6. Given 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 $\qquad$
Solve:
From $\frac{\tan A+\tan C}{1-\tan A \tan C}=-\tan B \Rightarrow \tan A \tan C-1=\frac{\tan A+\tan C}{\tan B}=2+2 \sqrt{2}$
$\Rightarrow \tan A \tan C=3+2 \sqrt{2}$, so $\tan B=\frac{\tan A+\tan C}{\tan A \tan C-1} \geqslant \frac{2 \sqrt{\tan A \tan C}}{\tan A \tan C-1}$ $=\frac{2 \sqrt{3+2 \sqrt{2}}}{2+2 \sqrt{2}}=\frac{2(1+\sqrt{2})}{2+2 \sqrt{2}}=1 \Rightarrow B \geqslant \frac{\pi}{4}$, equality holds when $A=C$.