§ Geometric proof of
Let's concentrate on the
e^x >= 1+x,
e^(-x) >= 1-x
e^x >= 1 + x part.
- The tangent of
x = 0 is
1 + x, since the taylor series of
e^x truncated upto
1 + x.
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.
e^(-x), working through the math:
1 -x is tangent at
(e^(-x))'' = -(e^(-x))' e^(-x) which is again positive everywhere, and hence,
e^(-x) is strongly convex.