3. In $\triangle A B C$, $\frac{2}{a}=\frac{1}{b}+\frac{1}{c}$. Then $\angle A()$.
(A) must be an acute angle
(B) must be a right angle
(C) must be an obtuse angle
(D) none of the above
3. (A).

Proof by contradiction. If $\angle .4$ is not an acute angle, then $\angle .4 \geqslant 90^{\circ}$. Given $a>b$ and $a>c$, we have $\frac{1}{a}<\frac{1}{b}, \frac{1}{a}<\frac{1}{c}$. Therefore, $-\frac{1}{a}+\frac{1}{a}<\frac{1}{b}+\frac{1}{c}$, which simplifies to $\frac{2}{a}<\frac{1}{b}+\frac{1}{c}$, contradicting the given condition. Hence, $\angle A-$ must be an acute angle.