Assume without loss of generality that EF > CE (meaning the proof would work if it was the other way around, you'd just have to switch around a few terms)

Now CF + CE > EF, since the shortest distance between two points is a straight line. (Or if you want to get technical, this is derived from the Triangle Inequality)

This means that CF > EF - CE. But we have that BC > CF (see diagram below). So BC > EF - CE

Also, BH + CH > BC, again since the shortest distance between two points is a straight line (or by the triangle inequality).

This means that BH + CH > EF - CE as desired

QED

Note that each pair of uppercase letters represent the magnitude of the length of a side. we could get a lot more technical in this problem by introducing absolute value signs into it, which kind of makes it resemble the triangle inequality even more. you could probably even use vectors to represent the sides if you wanted. But i didn't think that much fanciness was necessary