A correct solution came from Bradley student Rob Andry.
Other correct solutions were submitted by
Lou Cairoli, Bryan Fluhrer, Ron Welch, Paul Lee, Cee Ann Franklin, Juan C
Marivela and Ahmed AboAdham.
EA+AB is bigger than BC+CD+DE.
Extend BC and DE until they intersect. Call this intersection F. Then ABFE is a parallelogram. BC+CD+DE is less than BF+FE, which is equal to EA+AB.