Prove by Mean Value Theorem $\ln(1+x) \ge\frac{x}{1+x}$ for $x>-1$

Question

Prove for $\;x>-1\,,\,\;\ln(1+x)\geqslant\dfrac{x}{1+x}\,.$

I have proved that for every $\,x>0\,,\;\ln(1+x)\geqslant\dfrac{x}{1+x}\,.$

However, I am not sure if it is suitable for $\,x>-1\,.$

MY WORK

Taking $f(x)=\ln(1+x)$, which is differentiable on $(0,x)$ and continuous on $[0,x]$. By using Mean Value Theorem, there exist $c$ in $(0,x)$ such that $$f'(c)=\dfrac{\ln(1+x)}{x}=\dfrac{1}{1+c}$$ $$\ln(1+x)=\frac{x}{1+c}$$

Since $0<c<x$, $$\dfrac{x}{1+c}>\dfrac{t}{1+t}$$

therefore, $$\ln(1+x)>\dfrac{x}{1+x}$$ for $x>0$

Some help would be much appreciated!

Answer 1

For $x<0$, $$ln(1+x)=\frac{x}{1+c}=-\frac{|x|}{|1+c|}\ge -\frac{|x|}{|1-|x||}=\frac{x}{1+x}$$ where $c\in (x,0)$.