$\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=k$
$\Rightarrow \mathrm{a}=\mathrm{k} \sin \mathrm{A}, \mathrm{b}=\mathrm{k} \sin \mathrm{B}, \mathrm{c}=\mathrm{k} \sin \mathrm{C}$
$\therefore \frac{\mathrm{a}}{\mathrm{b}+\mathrm{c}}=\frac{\sin \mathrm{A}}{\sin \mathrm{B}+\sin \mathrm{C}}=\frac{2 \sin \frac{\mathrm{A}}{2}+\cos \frac{\mathrm{A}}{2}}{2 \sin \frac{\mathrm{B}+\mathrm{C}}{2}+\cos \frac{\mathrm{B}-\mathrm{C}}{2}}$
$\frac{\mathrm{a}}{\mathrm{b}+\mathrm{c}}=\frac{\sin \frac{\mathrm{A}}{2}+\cos \frac{\mathrm{A}}{2}}{\cos \frac{\mathrm{A}}{2}+\cos \frac{\mathrm{B}-\mathrm{C}}{2}}$
$\therefore \sin \frac{\mathrm{A}}{2}=\frac{\mathrm{a}}{\mathrm{b}+\mathrm{c}} \cos \left(\frac{\mathrm{B}-\mathrm{C}}{2}\right)$
$\because \frac{\mathrm{B}-\mathrm{C}}{2}$ in acuite
$\therefore \quad \cos \left(\frac{\mathrm{B}-\mathrm{C}}{2}\right) 2 \leq 1$
$\therefore \sin \frac{A}{2} \leq \frac{a}{b+c}$ ...(i)
We have $\mathrm{GM} \leq \mathrm{AM}$
$\begin{aligned} & 2 \sqrt{b c} \leq b+c \\ & \Rightarrow \frac{1}{b+c} \leq \frac{1}{2 \sqrt{b c}}\end{aligned}$
from (i)
$\sin \frac{\mathrm{A}}{2} " \frac{\mathrm{a}}{2 \sqrt{\mathrm{bc}}}$