First we note that when , and therefore that
We want to compare and to show that does not surpass on .
To do this, we may use a proof by contradiction. It can be easily shown that there are for which . Suppose that for some ,
By the Intermediate Value Theorem, there must be a number between and such that
If this is so, then we must have
By the Mean Value Theorem, there must be a number between and such that the derivative of at , , vanishes.
However, solving for , we see that
provided that , contradicting the vanishing of where was supposed to lie between and .