TITLE: Showing that a complex number $z$ satisfies $|z|\leq 1$ given a certain condition
QUESTION [5 upvotes]: If $z$ is a complex number such that $$|z-\varepsilon|\leq1\quad\mbox{ and}\quad |z-\varepsilon^{2}|\leq1$$ 
  where $\varepsilon \neq 1$ is the 3th root of unity. 
  Prove that $|z| \leq 1$.

I don't know how to start. Any ideas?

REPLY [10 votes]: This picture may give a geometric idea for a proof: