If $\left|z+\frac{2}{z}\right |=2$, then prove that max value of $|z|$ is $\sqrt 3 + 1$ (Corrected version) Then the expression will be 
$$|z|+\frac{2}{|z|}\ge 2$$

Solving the quadratic inequality 

$$|z|\ge1+i$$ and $$|z|\le1-i$$

I don’t know how to solve further. Please help

Thanks! 

this is the corrected version of an identical question asked by me. I am sorry for the inconvenience caused
$|z^{2}+2|=2|z|$ gives $|z|^{2}-2 \leq 2|z|$ or $(|z|-1)^{2} \leq 3$. Hence $|z| \leq 1+\sqrt 3$. I will let you find a value of $z$ for which equality holds. 

You have to choose $z$ to be $(1+\sqrt 3)e^{i\theta}$ for  some $\theta$.
Let $z=re^{i\theta}$. Then, $\left|z+\frac{2}{z}\right |=\left|r+\frac{2}{r}e^{-i2\theta}\right |=2$ leads to

$$\left(r+\frac2r \cos 2\theta \right)^2+\frac{4}{r^2}\sin^2 2\theta= 4$$

or,

$$r^2+\frac{4}{r^2}=8\sin^2\theta\le 8$$

Rearrange the inequality to get,

$$(r^2 -4 )^2\le 12$$

which leads to

$$r\le \sqrt{4+2\sqrt3}=1+\sqrt3$$