suppose $AE \cap CD =F $, the incircle touches $AB,BC,CD,EA$ at $X,Y,Z,W$ respectively,and let $r$ be the radius of the incircle. Since $AX=AW$, we have $\tan \angle XAC=\dfrac {r} {AX} =\dfrac {r} {AW}= \tan \angle WAC$ which implies $ \angle BAC=\angle FAC$ similarly, $ \angle BCA=\angle FCA$, so with $AC=AC$, we have $\triangle BAC \cong \triangle FAC$,which implies $AB=AF,BC=CF$ then $DE<EF+DF \Rightarrow AE+DE+DC<AE+EF+DF +DC=AF+FC=AB+BC$. Q.E.D.

The center of the inscribed circle lie on AC we know that intersection of bisectors gives the center of inscribed circle , so angles ∠BAC=∠CAE=x and from the other side ∠DCA=∠ACB=y Let extension of AE and CD intersect on L and ∠ABC=∠ALC because other two angles are equals to each other one By similarity of two triangles (ABC)~(ALC) AB/AL=BC/BL=AC/AC=1 k=1 Therefore AL=AB, CL=CB, AL=AE+EL, CL=CD+DL ED+DL>ED by (EDL) triangle inequality AB+BC=AL+CL=AE+EL+CD+DL>AE+ED+CD