Hi

There might be a better way to do this, but IŽll give one solution.

Note: is the first two terms of the maclaurin series of .

Anyway, let , then we want to show that for all .

Solution: is continous and differentiable for all , with . (I have already simplified the derivative here).

Note tha we are only interested in , so we donŽt have to worry about the denominator of the derivative becoming zero.

Looking at we see that the only stationary point is . And for all we have that .

And because , and the function is strictly decreasing for all , it immediately follows that .

Now for the right inequality:

You could do the same thing for , and show that .