Call $ \frac{MA}{MB}=t $. By Ptolemy's theorem, we have $ MC.AB=MA.BC+MB.AC \Rightarrow MC=MA+MB\sqrt{2} \Rightarrow \frac{MC}{MB}=t+\sqrt{2} $. Again by Ptolemy, $ MD.AB=MA.BD+MB.AD \Rightarrow \frac{MD}{MA} =\frac{1}{t}+\sqrt{2} $. The inequality then just $ (t+\sqrt{2})(t^{-1}+\sqrt{2})>3\sqrt{3} $ which is $ t+t^{-1}>\frac{3(\sqrt{3}-1)}{\sqrt{2}} $ which is obviously true since by AM-GM, $ t+t^{-1}\ge2>\frac{3(\sqrt{3}-1)}{\sqrt{2}} $