7. Let $\triangle A B C$ be an acute-angled triangle with the lengths of the sides opposite to its internal angles being $a$, $b$, and $c$ respectively. If $2 a^{2}=2 b^{2}+c^{2}$, then the minimum value of $\tan A+\tan B+\tan C$ is $\qquad$.
7.6.
$$
\begin{array}{l}
\text { Given } 2 a^{2}=2 b^{2}+c^{2} \\
\Rightarrow 2 \sin ^{2} A=2 \sin ^{2} B+\sin ^{2} C \\
\Rightarrow \cos 2 B-\cos 2 A=\sin ^{2} C \\
\Rightarrow 2 \sin (A-B) \cdot \sin (A+B)=\sin ^{2} C \\
\Rightarrow 2 \sin (A-B)=\sin C=\sin (A+B) \\
\Rightarrow \sin A \cdot \cos B=3 \cos A \cdot \sin B \\
\Rightarrow \tan A=3 \tan B .
\end{array}
$$

Then $\tan A+\tan B+\tan C$
$$
\begin{array}{l}
=\tan A+\tan B-\tan (A+B) \\
=\tan A+\tan B-\frac{\tan A+\tan B}{1-\tan A \cdot \tan B} \\
=4 \tan B-\frac{4 \tan B}{1-3 \tan ^{2} B} .
\end{array}
$$

Let $\tan B=x>0$.
Thus, $\tan A+\tan B+\tan C=4 x-\frac{4 x}{1-3 x^{2}}$
$$
=\frac{12 x^{3}}{3 x^{2}-1} \triangleq f(x) \text {. }
$$

Notice that, $f^{\prime}(x)=\frac{36 x^{2}\left(x^{2}-1\right)}{\left(3 x^{2}-1\right)^{2}}$.
Therefore, $f(x)$ is monotonically decreasing on the interval $(0,1]$ and monotonically increasing on the interval $[1,+\infty)$.

Hence, $f_{\min }(x)=f(1)=6$, which means the minimum value of $\tan A+\tan B+$ $\tan C$ is 6.