Prove that for every $x,y \in \mathbb{R}$, $\left| \ln \left(\frac{x+\sqrt{a^2+x^2}}{y+\sqrt{a^2+y^2}}\right)\right| \leq \frac{|x-y|}{a}$

Question

Let be $a > 0$. I have to prove that for every $x,y \in \mathbb{R}$,

$$\left| \ln \left(\frac{x+\sqrt{a^2+x^2}}{y+\sqrt{a^2+y^2}}\right)\right| \leq \frac{|x-y|}{a}$$

I think we use here Lagrange but I do not know how.

Any help?

Answer 1

this is the solution,we need to prove that the derivative of this function is bounded above and we need Logarithmic identities of quotinent to construct such function

Answer 2

Let f(x)=ln(x+$\sqrt{a^2+x^2})\,\,\forall\,x\in R$ $$f'(y)=\lim_{x \to y}\vert \frac{f(x)-f(y)}{x-y}\vert$$ $$=\vert \frac{1+\frac{2y}{2\sqrt{a^2+y^2}}}{y+\sqrt{a^2+y^2}} \vert$$ $$=\vert \frac{1}{\sqrt{a^2+y^2}}\vert \le \vert \frac{1}{a} \vert$$ $$\therefore \vert f(x)-f(y) \vert \le \vert \frac{x-y}{a} \vert $$