With $z\in \mathbb C$ find the maximum value for $\lvert z\rvert$ such that $\lvert z+\frac{1}{z}\rvert=1$ 
With $z\in \mathbb C$ find the maximum value for |z| such that 
  $$\left\lvert z+\frac{1}{z}\right\rvert=1.$$
  Source: List of problems for math-contest training.

My attempt: it is easy to see that the given condition is equivalent
$$\lvert z^2+1\rvert=\lvert z\rvert$$ 
and if $z=a+bi$,
\begin{align*}
\lvert z\rvert&=\lvert a^2-b^2+1+2ab i\rvert=\sqrt{(a^2-b^2+1)^2+4 a^2b^2}\\
&=\sqrt{(a^2-b^2)^2+2(a^2-b^2)+1+4a^2b^2}\\
&=\sqrt{(a^2+b^2)^2+2(a^2-b^2)+1}\\
&=\sqrt{\lvert z\rvert^4+2(a^2-b^2)+1}
\end{align*}
I think it is not leading to something useful... the approach I followed is probably not useful.
Hints and answers are welcomed.

Since $$|z| = |z^2-(-1)|\geq |z|^2-|-1|$$
Write $r=|z|$ and we have $$r^2-r-1\leq 0\implies r\in [0,r_2]$$ where $$r_{1,2} ={1\pm \sqrt{5}\over 2} $$ 
So $r\leq \displaystyle{1+ \sqrt{5}\over 2}$
