Solution:

Denote by $X$ and $Y$ the points of intersection of the segments $AB$ and $CD$ with the line through $M$ parallel to $BC$. Similarly, let $U$ and $V$ denote the points of intersection of the segments $AD$ and $BC$ with the line through $M$ parallel to $AB$.

Then $MA < AU + UM = XM + UM$, $MB < MX + XB = MX + MV$, $MC < MV + VC = MV + MY$, and $MD < MY + YD = MY + MU$.

Hence
$$MA + MB + MC + MD < 2(XM + YM) + 2(UM + VM) = 2AD + 2AB = AB + BC + CD + DA.$$