Greatest value of $|z|$ given that $\left|z- \frac 4z \right| = 2$. Why does using triangle inequality in this way not work? 
Greatest value of $|z|$ given that $\left|z- \dfrac 4z \right| = 2$ is? 

This has been asked here before but my question is about a specific method that doesn't seem to work. 
The method is: 
$\left|z- \dfrac 4z \right|\le |z|+ \left|\dfrac{4}{z}\right|\implies 2 \le |z|+ \dfrac{4}{|z|}$
$\implies |z|^2 - 2|z| +4\ge 0$ since $|z| > 0$ (here)
which is true for all values of $|z|$. 
Hence $|z| \in (0, \infty)$. 
Then why is the maximum given by $|z| =1 + \sqrt 5$, in other words, what's the fault in this method ?
A user has also asked that in one of the comments but hasn't received a reply there. 
 A: You can apply the triangle inequality this way:
$2 \geq |z-\frac{4}{z}| \geq |z|-|\frac{4}{z}|$, whence $|z|^2-2|z|-4 \leq 0$ whence $|z| \leq 1 + \sqrt{5}$ and note(very important!) that equality is indeed attained for $z=1+\sqrt{5}$
A: I was solving a similar problem in this question
edit 
Assume $z\neq 0.\;$
$\left|z-\frac{4}{z}\right|$ is the distance of points representing $z$ and $\frac{4}{z}.$ The maximum of $|z|$ is achieved when $0, z, \frac 4z$ are collinear (see bellow) and we have to take the difference, not the sum of absolute values.
Solution


*

*If $z$ is real, then $z, \frac 4z$ lie on the same half-line starting in $0$ and we have 
$$\left|z-\frac 4z \right|=|z|-\frac {4}{|z|}\quad \text{or} \quad \left| z-\frac 4z \right|=\frac{4}{|z|}-|z|,$$
and so $$2=|z|-\frac {4}{|z|} \quad \text{or} \quad 2=\frac{4}{|z|}-|z|.$$
Multiplying  by $|z|$ and solving the quadratic equations gives positive solutions $|z|=\sqrt 5 +1\;$ from the first and $|z|=\sqrt 5 -1\;$ from the second one. Note that $\sqrt 5 -1=\frac{4}{\sqrt 5 +1}.$


*

*If $z=ib, b\in \mathbb{R},$ then $0$ lies on the segment with bounds $z, \frac 4z$. The equation to solve is then $2=|z|+\frac {4}{|z|}$ and doesn't have solution.

*In all other cases, by triangle inequality is $|z|<\sqrt 5 +1.$
The maximal value of $|z|$ is $\sqrt5 + 1,$ the only convenient numbers are $z=\pm(\sqrt 5 +1).$ 
