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 .