Is there a simple, intuitive way to see that $f(x)=x-\sqrt{x^2-1}<1$ if $x>1$ Is there a simple intuitive way to show that $f(x)=x-\sqrt{x^2-1}<1$ if $x>1$?
I sense it could be done more simple than this:
1 - take the derivative
$f'(x)=1-\frac{x}{\sqrt{x^2-1}}<0$ if $x>1$ so the slope of $f(x)$ in $(1,\infty)$ is negative.
2 - conclude
Since $f(1)=1$ and the derivative is negative we have showed that $f(x)=x-\sqrt{x^2-1}<1$ if $x>1$
Question: is there a simple, intuitive way to see that $f(x)=x-\sqrt{x^2-1}<1$ if $x>1$?

$f(x)=x-\sqrt{x^2-1}= \frac{(x-\sqrt{x^2-1})(x+\sqrt{x^2-1})}{x+\sqrt{x^2-1}}=\frac{x^2-x^2+1}{x+\sqrt{x^2-1}}=\frac{1}{x+\sqrt{x^2-1}}$
From this, it should be easy to deduct.
