<s>[INST] 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 $x>1$?
 [/INST] For $x\gt 1$ one has $2x\gt 2$ thus $2x-2\gt 0$. This means
$x^2-1\gt x^2-1-(2x-2)$$
But the r.h.s of the previous inequality is $(x-1)^2$ so we have $(x^2-1)\gt (x-1)^2$.
Now take the square root and bear in mind that $\sqrt{(x-1)^2}=x-1$ because $x\gt 1$ we get
$\sqrt{x^2-1}\gt x-1$ and so
$$-\sqrt{x^2-1}\lt -(x-1)$$
Now add $x$ to get the inequality expected
$$x-\sqrt{x^2-1}\lt x-x+1=1$$
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