9. Let $F$ be the left focus of the ellipse $E: \frac{x^{2}}{3}+y^{2}=1$. A line $l$ passing through point $F$ with a positive slope intersects $E$ 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 $A M, B N$ intersect the $x$-axis at points $M$ and $N$. Find the minimum value of $|M N|$.
Given the line $AB: x = ny - \sqrt{2}$, with points $A(x_1, y_1)$ and $B(x_2, y_2)$, then the line $AM: y - y_1 = -n(x - x_1)$,
let $y = 0$, we get $x_M = x_1 + \frac{y_1}{n}$,
similarly, we get $x_N = x_2 + \frac{y_2}{n}$. Solving the system of equations
$$
\left\{\begin{array}{l}
x = ny - \sqrt{2}, \\
x^2 + 3y^2 = 3
\end{array} \Rightarrow (n^2 + 3)y^2 - 2\sqrt{2}ny - 1 = 0 \right. \text{. }
$$

Then $x_M - x_N = x_1 - x_2 + \frac{y_1}{n} - \frac{y_2}{n} = \left(n + \frac{1}{n}\right)(y_1 - y_2) \Rightarrow |MN| = \left(n + \frac{1}{n}\right)|y_1 - y_2|$
$= \frac{n^2 + 1}{n} \cdot \frac{\sqrt{8n^2 + 4(n^2 + 3)}}{n^2 + 3} = \frac{2\sqrt{3}(n^2 + 1)\sqrt{n^2 + 1}}{n(n^2 + 3)} = \frac{2\sqrt{3}\left(1 + \frac{1}{n^2}\right)\sqrt{1 + \frac{1}{n^2}}}{1 + \frac{3}{n^2}}$
$\xlongequal{t = \sqrt{1 + \frac{1}{n^2} > 1}} \frac{2\sqrt{3}t^3}{3t^2 - 2} = \frac{2\sqrt{3}}{\frac{3}{t} - \frac{2}{t^3}} \xlongequal{u = \frac{1}{t}} \frac{2\sqrt{3}}{3u - 2u^3}$, let $\varphi(u) = 3u - 2u^3$,
$\varphi'(u) = 3 - 6u^2 \Rightarrow \varphi(u)$ is monotonically increasing on $\left(0, \frac{\sqrt{2}}{2}\right)$ and monotonically decreasing on $\left(\frac{\sqrt{2}}{2}, 1\right)$,
thus $\varphi(u)_{\max} = \varphi\left(\frac{\sqrt{2}}{2}\right) = \sqrt{2}$. Therefore, the minimum value of $|MN|$ is $\sqrt{6}$, with equality when $n = 1$.