Proof of complex inequality: $|z-1|\le|\sqrt{z^2-1}|\le|z+1|$ for $\operatorname{Re}(z)>0$ Hello I'm trying to prove some inequality of complex number but I'm getting stuck: the inequality is as follows:

$$|z-1|\le|\sqrt{z^2-1}|\le|z+1|\quad\mbox{ for } \operatorname{Re}(z)>0$$

Now what I do is as follow:


*

*$|z-1|^2=|z|^2-2\operatorname{Re}(z)+1 \implies |z-1|^2\le|z|^2+1 $ (and now I'm stuck).

*$|z+1|^2=|z|^2+2\operatorname{Re}(z)+1 \implies |z+1|^2\ge|z|^2+1 $ (and now I'm stuck)


I feel I'm close but I lack the transition from what I get to where in need to go.
Hope someone will be able to enlighten me in that matter.
 A: You're actually on the way to an answer. What confused me about the question, and may be confusing you as well, is the use of the square root symbol, when the square root of a complex number is not uniquely defined.
But if $w$ is either of the two square roots of $z^2 - 1$ (they are "both" equal to $0$ if $z = 1$, of course!), then $w^2 = z^2 - 1 = (z - 1)(z + 1)$, therefore $|w|^2 = |w^2| = |z - 1||z + 1|$; and what we have to prove is $|z - 1| \leqslant |w| \leqslant |z + 1|$.
But you have proved $|z - 1| \leqslant |z + 1|$ (because $\operatorname{Re}(z) \geqslant 0$ - strict inequality not needed). Therefore ...?
A: We have $|z-1|^2=|z|^2-2Re(z)+1$ and $|z+1|^2=|z|^2+2Re(z)+1$; and because $Re(z)\gt0$, it can be directly stated that $|z-1|^2\lt|z+1|^2$, whence $|z-1|\lt|z+1|$.
If $w$ is a square root of $z^2-1$, the required inequality is equivalent to:
$$
|z-1|^2\le|w|^2\le|z+1|^2.
$$
Because $|w|^2 = |w^2| = |z^2 - 1| = |(z - 1)(z + 1)|$, this inequality is in turn equivalent to:
$$
|z-1|^2\le|z-1||z+1|\le|z+1|^2.
$$
$\rightarrow (\sqrt{(z-1)(z^*-1)})^2\le\sqrt{(z-1)(z^*-1)}\sqrt{(z+1)(z^*+1)}\le(\sqrt{(z+1)(z^*+1)})^2$
Now because $Re(z)\gt0$ and the fact that $|z-1|\lt|z+1|$ which means $\sqrt{(z-1)(z^*-1)}\lt\sqrt{(z+1)(z^*+1)}$ we can directly state that the inequality holds.
side note, if $z=1$ then $w=0$ and the inequality also holds {$0\le0\le2$}.
