Inequality $|\ln(\frac{x+\sqrt{1+x^2}}{y+\sqrt{1+y^2}})| \leq |x-y|$ So I want to prove that for any $x,y \in \mathbb{R}$ the following inequality is true:
$$|\ln(\frac{x+\sqrt{1+x^2}}{y+\sqrt{1+y^2}})| \leq |x-y|$$
So I know it's true if $x=y$, since it's $0\leq0$.
But I don't know how to evaluate for $x \neq y$.
I found this while going through calculus, so it should probably have a solution that uses calculus basics, I could be wrong though.  
Any help would be appreciated.
 A: One may observe that
$$
\left(\ln \left(x+\sqrt{1+x^2}\right)\right)'=\frac1{\sqrt{1+x^2}}, \quad x \in \mathbb{R},
$$ then, one may write
$$
\begin{align}
\left|\ln \left(\frac{x+\sqrt{1+x^2}}{y+\sqrt{1+y^2}}\right)\right|&=\left|\ln \left(x+\sqrt{1+x^2}\right)-\ln \left(y+\sqrt{1+y^2}\right)\right|
\\\\&= \left|\int_x^y\frac{dt}{\sqrt{1+t^2}}\right|
\\\\&\le \left|\max_{t \in\mathbb{R}} \frac1{\sqrt{1+t^2}}\right|\cdot |x-y|
\\\\&\le |x-y|.
\end{align}
$$
A: Let $f(x)=\mbox{arcsinh}(x)=\ln(x+\sqrt{1+x^2})$ (this is the Inverse Hyperbolic Sine). Then 
$$\left|\ln\left(\frac{x+\sqrt{1+x^2}}{y+\sqrt{1+y^2}}\right)\right|=|f(x)-f(y)|.$$
Now $0<f'(x)=1/\sqrt{1+x^2}\leq 1$, and by the Mean Value Theorem there is a $c$ between $x$ and $y$ such that
$$|f(x)-f(y)|=|f'(c)||x-y|\leq |x-y|.$$
A: Set $x=\sinh(a)$, $y=\sinh(b)$ and your inequality will become
$$ \left|a-b\right|\leq \left|\sinh(a)-\sinh(b)\right| \tag{1}$$
that is a straightforward consequence of Lagrange's theorem, since
$$ \frac{\sinh(a)-\sinh(b)}{a-b}=\cosh(\xi),\qquad \xi\in(a,b) \tag{2} $$
and $\cosh(\xi)\geq 1$.
