9. (This question is worth 16 points) Let $O$ be the circumcenter of acute $\triangle A B C$, and the areas of $\triangle B O C, \triangle C O A, \triangle A O B$ form an arithmetic sequence. Find the minimum value of $\tan A+2 \tan C$.

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Solution: $2 \sqrt{6}$
The areas of $\triangle B O C, \triangle C O A, \triangle A O B$ form an arithmetic sequence,
i.e., $R^{2} \sin 2 B=\frac{1}{2} R_{2} \sin 2 A+\frac{1}{2} R^{2} \sin 2 C$
which is equivalent to $2 \sin 2 B=\sin 2 A+\sin 2 C=2 \sin (A+C) \cos (A-C)=2 \sin B \cos (A-C)$
Thus, $2 \cos B=\cos (A-C)$
This implies $2 \cos (A+C)+\cos (A-C)=0$
i.e., $\sin A \sin C=3 \cos A \cos C$
Therefore, $\tan A \tan C=3$
Since $\triangle A B C$ is an acute triangle, we have $\tan A+2 \tan C \geq 2 \sqrt{2 \tan A \tan C}=2 \sqrt{6}$
Equality holds when: $\tan A=\sqrt{6}, \tan B=\frac{3 \sqrt{6}}{4}, \tan C=\frac{\sqrt{6}}{2}$