9. Given $F_{1} 、 F_{2}$ are the two foci of the ellipse $\frac{x^{2}}{4}+y^{2}=1$, $A 、 B$ are the left vertex and the upper vertex of the ellipse, respectively, and point $P$ lies on the line segment $A B$. Then the minimum value of $\overrightarrow{P F_{1}} \cdot \overrightarrow{P F_{2}}$ is
9. $-\frac{11}{5}$.

Let $P(x, y), F_{1}(-c, 0), F_{2}(c, 0)$. Then
$$
\begin{array}{l}
\overrightarrow{P F_{1}}=(-c-x,-y), \overrightarrow{P F_{2}}=(c-x,-y) \\
\Rightarrow \overrightarrow{P F_{1}} \cdot \overrightarrow{P F_{2}}=x^{2}+y^{2}-c^{2} .
\end{array}
$$

When $\overrightarrow{O P} \perp A D$, $x^{2}+y^{2}$ reaches its minimum value. Using the area method, the minimum value of $x^{2}+y^{2}$ is $\frac{a^{2} b^{2}}{a^{2}+b^{2}}$.
$$
\begin{array}{l}
\text { Therefore, }\left(\overrightarrow{P F_{1}} \cdot \overrightarrow{P F_{2}}\right)_{\text {min }}=\frac{a^{2} b^{2}}{a^{2}+b^{2}}-c^{2} \\
=\frac{4}{4+1}-3=-\frac{11}{5} .
\end{array}
$$