I am looking for a proof of the inequality as follows:

Let $A_1A_2....A_n$ be the regular polygon incribed in a circle $(O)$ with radius $R$. Let $B_1B_2....B_n$ be a polygon incribed the circle $(O)$. We let $x_{ij}=A_iA_j$ and $y_{ij}=B_{i}B_{j}$. Let $f(x)=x^m$ (where $m=1, m=2$), I conjecture that:

$$\sum_{i<j} f(x_{ij}) \ge \sum_{i<j} f(y_{ij})$$

The case $ m = 1 $ was proved in our paper in here


  • $n=3$, let $ABC$ be a triangle with sidelength $a, b, c$ then we have the inequality as follows:

$$a^2+b^2+c^2 \le (\sqrt{3}R)^2+(\sqrt{3}R)^2+(\sqrt{3}R)^2=9R^2$$


$$a+b+c \le 3\sqrt{3}R$$

  • $n=4$, let $ABCD$ be a cyclic quadrilateral with sidelength $a=AB$, $b=BC$, $c=CD$, $d=DA$, $e=AC$, $f=BD$ then we have the inequality as follows:

$$a^2+b^2+c^2+d^2+e^2+f^2 \le (\sqrt{2}R)^2+(\sqrt{2}R)^2+(\sqrt{2}R)^2+(\sqrt{2}R)^2+(2R)^2+(2R)^2=16R^2$$


$$a+b+c+d+e+f \le 4(\sqrt{2}+1)R$$

See also:

  • Dear Mr Dao, I think that $y_{ij}=B_iB_j,$ thank for nice question. – Tran Quang Hung Jul 7 at 3:20
  • @TranQuangHung You are right, Thank You very much. – Đào Thanh Oai Jul 7 at 3:23
up vote 1 down vote accepted

As for the sum of squares (m=2), denoting the vectors $\overline{OB_i}=b_i$ we get $\sum_{i<j} y_{ij}^2=\sum_{i<j} (b_i-b_j)^2=n\sum_i b_i^2-(\sum b_i)^2=n^2R^2-(\sum b_i)^2$ that is maximal if and only if $\sum b_i=0$ --- so, in particular, for a regular polygon.

  • I am sorry, I don’t think the your proof is true. Because because case n=3, 4. The factor of $R^2$ is not $n$ – Đào Thanh Oai Jul 7 at 6:38
  • 1
    Let's remark that sum(b_i)=0 in rectangle for example. Which is not regular. For more details,see our 4217 from "Crux Mathematicorum"-Marinescu and Giugiuc. – user120123 Jul 7 at 6:44
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    ah, $n^2$, not $n$, fixed – Fedor Petrov Jul 7 at 7:24
  • So now, I believe the inequality true with any $ 1 \le m \le 2$ what do You think? Do You have a proof? – Đào Thanh Oai Jul 7 at 8:10
  • 1
    Ah, your $m$ is not integer. This requires extra thinking. – Fedor Petrov Jul 7 at 14:28

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