Showing $\frac{x}{1+x}<\log(1+x)<x$ for all $x>0$ using the mean value theorem [duplicate]

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 $$x_0>0 \Rightarrow \frac{1}{x_0+1}<1.$$ $$\Rightarrow 1 > \frac{1}{x_0+1}= \frac{ \log(x+1)}{x} \Rightarrow x> \log(x+1)$$

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 $y>1$.

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}

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 $h(x)<h(0)$.