Is there always a positive $x$ that satisfies $\cos(n_1x)\leq0$, $\cos(n_2x)\leq0$, $\cos(n_3x)\leq0$ for given distinct positive integers $n_i$?

Solution 1:

For each set of positive integers $n_1,n_2,n_3,$
$$\cos(n_1x),\cos(n_2x),\cos(n_3 x)\leq0$$ has a real solution.

Observation: if the statement specifies greater than three positive integers instead, then it is false: a counterexample is $n_4=4.$

Proof (but missing Case 2)

It suffices to show that for ascendingly-ordered positive integers $a,b,c,$ $$\cos\left(\frac\pi2ax\right),\cos\left(\frac\pi2bx\right),\cos\left(\frac\pi2cx\right)\leq0$$ has a real solution.

Noting that $$\cos\left(\frac\pi2nx\right)\leq0\\\iff\\ x\in\ldots\cup\left[-\frac3n,-\frac1n\right]\cup\left[\frac1n,\frac3n\right]\cup\left[\frac5n,\frac7n\right]\cup\left[\frac9n,\frac{11}n\right]\cup\ldots,$$ there are four possible cases:

1. $c\leq 3a:$

   \begin{align}&a<b &\text{and} &&b<c &&\text{and} &&a<c\leq3a<3c\\ &\frac1b<\frac1a &\text{and} &&\frac3c<\frac3b &&\text{and} &&\frac1c<\frac1a\leq\frac3c<\frac3a \end{align} $$[\frac1a,\frac3c]\subseteq[\frac1a,\frac3a]\cap[\frac1b,\frac3b] \cap[\frac1c,\frac3c] \\\cos\left(\frac\pi2ax\right), \cos\left(\frac\pi2bx\right), \cos\left(\frac\pi2cx\right)\leq0 \quad\text{ on } \left[\frac1a,\frac3c\right].$$

2. $\displaystyle b<3a<c<\frac{2ab}{3a-b}:$

   This section needs to be filled in; for convenience: Desmos links 1 & 2. I have duplicated the next four lines from Case 3 only so that this Answer is at least a proof of the given statement were it to require only $\mathbf{\mathit{\cos(n_1x),\cos(n_2x)}}$ to be nonpositive.

   $$a<b<3a<3b \\\frac1b<\frac1a<\frac3b<\frac3a \\ [\frac1a,\frac3b]\subseteq[\frac1a,\frac3a]\cap[\frac1b,\frac3b] \\\cos\left(\frac\pi2ax\right), \cos\left(\frac\pi2bx\right)\leq0 \quad\text{ on } \left[\frac1a,\frac3b\right].$$

3. $b<3a<c$ and $\displaystyle c\geq\frac{2ab}{3a-b}:$

   $$a<b<3a<3b \\\frac1b<\frac1a<\frac3b<\frac3a \\ [\frac1a,\frac3b]\subseteq[\frac1a,\frac3a]\cap[\frac1b,\frac3b] \\\cos\left(\frac\pi2ax\right), \cos\left(\frac\pi2bx\right)\leq0 \quad\text{ on } \left[\frac1a,\frac3b\right].$$

   Since $\displaystyle \frac3b-\frac1a=\frac{3a-b}{ab}\geq\frac2c,$ which is the size of the smallest interval on which $\displaystyle\cos\left(\frac\pi2cx\right)$ must be somewhere nonpositive, some point in $\displaystyle\left[\frac1a,\frac3b\right]$ in fact contains a solution for $\cos\left(\frac\pi2ax\right),\cos\left(\frac\pi2bx\right),\cos\left(\frac\pi2cx\right)\leq0.$

4. $b\in\big[(2m{+}1)a,(2m{+}3)a\big]$ and $m\in\mathbb Z^+:$

   \begin{align}b&\leq(2m+3)a &\text{and} &&b&\geq(2m+1)a\\ \frac1a&\leq\frac{2m+3}b &\text{and} &&\frac3a&\geq\frac{6m+3}b\\&\leq\frac{4m+1}b &&&&>\frac{4m+3}b\\ &\frac1a\leq\frac{4m+1}b<\frac{4m+3}b<\frac3a\end{align} $$\left[\frac{4m+1}b,\frac{4m+3}b\right]\subseteq \left[\frac1a,\frac3a\right]\cap\left[\frac{4m+1}b,\frac{4m+3}b\right]\\ \cos\left(\frac\pi2ax\right),\cos\left(\frac\pi2bx\right)≤0 \quad\text{ on } \left[\frac{4m+1}b,\frac{4m+3}b\right].$$

   Since $\displaystyle\frac{4m+3}b-\frac{4m+1}b=\frac2b>\frac2c,$ which is the size of the smallest interval on which $\displaystyle\cos\left(\frac\pi2cx\right)$ must be somewhere nonpositive, some point in $\displaystyle\left[\frac{4m+1}b,\frac{4m+3}b\right]$ in fact contains a solution for $\cos\left(\frac\pi2ax\right),\cos\left(\frac\pi2bx\right),\cos\left(\frac\pi2cx\right)\leq0.$