Show that for every and ,
the inequality is clearly true for x = 0. so we assume that x > 0. so the claim is:
suppose for now that a > 1. since we have thus: hence which obviously holds
for a = 1 as well. therefore: which proves (1). Q.E.D.
A simple proof for the case .
By the Taylor's expansion, we have since and by the binomial expansion, we have
Using the fact that and the inequalities above, we get the resired result since is increasing in when