Four. (20 points) Given the ellipse $C: \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>$ $b>0$ ). As shown in Figure 2, $F_{1}$ and $F_{2}$ are its left and right foci, and $P$ is any point on the ellipse $C$ (not an endpoint of the major axis). $PD$ is the angle bisector of $\angle F_{1} P F_{2}$ in the triangle $\triangle F_{1} P F_{2}$. Perpendiculars are drawn from $D$ to $P F_{1}$ and $P F_{2}$, with $M$ and $N$ being the feet of these perpendiculars. Try to find: the maximum value of $\frac{S_{\triangle D M N}}{S_{\triangle P F_{1} F_{2}}}$.
Let $\angle F_{1} P F_{2}=\theta$, it is easy to know that $\angle M D N=180^{\circ}-\theta$,
$|D M|=|D N|=|P D| \sin \frac{\theta}{2}$.
From $S_{\triangle P F_{1} F_{2}}=S_{\triangle P F_{1} D}+S_{\triangle P F_{2} D}$, we get
$$
\begin{array}{l}
\frac{1}{2}\left|P F_{1}\right| \cdot\left|P F_{2}\right| \sin \theta \\
=\frac{1}{2}\left|P F_{2}\right| \cdot|P D| \sin \frac{\theta}{2}+\frac{1}{2}\left|P F_{2}\right| \cdot|P D| \sin \frac{\theta}{2} .
\end{array}
$$

Then $|P D| \sin \frac{\theta}{2}=\frac{\left|P F_{1}\right| \cdot\left|P F_{2}\right| \sin \theta}{\left|P F_{1}\right|+\left|P F_{2}\right|}$.
Thus, $\frac{S_{\triangle D M}}{S_{\triangle P F_{1} F_{2}}}=\frac{\frac{1}{2}|D M| \cdot|D N| \sin \left(180^{\circ}-\theta\right)}{\frac{1}{2}\left|P F_{1}\right| \cdot\left|P F_{2}\right| \sin \theta}$
$=\frac{|D M| \cdot|D N|}{\left|P F_{1}\right| \cdot\left|P F_{2}\right|}=\frac{|P D|^{2} \sin ^{2} \frac{\theta}{2}}{\left|P F_{1}\right| \cdot\left|P F_{2}\right|}$
$=\frac{\left|P F_{1}\right| \cdot\left|P F_{2}\right| \sin ^{2} \theta}{\left(\left|P F_{1}\right|+\left|P F_{2}\right|\right)^{2}}$
$=\frac{1}{4 a^{2}}\left|P F_{1}\right| \cdot\left|P F_{2}\right|(1+\cos \theta)(1-\cos \theta)$.
In $\triangle P F_{1} F_{2}$,
$$
\begin{array}{l}
\left|F_{1} F_{2}\right|^{2}=\left|P F_{1}\right|^{2}+\left|P F_{2}\right|^{2}-2\left|P F_{1}\right| \cdot\left|P F_{2}\right| \cos \theta \\
=\left(\left|P F_{1}\right|+\left|P F_{2}\right|\right)^{2}-2\left|P F_{1}\right| \cdot\left|P F_{2}\right|(1+\cos \theta) .
\end{array}
$$

Thus, $1+\cos \theta=\frac{(2 a)^{2}-(2 c)^{2}}{2\left|P F_{1}\right| \cdot\left|P F_{2}\right|}=\frac{2 b^{2}}{\left|P F_{1}\right| \cdot\left|P F_{2}\right|}$.
Then $\frac{S_{\triangle D Q N}}{S_{\triangle P F_{1} F_{2}}}=\frac{1}{4 a^{2}}\left|P F_{1}\right| \cdot\left|P F_{2}\right| \cdot \frac{2 b^{2}}{\left|P F_{1}\right| \cdot\left|P F_{2}\right|}$.
$$
\begin{array}{l}
\left(2-\frac{2 b^{2}}{\left|P F_{1}\right| \cdot\left|P F_{2}\right|}\right) \\
= \frac{b^{2}}{a^{2}}\left(1-\frac{b^{2}}{\left|P F_{1}\right| \cdot\left|P F_{2}\right|}\right) \\
\leqslant \frac{b^{2}}{a^{2}}\left[1-\frac{b^{2}}{\left(\frac{\left|P F_{1}\right|+\left|P F_{2}\right|}{2}\right)^{2}}\right] \\
= \frac{b^{2}}{a^{2}}\left(1-\frac{b^{2}}{a^{2}}\right)=\frac{b^{2} c^{2}}{a^{4}} .
\end{array}
$$

When and only when $\left|P F_{1}\right|=\left|P F_{2}\right|$, i.e., $P$ is the endpoint of the minor axis, the equality holds.
Therefore, the maximum value sought is $\frac{b^{2} c^{2}}{a^{4}}$.