<s>[INST] How to prove $\cos(\frac{B-C}2)\ge \sqrt{\frac{2r}{R}}$? 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? 
 [/INST] It is known that $$\frac{r}{4R}= \sin \frac{A}{2}\cdot\sin \frac{B}{2}\cdot\sin \frac{C}{2}.$$
and consequently$$\frac{2r}{R}= 8\sin \frac{A}{2}\cdot\sin \frac{B}{2}\cdot\sin \frac{C}{2}.$$
$$\cos \frac{B-C}{2}\geq\sqrt{\frac{2r}{R}}\Leftrightarrow\cos^2 \frac{B-C}{2}\geq\frac{2r}{R}\Leftrightarrow\cos^2 \frac{B-C}{2}\geq 8\sin \frac{A}{2}\cdot\sin \frac{B}{2}\cdot\sin \frac{C}{2}\Leftrightarrow\cos^2 \frac{B-C}{2}\geq 4\sin \frac{A}{2}(\cos \frac{B-C}{2}-\cos \frac{B+C}{2})\Leftrightarrow$$
$$\cos^2 \frac{B-C}{2}\geq 4\sin \frac{A}{2}(\cos \frac{B-C}{2}-\sin \frac{A}{2})\Leftrightarrow(\cos \frac{B-C}{2}-\sin \frac{A}{2})^2\geq 0.$$
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