Let $F$ be the reflection of $D$ about $AC$. We have that $\angle FCB = \angle ECB - \angle ECF = \angle CEB - \angle ECD = \angle CDB = \angle CBE$ $BE=CD=CF$ and $BC=BC \implies \triangle BCE \equiv \triangle CBF$ So, $BF = CE \leq DC \implies AD+DC \geq AF+FB \geq AB$ So, $AD+DC \geq AB$.

Note that if $AC \geq AB$ or $AD \geq AE$ we are done, as then $AD+DC > AC \geq AB$ and $AD+DC \geq AE+BE =AB$. Now, assume that $AC<AB$ and $AD<AE$. As $AC<AB$ we obtain $\angle ACB>\angle ABC$, that is $\angle BEC>\angle ABD+\angle DBC$, hence $\angle ABD<\angle BEC-\angle DBC=\angle BEC-\angle BDC=\angle ACD$, that is $\angle ABD<\angle ACD$. Moreover, $AD<AE$ and so $\angle ADB>\angle AED=\angle BEC=\angle ACB$, that is $\angle ADB>\angle ACB$. Now, since $\angle ADB>\angle ACB$, circle $(ADB)$ intersects $AC$ at an interior point $R$, implying $\angle ABD=\angle ARD=\angle ACD+\angle RDC>\angle ACD,$ a contradiction.

Let $D'$ be the reflection of $D$ over $AC$. $AD'B$ is either a line or a triangle, so $AD+BD'=AD' + BD' \ge AB$. We want to prove that, $BD'\le CD$. It is easy to check that $BCED'$ is an isosceles trapezoid as $BE=CD=CD'$ and $\angle CBE=\angle CDE=\angle CD'E\implies BCED' \text{ cyclic}$. Finally, $\angle DEC = \angle EBC+\angle BCE > \angle CDE\implies CD > CE = BD'$ as desired.

Very nice problem : Let $D'$ be the reflection of $D$ about $AC$. Now since we know $\angle CBD<\frac{\pi}{3}$ simple angle chasing gives $\angle{CBD}=\angle{CDB}>\angle{BCD}$. So $CD>BD$ and we can simply use the triangle inequality on $ABD$.