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.

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)
