Using the power of the point in triangle $ABC$, we have to find the minimum of the expression $\frac{AM\cdot BM \cdot CM}{R^2-OM^2}$. Let $A'B'C'$ be the pedal triangle of triangle $ABC$ - triangle with vertices that are projections of $M$ onto triangle's sides. By the Law of Sines, the lengths of the sides of triangle $A'B'C'$ are equal to $AM \sin A$, $BM \sin B$, $CM \sin C$. Hence \[4R_{A'B'C'}K_{A'B'C'} = AM\cdot BM \cdot CM \cdot \frac{abc}{8R^3}.\] Also, we know that the area of the pedal triangle is given by $K_{A'B'C'} =\frac{R^2-OM^2}{4R^2}\cdot K_{ABC}$. Therefore \[\frac{AM\cdot BM \cdot CM}{R^2-OM^2} = 2R_{A'B'C'} \geq 2r,\] because $R_{A'B'C'} \geq r$ and equality occurs when $M \equiv I$, the incenter of triangle $ABC$,