A sharp lower bound on the polygonal isoperimetric deficit (Q2802128)

The isoperimetric inequality for polygons states that if \(P\) is an \(n\)-gon, \(n\geq 3\), with perimeter \(L(P)\) and area \(|P|\), then the isoperimetric deficit NEWLINE\[NEWLINE \delta(P):=L(P)^2-4n\tan\frac{\pi}{n}|P|\geq 0, NEWLINE\]NEWLINE with equality if and only if \(P\) is the convex and regular \(n\)-gon.NEWLINENEWLINEIn the paper under review the author establishes a lower bound for \(\delta(P)\) in terms of {\parindent=6mm \begin{itemize}\item[(1)] the variance \(\sigma_r^2(P)\) of the radii of \(P\) (i.e., the distances between the vertices and the barycenter), \item[(2)] the variance \(\sigma_a^2(P)\) of the barycentric angles of \(P\) (i.e., the angles generated by the vertices and the barycenter), and \item[(3)] the area of \(P\). NEWLINENEWLINE\end{itemize}} More precisely, he shows that if \(P\) is a convex \(n\)-gon, \(n\geq 3\), then there exists a constant \(c_n>0\) such that \(c_n\delta(P)\geq\sigma_r^2(P)+|P|\sigma_a^2(P)\).NEWLINENEWLINEBesides the undoubted interest of this result, one of the goals of the paper is that it is proved using a technique based on a suitably defined Taylor expansion of the deficit, as well as the so-called circulant matrix method. This interesting method was introduced in [the author and \textit{L. Nurbekyan}, Adv. Math. 276, 62--86 (2015; Zbl 1316.35007)], and it is based on viewing a large class of polygons as points in the (\(2n\))-dimensional Euclidean space satisfying certain constraints.