$\DeclareMathOperator\Area{Area}\DeclareMathOperator\cotg{cotg}$I am looking for a proof (or a reference) of an inequality related to area and the sidelengths of a polygon as follows:

Let $A_1A_2\cdots A_n$ be an arbitrary polygon, then:

$$\Area(A_1A_2\cdots A_n) \le \frac{1}{4}\cotg{\frac{\pi}{n}} \sum_{i=1}^nA_iA_{i+1}^2$$

Geometric meanings: $$\Area(A_1A_2\cdots A_n) \le \frac{\Area(1)+\Area(2)+\dotsb +\Area(n)}{n}.$$

Figure illustrating geometric meaning of the inequality

PS: I found this inequality long time ago, that time I thought this is old inequality. But today, I think this is new because I can not see any reference for the inequality.


2 Answers 2


Presumably the indices $i$ in $A_i$ are taken mod $n$, so "$A_{n+1}$" is to be identified with $A_1$. This must be a known isoperimetric inequality, but it's easier to prove than to find in the literature.

Fix the area $\cal A$ of the $n$-gon. By a standard compactness argument there exists an $n$-gon $A_1 A_2 \ldots A_n$ that minimizes $\sum_{i=1}^n (A_i A_{i+1})^2$. We first show that this polygon is convex (but possibly with $A_{i-1} A_i A_{i+1}$ collinear for some $i$). Indeed if it is not we can replace it by the convex hull, with each side $A_j A_k$ of the convex hull divided into $k-j$ equal subsegments; this both increases the area and decreases $\sum_{i=1}^n (A_i A_{i+1})^2$, so we can shrink the polygon back to area $\cal A$ and make the sum of its sides' squares even smaller.

Given convexity, fix all but one of the vertices, say $A_2$. Then $A_2$ is limited to a line parallel to $A_1 A_3$, and we readily see (as by choosing coordinates that make $A_1,A_3 = (\pm 1, 0)$ ) that $(A_1 A_2)^2 + (A_2 A_3)^2$ is minimized when $(A_1 A_2) = (A_2 A_3)$. Thus the minimizing $n$-gon has all sides equal, say with each $(A_i A_{i+1}) = s$; and then $ns^2$ is minimized when $ns$ is $-$ but $ns$ is the perimeter, and the usual isoperimetric inequality for $n$-gons then finishes the proof that the area-$\cal A$ polygon with the smallest $\sum_{i=1}^n (A_i A_{i+1})^2$ is regular.


Cauchy–Schwarz tells you that $$\sum_{i=1}^n \lvert A_iA_{i+1}\rvert^2\geq \frac{P^2}{n}$$ where $P$ is the perimeter of the polygon. Then we need the inequality $$P^2\geq 4n\tan(\pi/n)A$$ which is the classical isoperimetric inequality for polygons with many proofs in the literature, analytic, geometric and algebraic. See Fan, Taussky, and Todd - An algebraic proof of the isoperimetric inequality for polygons article and its references, for example.


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