Let $ABCD$ be a convex quadrilateral with the lengths $a, b, c, d$ and the area $S$. The main result in [our paper](https://ijgeometry.com/wp-content/uploads/2018/04/81-86.pdf) 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](https://en.wikipedia.org/wiki/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](https://en.wikipedia.org/wiki/Hadwiger%E2%80%93Finsler_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](https://ijgeometry.com/wp-content/uploads/2018/04/81-86.pdf), $k=\frac{\sqrt{3}-1}{\sqrt{3}}$, $k=tan{\frac{\pi}{4}}-tan{\frac{\pi}{6}}$


\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:**


* [Weitzenböck's inequality](https://en.wikipedia.org/wiki/Weitzenb%C3%B6ck%27s_inequality)
* [Hadwiger–Finsler inequality](https://en.wikipedia.org/wiki/Hadwiger%E2%80%93Finsler_inequality)
* [Isoperimetric inequality](https://en.wikipedia.org/wiki/Isoperimetric_inequality).



Strengthened version of Isoperimetric inequality with n-polygon
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 $c\approx0$. 


