One, (50 points) Let $D$ be a moving point on side $BC$ of a given $\triangle ABC$, and $P$ divides $AD$ in a fixed ratio $\lambda$. Let $BP$ intersect $CD$ at $E$, and $CP$ intersect $AB$ at $F$. Find the maximum value of the area of $\triangle AEF$.

---

Translate the above text into English, preserving the original text's line breaks and formatting. Directly output the translation result.
$$
\text { I. Let } \frac{A F}{F B}=u, \frac{A E}{E C}=v \text {. }
$$

By Ceva's Theorem, we have
$$
\begin{array}{l}
\frac{B D}{D C} \cdot \frac{C E}{E A} \cdot \frac{A F}{F B}=1 \\
\Rightarrow \frac{B D}{D C}=\frac{E A}{C E} \cdot \frac{F B}{A F}=\frac{v}{u} \\
\Rightarrow \frac{B C}{D C}=\frac{C D+D B}{D C}=1+\frac{v}{u} .
\end{array}
$$

By the intercept theorem in $\triangle A B D$ cut by line $C P F$, we get
$$
\begin{array}{l}
\frac{A P}{P D} \cdot \frac{D C}{C B} \cdot \frac{B F}{F A}=1 \\
\Rightarrow \frac{A P}{P D}=\frac{C B}{D C} \cdot \frac{A F}{F B}=\left(1+\frac{v}{u}\right) u=u+v \\
\Rightarrow u+v=\lambda . \\
\text { Therefore, } \frac{S_{\triangle A B C}}{S_{\triangle A E F}}=\frac{A B \cdot A C}{A F \cdot A E}=\frac{A F+F B}{A F} \cdot \frac{A E+E C}{A E} \\
=\left(1+\frac{1}{u}\right)\left(1+\frac{1}{v}\right)=1+\frac{\lambda+1}{u v} .
\end{array}
$$

By $\lambda=u+v \geqslant 2 \sqrt{u v}$
$\Rightarrow u v \leqslant \frac{\lambda^{2}}{4} \Rightarrow \frac{1}{u v} \geqslant \frac{4}{\lambda^{2}}$
$\Rightarrow \frac{S_{\triangle A B C}}{S_{\triangle A E F}}=1+\frac{\lambda+1}{u v} \geqslant 1+\frac{4 \lambda+4}{\lambda^{2}}=\left(\frac{\lambda+2}{\lambda}\right)^{2}$
$\Rightarrow S_{\triangle A E F} \leqslant\left(\frac{\lambda}{\lambda+2}\right)^{2} S_{\triangle A B C}$.
Equality holds if and only if $u=v$.
At this point, $\frac{B D}{D C}=\frac{v}{u}=1$, meaning $D$ is the midpoint of $B C$.
Thus, the maximum area of $\triangle A E F$ is $\left(\frac{\lambda}{\lambda+2}\right)^{2} S_{\triangle A B C}$.