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:**

- Hadwiger–Finsler inequality, $k=1 \approx 1.00550827956352=tan{\frac{\pi}{3}}-tan{\frac{\pi}{5}}$:

\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:**