let suppose $AB+CD>\max(AM+DM,BN+CN)$and let symetry of $B$ to $N$ , $S$. So $ABDS$ is parallelogram. So $AB=DS$ and $NS=BN$. if $D$ is in triangle of $NCS$ we have $AB+CD=DS+CD<NS+CN=BN+CN$. So $D$ must be outside of triangle $NCS$. So $\angle{SDA}+\angle{ADC}=\angle{BAC}+\angle{ADC}< 180^{\circ}$. Similarly we have $\angle{ABC}+\angle{BCD}< 180^{\circ}$. So we have $\angle{ABC}+\angle{BCD}+\angle{BAC}+\angle{ADC}< 360^{\circ}$. This is contradiction. So it is not possible.

WLOG let's assume that $P = AB \cap CD$ is on rays $BA$ and $CD$. The case where $AB || CD$ can be handled with continuity. Then, we claim that $BN + NC > AB + CD.$ Indeed, we have that $BN + NP > BP$ and $CN + NP > CP$ by the Triangle Inequality, and so adding gives: $$BN + NC > BP + CP - 2NP = AB + CD + (AP + PD - 2NP).$$ However, it's well known that $AP + PD > 2NP$ (construct paralellogram $PAXD$), and so we're done. $\square$

Let $B'$ be reflection of $B$ across $N$ and $A'$ be reflection of $A$ across $M$. Now we have $DC + CA' > DB + BA', CD + DB' > CA + AB'$ so both $D,C$ are out of $B'AC,A'BC$ so $\angle B'DA + \angle ADC , \angle DCB + \angle BCA' < 180$ But then sum of angles of $ABCD$ is less than $360$ so contradiction so one of $D$ or $C$ lies in triangles which implies $AB+CD <\max(AM+DM,BN+CN)$