Prove that $1+x\log(x+\sqrt{1+x^2})>\sqrt{1+x^2}$ for all $x\in\mathbb{R}$ I tried to approach this problem the most obvious way. I called to life $f(x) =1+x\log(x+\sqrt{1+x^2})-\sqrt{1+x^2}$ and I wanted to show that f(x) in the limit in $-\infty$ is positive (but it diverges to $\infty$ what may be a problem already, is it?) and it is increasing function. I found out that $f'(x)=\log(x+\sqrt{1+x^2})$ so f'(x) is well defined everywhere since $x+\sqrt{1+x^2}$ is always positive but for $x<0$ we have $f'(x)<0$ so my idea of a proof collapses. Does anyone have any hints how to solve this problem?
 A: You can show your inequality (except for $x= 0$) using hyperbolic functions when you recognize that


*

*$\ln{(x + \sqrt{1+x^2})} = \sinh^{-1}x$


So, set


*

*$x = \sinh y$ and note that

*$\sqrt{1+\sinh^2y}= \cosh y$


Now, your inequality turns into
$$1 + y\sinh y \gt \cosh y \Longleftrightarrow y\sinh y > \cosh y - 1$$
Now, use the series expansions


*

*$\sinh y = \sum_{n=0}^\infty \frac{y^{2n+1}}{(2n+1)!}$

*$\cosh y = \sum_{n=0}^\infty \frac{y^{2n}}{(2n)!}$


A short calculation shows
$$y\sinh y = \sum_{n=\color{blue}{1}}^\infty \frac{y^{2n}}{(2n-1)!} \gt \sum_{n=\color{blue}{1}}^\infty \frac{y^{2n}}{(2n)!} = \cosh y - 1 \mbox{ for all } y \color{blue}{\neq} 0$$
A: You don't need to show it's an increasing function, you have shown that $f$ is increasing (since $x>0 \implies x+\sqrt{1+x^2}>1$ ) for $x>0$ and decreasing for $x<0$  and $f'(0)=0$ this implies that $f$ attains it's minimum value for $x=0$ or equivilently $f(x)\geq{}f(0)=0,\forall{}x\in{}\mathbb{R}$.
A: You are almost there. For $x>0$ the function is increasing and for $x<0$ the function is decreasing. So for both the cases you get what you want. That is for $x>0: f(x)>f(0)$ by increasingness and for $x<0: f(x)>f(0)$ by decreasingness.
