Why Is ln(1+x) Greater Than x/(2+x) for x > 0?

[Lisa91]

How to prove that for $$x>0$$
$$\ln(1+x) > \frac{x}{2+x}$$ is true?

 

[Fernando Revilla]

How to prove that for $$x>0$$
$$\ln(1+x) > \frac{x}{2+x}$$ is true?

Yes, it is true. One way to prove it: denote, $f(x)=\ln(1+x) - \dfrac{x}{2+x}$, then $f'(x)=\ldots=\dfrac{x^2+2x+2}{(1+x)(2+x)^2}>0$ for all $x>0$. This means that $f$ is strictly increasing in $(0,+\infty)$. On the other hand,

$\displaystyle\lim_{x\to 0^+}f(x)=\displaystyle\lim_{x\to 0^+}\left(x+o(x)-\frac{x}{2+x}\right)=0$.