{1}{1-x^{2}}+\frac{1}{1-y^{2}} & =\left(1+x^{2}+x^{4}+x^{6}+\cdots\right)+\left(1+y^{2}+y^{4}+y^{6}+\cdots\right) \\
& =2+\left(x^{2}+y^{2}\right)+\left(x^{4}+y^{1}\right)+\left(x^{6}+y^{6}\right)+\cdots \\
& \geqslant 2+2 x y+2 x^{2} y^{2}+2 x^{3} y^{3}+\cdots=\frac{2}{1-x y} .
\end{aligned}
$$

Explanation (1) Similar to the above solution, the conclusion can be generalized to a more general case:
Let $0<a_{i}<1\left(i=1,2, \cdots, n, n \in \mathbf{N}^{*}\right)$, then $\sum_{i=1}^{n} \frac{1}{1-a_{i}^{n}} \geqslant \frac{n}{1-a_{1} a_{2} \cdots a_{n}}$.
(2) For some fractional inequality problems, if each term appears in the form of $\frac{a}{b \pm c}$ or $\frac{a^{2}}{b \pm c}$, if these terms can be appropriately transformed, or if appropriate auxiliary quantities are introduced, they can be transformed into the form $\frac{a_{1}}{1-q}(|q|<1)$, then the method introduced above can be considered for solving.
However, if the minimum value is to be found and $x \neq y$, please refer to the following problem:
(2003 National High School Mathematics Competition) Given $x 、 y \in(-2,2)$, and $x y=-1$, find the minimum value of the function $u=\frac{4}{4-x^{2}}+$ $\frac{9}{9-y^{2}}$.
Solution Since $x 、 y \in(-2,2)$, it follows that $\left(\frac{x}{2}\right)^{2}<1,\left(\frac{y}{3}\right)^{2}<1$.
Since $x y=-1$, we have
$$
\begin{aligned}
u & =\frac{4}{4-x^{2}}+\frac{9}{9-y^{2}}=\frac{1}{1-\left(\frac{x}{2}\right)^{2}}+\frac{1}{1-\left(\frac{y}{3}\right)^{2}}=\sum_{k=1}\left(\frac{x}{2}\right)^{2 k}+\sum_{k=0}\left(\frac{y}{3}\right)^{2 k} \\
& =\sum_{k=0}\left[\left(\frac{x}{2}\right)^{2 k}+\left(\frac{y}{3}\right)^{2 k}\right] \geqslant 2 \sum_{k=1}\left|\left(\frac{x}{2}\right)^{k}\right|\left|\left(\frac{y}{3}\right)^{k}\right| \\
& =2 \sum_{k=0} \frac{1}{6^{k}}=2 \cdot \frac{1}{1-\frac{1}{6}}=\frac{12}{5} .
\end{aligned}
$$

The function $u$ achieves its minimum value $\frac{12}{5}$ when and only when $\left|\frac{x}{2}\right|=\left|\frac{y}{3}\right|$. At this time, solving
$$
\left\{\begin{array}{l}
\left|\frac{x}{2}\right|=\left|\frac{y}{3}\right|, \\
x y=-1,
\end{array}\right.
$$

yields $\left\{\begin{array}{l}x=\frac{\sqrt{6}}{3}, \\ y=-\frac{\sqrt{6}}{2},\end{array}\right.$ or $\quad\left\{\begin{array}{l}x=-\frac{\sqrt{6}}{3}, \\ y=\frac{\sqrt{6}}{2} .\end{array}\right.$