Prove that $\operatorname{Arsinh}(x) \ge \ln(1+x)$ for $x>-1$ Prove that $\operatorname{Arsinh}(x) \ge \ln(1+x)$ for $x>-1$.
I have solved similar inequalities for other trigonometric functions, but for this one I have no idea where to start, other than the fact that the plot of the functions makes it obvious.
For other examples, I was using the derivatives of various related functions and facts like "its derivative is $>0$".
Some indications would be welcome.

Defining $$f(x)=\operatorname{arsinh}(x)-\ln(x+1),$$ then we get by differentiating with respect to $x$: $$f'(x)=\frac{1}{\sqrt{1+x^2}}-\frac{1}{x+1}.$$
