6. In $\triangle A B C$,
$$
\tan A 、(1+\sqrt{2}) \tan B 、 \tan C
$$

form an arithmetic sequence. Then the minimum value of $\angle B$ is $\qquad$
6. $\frac{\pi}{4}$.

From the problem, we know
$$
\begin{array}{l}
2(1+\sqrt{2}) \tan B=\tan A+\tan C . \\
\text { Also, } \angle A+\angle B+\angle C=\pi \text {, so } \\
\tan B=\tan (\pi-(A+C)) \\
=-\frac{\tan A+\tan C}{1-\tan A \cdot \tan C} \\
\Rightarrow \tan A \cdot \tan B \cdot \tan C \\
=\tan A+\tan B+\tan C .
\end{array}
$$

Substituting equation (1) into equation (2) gives
$$
\begin{array}{l}
\tan A \cdot \tan C=1+2(1+\sqrt{2}) \\
=3+2 \sqrt{2}=(1+\sqrt{2})^{2} .
\end{array}
$$

Thus, $\tan A, \tan C > 0$.
Using the AM-GM inequality, we get
$$
\begin{array}{l}
2(1+\sqrt{2}) \tan B=\tan A+\tan C \\
\geqslant 2 \sqrt{\tan A \cdot \tan C}=2(1+\sqrt{2}) \\
\Rightarrow \tan B \geqslant 1 .
\end{array}
$$

Since $0<\angle B<\pi$, the minimum value of $\angle B$ is $\frac{\pi}{4}$.