2. The three sides of $\triangle A B C$, $a$, $b$, and $c$, satisfy $b+$ $c \leqslant 2 a, a+c \leqslant 2 b$. Then the range of $\frac{b}{a}$ is
(A) $(0, \infty)$
(B) $\left(\frac{\sqrt{5}-1}{2}, \frac{\sqrt{5}+1}{2}\right)$
(C) $\left(\frac{1}{2}, 2\right)$
(D) $(0,2)$
2. (C).

From the conditions, we know that $c$ is the smallest (otherwise, if $b < c$, then $b^2 < c^2$, which contradicts $b^2 > c^2$. If $a < c$, then $a^2 < c^2$, which contradicts the given conditions).
From $\frac{b}{a}+\frac{c}{a} \leqslant 2$ and $1+\frac{c}{a} \leqslant 2 \cdot \frac{b}{a}$, we have
$$
\frac{1}{2}+\frac{c}{2 a} \leqslant \frac{b}{a} \leqslant 2-\frac{c}{a} \text {. }
$$

Since $c$ is the smallest, we can let $c \rightarrow 0$ to get $\frac{1}{2}<\frac{b}{a}<2$.