Inequality involving circumradii Let $ABC$ be a triangle and $M$ a point on the side $BC$. Let $R_1$,$R_2$, and $R$ be the circumradii of the triangles $ABM, ACM$, and $ABC$. Show that $\max\{R_1,R_2\} \geq R\cos\frac A 2$.
 A: In $\Delta ABC$ we have:
$$\frac{a}{\sin A}=2R$$
$\Rightarrow 2R\cdot\sin A=a\Rightarrow2R(2\sin\frac{A}{2}\cdot\cos\frac{A}{2})=a
\Rightarrow R\cdot\cos\frac{A}{2}=\dfrac{1}{4}\cdot\dfrac{a}{\sin\frac{A}{2}}$
Now assume $AM$ divides $\angle A$ into $\angle A_1$ and $\angle A_2$, also consider
$|\alpha|\lt\dfrac{A}{2}\lt 90^\circ$
$\max\{R_1,R_2\}\ge\dfrac{1}{2}(R_1+R_2)=\dfrac{1}{4}(\dfrac{BM}{\sin A_1}+\dfrac{CM}{\sin A_2})\ge\dfrac12(\dfrac{BM+CM}{\sin A_1+\sin A_2})$
$=\dfrac12\cdot\dfrac{a}{\sin(\frac{A}{2}-\alpha)+\sin(\frac{A}{2}+\alpha)}
\ge \dfrac12\cdot\dfrac{a}{2\sin\frac{A}{2}\cos\alpha}\ge\dfrac14\cdot\dfrac{a}{\sin\frac{A}{2}}=R\cdot\cos\frac{A}{2}$

P.S: Proof of the used inequality: Suppose $a=\sin A_1$, $b=BM$, $c=\sin A_2$, $d=CM$ $(b,d,b+d\neq0)$ 
By using Chebyshev's sum inequality (since $\sin A_1\ge\sin A_2 \equiv BM\ge CM$) on the third step we have:
$\dfrac{a}{b}+\dfrac{c}{d}=\dfrac{ad+bc}{bd}\ge\dfrac{(a+c)(b+d)}{2bd}
=\dfrac{(a+c)(b+d)^2}{(b+d)\cdot2bd}=\dfrac{(a+c)(b^2+d^2+2bd)}{(b+d)\cdot2bd}
\ge\dfrac{(a+c)\cdot4bd}{(b+d)\cdot2bd}=2\cdot\dfrac{a+c}{b+d}$
