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 $$x_0>0 \Rightarrow  \frac{1}{x_0+1}<1.$$ $$\Rightarrow 1 > \frac{1}{x_0+1}= \frac{ \log(x+1)}{x} \Rightarrow x> \log(x+1)$$
 A: By Definition of log(which already a kind of mean value theorem ) we have
For any $x>0$ We have 
$$\frac{x}{x+1} =\int_{0}^x\frac{dt}{x+1} \le \int_{0}^x\frac{dt}{t+1} =\color{red}{\ln(x+1 )}=\int_{0}^x\frac{dt}{t+1}  \le \int_{0}^x\frac{dt}{1}  = x $$
Thus,
$$\frac{x}{x+1} \le \ln(x+1 ) \le   x $$
A: Substitute $x=y-1$ to get $$1-y^{-1}<\log y<y-1$$ for all $y>1$. Now note  $$\int_1^y t^{-2}dt <\int_1^y t^{-1}dt <\int_1^y dt $$ for $y>1$.
A: 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}
A: I wanted to ask if there are other ways to solve this and wrote my solution on the way, this question didn't turn up at first when I was searching. 
Well here's another way to solve this that doesn't use MVT: 
Define $g(x)=\ln(1+x)-x, \ g'(x)=\frac1 {1+x} -1$
$g'(x)=0\implies x=0$ and after checking it is a maximum. so it holds that $g(x)<g(0)$.
Define: $h(x)=\frac x {1+x}-\ln(1+x),\ h'(x)=-\frac x {(1+x)^2}$
$h'(x)=0 \implies x=0$ and it's also a maximum so $h(x)<h(0)$.
A: Sounds like you're not asking for a solution (which is great!) but just confirmation that what you have so far is correct.  Yes (after some edits), so far so good, except how do you justify that $f(0) = 0$ and $f'(x) < 0$ for $x > 0$ implies that $f(x) < 0$ for $x > 0$?  What's the key part of your question that you haven't used yet?
