This might be suitable for middle school. It uses very little geometry even. Draw circle with center $B$ and radius $BD$, intersecting $AB$, $CB$, in $F$, $G$, and draw circle with radius $BA$, passing through $C$ and intersecting $EB$ at $H$. Draw a line through $F$, $G$ parallel to $AE$, and join $GH$.
$H$ lies above line $FG$. For considering $D$ and $C$ as pendulums suspended from $B$, then if $D$ in rising to $F$ moves through projected horizontal distance $DA$, then since $C$ is rising more steeply, and hence is less horizontal in its movement, it must rise higher than $F$ to move through a projected horizontal distance $CE=DA$.
Hence $$EH>CG=AF$$(Even the portion of $EH$ between the parallels is greater than $CG$.)
But $$EH=EB-CB$$ and $$AF=AB-DB$$Therefore$$EB-CB>AB-DB$$making$$EB+DB>AB+CB$$