To prove: circles $U,V$ have sum of radii $\ge MS + \frac{AB-AM-BM}{2}$ With $rs=A$ this can be solved with a lot of calculus in function of boring angles, it seems. But can smb give a nice sol. to complete it?

Denote $\frac{AB}{2}$ by $R,$ radius of $U$ and $V$ by $r_1$ and $r_2$, centers of $U$ and $V$ by $O_{1}$ and $O_{2}$. Let $X = U \cap AB, Y=V\cap AB, \varphi = \angle{ O_{1}SM }$ $(\varphi \in (0, \frac{\pi}{2})\implies \sin\varphi > 0, \cos\varphi > 0)$ $R-r_{1}=SO_{1} =\frac{r_{1}}{\sin\varphi}\implies r_{1}=\frac{R\cdot \sin\varphi}{1+\sin\varphi}\implies SX=\frac{r_{1}}{\tan \varphi}=\frac{R\cdot \cos\varphi}{1+\sin\varphi}.$ Symilary $r_2 = \frac{R\cdot \sin(\frac{\pi}{2} - \varphi)}{1+\sin(\frac{\pi}{2} - \varphi)} = \frac{R\cdot \cos\varphi}{1+\cos\varphi}$ and $SY = \frac{R\cdot \cos(\frac{\pi}{2} -\varphi)}{1+\sin(\frac{\pi}{2} -\varphi)} = \frac{R\cdot \sin\varphi}{1+\cos\varphi}.$ By rearrangement inequality $\frac{R\cdot \sin\varphi}{1+\cos\varphi} + \frac{R\cdot \cos\varphi}{1+\sin\varphi}\geq \frac{R\cdot \sin\varphi}{1+\sin\varphi} + \frac{R\cdot \cos\varphi}{1+\cos\varphi}$ $\implies XY = SX+SY \geq r_{1}+r_{2}$ it needed. Or another way to prove it inequality: $\frac{R\cdot \sin\varphi}{1+\cos\varphi} - \frac{R\cdot \cos\varphi}{1+\cos\varphi} + \frac{R\cdot \cos\varphi}{1+\sin\varphi}- \frac{R\cdot \sin\varphi}{1+\sin\varphi} = R (\sin\varphi - \cos\varphi)(\frac{1}{1+\cos\varphi} - \frac{1}{1+\sin\varphi})=\frac{R(\sin\varphi-\cos\varphi)^2}{(1+\sin\varphi)(1+\cos\varphi)}\geq 0.$