You should have noticed that GH ≤ (AF + FE + BC + CD)/2, with equality only holding when ABDE is a trapezium (trapezoid) with AE // BD, and the points A, F and E collinear, as well as B, C and D collinear. (In other words, equality only holds when the hexagon degenerates into a trapezium).
Proof: From the 1st result about a quadrilateral, we have GH ≤ (AE + BD)/2. But from the triangle inequality we have AE ≤ AF + FE and BD ≤ BC + CD. By substitition into the 1st inequality the result follows.
Further generalization: From the triangle inequality, the result obviously generalizes by the addition of any number of sides on AE and BD. In this way, it not only generalizes to an octagon, decagon, etc. but also to polygons with an odd number of sides such as a pentagon, septagon, etc.
Modified by Michael de Villiers, 14 June 2014.