10. (20 points) As shown in Figure 1, the ellipse $E: \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ $(a>b>0)$, $F_{1}$ and $F_{2}$ are the foci, $P$ is any point on the ellipse $E$ (not the endpoint of the major axis), $D$ is the intersection of the angle bisector of $\angle F_{1} P F_{2}$ with the $x$-axis, and perpendiculars are drawn from point $D$ to $P F_{1}$ and $P F_{2}$, with $M$ and $N$ being the feet of these perpendiculars. Find the maximum value of $\frac{S_{\triangle D M N}}{S_{\triangle F_{1} P F_{2}}}$.
10. Let $\angle F_{1} P F_{2}=\theta$. Then
$$
\begin{array}{l}
\angle M D N=180^{\circ}-\theta, \\
|D M|=|D N|=|P D| \sin \frac{\theta}{2} .
\end{array}
$$

Let $\left|P F_{1}\right|=m,\left|P F_{2}\right|=n$, then $m+n=2 a$.
In $\triangle F_{1} P F_{2}$, by the cosine rule we get
$$
\begin{aligned}
4 c^{2} & =m^{2}+n^{2}-2 m n \cos \theta \\
\Rightarrow & 4 c^{2}=(m+n)^{2}-2 m n(1+\cos \theta) \\
& =4 a^{2}-2 m n(1+\cos \theta) \\
\Rightarrow & 2 m n(1+\cos \theta)=4 a^{2}-4 c^{2}=4 b^{2} \\
\Rightarrow & \cos \theta=\frac{2 b^{2}}{m n}-1 .
\end{aligned}
$$
$$
\begin{array}{l}
\text { By } S_{\triangle F_{1} P F_{2}}=S_{\triangle P F_{1} D}+S_{\triangle P F_{2} D} \\
\Rightarrow \frac{1}{2} m n \sin \theta \\
=\frac{1}{2} m|P D| \sin \frac{\theta}{2}+\frac{1}{2}|P D| n \sin \frac{\theta}{2} \\
\Rightarrow|P D| \sin \frac{\theta}{2}=\frac{m n \sin \theta}{m+n}=\frac{m n \sin \theta}{2 a} \text {. } \\
\text { Hence } \frac{S_{\triangle D M N}}{S_{\triangle F_{1} P F_{2}}}=\frac{\frac{1}{2}|D M||D N| \sin \left(180^{\circ}-\theta\right)}{\frac{1}{2} m n \sin \theta} \\
=\frac{|D M||D N|}{m n}=\frac{m n \sin ^{2} \theta}{(2 a)^{2}} \\
=\frac{m n}{4 a^{2}}\left(1-\cos ^{2} \theta\right)=\frac{m n}{4 a^{2}}\left(1-\left(\frac{2 b^{2}}{m n}-1\right)^{2}\right) \\
=\frac{b^{2}}{a^{2}}\left(1-\frac{b^{2}}{m n}\right) \\
\leqslant \frac{b^{2}}{a^{2}}\left(1-\frac{b^{2}}{\left(\frac{m+n}{2}\right)^{2}}\right)=\frac{b^{2} c^{2}}{a^{4}}, \\
\end{array}
$$

When and only when $m=n$, i.e., $\left|P F_{1}\right|=\left|P F_{2}\right|$, the equality holds.
Therefore, the maximum value of $\frac{S_{\triangle D M N}}{S_{\triangle F_{1} P F_{2}}}$ is $\frac{b^{2} c^{2}}{a^{4}}$.