Why is $|-a+\sqrt{a^2-1}|<1<|-a-\sqrt{a^2-1}|$ (where $a>1$) true? Why does $|-a+\sqrt{a^2-1}|<1<|-a-\sqrt{a^2-1}|$ (where $a>1$) hold? I understand that $a>1 \implies 1<|-a-\sqrt{a^2-1}|$ and that $|-a+\sqrt{a^2-1}|<|-a-\sqrt{a^2-1}|$
But I can't see why $a>1 \implies |-a+\sqrt{a^2-1}|<1$.
Does anyone see why? Thank you.

We define $f(a)=-a+\sqrt{a^2-1}$. You can easily see that its derivative is defined for $a \in [1,+\infty[$ and $f'(a)=\frac{a-\sqrt{a^2-1}}{\sqrt{a^2-1}} >0$ in this interval (just consider the numerator).
Then $f$ is increasing on the interval $[1,+\infty[$, $f(1)=-1$ and $\lim_{x\to \infty} f(x)=0$ (as $\sqrt{a^2-1} \sim \sqrt{a^2}=a$ for sufficiently large considerations).
So $|f(x)|<1$ on the interval you want.
