Prove :
Hi
There might be a better way to do this, but IŽll give one solution.
Note: $\displaystyle x-\frac{x^{2}}{2}$ is the first two terms of the maclaurin series of $\displaystyle ln(1+x) $ .
Anyway, let $\displaystyle f(x)=x-\frac{x^{2}}{2}-ln(1+x) $ , then we want to show that $\displaystyle f(x)<0 $ for all $\displaystyle x>0 $.
Solution: $\displaystyle f $ is continous and differentiable for all $\displaystyle x>-1 $ , with $\displaystyle f'(x)=\frac{-x^{2}}{1+x} $ . (I have already simplified the derivative here).
Note tha we are only interested in $\displaystyle x>0 $, so we donŽt have to worry about the denominator of the derivative becoming zero.
Looking at $\displaystyle f'(x) $ we see that the only stationary point is $\displaystyle x=0 $ . And for all $\displaystyle x>0$ we have that $\displaystyle f'(x)<0 $ .
And because $\displaystyle f(0)=0 $ , and the function is strictly decreasing for all $\displaystyle x>0 $ , it immediately follows that $\displaystyle f(x)<0 $ .
Now for the right inequality:
You could do the same thing for $\displaystyle g(x)=x-ln(1+x) $ , and show that $\displaystyle g(x)>0 $ .
Let $\displaystyle f(x)=\ln(1+x)-x, \ f0,\infty)\to\mathbf{R}$
$\displaystyle f'(x)=-\frac{x}{1+x}<0, \ \forall x>0$, so f is strictly decreasing.
Then, $\displaystyle x>0\Rightarrow f(x)<f(0)=0\Rightarrow\ln(1+x)<x$
Try to prove the other inequality in a similar way by using the function
$\displaystyle g(x)=x-\frac{x^2}{2}-\ln(1+x), \ x>0$