Prove $|\sqrt{x^2+1}-\sqrt{y^2+1}| < |x-y|$ when $x\neq y$

Question

I am trying to prove the basic formula for contraction mapping as an extension of an undergraduate homework problem I had, and have found this to be true using a calculator but can't figure out how to do it formally or even intuitively.

$|\sqrt{x^2+1}-\sqrt{y^2+1}| < |x-y|$ when $x\neq y$

Answer 1

For $x \ne y$, we have $\sqrt{x^2+1} - \sqrt{y^2+1} = \dfrac{x^2-y^2}{\sqrt{x^2+1} + \sqrt{y^2+1}} = \dfrac{(x-y)(x+y)}{\sqrt{x^2+1} + \sqrt{y^2+1}} $ so we need to show that: $$ \left| \frac{(x-y)(x+y)}{\sqrt{x^2+1} + \sqrt{y^2+1}} \right| < |x-y| \iff |x+y| < \left| \sqrt{x^2+1} + \sqrt{y^2+1} \right|$$ Which is clearly true.

Answer 2

Elementary Calculus also works. The derivative of $f(t)=\sqrt{t^2+1}$ is $f'(t)=\frac{t}{\sqrt{t^2+1}}$. If $x\neq y$, then without loss of generality we have $|x|<|y|$, and we have $|f(x)-f(y)|=|f(|x|)-f(|y|)|$, so by the Mean Value Theorem there exists some $|x|<t<|y|$ such that $$|f(|x|)-f(|y|)|=||x|-|y||\cdot |f'(t)|<|x-y|$$ whereby the inequalities $||x|-|y||\leq |x-y|$ and $\frac{t}{\sqrt{t^2+1}}<1$ (for all $t> 0$) were used.