9. (16 points) Given the ellipse $C: \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$, points $F_{1}, F_{2}$ are its left and right foci, respectively, and $A$ is the right vertex. A line $l$ passing through $F_{1}$ intersects the ellipse at points $P, Q$, and $\overrightarrow{A P} \cdot \overrightarrow{A Q}=\frac{1}{2}(a+c)^{2}$. Try to find the minimum value of the eccentricity of the ellipse $C$.
9. According to the problem, we can set the line $l: x=m y-c$, and combine the equations
$$
\left\{\begin{array}{l}
x=m y-c, \\
\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1 .
\end{array}\right.
$$

Eliminating $x$, we get
$$
\left(a^{2}+b^{2} m^{2}\right) y^{2}-2 b^{2} c m y-b^{4}=0 .
$$

Let $P\left(x_{1}, y_{1}\right), Q\left(x_{2}, y_{2}\right)$, then
$$
y_{1}+y_{2}=\frac{2 b^{2} c m}{a^{2}+b^{2} m^{2}}, \quad y_{1} y_{2}=\frac{-b^{4}}{a^{2}+b^{2} m^{2}},
$$

Thus
$$
\begin{array}{l}
x_{1}+x_{2}=m\left(y_{1}+y_{2}\right)-2 c=\frac{-2 a^{2} c}{a^{2}+b^{2} m^{2}}, \\
x_{1} x_{2}=\left(m y_{1}-c\right)\left(m y_{2}-c\right)=m^{2} y_{1} y_{2}-m c\left(y_{1}+y_{2}\right)+c^{2}=\frac{a^{2}\left(-b^{2} m^{2}+c^{2}\right)}{a^{2}+b^{2} m^{2}} .
\end{array}
$$

Also, $A(a, 0)$, so $\overrightarrow{A P}=\left(x_{1}-a, y_{1}\right), \overrightarrow{A Q}=\left(x_{2}-a, y_{2}\right)$, thus
$$
\begin{aligned}
\overrightarrow{A P} \cdot \overrightarrow{A Q} & =\left(x_{1}-a\right)\left(x_{2}-a\right)+y_{1} y_{2}=x_{1} x_{2}-a\left(x_{1}+x_{2}\right)+a^{2}+y_{1} y_{2} \\
& =\frac{a^{2}(a+c)^{2}-b^{4}}{a^{2}+b^{2} m^{2}} .
\end{aligned}
$$

By the problem, we have
$$
\frac{a^{2}(a+c)^{2}-b^{4}}{a^{2}+b^{2} m^{2}}=\frac{1}{2}(a+c)^{2},
$$

Simplifying, we get
$$
\frac{a^{2}-(a-c)^{2}}{a^{2}+b^{2} m^{2}}=\frac{1}{2},
$$

which means
$$
m^{2}=\frac{a^{2}-2(a-c)^{2}}{b^{2}} .
$$

Since $m^{2} \geqslant 0$, we have $a^{2}-2(a-c)^{2} \geqslant 0 \Rightarrow a \geqslant \sqrt{2}(a-c) \Rightarrow \frac{c}{a} \geqslant 1-\frac{\sqrt{2}}{2}$. Therefore, the minimum value of the eccentricity of the ellipse $C$ is $1-\frac{\sqrt{2}}{2}$.