# Show there exists $x\in[0,2\pi]$ such that the following inequality holds

Let $a_1,a_2,a_3$ be given numbers in $[0,2\pi]$. I am interested in showing there exists $x\in[0,2\pi]$ such that $\dfrac{1}{1-\cos(x-a_1)}+\dfrac{1}{1-\cos(x-a_2)}+\dfrac{1}{1-\cos(x-a_3)}\leq 3$.

I cannot think of a good upper bound for this sum. I am not knowledgeable in more sophisticated trigonometric inequalities which may help in giving an easier upper bound. I tried the derivative test but all it gives is that there is a minimum, nothing more.

I did some numerical plotting and every time I see there exists an $x$ like this. Please help me prove this.

I am not sure if the dual inequality would be easier: $\sum_{i=1}^3\dfrac{\cos(x-a_i)}{1-\cos(x-a_i)}\leq 0$

This is not true. Consider $a_1=0$, $a_2=2\pi/3$, and $a_3=4\pi/3$. For any $x$, there is some $i$ such that $x$ is within $\pi/3$ of $a_i$. We then have $\cos(x-a_i)\in [1/2,1]$ so $\frac{1}{1-\cos(x-a_1)}\geq 2$. Since cosine cannot be less than $-1$, the other two terms are each at least $1/2$, so the sum of all three terms is at least $3$. Moreover, it is impossible to have equality, since then $\cos(x-a_j)$ would have to be $-1$ for two different values of $j$ which is impossible.
• Well, if you want the sum to be small, you want each $\cos(x-a_i)$ to be small, so you want $x$ to be far away from all of the $a_i$. So it is natural to guess that the hardest case would be when the $a_i$ are equally spaced, so that $x$ is forced to be close to some of them. – Eric Wofsey Jun 7 '18 at 7:19