calculus - how to prove that $ln(1+x)< x$

I want to prove that: $\ln(x+1)< x$.

My idea is to define: $f(x) = \ln(x+1) - x$, so:

$f'(x) = \dfrac1{1+x} - 1 = \dfrac{-x}{1+x} < 0, \text{ for }x >0$.

Which leads to $f(x)

Is that a valid proof? Any other ideas?

Thanks.

Answer

I think your approach is correct but you need to add some more details. Based on your approach let $f(x) = \log(1 + x) - x$ so that $f(0) = 0$. Clearly $$f'(x) = -\frac{x}{1 + x}$$ and hence $f'(x) > 0$ if $-1 < x < 0$ and $f'(x) < 0$ if $x > 0$. It follows that that $f(x)$ in increasing in $(-1, 0]$ and decreasing in $[0, \infty)$. Thus we have $f(x) < f(0)$ if $-1 < x < 0$ and $f(x) < f(0)$ if $x > 0$. It thus follows that $f(x) \leq f(0) = 0$ for all $x > -1$ and there is equality only when $x = 0$. So we can write $$\log(1 + x) \leq x$$ for all $x > -1$ and there is equality only when $x = 0$.

Note: We have considered $x > -1$ because $\log(1 + x)$ is not defined if $x \leq -1$.