You don't need to separate out the terms where k=l.

Here's how to solve.

First square the absolute value term as the other anon suggested. The square of the aboslute value is monotonically increasing preserving order, so [math]\alpha_n[/math] is also just squared.

[math]\displaystyle \alpha_n^2 = \underset{(\epsilon_1,\dots,\epsilon_n) \in \{-1,1\}^n }{\inf} \underset{ \theta \in [0,2 \pi] }{\sup} \left| \sum_{k=1}^n \epsilon_k e^{i k \theta} \right|^2 [/math]

Then some algebra. Don't know why I bothered to write all this out.

[math]\displaystyle \left| \sum_{k=1}^n \epsilon_k e^{i k \theta} \right|^2[/math]

[math]\displaystyle \left( \sum_{k=1}^n \epsilon_k e^{i k \theta} \right) \left( \sum_{k=1}^n \epsilon_k e^{-i k \theta} \right)[/math]

[math]\displaystyle \left( \sum_{k=1}^n \epsilon_k \cos(k \theta) + i \sum_{k=1}^n \epsilon_k \sin(k \theta) \right) \left( \sum_{k=1}^n \epsilon_k \cos(k \theta) - i \sum_{k=1}^n \epsilon_k \sin(k \theta) \right) [/math]

[math]\displaystyle \left( \sum_{k=1}^n \epsilon_k \cos(k \theta) \right)^2 + \left( \sum_{k=1}^n \epsilon_k \sin(k \theta) \right)^2 [/math]

[math]\displaystyle \sum_{k=1}^n \sum_{l=1}^n \epsilon_k \epsilon_l \cos(k \theta) \cos(l \theta) + \sum_{k=1}^n \sum_{l=1}^n \epsilon_k \epsilon_l \sin(k \theta) \sin(l \theta) [/math]

[math]\displaystyle \sum_{k=1}^n \sum_{l=1}^n \epsilon_k \epsilon_l \cos((k-l) \theta) [/math]

Now we can establish an inequality by noting the supremum is greater than the average.

The average of [math]\cos(n \theta)[/math] over [math]\theta \in [0,2 \pi][/math] is [math]\mathrm{sinc}(2 \pi n)[/math] so all but the [math]l=k[/math] term of inner sum vanish. All the epsilons are just -1 or 1 so

[math]\displaystyle \underset{ \theta \in [0,2 \pi] }{\sup} \left| \sum_{k=1}^n \epsilon_k e^{i k \theta} \right|^2 >= \sum_{k=1}^n \epsilon_k^2 = n[/math]

The infimum of a finite number of a constant is obviously just the constant so [math]\alpha_n >= \sqrt{n}[/math]