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Let $ABCD$ be a convex quadrilateral with the lengths $a, b, c, d$ and the area $S$. The main result in our paper equivalent to:

\begin{equation} a^2+b^2+c^2+d^2 \ge 4S + \frac{\sqrt{3}-1}{\sqrt{3}}\sum{(a-b)^2}\end{equation}

where $\sum{(a-b)^2}=(a-b)^2+(a-c)^2+(a-d)^2+(b-c)^2+(b-d)^2+(c-d)^2$ and note that $\frac{\sqrt{3}-1}{\sqrt{3}}$ is the best constant for this inequality.

Similarly with the same form above apply to n-polygon, I propose a conjecture that:

Let a convex polygon $A_1A_2...A_n$ with the lengths are $a_1, a_2, ...a_n$ and area $S$ we have:

\begin{equation} \sum_i^n{a_i^2} \ge 4\tan{\frac{\pi}{n}}S + k\sum_{i < j}{(a_i-a_j)^2}\end{equation}

I guess that $k=tan{\frac{\pi}{n}}-tan{\frac{\pi}{n+2}}$

I am looking for a proof of the inequality above.

Note that using Isoperimetric inequality we can prove that:

\begin{equation} \sum_i^n{a_i^2} \ge 4\tan{\frac{\pi}{n}}S\end{equation}

Case n=3:

\begin{equation}a^{2} + b^{2} + c^{2} \geq 4 \sqrt{3}S+ (a - b)^{2} + (b - c)^{2} + (c - a)^{2}\end{equation}

Case n=4: our paper, $k=\frac{\sqrt{3}-1}{\sqrt{3}}$, $k=tan{\frac{\pi}{4}}-tan{\frac{\pi}{6}}$

\begin{equation} a^2+b^2+c^2+d^2 \ge 4S + \frac{\sqrt{3}-1}{\sqrt{3}}\sum{(a-b)^2}\end{equation}

See also:

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The conjectured inequality, with $k=\tan{\frac{\pi}{n}}-\tan{\frac{\pi}{n+2}}$, is false for $n=3$. More specifically, the constant factor $k=1$ is optimal in the Hadwiger--Finsler inequality: e.g., consider $a=b=1$ and $c\approx0$.

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  • $\begingroup$ I will chesk by my computer now. Thank You very much. $\endgroup$ – Đào Thanh Oai Jul 10 '18 at 20:50
  • $\begingroup$ I check in long time, but I can not find the optimal formular of $k$ $\endgroup$ – Đào Thanh Oai Jul 11 '18 at 2:25
  • $\begingroup$ So far, we don't even seem to have a good expression for the optimal $k$ covering the cases of $n=3$ and $n=4$. $\endgroup$ – Iosif Pinelis Jul 11 '18 at 11:44
  • $\begingroup$ I think optimal of $k \approx \tan\frac{\pi}{n}- \tan\frac{\pi}{n+2}$ $\endgroup$ – Đào Thanh Oai Jul 11 '18 at 12:16

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