1. ## Demonstration

Prove :

2. 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 $\displaystyle g(x)>0$ .

3. 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

$\displaystyle g(x)=x-\frac{x^2}{2}-\ln(1+x), \ x>0$