
> 
> **Problem**: Given three positive integers $0 < n\_1 < n\_2 < n\_3$ such that
> $$n\_1 + n\_2 \ne n\_3, \quad n\_2 \ne 2n\_1, \quad n\_3 \ne 2n\_1, \quad n\_3 \ne 2n\_2,$$
> is there always a real number $x$ such that
> $$\cos n\_1 x + \cos n\_2 x + \cos n\_3 x < -2?$$
> 
> 
> 


This is a follow-up of the question in MO: [Is there always a real $x$ such that $\cos n\_1 x + \cos n\_2 x + \cos n\_3 x < -2$?](https://mathoverflow.net/questions/410115/is-there-always-a-real-x-such-that-cos-n-1-x-cos-n-2-x-cos-n-3-x-2)


I checked that (brute force), the statement is true for all positive integers $1\le n\_1 < n\_2 < n\_3 \le 1000$
with $n\_1 + n\_2 \ne n\_3, \, n\_2 \ne 2n\_1, \, n\_3 \ne 2n\_1, \, n\_3 \ne 2n\_2$.
For $n\_3 > 1000$, I randomly generate some $n\_1, n\_2, n\_3$ without finding a counterexample.


I also found a related problem in MO:
[The maximum of a real trigonometric polynomial](https://mathoverflow.net/questions/35538/the-maximum-of-a-real-trigonometric-polynomial).  

Perhaps the problem here is easier to deal with.


My idea is to consider the case $\cos n\_1 x = -1$ or
$\cos n\_2 x = -1$ or $\cos n\_3 x = -1$.  

For example, $\cos n\_3 x = -1$ leads to $x = \frac{(2k\_3 + 1)\pi}{n\_3}$ where $0\le k\_3 \le n\_3; k\_3\in \mathbb{Z}$;  

We consider $f(k\_3) = \cos \frac{(2k\_3 + 1)n\_1\pi}{n\_3} + \cos\frac{(2k\_3 + 1)n\_2\pi}{n\_3} - 1$. However, I failed to go proceed.


I also tried non-negative trigonometric polynomials
and found this article: "Extremal Positive Trigonometric Polynomials", [https://www.dcce.ibilce.unesp.br/~dimitrov/papers/main.pdf](https://www.dcce.ibilce.unesp.br/%7Edimitrov/papers/main.pdf)  

Theorem 4 gives the necessary and sufficient condition for
$f(\theta) = a\_0/2 + \sum\_{k=1}^n a\_k \cos k \theta$ to be non-negative on $\mathbb{R}$.



(update) Is there always a real $x$ such that $\cos n_1 x + \cos n_2 x + \cos n_3 x < -2$?
In principle this problem can be resolved numerically in finite time, by exploiting the dichotomy between structure (small linear relations between the frequencies $n\_1,n\_2,n\_3$) and randomness (equidistribution), though I do not know if the approach below can actually be implemented in a feasible amount of time. (One could in fact use quantitative equidistribution theorems on tori for this, such as Proposition 1.1.17 of [this book of mine](https://terrytao.wordpress.com/books/higher-order-fourier-analysis/), but I will take a more explicit approach here as it will likely give better numerical constants.)


We first claim that if $\cos(n\_1 x) + \cos(n\_2 x) + \cos(n\_3 x) \geq -2$ for all $x$ then there must be a non-trivial linear relation $a\_1 n\_1 + a\_2 n\_2 + a\_3 n\_3 = 0$ between the $n\_1,n\_2,n\_3$ with integers $a\_1,a\_2,a\_3$ with $|a\_1| + |a\_2| + |a\_3| \leq C$ for some effectively computable absolute constant $C$. Indeed, use the Weierstrass approximation theorem to find a polynomial $P: {\bf R} \to {\bf R}$ such that
$$ \int\_0^1 \int\_0^1 \int\_0^1 P( \cos(2\pi \theta\_1) + \cos(2\pi \theta\_2) + \cos(2\pi \theta\_3) )\ d\theta\_1 d\theta\_2 d\theta\_3 > \sup\_{-2 \leq x \leq 3} P(x)$$
(this can be done by choosing $P$ to be large and positive on $[-3,-2]$ and small on $[-2,3]$). Then we have
$$ \int\_0^1 \int\_0^1 \int\_0^1 P( \cos(2\pi \theta\_1) + \cos(2\pi \theta\_2) + \cos(2\pi \theta\_3) )\ d\theta\_1 d\theta\_2 d\theta\_3 
\neq \int\_0^1 P( \cos(2\pi n\_1 \theta) + \cos(2\pi n\_2 \theta) + \cos(2\pi n\_3 \theta) )\ d\theta.$$
But expanding out the polynomial and extracting the Fourier coefficients we see that the LHS and RHS in fact agree unless there is a non-trivial collision $a\_1 n\_1 + a\_2 n\_2 + a\_3 n\_3 = 0$ with $|a\_1| + |a\_2| + |a\_3| \leq \mathrm{deg} P$. In order to make this algorithm run in as feasible a time as possible it is desirable to get $\mathrm{deg} P$ as small as one can; presumably this can be done numerically since for each fixed choice of degree, one has to solve a linear program in the coefficients of $P$.


Next, once one has restricted to the case $a\_1 n\_1 + a\_2 n\_2 + a\_3 n\_3=0$ for some fixed $a\_1,a\_2,a\_3$, one can perform a similar argument to see that either one has $\cos(2\pi \theta\_1) + \cos(2\pi \theta\_2) + \cos(2\pi \theta\_3) \geq -2$ on the entire hyperplane $\{ (\theta\_1,\theta\_2,\theta\_3): a\_1 \theta\_1 + a\_2 \theta\_2 + a\_3 \theta\_3 =0\}$ (in which the answer to your question is negative), or else there must be a second constraint $b\_1 n\_1 + b\_2 n\_2 + b\_3 n\_3 = 0$ with $(b\_1,b\_2,b\_3)$ independent of $(a\_1,a\_2,a\_3)$ and with $|b\_1|+|b\_2|+|b\_3|$ also bounded. After reducing $n\_1,n\_2,n\_3$ to lowest terms, this leaves one with an explicit finite list of candidate triples $(n\_1,n\_2,n\_3)$ for which the question can be decided by case-by-case check. (To reduce the number of cases, one can normalise $0 \leq a\_1 \leq a\_2 \leq a\_3$ (with $n\_1,n\_2,n\_3$ now arbitrary integers) rather than $0 < n\_1 < n\_2 < n\_3$.)


