1
$\begingroup$

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: $\sum_{i=1}^3\dfrac{\cos(x-a_i)}{1-\cos(x-a_i)}\leq 0$

$\endgroup$
0
1
$\begingroup$

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.

$\endgroup$
3
  • $\begingroup$ Thank you! How did you see this? $\endgroup$ – Andrew Richards Jun 7 '18 at 7:03
  • 1
    $\begingroup$ 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. $\endgroup$ – Eric Wofsey Jun 7 '18 at 7:19
  • $\begingroup$ Thank you for sharing your method of thinking. This will be helpful for me. $\endgroup$ – Andrew Richards Jun 7 '18 at 7:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.