Prove that $\operatorname{Arsinh}(x) \ge \ln(1+x)$ for $x>-1$

Question

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.

Answer 1

Use the definition of $\operatorname{arcsinh} x$:

$$\operatorname{arcsinh} x = \ln (x + \sqrt{1 + x^2})$$

Answer 2

Note that $$\sinh^{-1}(x) = \ln(x+\sqrt{x^2+1}).$$ Using $x^2+1 \ge 1$, we get $$\sinh^{-1}(x) = \ln(x+\sqrt{x^2+1})\ge\ln(x+1).$$

Answer 3

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

Answer 4

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 $x\in(-1,0)$?

Answer 5

$\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.