3. In $\triangle A B C$, $\lg \operatorname{tg} A+\lg \operatorname{tg} C=$ $2 \lg \operatorname{tg} B$. Then the range of $\angle B$ is () .
(A) $0<\angle B \leqslant \frac{\pi}{3}$
(B) $\frac{\pi}{3} \leqslant \angle B<\frac{\pi}{2}$
(C) $0<\angle B \leqslant \frac{\pi}{6}$
(D) $\frac{\pi}{6} \leqslant \angle B<\frac{\pi}{2}$
3. (B).

From the given, we have $\operatorname{tg}^{2} B=\operatorname{tg} A \operatorname{tg} C \leqslant\left(\frac{\operatorname{tg} A+\operatorname{tg} C}{2}\right)^{2}$. Therefore, $2 \operatorname{tg} B \leqslant \operatorname{tg} A+\operatorname{tg} C=\operatorname{tg}(A+C)(1-\operatorname{tg} A \operatorname{tg} C)=$ $-\operatorname{tg} B \cdot\left(1-\operatorname{tg}^{2} B\right)$. Then $\operatorname{tg}^{2} B \geqslant 3, \operatorname{tg} B \geqslant \sqrt{3}$.