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.
 A: Note that
$$\sinh^{-1}(x) = \ln(x+\sqrt{x^2+1}).$$
Using $x^2+1 \ge 1$, we get
$$\sinh^{-1}(x) = \ln(x+\sqrt{x^2+1})\ge\ln(x+1).$$
A: Let $f\colon(-1,+\infty)\longrightarrow\mathbb R$ be the function defined by $f(x)=\operatorname{arcsinh}(x)-\ln(1+x)$. It is clear that $f(0)=0-0=0$. On the other hand$$\bigl(\forall x\in(-1,+\infty)\bigr):f'(x)=\frac1{\sqrt{1+x^2}}-\frac1{1+x}.$$But, if $x\geqslant0$, then $(1+x)^2=1+x^2+2x\geqslant1+x^2$, and therefore $1+x\geqslant\sqrt{1+x^2}$, from which it follows that $f'(x)\geqslant0$. Since $f(0)=0$,$$\bigl(\forall x\in[0,+\infty)\bigr):f(x)\geqslant0.$$Can you deal with the case $x\in(-1,0)$?
A: $\begin{array}\\
\sinh(\ln(1+x))
&=\frac12(e^{\ln(1+x)}-e^{-\ln(1+x)})\\
&=\frac12(1+x-\frac1{1+x})\\
&=\frac12(\frac{(1+x)^2-1}{1+x})\\
&=\frac{2x+x^2}{2(1+x)}\\
&=\frac{2x+2x^2-x^2}{2(1+x)}\\
&=x-\frac{x^2}{2(1+x)}\\
&\le x
\qquad\text{for } x > -1\\
\end{array}
$
Since
$\sinh(x)$
and its inverse are monotonic increasing.
the result follows. 
A: Use the definition of $\operatorname{arcsinh} x$:
$$\operatorname{arcsinh} x = \ln (x + \sqrt{1 + x^2})$$
A: Defining $$f(x)=\operatorname{arsinh}(x)-\ln(x+1),$$ then we get by differentiating with respect to $x$: $$f'(x)=\frac{1}{\sqrt{1+x^2}}-\frac{1}{x+1}.$$
