(3) Given that the three sides $a, b, c$ of $\triangle ABC$ form a geometric sequence, and the angles opposite to $a, b, c$ are $A, B, C$ respectively. Then the range of $\sin B + \cos B$ is ( ).
(A) $\left(1,1+\frac{\sqrt{3}}{2}\right]$
(B) $\left[\frac{1}{2}, 1+\frac{\sqrt{3}}{2}\right]$
(C) $(1, \sqrt{2}]$
(D) $\left[\frac{1}{2}, \sqrt{2}\right]$
(3) $\mathrm{C}$ Hint: $a c=b^{2}=a^{2}+c^{2}-2 a c \cos B \geqslant 2 a c-2 a c \cos B$
$$
\Rightarrow \cos B \geqslant \frac{1}{2} \Rightarrow 0<B \leqslant \frac{\pi}{3} \text {, }
$$

Thus,
$$
\sin B+\cos B=\sqrt{2} \sin \left(B+\frac{\pi}{4}\right) \in(1, \sqrt{2}] .
$$

Therefore, the answer is C.