7. In $\triangle A B C$, $M$ is the midpoint of side $B C$, and $N$ is the midpoint of segment $B M$. If $\angle A=\frac{\pi}{3}, S_{\triangle A B C}=\sqrt{3}$, then the minimum value of $\overrightarrow{A M} \cdot \overrightarrow{A N}$ is $\qquad$
7. $\sqrt{3}+1$

From the conditions, we have
$$
\begin{array}{l} 
\overrightarrow{A M}=\frac{1}{2}(\overrightarrow{A B}+\overrightarrow{A C}), \overrightarrow{A N}=\frac{3}{4} \overrightarrow{A B}+\frac{1}{4} \overrightarrow{A C} . \\
\text { Therefore, } \overrightarrow{A M} \cdot \overrightarrow{A N}=\frac{1}{8}\left(3|\overrightarrow{A B}|^{2}+|\overrightarrow{A C}|^{2}+4 \overrightarrow{A B} \cdot \overrightarrow{A C}\right) . \\
\text { Given } \overrightarrow{A B} \cdot \overrightarrow{A C}=|\overrightarrow{A B}||\overrightarrow{A C}| \cos A \\
=2 S_{\triangle A B C} \cdot \cot A=2 \\
\Rightarrow|\overrightarrow{A B}||\overrightarrow{A C}|=4 \\
\Rightarrow \overrightarrow{A M} \cdot \overrightarrow{A N} \geqslant \frac{\sqrt{3}}{4}|\overrightarrow{A B}||\overrightarrow{A C}|+\frac{1}{2} \overrightarrow{A B} \cdot \overrightarrow{A C}=\sqrt{3}+1 . \\
\quad \text { When }|\overrightarrow{A B}|=\frac{3}{4},|\overrightarrow{4 C}|=2 \times \sqrt[4]{3} \text {, } \overrightarrow{A M} \cdot \overrightarrow{A N} \text { is }
\end{array}
$$

the minimum value is $\sqrt{3}+1$.