Consider the case $AB+CD\ge AD+BC$. The other case can be done similarly. Let $C'\in DC$ such that $KC'\parallel LC$ and $B'\in AB$ such that $B'C'\parallel BC$ (see the first figure). Then the quadrilateral $AB'C'D$ is circumscribed and $KB'$ bisects the angle $AB'C$ thus $B'K\parallel BL$. Also let the line through $C'$ parallel to $KL$ intersect $CL, CB$ at $J, M$ respectively and let the line through $B'$ parallel to $KL$ intersect $BL, BC$ at $I, N$ respectively. Then $$2KL \geq |AB-BC+CD-DA|\iff$$$$2KL\ge BB’+B’C’+C’C+CB\iff$$$$2KL\ge BB’+CC’-NB-CM.\quad (*)$$Now note that we can merge the triangles $C’CM$ and $B’BM$ along the sides $C’M$ and $B’N$ (see the second figure). Moreover the points $I$ and $J$ will be merged together in the incenter of the resulted triangle. Hence if $P$ is the touchpoint of $B’B$ with the incircle then $$(*)\iff KL\ge B’P$$which is obvious as $B’J=KL$.

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