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

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  • $\begingroup$ What a surprise to find you here! $\endgroup$
    – user164469
    Commented Dec 6, 2021 at 7:49
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    $\begingroup$ @haidangel Well, nice to see you here. $\endgroup$
    – River Li
    Commented Dec 6, 2021 at 9:28

1 Answer 1

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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.

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