Example 1.3.1 (Shanghai 2000) As shown in the figure, $ABCD$ is a square with a side length of 1, $U, V$ are points on $AB, CD$ respectively, $AV \cap DU=P, BV \cap CU=Q$, find the maximum value of the area $S_{PUQV}$.
Analyzing the problem, let $A U=a, D V=b$, expressing $S_{P V Q V}$ as a function of $a$ and $b$ is one direction to solve the problem. The right trapezoids $A U V D$ and $B U V C$ are two basic figures. Draw $P E \perp A D, Q F \perp B C$, and let $P E=x, Q F=y$. $x$ and $y$ can serve as intermediary variables in the process of calculating the area of quadrilateral $P U Q V$.

Solution: Connect $U V$, draw $P E \perp A D, Q F \perp B C$. Studying the basic figure, the right trapezoid $A U V D$, we get: $S_{U P V}=S_{A P D}=\frac{1}{2} x$, here let $P E=x, Q F=y, A U=a, D V=b$. Then $S_{P C Q V}=\frac{1}{2}(x+y)$.
$$
\text { and } \frac{x}{a}+\frac{x}{b}=D E+A E=1, x=\frac{a b}{a+b} \text {. }
$$

Similarly, $y=\frac{(1-a)(1-b)}{(1-a)+(1-b)}$.
$$
\begin{aligned}
\therefore S_{\text {PUQV }} & =\frac{1}{2}\left[\frac{a b}{a+b}+\frac{(1-a)(1-b)}{2-a-b}\right]=\frac{2(a+b)-2 a^{2}-2 b^{2}}{4(a+b)(2-a-b)} \\
& \leqslant \frac{2(a+b)-a^{2}-b^{2}-2 a b}{4(a+b)(2-a-b)}=\frac{1}{4} .
\end{aligned}
$$

The equality holds if and only if $a=b$.
Explanation: By studying the basic figures, we can also get: $S_{P U Q V}=\frac{1}{2}\left[1-S_{1}-S_{2}-S_{3}-S_{4}\right]$.
Studying the extreme case when $U$ and $V$ are at the midpoints of the sides, we get $S_{P \cup Q V}=\frac{1}{4}$, which suggests that this might be the maximum value of $S$.

Thus, at this step: $\frac{2(a+b)-2 a^{2}-2 b^{2}}{4(a+b)(2-a-b)}$, we can use the method of approximation to ensure that the numerator is no greater than one-fourth of the denominator, i.e., to prove: $2(a+b)-2 a^{2}-2 b^{2} \leqslant(a+b)(2-a-b)$.