Riding the Angle Bisector

Drop perpendiculars from $X$ to sides $AB$ and $AC$ at $P$ and $Q$ respectively:

Then $APXQ$ is a kite, with $AP=AQ=t$ (say) and $PX=QX=t \tan \frac{A}{2}$ .

Also, $BP=c-t$ and $CQ=b-t$ .

By Pythagoras, $BX^2=(c-t)^2+ t^2 \tan^2 \frac{A}{2}=k^2 t^2 - 2ct+c^2$

where $k=\sec \frac{A}{2}$ , and similarly

$CX^2=k^2 t^2 - 2bt+b^2$

We are interested in the max/min values of $f(t)=\frac{k^2 t^2 - 2ct+c^2}{k^2 t^2 - 2bt+b^2}$

Setting $f'(t)=0$ and tidying up, we find at the max/min, $t^2-(b+c)t+\frac{bc}{k^2}=0$

After solving and substituting, and some messy algebra, we find $Mm=\boxed{\frac{c}{b}}$ .

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I suspect there's a better approach once $f(t)$ is found - the function is not complicated, and nor is the final result; calculus is probably not the best method.