# 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.

• Let $f(x)=\ln(x+\sqrt{1+x^2})$, then observe that $f'(x)=1/\sqrt{1+x^2}<1$. – Cave Johnson Aug 4 '16 at 8:16

## 3 Answers

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 $$|f(x)-f(y)|=|f'(c)||x-y|\leq |x-y|.$$

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}

• Thank you as well, for giving an interesting approach(integral). – MathIsTheWayOfLife Aug 4 '16 at 9:41

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 $\cosh(\xi)\geq 1$.