TITLE: Is there always a real $x$ such that $\cos n_1 x + \cos n_2 x + \cos n_3 x < -2$?
QUESTION [5 upvotes]: Problem: Given three positive integers $0 < n_1 < n_2 < n_3$. Is there always a real number $x$ such that
$$\cos n_1 x + \cos n_2 x + \cos n_3 x < -2?$$

REPLY [17 votes]: The answer is negative. If $n_3=n_1+n_2$, then
$$\cos n_1 x + \cos n_2 x + \cos n_3 x\geq -2.$$
Indeed, the left-hand side equals
$$(1+\cos n_1 x)(1+\cos n_2 x)-1-\sin n_1 x\sin n_2 x,$$
where $1+\cos n_j x\geq 0$, and $\sin n_1 x\sin n_2 x\leq 1$.
Furthermore, if $n_2=2 n_1$ and $n_3=3 n_1$, then
$$\min_{x\in\mathbb{R}}\,(\cos n_1 x + \cos n_2 x + \cos n_3 x)=-\frac{17+7\sqrt{7}}{27}\approx-1.31557.$$
More generally, the minimum of the Dirichlet kernel is known to great precision, see e.g. here.