9. (16 points) Let $F$ be the left focus of the ellipse $E: \frac{x^{2}}{3}+y^{2}=1$. A line $l$ with a positive slope passes through point $F$ and intersects the ellipse $E$ at points $A$ and $B$. Lines $A M$ and $B N$ are drawn through points $A$ and $B$ respectively, such that $A M \perp l, B N \perp l$, and lines $A M$ and $B N$ intersect the $x$-axis at points $M$ and $N$. Find the minimum value of $|M N|$.
9. Let the straight line $l$ passing through the left focus $F$ of the ellipse $E$ have an inclination angle $\alpha$.
According to the problem, we know $\frac{\pi}{2}>\alpha>0$.
As shown in Figure 1, let $F^{\prime}$ be the right focus of the ellipse $E$.

In the right triangle $\triangle M A F$,
$\cos \angle M F A=\frac{|A F|}{|M F|}$
$\Rightarrow|M F|=\frac{|A F|}{\cos \angle M F A}$.
Similarly, in the right triangle $\triangle N B F$,
$|N F|=\frac{|B F|}{\cos \angle N F B}$.
Then $|M N|=|M F|+|N F|$
$$
\begin{array}{l}
=\frac{|A F|}{\cos \angle M F A}+\frac{|B F|}{\cos \angle N F B} \\
=\frac{|A B|}{\cos \alpha} .
\end{array}
$$

Connecting $B F^{\prime}$, in $\triangle F B F^{\prime}$, let $|B F|=x$. Then
$$
\left|B F^{\prime}\right|=2 \sqrt{3}-x \text {. }
$$

By the cosine rule,
$$
\begin{array}{l}
\left|B F^{\prime}\right|^{2}=|B F|^{2}+\left|F F^{\prime}\right|^{2}-2|B F|\left|F F^{\prime}\right| \cos \alpha \\
\Rightarrow(2 \sqrt{3}-x)^{2}=x^{2}+(2 \sqrt{2})^{2}-2 x \cdot 2 \sqrt{2} \cos \alpha \\
\Rightarrow x=\frac{1}{\sqrt{3}-\sqrt{2} \cos \alpha} \\
\Rightarrow|B F|=\frac{1}{\sqrt{3}-\sqrt{2} \cos \alpha} .
\end{array}
$$

Similarly, $|A F|=\frac{1}{\sqrt{3}+\sqrt{2} \cos \alpha}$.
From equations (2) and (3),
$$
\begin{array}{l}
|A B|=\frac{1}{\sqrt{3}-\sqrt{2} \cos \alpha}+\frac{1}{\sqrt{3}+\sqrt{2} \cos \alpha} \\
=\frac{2 \sqrt{3}}{3-2 \cos ^{2} \alpha} .
\end{array}
$$

From equation (1),
$$
\begin{array}{l}
|M N|=\frac{2 \sqrt{3}}{\left(3-2 \cos ^{2} \alpha\right) \cos \alpha} . \\
\text { Let } f(\alpha)=\left(3-2 \cos ^{2} \alpha\right)^{2} \cdot 4 \cos ^{2} \alpha \text {, i.e., } \\
f(\alpha)=\left(3-2 \cos ^{2} \alpha\right)\left(3-2 \cos ^{2} \alpha\right) \cdot 4 \cos ^{2} \alpha .
\end{array}
$$

By the AM-GM inequality,
$$
\begin{array}{l}
\frac{\left(3-2 \cos ^{2} \alpha\right)+\left(3-2 \cos ^{2} \alpha\right)+4 \cos ^{2} \alpha}{3} \\
\geqslant \sqrt[3]{\left(3-2 \cos ^{2} \alpha\right)^{2} \cdot 4 \cos ^{2} \alpha} .
\end{array}
$$

Thus, $\left(3-2 \cos ^{2} \alpha\right)^{2} \cdot 4 \cos ^{2} \alpha \leqslant 8$, with equality holding if and only if
$$
3-2 \cos ^{2} \alpha=4 \cos ^{2} \alpha \Rightarrow \alpha=\frac{\pi}{4} \text {. }
$$

Therefore, when and only when $\alpha=\frac{\pi}{4}$, $(f(\alpha))_{\text {max }}=8$. At this time, $|M N|_{\text {min }}=\sqrt{6}$.