§ Geometric proof of e^x >= 1+x, e^(-x) >= 1-x

Let's concentrate on the e^x >= 1 + x part.

1. The tangent of e^x at x = 0 is 1 + x, since the taylor series of e^x truncated upto x is 1 + x.
2. e^x is a strongly convex function, since (e^x)'' = e^x which is positive everywhere. Hence, e^x will always lie above its tangent.

Similarly for e^(-x), working through the math:

1. 1 -x is tangent at x=0 to e^(-x)
2. (e^(-x))'' = -(e^(-x))' e^(-x) which is again positive everywhere, and hence, e^(-x) is strongly convex.