<s>[INST] Showing $\frac{x}{1+x}<\log(1+x)0$ using the mean value theorem I want to show that $$\frac{x}{1+x}<\log(1+x)<x$$ for all $x>0$ using the mean value theorem. I tried to prove the two inequalities separately.
$$\frac{x}{1+x}<\log(1+x) \Leftrightarrow \frac{x}{1+x} -\log(1+x) <0$$
Let $$f(x) = \frac{x}{1+x} -\log(1+x).$$ Since $$f(0)=0$$ and $$f'(x)= \frac{1}{(1+x)^2}-\frac{1}{1+x}<0$$ for all $x > 0$,  $f(x)<0$ for all $x>0$.  Is this correct so far?
I go on with the second part:
Let $f(x) = \log(x+1)$. Choose $a=0$ and $x>0$ so that there is, according to the mean value theorem, an $x_0$ between $a$ and $x$ with 
$f'(x_0)=\frac{f(x)-f(a)}{x-a} \Leftrightarrow \frac{1}{x_0+1}=\frac{ \log(x+1)}{x}$.
Since $$x_0>0 \Rightarrow  \frac{1}{x_0+1}<1.$$ $$\Rightarrow 1 > \frac{1}{x_0+1}= \frac{ \log(x+1)}{x} \Rightarrow x> \log(x+1)$$
 [/INST] As a consequence of MVT, there is a $\xi\in(0,x)$, such that
$$
\log(1+x)=\log(1+x)-\log 1=x\cdot \left(\log(1+x)\right)'_{x=\xi}=x\cdot\frac{1}{1+\xi}<x.
$$
Let $y=\frac{x}{1+x}$. Then there is $\xi\in\big(0,y\big)$, such that
\begin{align}
\log(1+x)&=-\log\left(\frac{1}{1+x}\right)=\log 1-\log\left(1-\frac{x}{1+x}\right) \\&=
\log 1-\log\left(1-y\right) =
y\left(\log(1-y)\right)'_{y=\xi}=y\cdot\frac{1}{1-\xi}>y=\frac{x}{x+1.}
\end{align}
</s>