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 $x>1$?
 A: Notice that $(x-\sqrt{x^2-1})(x+\sqrt{x^2-1})=x^2-(x^2-1)=1$, and that $x+\sqrt{x^2-1}>x>1$, and therefore, we have that $x-\sqrt{x^2-1}=\frac{1}{x+\sqrt{x^2-1}}<\frac{1}{1}=1$
A: $f(x)=x-\sqrt{x^2-1}= \frac{(x-\sqrt{x^2-1})(x+\sqrt{x^2-1})}{x+\sqrt{x^2-1}}=\frac{x^2-x^2+1}{x+\sqrt{x^2-1}}=\frac{1}{x+\sqrt{x^2-1}}$
From this, it should be easy to deduct.
A: WLOG:
$x=\csc2y$ with $0\le2y\le\dfrac\pi2$
$$x-\sqrt{x^2-1}=\dfrac{1-\cos2y}{\sin2y}=\tan y$$
As $0\le y\le\dfrac\pi4,\tan y\le1$
A: $$x-\sqrt{x^2-1}<1$$
$$\iff (x-1) < \sqrt{x^2-1}$$
$$\iff x^2-2x+1 < x^2-1,\ \ \mathrm{for} \ \ x>1$$
$$\iff x>1$$
A: 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$$
A: $x-\sqrt{x^2-1}<1\implies x-1<\sqrt{x^2-1}\implies x-1<\sqrt{(x+1)(x-1)}$$\implies \sqrt{x-1}<\sqrt{x+1}$
Since these are algebraic operations perserving order, it is easy to see that the implications follow backwards assuming $x>1$.
A: The "horizontal" hyperbola $ \ x^2 \ - \ y^2 \ = \ 1 \ $ has the asymptotes $ \ y \ = \ \pm x \ $ .  In the first quadrant then, that "branch" of the hyperbola follows $ \ y \ = \ \sqrt{x^2 - 1} \ $ .  At $ \ x \ = \ 1 \ $ , the asymptote passes through $ \ ( 1, \ 1) \ $ , while the hyperbola has a vertex at $ \ (1, \ 0 ) \ $ .  For $ \ x \ > \ 1 \ $ , the branch of the hyperbola approaches the asymptote indefinitely, so $ \ x \ - \ \sqrt{x^2 - 1} \ $ approaches zero thenceforth, making the difference less than 1 everywhere in the interval. 
A: Substitute $x-1=y$. Now you need to show that for $y\geq 0$ the next inequality holds: $$y+1-\sqrt{(y+1)^2-1}<1$$
$$y<\sqrt{y^2+2y}$$
$$y=\sqrt{y^2}\leq \sqrt{y^2+2y}$$
