1
$\begingroup$

Four years ago, I proposed an inequality related to area and sides of a polygon.

After computer checking, I conjecture that the previous inequality can be strengthened as follows:

Let $A_1A_2\cdots A_n$ be an arbitrary polygon, then: $$4(\tan\frac{\pi}{n})Area(A_1A_2\cdots A_n) \le \sum_{i=1}^n(A_iA_{i+1})^2-\frac{1}{n-2}\sum_{i=1}^n(A_iA_{i+1}-A_{i+1}A_{i+2})^2$$

Question: Can you prove this conjectured inequality?

PS: I also found that for an equiangular convex pentagon, $Area(ABCDE)=\frac{1}{4}(\cot\frac{\pi}{5})(a^2+b^2+c^2+d^2+e^2-\frac{(a-b)^2+(b-c)^2+(c-d)^2+(d-e)^2+(e-a)^2}{\sqrt{5}})$

enter image description here

Improve this question
$\endgroup$
3
  • $\begingroup$ It's not clear to me whether you have a proof of the inequality, or whether it is a conjecture. $\endgroup$
    – Gerry Myerson
    Commented Jul 9, 2022 at 0:30
  • $\begingroup$ Thank you very much, I added word "conjecture" @GerryMyerson $\endgroup$
    – Đào Thanh Oai
    Commented Jul 9, 2022 at 0:32
  • 1
    $\begingroup$ Is the inequality sharp, and if so which polygons achieve equality? $\endgroup$
    – user44143
    Commented Jul 27, 2022 at 0:26

0

Reset to default

You must log in to answer this question.