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.
 A: the reciprocal of $|-a+\sqrt{a^2-1}|$ is $|-a-\sqrt{a^2-1}|$
as you have discovered $|-a-\sqrt{a^2-1}|>1$ or
its reciprocal$|-a+\sqrt{a^2-1}|<1$
A: $$\begin{align}\left|-a-\sqrt{a^2-1}\right|=\left|-\left(a+\sqrt{a^2-1}\right)\right|=\left|a+\sqrt{a^2-1}\right|>a>1\end{align}$$
$$\begin{align}\left|-a+\sqrt{a^2-1}\right|=\left|-\left(a-\sqrt{a^2-1}\right)\right|=\left|a-\sqrt{a^2-1}\right|=\dfrac{1}{a+\sqrt{a^2-1}}<1\end{align}$$
A: 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.
A: Observe that $$-a+\sqrt{a^2-1}<-a+\sqrt{a^2}=0$$ since $a>1>0$, which implies that $|-a+\sqrt{a^2-1}|=a-\sqrt{a^2-1}$. So, it remains to show that
\begin{align}&a-\sqrt{a^2-1}<1 \\\iff& a-1<\sqrt{a^2-1}\\\iff& (a-1)^2<a^2-1\\\iff& a^2-2a+1<a^2-1\iff1<a
\end{align} which holds by assumption. (Taking roots and squares may be tricky when we have inequalities, but here everything was positive throughout due to $a>1$, so we didn't have to worry about it)
A: $|(-a+\sqrt{a^2-1} ) \frac{-a-\sqrt{a^2-1}} {-a-\sqrt{a^2-1}} |= |\frac{1} {a+\sqrt{a^2-1}}|$ but $a+\sqrt{a^2-1}>1 $ with $a>1$
