3. Given that the three sides $a$, $b$, and $c$ of $\triangle ABC$ form a geometric sequence, and the angles opposite to $a$, $b$, and $c$ are $\angle A$, $\angle B$, and $\angle C$ respectively. Then the range of $\sin B + \cos B$ is ( ).
(A) $\left[\frac{1}{2}, 1+\frac{\sqrt{3}}{2}\right]$
(B) $(1, \sqrt{2}]$
3. B.

From the known information and the cosine rule, we have
$$
\begin{array}{l}
a c=b^{2}=a^{2}+c^{2}-2 a c \cos B \\
\geqslant 2 a c-2 a c \cos B .
\end{array}
$$

Then $\cos B \geqslant \frac{1}{2}$.
Therefore, $0<\angle B \leqslant \frac{\pi}{3}$.
Hence, $\sin B+\cos B$
$$
=\sqrt{2} \sin \left(B+\frac{\pi}{4}\right) \in(1, \sqrt{2}] .
$$