For any triangle $ABC$, prove that: $$\cos(\frac{B-C}2)\ge \sqrt{\frac{2r}{R}}$$
I have tried many approaches but none seems to work. I noted that $\cos(\frac{B-C}2)=\frac{AM}{2R}$, where $M$ is point of intersection of the circumcircle and bisector of $\angle A$, but it wasn't much helpful. Although, Euler's inequality seems to be helpful, it isn't. And expanding out $\cos(\frac{B-C}2)$ gets very complicated.
Does anyone have some elegant methods?
It is known that $$\frac{r}{4R}= \sin \frac{A}{2}\cdot\sin \frac{B}{2}\cdot\sin \frac{C}{2}.$$ and consequently$$\frac{2r}{R}= 8\sin \frac{A}{2}\cdot\sin \frac{B}{2}\cdot\sin \frac{C}{2}.$$ $$\cos \frac{B-C}{2}\geq\sqrt{\frac{2r}{R}}\Leftrightarrow\cos^2 \frac{B-C}{2}\geq\frac{2r}{R}\Leftrightarrow\cos^2 \frac{B-C}{2}\geq 8\sin \frac{A}{2}\cdot\sin \frac{B}{2}\cdot\sin \frac{C}{2}\Leftrightarrow\cos^2 \frac{B-C}{2}\geq 4\sin \frac{A}{2}(\cos \frac{B-C}{2}-\cos \frac{B+C}{2})\Leftrightarrow$$ $$\cos^2 \frac{B-C}{2}\geq 4\sin \frac{A}{2}(\cos \frac{B-C}{2}-\sin \frac{A}{2})\Leftrightarrow(\cos \frac{B-C}{2}-\sin \frac{A}{2})^2\geq 0.$$