$\left |\sum_{k=1}^n\epsilon_ke^{ik\theta}\right |\ge \sqrt{n}$

Question

I want to prove that
$\inf_{(\epsilon_1,...,\epsilon_n)\in \{-1,1\}^n}\left (\sup_{0\le \theta\le 2\pi}\left ( \left |\sum_{k=1}^n\epsilon_ke^{ik\theta}\right |\right )\right )\ge \sqrt{n}$
which is of course the same as proving that for any $\epsilon_1,...,\epsilon_n\in \{-1,1\}$ there exists $\theta\in [0,2\pi]$ such that $\left |\sum_{k=1}^n\epsilon_ke^{ik\theta}\right |\ge \sqrt{n}$.
I have no idea how to do this, any help would be appreciated. Thank you in advance!

Answer 1

$\frac 1 {2\pi} \int_0^{2\pi} |\sum\limits_{k}^n \epsilon_k e^{ik\theta}|^{2}d\theta=\sum\limits_{k}^n |\epsilon_k|^{2}=n$. Hence there must be at least one value of $\theta \in [0,2\pi]$ such that $|\sum\limits_{k}^n \epsilon_k e^{ik\theta}|^{2} >n$.