Example 3. On the lower base $AB$ of trapezoid $ABCD$, there are two fixed points $M, N$, and on the upper base $CD$, there is a moving point $P$. Let $E = DN \cap AP, F = DN \cap MC, G = MC \cap PB$, and $DP = \lambda DC$. For what value of $\lambda$ is the area of quadrilateral $PEFG$ maximized? (88 China Training Team Selection Test)
Solve: Because
$$
S_{\mathrm{PE}}{ }^{\mathbb{F G}}=S_{\mathrm{ABP}}-S_{\mathrm{ANE}}-S_{\mathrm{MBG}}+S_{\mathrm{MNE}},
$$

and among them, $S_{A B P}$ and $S_{\mathrm{ANF}}$ are constants, so $S_{\mathrm{PEFG}}$ is maximized if and only if $S_{\mathrm{ANE}}+S_{\mathrm{MBG}}$ takes the minimum value.

Let $A B=a, C D=b, M N=c$, and set $A N=\mu(a+c)$, thus $M B=(1-\mu)(a+c)$. Suppose the height of the trapezoid is 1, it is easy to see that
$$
\begin{aligned}
S_{A N E} & =\frac{1}{2} \frac{A N^{2}}{A N+D P}=\frac{1}{2} \frac{\mu^{2}(a+c)^{2}}{\mu(a+c)+\lambda b}, \\
S_{\mathrm{MBG}} & =\frac{1}{2} \frac{M B^{2}}{M B+P C} \\
& =\frac{1}{2} \frac{(1-\mu)^{2}(a+c)^{2}}{(1-\mu)(a+c)+(1-\lambda) b}
\end{aligned}
$$

Thus, we have
$$
\begin{array}{l}
S_{\mathrm{ANE}}+S_{\mathrm{MBG}} \\
=\frac{1}{2}(a+c)^{2}\left[\frac{\mu^{2}}{\mu(a+c)+\lambda b}\right. \\
\left.+\frac{(1-\mu)^{2}}{(1-\mu)(a+c)+(1-\lambda) b}\right] .
\end{array}
$$

By the Cauchy-Schwarz inequality,
$$
\begin{array}{l}
\frac{\mu^{2}}{\mu(a+c)+\lambda b}+\frac{(1-\mu)^{2}}{(1-\mu)(a+c)+(1-\lambda) b} \\
=\left[\frac{\mu^{2}}{\mu(a+c)+\lambda b}\right. \\
\left.+\frac{(1-\mu)^{2}}{(1-\mu)(a+c)+(1-\lambda) b}\right] \\
\cdot[\mu(a+c)+\lambda b+(1-\mu)(a+c) \\
+(1-\lambda) b] \frac{1}{a+b+c} \geqslant(\mu+1-\mu) \frac{1}{a+b+c} \\
=\frac{1}{a+b+c} .
\end{array}
$$

Combining (1) and (2), we get
$$
S_{\mathrm{ANE}}+S_{\mathrm{MBG}} \geqslant \frac{1}{2} \frac{(a+c)^{2}}{a+b+c} \text { (constant). }
$$

The equality holds if and only if the equality in (2) holds, which is true if and only if
$$
\frac{\mu}{\mu(a+c)+\lambda b}=\frac{1-\mu}{(1-\mu)(a+c)+(1-\lambda) b} .
$$

Solving this, we get $\lambda=\mu$, i.e., when $\lambda=\mu=A N:(A B+M N)$, $S_{\mathrm{PEFG}}$ takes the maximum value.