In $\triangle A B C$, the three sides $a, b, c$ 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) $(0,2)$
Given pipe:
From the conditions, $c$ is the smallest (otherwise, if $b < 2b$, it leads to a contradiction; if $a < 2a$, it contradicts the given information).
From $\frac{b}{a}+\frac{c}{a} \leqslant 2$ and $1+\frac{c}{a} \leqslant 2 \cdot \frac{b}{a}$, we have $\frac{1}{2}+\frac{c}{2a} \leqslant \frac{b}{a} \leqslant 2-\frac{c}{a}$.
Since $c$ is the smallest, we can let $c \rightarrow 0$, yielding $\frac{1}{2}<\frac{b}{a}<2$.
Therefore, the answer is (C).