Why must $|z|\gt 1$ be the necessary condition 
Question:- If $\left|z+\dfrac{1}{z} \right|=a$ where $z$ is a complex number and $a\gt 0$, find the greatest value of $|z|$.


My solution:- From triangle inequality we have 
$$|z|-\left|\dfrac{1}{z}\right|\le\left|z+\dfrac{1}{z} \right|\le|z|+\left|\dfrac{1}{z}\right| 
\implies |z|-\left|\dfrac{1}{z}\right|\le a\le|z|+\left|\dfrac{1}{z}\right|$$
Now on solving the inequalities separately, we get the following
$$\begin{equation}\tag{1}|z|-\left|\dfrac{1}{z}\right|\le a 
\implies \dfrac{a-\sqrt{a^2+4}}{2}\le|z|\le\dfrac{a+\sqrt{a^2+4}}{2}\end{equation}$$
$$\begin{equation}\tag{2}|z|+\left|\dfrac{1}{z}\right|\ge a 
\implies |z| \in \mathbb{R}-\left(\dfrac{a-\sqrt{a^2-4}}{2},\dfrac{a+\sqrt{a^2-4}}{2} \right)\end{equation}$$
From $(1)$ and $(2)$, we get $$\boxed{|z|_{max}=\dfrac{a+\sqrt{a^2+4}}{2}}$$

My problem with the question:-
The book from which I am solving tells to take note of the following point for the question.

$|z_1+z_2|\ge |z_1|-|z_2|$ and $|z_1+z_2|\ge |z_2|-|z_1|$. Here we have taken $|z|-\dfrac{1}{|z|}$ since we have to find the greatest value of $|z|$ and hence we take the case $|z| \gt 1$

Now all this does is that make the bound tighter nothing else, so why the need of the specific condition $|z|\gt 1$ and also why, only $|z|-\dfrac{1}{|z|}$ provides the maximum value. From, this what I mean to ask is how can we tell even before solving for $|z|$ that $|z|-\dfrac{1}{|z|}$ provides the maximum value


why the need of the specific condition $|z|\gt 1$

$$a=\left|z+\frac 1z\right|\ge|z|-\frac{1}{|z|}\tag1$$
If $0\lt |z|\le 1$, then $-\frac{1}{|z|}\le -1$, so
$$|z|-\frac{1}{|z|}\le 1-1=0\tag2$$
From $(1)$, we have
$$a=\left|z+\frac 1z\right|\ge |z|-\frac{1}{|z|}=(\text{non-positive})$$
which is true since $a\gt 0$, so in this case the maximum value of $|z|$ is $1$. Now, of course, we are interested in the case when $|z|\gt 1$. (so, I think that the book does not say that $|z|\gt 1$ is the necessary condition, and that the book implies that the case $0\lt |z|\le 1$ is trivial.)

why, only $|z|-\dfrac{1}{|z|}$ provides the maximum value.

If we take
$$a\ge \frac{1}{|z|}-|z|$$
we have
$$|z|^2+a|z|-1\ge 0$$
which is not useful to find the maximum value of $|z|$.
