Let $G$ be the projection of $D$ onto $BC$. First we notice that $\angle ABM = \angle ABD = \angle EBD = \angle EGD$. Analogously $\angle ACN = \angle FGD$. So by law of sines we get $MN = 2R \sin(\angle NAM) = 2R \sin(\angle ABM + \angle ACN) = 2R\sin \angle EGF$. Let $\angle EGF = \theta$. Then we need to prove that $\frac{EF}{2R\sin\theta} \geq \frac{r}{R},$ in other words $EF \geq 2r\sin\theta$, but if $r'$ is the radius of the circumcircle of $EFG$, then by law of sines: $EF = 2r'\sin\theta$, but $r' \geq r$, because the incircle is the smallest circumcircle of a pedal triangle.