11. (20 points) As shown in Figure 1, the ellipse $E: \frac{x^{2}}{4}+y^{2}=1$ has its right focus at $F$, and the circle $\odot O: x^{2}+y^{2}=1$ has a tangent line $AB$ that intersects the ellipse $E$ at points $A$ and $B$ ($A$ and $B$ are both on the right side of the $y$-axis). Find the maximum value of the inradius of $\triangle ABF$.
11. Let point $A\left(x_{1}, y_{1}\right), B\left(x_{2}, y_{2}\right)$, and the inradius be $r$.
(1) If the slope of line $A B$ does not exist, $l_{A B}: x=1$, $A B=\sqrt{3}$, at this time, $r=\frac{3-\sqrt{3}}{4}$.
(2) If the slope of line $A B$ exists, let $l_{A B}: y=k x + m$. Without loss of generality, assume $k>0$, then $m<0$.
From the conditions we know
$\frac{|m|}{\sqrt{1+k^{2}}}=1 \Rightarrow m^{2}=1+k^{2}$.
Solving the system $\left\{\begin{array}{l}\frac{x^{2}}{4}+y^{2}=1, \\ y=k x+m\end{array}\right.$ and eliminating $y$ gives
$$
\begin{array}{l}
\left(1+4 k^{2}\right) x^{2}+8 k m x+4 m^{2}-4=0 \\
\Rightarrow x_{1}+x_{2}=-\frac{8 k m}{1+4 k^{2}}, x_{1} x_{2}=\frac{4 m^{2}-4}{1+4 k^{2}} \\
\Rightarrow\left|x_{1}-x_{2}\right|=\sqrt{\left(x_{1}+x_{2}\right)^{2}-4 x_{1} x_{2}} \\
=\frac{4 \sqrt{4 k^{2}-m^{2}+1}}{1+4 k^{2}} .
\end{array}
$$

From equations (1) and (2), we get $\left|x_{1}-x_{2}\right|=\frac{4 \sqrt{3} k}{1+4 k^{2}}$.
Thus, $A B=\sqrt{1+k^{2}}\left|x_{1}-x_{2}\right|$
$=\sqrt{1+k^{2}} \frac{4 \sqrt{3} k}{1+4 k^{2}}$.
Also, $A F+B F=a-e x_{1}+a-e x_{2}$
$=2 a-e\left(x_{1}+x_{2}\right)=4-\sqrt{1+k^{2}} \frac{4 \sqrt{3} k}{1+4 k^{2}}$.
Therefore, the perimeter of $\triangle A B F$ is $A B+A F+B F=4$ (a constant).
On one hand, by the triangle area formula we get
$S=\frac{1}{2} \times 4 r=2 r$.
On the other hand,
$$
\begin{aligned}
S & =\frac{1}{2} A B h=\frac{1}{2} \sqrt{1+k^{2}} \frac{4 \sqrt{3} k}{1+4 k^{2}} \cdot \frac{|\sqrt{3} k+m|}{\sqrt{1+k^{2}}} \\
& =\frac{2 \sqrt{3} k}{1+4 k^{2}}|\sqrt{3} k+m| .
\end{aligned}
$$

Thus, $r=\frac{\sqrt{3} k}{1+4 k^{2}}|\sqrt{3} k+m|$
$=\frac{\sqrt{3} k}{1+4 k^{2}} \left|\sqrt{3} k-\sqrt{1+k^{2}}\right|$
$=\frac{\sqrt{3}}{4+\frac{1}{k^{2}}}\left(\sqrt{3}-\sqrt{1+\frac{1}{k^{2}}}\right)$
$\Rightarrow \lim _{k \rightarrow+\infty} r(k)=\frac{\sqrt{3}(\sqrt{3}-1)}{4}$.
In summary, the maximum value of the inradius of $\triangle A B F$ is $\frac{3-\sqrt{3}}{4}$.