# On some discrete Bonnesen-style isoperimetric inequalities

Chunna Zeng and Xu Dong  School of Mathematics and Statistics, Chongqing
Normal University, Chongqing 401332, People’s Republic of China
<EMAIL_ADDRESS>School of Mathematics and Statistics, Chongqing Normal
University, Chongqing 401332, People’s Republic of China
<EMAIL_ADDRESS>
###### Abstract.

This article deals with the sharp discrete isoperimetric inequalities in
analysis and geometry for planar convex polygons. First, the analytic
isoperimetric inequalities based on Schur convex function are established. In
the wake of the analytic isoperimetric inequalities, Bonnesen-style
isoperimetric inequalities and inverse Bonnesen-style inequalities for the
planar convex polygons are obtained.

This work is supported in part by the Major Special Project of NSFC (Grant No.
12141101), the Young Top-Talent program of Chongqing (Grant No.
CQYC2021059145), NSF-CQCSTC (Grant No. cstc2020jcyj-msxmX0609) and Technology
Research Foundation of Chongqing Educational committee (Grant No.
KJQN201900530, KJZD-K202200509).

Keywords: Schur-convex (concave) functions, convex polygons, analytic
isoperimetric inequality, discrete Bonnesen-style isoperimetric inequality.

##  1\. Introduction

The isoperimetric problem is an ancient problem in geometry that has a
significant impact on various branches of mathematics. The problem deals with
finding a closed curve or surface that encloses the maximum area or volume for
a given perimeter or boundary length. The geometric inequalities derived from
the isoperimetric problem have far-reaching effects and connections to other
areas of mathematics. In the 1950s, a connection was discovered between the
isoperimetric problem and the Sobolev embedding problem. The link between the
two problems provided insights into the relationship between geometric and
functional analysis. In the 1970s, the Aleksandrov-Fenchel inequality, which
is a generalization of the isoperimetric inequality, was found to be closely
related to the Hodge index theorem in algebraic geometry. In recent decades,
there has been extensive research on convex geometry related to the
isoperimetric problem. This research has established important connections
among various fields, including functional analysis, harmonic analysis, affine
geometry, partial differential equations, and information theory.

The complete mathematical proof of isoperimetric inequality was not
established until the variational method based on calculus appeared in the
19th century. After this breakthrough, various proofs for the isoperimetric
inequality were developed. In 1962, Hurwitz proposed two ingenious methods to
prove the isoperimetric inequalities. He employed Fourier analysis in his
proofs, one for convex curves and the other for general curves. These methods
were detailed in his works [3, Section 4.2] and [5, Page 392-394],
respectively. The planar isoperimetric inequality can be derived as an
immediate consequence of Poincaré formula [1, Section 23], which can be
founded in references such as [4, Chapter 7, Section 7] and [8, Page
1183-1185]. Zhang demonstrated that the Sobolev inequality, a key result in
functional analysis, is a special form of the isoperimetric inequality
([20].Zhou utilized the fundamental kinematic formula to establish Bonnesen-
type isoperimetric inequalities and other new inequalities in integral
geometry ([10, 21, 22]). It is worth mentioning that Zhang [17] introduced the
discrete Wirtinger inequality to obtain some new analytical isoperimetric
inequalities.

The description of classical isoperimetric inequality in the Euclidean plane
$\mathbb{R}^{2}$ is: assume that $K$ is a domain with length $L$ and area $A$,
then

$L^{2}-4\pi A\geq 0,$

where the equality holds if and only if $K$ is a disc.

In 1920, Bonnesen discovered a series of inequalities in $\mathbb{R}^{2}$ as
follows ([10, 8, 9, 15])

$\Delta\left(K\right)=L^{2}-4\pi A\geq B_{K},$ (1.1)

where $B_{k}$ is non-negative with geometric significance and vanishes only
when $K$ is a disc.

Denote by $\Delta\left(K\right)=L^{2}-4\pi A$ the isoperimetric deficit of
$K,$ and $B_{K}$ measures the “deviation” between $K$ and a disc. Many
$B_{K}$’s for planar domains are found by mathematicians. For instance,
Bonnesen, Zhang, Zhou etc. obtained a series of Bonnesen-type isoperimetric
inequalities ([8, 9, 16]): assume that $K$ is the domain of length $L$ and
area $A.$ Let $r_{K}$ and $R_{K}$ be the maximum inscribed radius and minimum
circumscribed radius, respectively. Then

$\pi r^{2}-Lr+A\leq 0;$ $\frac{L-\sqrt{L^{2}-4\pi A}}{2\pi}\leq
r_{K}\leq\frac{L}{2\pi}\leq R_{k}\leq\frac{L-\sqrt{L^{2}+4\pi A}}{2\pi};$

and

$L^{2}-4\pi A\geq A^{2}\left(\frac{1}{R_{K}}-\frac{1}{r_{K}}\right)^{2};\ \ \
\ L^{2}-4\pi A\geq L^{2}\left(\frac{R_{K}-r_{K}}{R_{K}+r_{K}}\right)^{2};$

$L^{2}-4\pi A\geq A^{2}\left(\frac{1}{R_{K}}-\frac{1}{r}\right)^{2};\ \ \ \
L^{2}-4\pi A\geq L^{2}\left(\frac{r-R_{K}}{r+R_{K}}\right)^{2};$

$L^{2}-4\pi A\geq A^{2}\left(\frac{1}{r}-\frac{1}{r_{K}}\right)^{2};\ \ \ \
L^{2}-4\pi A\geq L^{2}\left(\frac{r_{K}-r}{r_{K}+r}\right)^{2}.$

With equalities hold when and only when $K$ is a disc.

Comparing these $B_{K}$’s and determining the best lower bound poses a
challenge, and mathematicians are still pursuing to discover these unknown
invariants of geometrical significance.

Rencently, mathematicians have turned their attention towards the discrete
isoperimetric problem, such as polygons or polyhedrons. In fact, by
establishing a series of analytic inequalities that related to solutions of
nonlinear second-order differential equalities, Zhang [17, Section 3] obtained
some discrete isoperimetric inequalities of polygons in $\mathbb{R}^{2}$.

Assume that $H_{n}$ is an $n$-sided planar convex polygon. Denote by $L_{n}$
and $A_{n}$ length and area of $H_{n}$, respectively. Then

$L_{n}^{2}-4d_{n}A_{n}\geq 0,\ \ \ \ d_{n}=n\tan{\frac{\pi}{n}}.$ (1.2)

The quantity $L_{n}^{2}-4d_{n}A_{n}$ is called the isoperimetric deficit of
$H_{n}$. The sharp discrete Bonnesen-style isoperimetric inequality has the
following form

$L_{n}^{2}-4d_{n}A_{n}\geq U_{n},$ (1.3)

where $U_{n}$ is non-negative with geometric significance and vanishes only
when $H_{n}$ is a regular polygon. $U_{n}$ measures the “deviation” of $H_{n}$
from “regularity”.

Utilizing analytic inequalities such as the discrete Wirtinger inequality and
Schur-convex (concave) function have been proved to be an effective method for
solving discrete isoperimetric problem (see [13, 17, 18, 19]). However, the
discrete Bonnesen-style isoperimetric inequalities remain largely unexplored,
with only a few inequalities discovered in $\mathbb{R}^{2}$. As noted in
Zhang’s paper [17], a convex polygon that is inscribed in a circle is called
as “cyclic” and such polygons always enclose the largest area. Therefore, in
studying geometric inequalities for planar convex polygon, one only needs to
consider cyclic polygons.

Assume that ${\Lambda}_{n}$ is an $n$-sided planar convex polygon with
perimeter $L_{n}$ and area $A_{n}$, which is inscribed in a circle with radius
$R$. Meanwhile, there exist a regular polygon with perimeter $L_{n}^{\ast}$
and area $A_{n}^{\ast}$ inscribed in the same circle ${\Lambda}_{n}$. Zhang
[18] achieved the following geometric inequality for planar convex polygon in
$\mathbb{R}^{2}$

$L_{n}^{2}-4d_{n}A_{n}\geq\left(L_{n}^{\ast}-L_{n}\right)^{2},\ \ \ \
d_{n}=n\tan{\frac{\pi}{n}},$ (1.4)

where equality holds when and only when $\Lambda_{n}$ is a regular polygon.

In this paper, inspired by Zhang, Qi and Ma’s work ([19, 13]), by establishing
two analytic isoperimetric inequalities (Theorem 1, 2), we derive a series of
discrete Bonnesen-style isoperimetric inequalities (Theorem 3, 4) and their
reverse forms (Theorem 5, 6). In Section five, we make up for the shortcomings
in the proof of Ma’s theorem in [6], and present a new proof of Zhang and Ma’s
results ([18, 6])(Theorem 7). Meanwhile, we suppose that on surfaces of
constant curvature, Schur convexity also will be a convenient and excellent
idea to explore geometric inequalities on surfaces of constant curvature.

##  2\.  isoperimetric inequalities in analysis

The Schur-convex (concave) function was introduced by Schur in 1923 ([11]) and
has many important applications in analytic inequalities. In this section, we
review some basic facts about Schur-convex (concave) function and give two
important analytic inequalities.

An $n\times n$ matrix $P=\left[p_{ij}\right]$ is called a doubly stochastic
matrix if $p_{ij}\geq 0$ for $1\leq i$, $j\leq n$, and

$\sum_{j=1}^{n}p_{ij}=1,\ \ i=1,2,\cdots,n;\ \ \ \ \ \sum_{\
i=1}^{n}p_{ij}=1,\ \ j=1,2,\cdots,n.$

For instance, $P=\left[p_{ij}\right]$ with $p_{ij}=\frac{1}{n}$, $1\leq i$,
$j\leq n$ is a doubly stochastic matrix.

Let $I$ be an open interval of the real number line R, and $I^{n}=I\times
I\times\cdots\times I$. A real function $f:I^{n}\rightarrow\textbf{R}$ ($n\geq
2$) is Schur-convex if for any doubly stochastic matrix $P$ and $x\in I^{n}$,
then

$f(Px)\leq f(x).$ (2.1)

Furthermore, $f$ is called strictly Schur-convex if (2.1) is strict.
Similarly, $f$ is strictly Schur-concave if (2.1) is inverse and strict.

A real function $f:I^{n}\rightarrow\textbf{R}$ ($n\geq 2$) is symmetric if for
any permutation matrix $T$ and $x\in I^{n}$

$f(Tx)=f(x).$ (2.2)

Every Schur-convex function is symmetric, but not every symmetric function can
be a Schur-convex function. The following lemma is defined as the Schur’s
condition.

######  Lemma 1.

_([14])_ Assume that the real function
$f\left(x\right)=f\left(x_{1},x_{2},\cdots,x_{n}\right)$ with symmetry has
continuous partial derivatives on $I^{n}$. Then $f:I^{n}\rightarrow\textbf{R}$
is Schur-convex (concave) when and only when

$\left(x_{i}-x_{j}\right)\left(\frac{\partial f}{\partial
x_{i}}-\frac{\partial f}{\partial x_{j}}\right)\geq 0\left(\leq 0\right).$
(2.3)

Furthermore $f$ is strictly Schur-convex if inequality (2.3) is strict for
$x_{i}\neq x_{j}$, $1\leq i$, $j\leq n$.

Due to the symmetry of $f(x)$, the Schur’s condition (2.3) can be rewritten as
([7, Page 57])

$\left(x_{1}-x_{2}\right)\left(\frac{\partial f}{\partial
x_{1}}-\frac{\partial f}{\partial x_{2}}\right)\geq 0\left(\leq 0\right),$
(2.4)

and $f$ is strictly Schur-convex if (2.4) is strict for $x_{1}\neq x_{2}$.

The following notations will be often used in the latter part of this article.

$I=\left(0,l\right);\ \ \
H_{n}=\Big{\\{}\mathrm{\Theta}=\left(\theta_{1},\cdots,\theta_{n}\right)\in\mathbb{R}^{n},~{}\sum_{i=1}^{n}\theta_{i}=ml\Big{\\}}\left(0<m<n\right);$
$D_{n}=I^{n}\cap H_{n};\ \ \
\mathrm{\Omega}=\left(\sigma,\cdots,\sigma\right)~{}where~{}\sigma=\frac{1}{n}\sum_{i=1}^{n}\theta_{i}=\frac{ml}{n}.$

######  Lemma 2.

If $f$: $I^{n}\rightarrow\textbf{R}$ is a Schur-concave function on $I^{n}$,
then $f\left(\mathrm{\Omega}\right)$ is a global maximum in $D_{n}$. If $f$ is
strictly Schur-concave on $I^{n}$, then $f\left(\mathrm{\Omega}\right)$ is the
unique global maximum in $D_{n}$.

###### Proof.

Because that $f$ is Schur-concave, then

$f\left(P\mathrm{\Theta}\right)\geq f\left(\mathrm{\Theta}\right),\ \ \
\Theta\in I^{n}$

for any doubly stochastic matrix $P$. Then for
$\mathrm{\Theta}=\left(\theta_{1},\cdots,\theta_{n}\right)$ in $D_{n}$, and

$P=\left[p_{ij}\right]\ \ with\ \ p_{ij}=\frac{1}{n},1\leq i,j\leq n.$

It yields

$f\left(\mathrm{\Omega}\right)=\ f\left(P\mathrm{\Theta}\right)\geq
f\left(\mathrm{\Theta}\right).$

If $f$ is strictly Schur-concave on $I^{n}$, then
$f\left(\mathrm{\Omega}\right)>f\left(\mathrm{\Theta}\right)$ for all
$\mathrm{\Theta}$ in $D_{n}$, $\mathrm{\Theta}\neq\mathrm{\Omega}$. ∎

######  Lemma 3.

_([19])_ If $f$: $I^{n}\rightarrow\textbf{R}$ is a Schur-convex function, then
$f\left(\mathrm{\Omega}\right)$ is a global minimum in $D_{n}$. If $f$ is
strictly Schur-convex on $I^{n}$, then $f\left(\mathrm{\Omega}\right)$ is
unique global minimum in $D_{n}$.

We obtain the following analytic inequality.

######  Theorem 1.

Let $f\left(\theta\right)$ be a positive, strictly convex function, then

$\left(\sum_{i=1}^{n}f\left(\theta_{i}\right)\right)^{2\alpha}-\left(nf\left(\sigma\right)\right)^{\alpha}\left(\sum_{i=1}^{n}f\left(\theta_{i}\right)\right)^{\alpha}\geq\left(f\left(\sigma\right)\right)^{\alpha}\left[\left(\sum_{i=1}^{n}f\left(\theta_{i}\right)\right)^{\alpha}-\left(nf\left(\sigma\right)\right)^{\alpha}\right],$
(2.5)

where $\alpha\in\mathbb{N}^{+}.$ The equality holds when and only when
$\theta_{1}=\theta_{2}=\cdots=\theta_{n}=\sigma$.

###### Proof.

Consider the following function

$F\left(\Theta\right)=\left(\sum_{i=1}^{n}f\left(\theta_{i}\right)\right)^{2\alpha}-\left(nf\left(\sigma\right)\right)^{\alpha}\left(\sum_{i=1}^{n}f\left(\theta_{i}\right)\right)^{\alpha}-\left(f\left(\sigma\right)\right)^{\alpha}\left[\left(\sum_{i=1}^{n}f\left(\theta_{i}\right)\right)^{\alpha}-\left(nf\left(\sigma\right)\right)^{\alpha}\right].$
(2.6)

It shows that $F\left(\Theta\right)$ is symmetric on $I^{n}$ and
$F\left(\mathrm{\Omega}\right)=0$. In order to prove Theorem 1, we need to
prove that $F\left(\Theta\right)\geq 0$ and $F\left(\Theta\right)$ is strictly
Schur-convex on $I^{n}$. Then by Lemma 3 it implies that
$F\left(\mathrm{\Theta}\right)$ has the unique global minimum
$F\left(\mathrm{\Omega}\right)$ in $D_{n}$. So the key of this proof is to
determine whether $F\left(\Theta\right)$ satisfies strict Schur-convexity on
$I^{n}$. By Lemma 1 and (2.4), in order to demonstrate that, it is necessary
to verify the following inequality

$\left(\theta_{1}-\theta_{2}\right)\left(\frac{\partial
F}{\partial\theta_{1}}-\frac{\partial F}{\partial\theta_{2}}\right)>0,\ \ if\
{\ \theta}_{1}\neq\theta_{2}.$ (2.7)

Let $P_{n}=\sum_{i=1}^{n}f\left(\theta_{i}\right)$, then

$F\left(\Theta\right)=\left(P_{n}\right)^{2\alpha}-\left(n^{\alpha}+1\right)\left(f\left(\sigma\right)\right)^{\alpha}\left(P_{n}\right)^{\alpha}+{n^{\alpha}\left(f\left(\sigma\right)\right)}^{2\alpha}.$

Therefore

$\frac{\partial
F}{\partial\theta_{i}}=2\alpha\left(P_{n}\right)^{2\alpha-1}f^{\prime}\left(\theta_{i}\right)-\alpha\left(n^{\alpha}+1\right)\left(f\left(\sigma\right)\right)^{\alpha}\left(P_{n}\right)^{\alpha-1}f^{\prime}\left(\theta_{i}\right)$

and

$\displaystyle\left(\theta_{1}-\theta_{2}\right)\left(\frac{\partial
F}{\partial\theta_{1}}-\frac{\partial F}{\partial\theta_{2}}\right)=$
$\displaystyle\left(\theta_{1}-\theta_{2}\right)\cdot\left(f^{\prime}\left(\theta_{1}\right)-f^{\prime}\left(\theta_{2}\right)\right)$
$\displaystyle\cdot\left[2\alpha\left(P_{n}\right)^{2\alpha-1}-\alpha\left(n^{\alpha}+1\right)\left(f\left(\sigma\right)\right)^{\alpha}\left(P_{n}\right)^{\alpha-1}\right].$

Since $f$ has strict convexity, then $f^{\prime\prime}>0$ and $f^{\prime}$ is
increasing on $\left(0,l\right).$ This leads to

$\left(\theta_{1}-\theta_{2}\right)\left(f^{\prime}\left(\theta_{1}\right)-f^{\prime}\left(\theta_{2}\right)\right)>0.$
(2.8)

Consider

$G\left(\mathrm{\Theta}\right)=2\alpha\left(P_{n}\right)^{2\alpha-1}-\alpha\left(n^{\alpha}+1\right)\left(f\left(\sigma\right)\right)^{\alpha}\left(P_{n}\right)^{\alpha-1}.$

Because that $f$ is strictly convex and by Jensen’s inequality it follows that

$\left(P_{n}\right)^{\alpha}>\left(nf\left(\sigma\right)\right)^{\alpha}.$
(2.9)

(2.9) is strict and it implies that

$\left(P_{n}\right)^{\alpha}+\left(P_{n}\right)^{\alpha}>\left(nf\left(\sigma\right)\right)^{\alpha}+\left(f\left(\sigma\right)\right)^{\alpha}=\left(n^{\alpha}+1\right)\left(f\left(\sigma\right)\right)^{\alpha},$

that is

${2\left(P_{n}\right)}^{\alpha}>\left(n^{\alpha}+1\right)\left(f\left(\sigma\right)\right)^{\alpha}.$
(2.10)

(2.10) is equivalent to

$2\alpha\left(P_{n}\right)^{2\alpha-1}>\alpha\left(n^{\alpha}+1\right)\left(f\left(\sigma\right)\right)^{\alpha}\left(P_{n}\right)^{\alpha-1}.$

Therefore

$G\left(\mathrm{\Theta}\right)=2\alpha\left(P_{n}\right)^{2\alpha-1}-\alpha\left(n^{\alpha}+1\right)\left(f\left(\sigma\right)\right)^{\alpha}\left(P_{n}\right)^{\alpha-1}>0.$
(2.11)

From (2.8) and (2.11), (2.7) is proved. So we complete the proof of Theorem 1.
∎

######  Remark 1.

_In the proof of Theorem 1, assume that $f\left(\theta\right)$ is a positive
convex (resp. concave) function, and taking $\alpha=1,$ then (2.9) becomes the
classical Jensen’s inequality: _

$\sum_{i=1}^{n}f\left(\theta_{i}\right)\geq\left(\leq\right)nf\left(\sigma\right),$
(2.12)

_with equalities when and only when
$\theta_{1}=\theta_{2}=\cdots=\theta_{n}=\sigma$. For example_

$\left(1\right)\ \ \sum_{i=1}^{n}{\tan\theta_{i}}\geq n\tan{\sigma},\ \ \ \ \
\ \ \ \ \ \ \ \left(2\right)\ \ \sum_{i=1}^{n}{\sin\theta_{i}}\leq
n\sin{\sigma},$

_where $\theta_{i}\in\left(0,\frac{\pi}{2}\right)$, $i=1,2,\cdots,n$; and
$\sum_{i=1}^{n}\theta_{i}=\pi$,
$\sigma=\frac{1}{n}\sum_{i=1}^{n}\theta_{i}=\frac{\pi}{n}$._

In particular, if $f\left(\theta\right)=\tan{\theta}$ and
$f\left(\theta\right)=\sec{\theta}$ for
$\theta\in\left(0,\frac{\pi}{2}\right)$, then

######  Corollary 1.

Let $\theta_{i}\in\left(0,\frac{\pi}{2}\right)$, $i=1,2,\cdots,n$; and
$\sum_{i=1}^{n}\theta_{i}=\pi$, then for $\alpha\in\mathbb{N}^{+}$

$\left(\sum_{i=1}^{n}\tan{\theta_{i}}\right)^{2\alpha}-\left(n\tan{\frac{\pi}{n}}\right)^{\alpha}\left(\sum_{i=1}^{n}\tan{\theta_{i}}\right)^{\alpha}\geq\left(\tan{\frac{\pi}{n}}\right)^{\alpha}\left[\left(\sum_{i=1}^{n}\tan{\theta_{i}}\right)^{\alpha}-\left(n\tan{\frac{\pi}{n}}\right)^{\alpha}\right];$
(2.13)

Specially, if $\alpha=1$, then

$\left(\sum_{i=1}^{n}\tan{\theta_{i}}\right)^{2}-\left(n\tan{\frac{\pi}{n}}\right)\left(\sum_{i=1}^{n}\tan{\theta_{i}}\right)\geq\tan{\frac{\pi}{n}}\left(\sum_{i=1}^{n}\tan{\theta_{i}}-n\tan{\frac{\pi}{n}}\right),$
(2.14)

both equalities hold when and only when
$\theta_{1}=\theta_{2}=\cdots=\theta_{n}=\frac{\pi}{n}$.

######  Corollary 2.

Let $\theta_{i}\in\left(0,\frac{\pi}{2}\right)$, $i=1,2,\cdots,n$; and
$\sum_{i=1}^{n}\theta_{i}=\pi$, then

$\left(\sum_{i=1}^{n}\sec{\theta_{i}}\right)^{2\alpha}-\left(n\sec{\frac{\pi}{n}}\right)^{\alpha}\left(\sum_{i=1}^{n}\sec{\theta_{i}}\right)^{\alpha}\geq\left(\sec{\frac{\pi}{n}}\right)^{\alpha}\left[\left(\sum_{i=1}^{n}\sec{\theta_{i}}\right)^{\alpha}-\left(n\sec{\frac{\pi}{n}}\right)^{\alpha}\right],$
(2.15)

where $\alpha\in\mathbb{N}^{+}.$ The equality holds when and only when
$\theta_{1}=\theta_{2}=\cdots=\theta_{n}=\frac{\pi}{n}$.

Notice that $f\left(x\right)=x^{2}$ on $\left(0,1\right)$ has strict
convexity, then

######  Corollary 3.

Let $\sum_{i=1}^{n}x_{i}=m$, $i=1,2,\cdots,n$, then

$\left(\sum_{i=1}^{n}x_{i}^{2}\right)^{2\alpha}-\left(\frac{m^{2}}{n}\right)^{\alpha}\left(\sum_{i=1}^{n}x_{i}^{2}\right)^{\alpha}\geq\left(\frac{m^{2}}{n^{2}}\right)^{\alpha}\left[\left(\sum_{i=1}^{n}x_{i}^{2}\right)^{\alpha}-\left(\frac{m^{2}}{n}\right)^{\alpha}\right].$
(2.16)

for $\alpha\in\mathbb{N}^{+}.$ The equality holds when and only when
$x_{1}=x_{2}=\cdots=x_{n}=\frac{m}{n}$ $\left(0<m<n\right)$.

######  Theorem 2.

Let $f\left(\theta\right)$ be a positive, strictly convex function, then

$\left(\sum_{i=1}^{n}f\left(\theta_{i}\right)\right)^{2\alpha}-\left(nf\left(\sigma\right)\right)^{\alpha}\left(\sum_{i=1}^{n}f\left(\theta_{i}\right)\right)^{\alpha}\leq\left(\sum_{i=1}^{n}f\left(\theta_{i}\right)\right)^{k\alpha}-\left(nf\left(\sigma\right)\right)^{k\alpha},$
(2.17)

where $\alpha,k\in\mathbb{N}^{+}$ and $k\geq 2$. The equality holds when and
only when $\theta_{1}=\theta_{2}=\cdots=\theta_{n}=\sigma$.

###### Proof.

Consider the following function

$F\left(\Theta\right)=\left(\sum_{i=1}^{n}f\left(\theta_{i}\right)\right)^{2\alpha}-\left(nf\left(\sigma\right)\right)^{\alpha}\left(\sum_{i=1}^{n}f\left(\theta_{i}\right)\right)^{\alpha}-\left.\left(\sum_{i=1}^{n}f\left(\theta_{i}\right)\right)^{k\alpha}+\left(nf\left(\sigma\right)\right)^{k\alpha}.\right.$
(2.18)

It is obviously that $F\left(\Theta\right)$ is symmetrical on $I^{n}$ and
$F\left(\mathrm{\Omega}\right)=0$. Lemma 2 implies that
$F\left(\mathrm{\Omega}\right)$ is the unique global maximum in $D_{n}$ when
$F\left(\Theta\right)$ is strictly Schur-concave. So the key is to determine
whether $F\left(\Theta\right)$ is strictly Schur-concave on $I^{n}$. By Lemma
1 and (2.4), in order to demonstrate that, it is necessary to verify the
following inequality

$\left(\theta_{1}-\theta_{2}\right)\left(\frac{\partial
F}{\partial\theta_{1}}-\frac{\partial F}{\partial\theta_{2}}\right)<0,\ \ if\
{\ \theta}_{1}\neq\theta_{2}.$ (2.19)

Let $P_{n}=\sum_{i=1}^{n}f\left(\theta_{i}\right)$, then

$F\left(\Theta\right)=\left(P_{n}\right)^{2\alpha}-\left(nf\left(\sigma\right)\right)^{\alpha}\left(P_{n}\right)^{\alpha}-\left(P_{n}\right)^{k\alpha}+\left(nf\left(\sigma\right)\right)^{k\alpha}.$

Therefore

$\displaystyle\frac{\partial F}{\partial\theta_{i}}$
$\displaystyle=2\alpha\left(P_{n}\right)^{2\alpha-1}f^{\prime}\left(\theta_{i}\right)-n\left(f\left(\sigma\right)\right)^{a}\alpha\left(P_{n}\right)^{\alpha-1}f^{\prime}\left(\theta_{i}\right)-k\alpha\left(P_{n}\right)^{k\alpha-1}f^{\prime}\left(\theta_{i}\right)$
$\displaystyle=f^{\prime}\left(\theta_{i}\right)\left[2\alpha\left(P_{n}\right)^{2\alpha-1}-n\left(f\left(\sigma\right)\right)^{a}\alpha\left(P_{n}\right)^{\alpha-1}-k\alpha\left(P_{n}\right)^{k\alpha-1}\right]$

and

$\displaystyle\left(\theta_{1}-\theta_{2}\right)\left(\frac{\partial
f}{\partial\theta_{1}}-\frac{\partial f}{\partial\theta_{2}}\right)=$
$\displaystyle\left(\theta_{1}-\theta_{2}\right)\cdot\left(f^{\prime}\left(\theta_{1}\right)-f^{\prime}\left(\theta_{2}\right)\right)$
(2.20)
$\displaystyle\cdot\left[2\alpha\left(P_{n}\right)^{2\alpha-1}-n\left(f\left(\sigma\right)\right)^{a}\alpha\left(P_{n}\right)^{\alpha-1}-k\alpha\left(P_{n}\right)^{k\alpha-1}\right].$

Since $f$ is strictly convex, then $f^{\prime\prime}>0$ and $f^{{}^{\prime}}$
is increasing. It leads to

$\left(\theta_{1}-\theta_{2}\right)\left(f^{\prime}\left(\theta_{1}\right)-f^{\prime}\left(\theta_{2}\right)\right)>0,$

and

$2\alpha\left(P_{n}\right)^{2\alpha-1}-n\left(f\left(\sigma\right)\right)^{a}\alpha\left(P_{n}\right)^{\alpha-1}-k\alpha\left(P_{n}\right)^{k\alpha-1}<0.$
(2.21)

From (2.20) and (2.21), (2.19) is obtained. So we complete the proof of
Theorem 2. ∎

In particular, assume that $f\left(\theta\right)=\tan{\theta}$ and
$f\left(\theta\right)=\csc{\theta}$ for
$\theta\in\left(0,\frac{\pi}{2}\right)$, then we have

######  Corollary 4.

Let $\theta_{i}\in\left(0,\frac{\pi}{2}\right)$, $i=1,2,\cdots,n;$ and
$\sum_{i=1}^{n}\theta_{i}=\pi$, then

$\left(\sum_{i=1}^{n}\tan{\theta_{i}}\right)^{2\alpha}-\left(n\tan{\frac{\pi}{n}}\right)^{\alpha}\left(\sum_{i=1}^{n}\tan{\theta_{i}}\right)^{\alpha}\leq\left(\sum_{i=1}^{n}\tan{\theta_{i}}\right)^{k\alpha}-\left(n\tan{\frac{\pi}{n}}\right)^{k\alpha},$
(2.22)

where $\alpha,k\in\mathbb{N}^{+}$ and $k\geq 2$. Especially, if $\alpha=1$ and
$k=3$, then

$\left(\sum_{i=1}^{n}\tan{\theta_{i}}\right)^{2}-\left(n\tan{\frac{\pi}{n}}\right)\left(\sum_{i=1}^{n}\tan{\theta_{i}}\right)\leq\left.\left(\sum_{i=1}^{n}\tan{\theta_{i}}\right)^{3}-\left(n\tan{\frac{\pi}{n}}\right)^{3},\right.$
(2.23)

both equalities hold when and only when
$\theta_{1}=\theta_{2}=\cdots=\theta_{n}=\frac{\pi}{n}$.

######  Corollary 5.

Let $f\left(\theta\right)=\csc{\theta}$,
$\theta_{i}\in\left(0,\frac{\pi}{2}\right)$, $i=1,2,\cdots,n$; and
$\sum_{i=1}^{n}\theta_{i}=\pi$, then

$\left(\sum_{i=1}^{n}\csc{\theta_{i}}\right)^{2\alpha}-\left(n\csc{\frac{\pi}{n}}\right)^{\alpha}\left(\sum_{i=1}^{n}\csc{\theta_{i}}\right)^{\alpha}\leq\left(\sum_{i=1}^{n}\csc{\theta_{i}}\right)^{k\alpha}-\left(n\csc{\frac{\pi}{n}}\right)^{k\alpha},$
(2.24)

where $\alpha,k\in\mathbb{N}^{+}$ and $k\geq 2$. Especially, when $\alpha=1$
and $k=3$, then

$\left(\sum_{i=1}^{n}\csc{\theta_{i}}\right)^{2}-\left(n\csc{\frac{\pi}{n}}\right)\left(\sum_{i=1}^{n}\csc{\theta_{i}}\right)\leq\left(\sum_{i=1}^{n}\csc{\theta_{i}}\right)^{3}-\left(n\csc{\frac{\pi}{n}}\right)^{3},$
(2.25)

both equalities hold when and only when
$\theta_{1}=\theta_{2}=\cdots=\theta_{n}=\frac{\pi}{n}$.

##  3\. Discrete Bonnesen-style isoperimetric inequalities

In this section, by using the analytic isoperimetric inequalities in Theorem
1, we establish some discrete Bonnesen isoperimetric inequalities.

######  Theorem 3.

Assume that ${\Gamma}_{n}$ is an $n$-sided planar convex polygon circumscribed
in a circle of radius $R$. Denote by $L_{n}$ and $A_{n}$ the perimeter and
area of ${\Gamma}_{n}$, then

$L_{n}^{2\alpha}-4^{\alpha}\left(d_{n}A_{n}\right)^{\alpha}\geq
2^{\alpha}R^{\alpha}\left(\tan{\frac{\pi}{n}}\right)^{\alpha}\left[L_{n}^{\alpha}-\left(L_{n}^{\ast}\right)^{\alpha}\right],\
\ \ d_{n}=ntan{\frac{\pi}{n}},$ (3.1)
$\left(\frac{A_{n}}{R^{2}}\right)^{2\alpha}-d_{n}^{\alpha}\left(\frac{L_{n}}{2R}\right)^{\alpha}\geq\left(\tan{\frac{\pi}{n}}\right)^{\alpha}\left[\left(\frac{L_{n}}{2R}\right)^{\alpha}-\left(\frac{L_{n}^{\ast}}{2R}\right)^{\alpha}\right],\
\ \ d_{n}=ntan{\frac{\pi}{n}},$ (3.2)

where $\alpha\in\mathbb{N}^{+}$, $L_{n}^{\ast}$ is the perimeter of the
regular $n$-sided convex polygon circumscribed in the same circle with
${\Gamma}_{n}$. Both equalities hold when and only when ${\Gamma}_{n}$ is a
regular polygon.

###### Proof.

Denote by $a_{i}$ and $\theta_{i}$ the length of the $i$th side of
${\Gamma}_{n}$, and the half of the central angle subtended by the $i$th
vertex $A_{i}$ of ${\Gamma}_{n}$, $i=1,2,\cdots,n$, respectively. Then

$L_{n}=\sum_{i=1}^{n}a_{i}=2R\sum_{i=1}^{n}{\tan\theta_{i}};\ \ \ \ \ \ \ \ \
\ L_{n}^{\ast}=2nR\tan{\frac{\pi}{n}}.$ (3.3) $\sum_{i=1}^{n}\theta_{i}=\pi;\
\ \ \ \ \ \ \
A_{n}=\frac{1}{2}\sum_{i=1}^{n}a_{i}R=R^{2}\sum_{i=1}^{n}{\tan\theta_{i}};{\ \
\ \ \ \ \ A}_{n}^{\ast}=nR^{2}\tan{\frac{\pi}{n}}.$ (3.4)

Substituting (3.3) and (3.4) into (2.13) , we obtain (3.1) and (3.2). ∎

######  Remark 2.

_The inequality ( 3.2) can be regard as the reverse inequality of (3.1). _

Assume that $\alpha=1$, we have the following special case.

######  Corollary 6.

Assume that ${\Gamma}_{n}$ is an $n$-sided planar convex polygon circumscribed
in a circle of radius $R$. Denote by $L_{n}$ and $A_{n}$ the perimeter and
area of ${\Gamma}_{n}$, then

$L_{n}^{2}-4d_{n}A_{n}\geq
2R\left(\tan{\frac{\pi}{n}}\right)\left[L_{n}-L_{n}^{\ast}\right],$ (3.5)
$\left(\frac{A_{n}}{R^{2}}\right)^{2}-d_{n}\left(\frac{L_{n}}{2R}\right)\geq\left(\tan{\frac{\pi}{n}}\right)\left(\frac{L_{n}}{2R}-\frac{L_{n}^{\ast}}{2R}\right),$
(3.6)

where $d_{n}=n\tan{\frac{\pi}{n}}$. Both equalities hold when and only when
${\Gamma}_{n}$ is a regular polygon.

Similarly, substituting (3.3) and (3.4) into (2.13), we obtain

######  Theorem 4.

Assume that ${\Gamma}_{n}$ is an $n$-sided planar convex polygon circumscribed
in a circle of radius $R$. Denote by $L_{n}$ and $A_{n}$ the perimeter and
area of ${\Gamma}_{n}$, then

$L_{n}^{2\alpha}-4^{\alpha}\left(d_{n}A_{n}\right)^{\alpha}\geq
4^{\alpha}\left(\tan{\frac{\pi}{n}}\right)^{\alpha}\left[A_{n}^{\alpha}-\left(A_{n}^{\ast}\right)^{\alpha}\right],\
\ \ \ d_{n}=n\tan{\frac{\pi}{n}},$ (3.7)
$\left(\frac{A_{n}}{R^{2}}\right)^{2\alpha}-d_{n}^{\alpha}\left(\frac{L_{n}}{2R}\right)^{\alpha}\geq\left(\tan{\frac{\pi}{n}}\right)^{\alpha}\left[\left(\frac{A_{n}}{R^{2}}\right)^{\alpha}-\left(\frac{A_{n}^{\ast}}{R^{2}}\right)^{\alpha}\right],\
\ \ \ d_{n}=n\tan{\frac{\pi}{n}},$ (3.8)

where $\alpha\in\mathbb{N}^{+}$, and $A_{n}^{\ast}$ denotes the area of the
regular $n$-sided convex polygon circumscribed in the same circle with
${\Gamma}_{n}$. Both equalities hold when and only when ${\Gamma}_{n}$ is a
regular polygon.

(3.8) can be regard as the reverse inequality of (3.7). When $\alpha=1$, we
have

######  Corollary 7.

Assume that ${\Gamma}_{n}$ is an $n$-sided planar convex polygon circumscribed
in a circle of radius $R$. Denote by $L_{n}$ and $A_{n}$ the perimeter and
area of ${\Gamma}_{n}$, then

$L_{n}^{2}-4d_{n}A_{n}\geq
4\left(\tan{\frac{\pi}{n}}\right)\left[A_{n}-A_{n}^{\ast}\right],$ (3.9)
$\left(\frac{A_{n}}{R^{2}}\right)^{2}-d_{n}\left(\frac{L_{n}}{2R}\right)\geq\left(\tan{\frac{\pi}{n}}\right)\left[\left(\frac{A_{n}}{R^{2}}\right)-\left(\frac{A_{n}^{\ast}}{R^{2}}\right)\right],$
(3.10)

where $d_{n}=n\tan{\frac{\pi}{n}}$. Both equalities hold when and only when
${\Gamma}_{n}$ is a regular polygon.

##  4\. Inverse Discrete Bonnesen-type isoperimetric inequalities

It is interesting and also difficult to study the reverse Bonnesen-style
isoperimetric inequalities. There are only a few reverse Bonnesen-style
isoperimetric inequalities of convex domains (see [2]). In 1993, Bottema
discovered a classic reverse Bonnesen-style isoperimetric inequality of oval
domains in $\mathbb{R}^{2}$([10]). The discrete cases are more complex and
difficult.

Assume that $K$ is a convex domain with $C^{2}$ smooth boundary in
$\mathbb{R}^{2}$, then ([10])

$\mathrm{\Delta}\left(K\right)=L^{2}-4\pi
A\leq\pi^{2}\left(\rho_{M}-\rho_{m}\right)^{2},$

where $\rho_{M}$ and $\rho_{m}$ are the maximum and minimum value of
$\rho\left(\partial K\right)$, respectively. The equality holds when and only
when $\partial K$ is a circle.

Mathematicians still continue to work hard on investigating the inverse
discrete Bonnesen-type isoperimetric inequalities. A natural question is that:
for an $n$-sided planar convex polygon if there is a geometric invariant
$D_{n}$ such that

$L_{n}^{2}-4d_{n}A_{n}\leq D_{n},\ \ \ d_{n}=n\tan{\frac{\pi}{n}},$

where $D_{n}$ satisfies

  1. (1)

$D_{n}\geq 0$;

  2. (2)

$D_{n}=0$ only when $H_{n}$ is regular.

In order to obtain the inverse discrete Bonnesen-style isoperimetric
inequalities, we use the analysis isoperimetric inequalities in Theroem 2.
Substituting (3.3) and (3.4) into (2.22), then we have

######  Theorem 5.

Assume that ${\Gamma}_{n}$ is an $n$-sided planar convex polygon circumscribed
in a circle of radius $R$. Denote by $L_{n}$ and $A_{n}$ the perimeter and
area of ${\Gamma}_{n}$, then

$L_{n}^{2\alpha}-\left(4d_{n}A_{n}\right)^{\alpha}\leq
L_{n}^{k\alpha}-\left(L_{n}^{\ast}\right)^{k\alpha},\ \ \
d_{n}=n\tan{\frac{\pi}{n}},$ (4.1)
$\left(\frac{A_{n}}{R^{2}}\right)^{2\alpha}-d_{n}^{\alpha}\left(\frac{L_{n}}{2R}\right)^{\alpha}\leq\left(\frac{L_{n}}{2R}\right)^{k\alpha}-\left(\frac{L_{n}^{\ast}}{2R}\right)^{k\alpha},\
$ (4.2)

where $\alpha,k\in\mathbb{N}^{+}$ and $k\geq 2$. Both equalities hold when and
only when ${\Gamma}_{n}$ is a regular polygon.

Specially, we can regard inequality (4.2) as the reverse inequality of (4.1).

In particular, when $\alpha=1$ and $k=2,3$, Theorem 5 can be expressed as
follows.

######  Corollary 8.

Assume that ${\Gamma}_{n}$ is an $n$-sided planar convex polygon circumscribed
in a circle of radius $R$. Denote by $L_{n}$ and $A_{n}$ the perimeter and
area of ${\Gamma}_{n}$, then

$L_{n}^{2}-4d_{n}A_{n}\leq L_{n}^{3}-(L_{n}^{\ast})^{3},$ (4.3)
$\left(\frac{A_{n}}{R^{2}}\right)^{2}-d_{n}\left(\frac{L_{n}}{2R}\right)\leq\left(\frac{L_{n}}{2R}\right)^{2}-\left(\frac{L_{n}^{\ast}}{2R}\right)^{2},$
(4.4)

where $d_{n}=n\tan{\frac{\pi}{n}}$. Both equalities hold when and only when
${\Gamma}_{n}$ is a regular polygon.

Substituting (3.3) and (3.4) into (2.22), then we have

######  Theorem 6.

Assume that ${\Gamma}_{n}$ is an $n$-sided planar convex polygon circumscribed
in a circle of radius $R$. Denote by $L_{n}$ and $A_{n}$ the perimeter and
area of ${\Gamma}_{n}$, then

$L_{n}^{2\alpha}-4^{\alpha}\left(d_{n}A_{n}\right)^{\alpha}\leq\frac{4^{\alpha}}{R^{2\left(k-1\right)\alpha}}\left[A_{n}^{k\alpha}-\left(A_{n}^{\ast}\right)^{k\alpha}\right],\
\ \ \ d_{n}=n\tan{\frac{\pi}{n}},$ (4.5)
$\left(\frac{A_{n}}{R^{2}}\right)^{2\alpha}-d_{n}^{\alpha}\left(\frac{L_{n}}{2R}\right)^{\alpha}\leq\left(\frac{A_{n}}{R^{2}}\right)^{k\alpha}-\left(\frac{A_{n}^{\ast}}{R^{2}}\right)^{k\alpha},\
$ (4.6)

where $\alpha,k$ and $k\geq 2$. Both equalities hold when and only when
${\Gamma}_{n}$ is a regular polygon.

In particular, when $\alpha=1$ and $k=2,3$, Theorem 5 also can be expressed as
follows.

######  Corollary 9.

Assume that ${\Gamma}_{n}$ is an $n$-sided planar convex polygon circumscribed
in a circle of radius $R$. Denote by $L_{n}$ and $A_{n}$ the perimeter and
area of ${\Gamma}_{n}$, then

$L_{n}^{2}-4d_{n}A_{n}\leq\frac{4}{R^{2}}\left[A_{n}^{2}-\left(A_{n}^{\ast}\right)^{2}\right],$
(4.7)
$\left(\frac{A_{n}}{R^{2}}\right)^{2}-d_{n}\left(\frac{L_{n}}{2R}\right)\leq\left(\frac{A_{n}}{R^{2}}\right)^{3}-\left(\frac{A_{n}^{\ast}}{R^{2}}\right)^{3},$
(4.8)

where $d_{n}=n\tan{\frac{\pi}{n}}$. Both equalities hold when and only when
${\Gamma}_{n}$ is a regular polygon.

##  5\. some notes

Many special analytic inequalities have interesting geometric consequences. In
this section, we first prove the following analytic inequalities (Theorem 7)
in which we set two different variables $\theta_{i}$ and $\psi_{i}$. As the
consequence of Theorem 7, we present new proofs of Zhang’s result [18] and
Ma’s result ([6]). Note that there is a defect in the proof of Ma’s main
theorem. For completeness, we give the new proofs of these isoperimetric style
inequalities.

Assume that $\theta_{i}$ and $\psi_{i}$ are real numbers in
$\left(0,l\right)$, $i=1,2,\cdots,n$,
$\sum_{i=1}^{n}\theta_{i}=ml=\sum_{i=1}^{n}\psi_{i}$, where $m$ is a positive
constant less then $n$. Then

######  Theorem 7.

Assume that $f\left(\theta\right)$ is a positive $C^{2}$-function on
$\left(0,l\right)$, and it satisfies

${f^{\prime}}^{2}\left(\theta\right)-f\left(\theta\right)f^{\prime\prime}\left(\theta\right)=\mu,\
\ \theta\in\left(0,l\right),$ (5.1)

where $\mu$ is a constant and
$f^{\prime}\left(\theta\right)f^{\prime\prime}\left(\theta\right)\neq 0$.

  1. (i)

    1. (1)

If $f^{\prime\prime}\left(\theta\right)<0$ and
$f^{\prime}\left(\theta\right)>0$ on $\left(0,l\right)$, then

$2\sum_{i=1}^{n}{f\left(\theta_{i}\right)f^{\prime}\left(\psi_{i}\right)}\geq\sum_{i=1}^{n}{f\left(\theta_{i}\right)f^{\prime}\left(\theta_{i}\right)}+\sum_{i=1}^{n}{f\left(\psi_{i}\right)f^{\prime}\left(\psi_{i}\right)};$
(5.2)

    2. (2)

If $f^{\prime\prime}\left(\theta\right)<0$ and
$f^{\prime}\left(\theta\right)<0$ on $\left(0,l\right)$, then

$2\sum_{i=1}^{n}{f\left(\theta_{i}\right)f^{\prime}\left(\psi_{i}\right)}\leq\sum_{i=1}^{n}{f\left(\theta_{i}\right)f^{\prime}\left(\theta_{i}\right)}+\sum_{i=1}^{n}{f\left(\psi_{i}\right)f^{\prime}\left(\psi_{i}\right)};$
(5.3)

  2. (ii)

    1. (1)

If $f^{\prime\prime}\left(\theta\right)>0$ and
$f^{\prime}\left(\theta\right)>0$ on $\left(0,l\right)$, then

$2\sum_{i=1}^{n}{f\left(\theta_{i}\right)f^{\prime}\left(\psi_{i}\right)}\leq\sum_{i=1}^{n}{f\left(\theta_{i}\right)f^{\prime}\left(\theta_{i}\right)}+\sum_{i=1}^{n}{f\left(\psi_{i}\right)f^{\prime}\left(\psi_{i}\right)};$
(5.4)

    2. (2)

If $f^{\prime\prime}\left(\theta\right)>0$ and
$f^{\prime}\left(\theta\right)<0$ on $\left(0,l\right)$, then

$2\sum_{i=1}^{n}{f\left(\theta_{i}\right)f^{\prime}\left(\psi_{i}\right)}\geq\sum_{i=1}^{n}{f\left(\theta_{i}\right)f^{\prime}\left(\theta_{i}\right)}+\sum_{i=1}^{n}{f\left(\psi_{i}\right)f^{\prime}\left(\psi_{i}\right)}.$
(5.5)

The equalities hold when and only when $\theta_{i}=\psi_{i}$.

###### Proof.

For the convenience, the following notations will be used in the latter of
this proof.

$\mathrm{\Theta}=\left(\theta_{1},\theta_{2},\cdots,\theta_{n}\right)\in\mathbb{R}^{n};\
\ \
\mathrm{\Psi}=\left(\psi_{1},\psi_{2},\cdots,\psi_{n}\right)\in\mathbb{R}^{n};$
$H_{n}=\Big{\\{}\mathrm{\Theta}\in\mathbb{R}^{n},\sum_{i=1}^{n}\theta_{i}=ml\Big{\\}}\quad\left(0<m<n\right);$

$I_{n}=\left(0,l\right)^{n}\subset\mathbb{R}^{n};\ \ \ D_{n}=H_{n}\cap I_{n}.$

We set

$\displaystyle G\left(\mathrm{\Theta},\mathrm{\Psi}\right)$
$\displaystyle=2\sum_{i=1}^{n}{f\left(\theta_{i}\right)f^{\prime}\left(\psi_{i}\right)}-\sum_{i=1}^{n}{f\left(\theta_{i}\right)f^{\prime}\left(\theta_{i}\right)}-\sum_{i=1}^{n}{f\left(\psi_{i}\right)f^{\prime}\left(\psi_{i}\right)}.$

Case (_i_). First, if $f^{\prime\prime}\left(\theta\right)<0$ and
$f^{\prime}\left(\theta\right)>0$ on $\left(0,l\right)$, then
$f^{\prime}\left(\theta\right)$ is a strict decreasing function on
$\left(0,l\right)$. Since $G\left(\mathrm{\Psi},\mathrm{\Psi}\right)=0$, we
should demonstrate that $G\left(\mathrm{\Theta},\mathrm{\Psi}\right)\geq 0$
with equality only when $\Theta=\Psi$. It suffices for us to verify the
following inequality

$G\left(\Phi,\Psi\right)>G\left(\Psi,\Psi\right)~{}for~{}any~{}\Phi=\left(\phi_{1},\phi_{2},\cdots,\phi_{n}\right)\neq\Psi~{}in~{}D_{n}.$

In fact the line segment joining $\Phi$ and $\Psi$ can be defined as the
following form

$\mathrm{\Theta}\left(t\right)=\left(\theta_{1}\left(t\right),\theta_{2}\left(t\right),\cdots,\theta_{n}\left(t\right)\right),$

in which

$\theta_{i}\left(t\right)=t\psi_{i}+\left(1-t\right)\phi_{i},\ \ \
i=1,2,\cdots,n,\ \ \ t\in\left[0,1\right].$

Then

$\theta_{i}^{\prime}\left(t\right)=\psi_{i}-\phi_{i},\ \ \ i=1,2,\cdots,n.$

Observe that

$\sum_{i=1}^{n}\left(\psi_{i}-\phi_{i}\right)=ml-ml=0.$ (5.6)

Furthermore, when $\psi_{i}\neq\phi_{i}$, we have

$\left(\psi_{i}-\phi_{i}\right)\left[f^{\prime}\left(\psi_{i}\right)-f^{\prime}\left(\theta_{i}\left(t\right)\right)\right]<0,\
\ \ i=1,2,\cdots,n,\ \ \ t\in\left[0,1\right].$ (5.7)

Notice that $f^{\prime}\left(\theta\right)$ is decreasing on
$\left(0,l\right)$. If $\psi_{i}>\phi_{i}$, then

$\theta_{i}\left(t\right)=t\psi_{i}+\left(1-t\right)\phi_{i}<t\psi_{i}+\left(1-t\right)\psi_{i}=\psi_{i},$

and
$f^{\prime}\left(\theta_{i}\left(t\right)\right)>f^{\prime}\left(\psi_{i}\right)$.
If $\psi_{i}<\phi_{i}$, then

$\theta_{i}\left(t\right)=t\psi_{i}+\left(1-t\right)\phi_{i}>t\psi_{i}+\left(1-t\right)\psi_{i}=\psi_{i},$

and
$f^{\prime}\left(\theta_{i}\left(t\right)\right)<f^{\prime}\left(\psi_{i}\right)$.

Hence, the inequality (5.8) also holds for $0\leq t\leq 1$

$\displaystyle\sum_{i=1}^{n}\left(\psi_{i}-\phi_{i}\right)\left[f^{\prime}\left(\psi_{i}\right)-f^{\prime}\left(\theta_{i}\left(t\right)\right)\right]f^{\prime}\left(\theta_{i}\left(t\right)\right)<0.$
(5.8)

Besides, from (5.1) and (5.6) we have

$\displaystyle\mu\sum_{i=1}^{n}\left(\psi_{i}-\phi_{i}\right)$
$\displaystyle=\sum_{i=1}^{n}[{f^{\prime}}^{2}\left(\theta_{i}\left(t\right)\right)-f\left(\theta_{i}\left(t\right)\right)f^{\prime\prime}\left(\theta_{i}\left(t\right)\right)]\left(\psi_{i}-\phi_{i}\right)=0.$

Thus

$\displaystyle\sum_{i=1}^{n}{\left(\psi_{i}-\phi_{i}\right)f\left(\theta_{i}\left(t\right)\right)f^{\prime\prime}\left(\theta_{i}\left(t\right)\right)}$
$\displaystyle=\sum_{i=1}^{n}{\left(\psi_{i}-\phi_{i}\right)\left[f^{\prime}\left(\theta_{i}\left(t\right)\right)\right]^{2}}.$
(5.9)

Now, differential $G\left(\mathrm{\Theta}\left(t\right),\mathrm{\Psi}\right)$
with respect to $t$ and by (5.9), we have

$\displaystyle
G^{\prime}\left(\mathrm{\Theta}\left(t\right),\mathrm{\Psi}\right)$
$\displaystyle=2\sum_{i=1}^{n}{f^{\prime}\left(\psi_{i}\right)f^{\prime}\left(\theta_{i}\left(t\right)\right)\left(\psi_{i}-\phi_{i}\right)}-\sum_{i=1}^{n}\left[2{f^{\prime}}^{2}\left(\theta_{i}\left(t\right)\right)-1\right]\left(\psi_{i}-\phi_{i}\right)$
(5.10)
$\displaystyle=2\sum_{i=1}^{n}{f^{\prime}\left(\psi_{i}\right)f^{\prime}\left(\theta_{i}\left(t\right)\right)\left(\psi_{i}-\phi_{i}\right)}-2\sum_{i=1}^{n}\left.{f^{\prime}}^{2}\left(\theta_{i}\left(t\right)\right)\right.\left(\psi_{i}-\phi_{i}\right)$
$\displaystyle=2\sum_{i=1}^{n}\left(\psi_{i}-\phi_{i}\right)\left[f^{\prime}\left(\psi_{i}\right)-f^{\prime}\left(\theta_{i}\left(t\right)\right)\right]f^{\prime}\left(\theta_{i}\left(t\right)\right)<0.$

This shows that $G\left(\mathrm{\Theta}\left(t\right),\mathrm{\Psi}\right)$ is
decreasing on $\left[0,1\right]$ about the variable $t$, thus

$G\left(\mathrm{\Theta}\left(0\right),\mathrm{\Psi}\right)=G\left(\mathrm{\Phi},\mathrm{\Psi}\right)>G\left(\mathrm{\Psi},\mathrm{\Psi}\right)=G\left(\mathrm{\Theta}\left(1\right),\mathrm{\Psi}\right).$

Since $\Phi$ is arbitrarily chosen, the proof of case (i)(1) is complete.

Second, if $f^{\prime}\left(\theta\right)<0$ then the inequality (5.8) is
reversed and
$G^{\prime}\left(\mathrm{\Theta}\left(t\right),\mathrm{\Psi}\right)>0$. This
shows that $G\left(\mathrm{\Theta}\left(t\right),\mathrm{\Psi}\right)$ is
increasing on $\left[0,1\right]$ about the variable $t$, thus

$G\left(\mathrm{\Theta}\left(0\right),\mathrm{\Psi}\right)=G\left(\mathrm{\Phi},\mathrm{\Psi}\right)<G\left(\mathrm{\Psi},\mathrm{\Psi}\right)=G\left(\mathrm{\Theta}\left(1\right),\mathrm{\Psi}\right).$

Then we complete the proof of case (i)(2).

Case (ii). First, if $f^{\prime\prime}\left(\theta\right)>0$ and
$f^{\prime}\left(\theta\right)>0$ on $\left(0,l\right)$. The proof is similar
to case (i)(1), then

$\sum_{i=1}^{n}\left(\psi_{i}-\phi_{i}\right)\left[f^{\prime}\left(\psi_{i}\right)-f^{\prime}\left(\theta_{i}\left(t\right)\right)\right]f^{\prime}\left(\theta_{i}\left(t\right)\right)>0.$

This shows that $G\left(\mathrm{\Theta}\left(t\right),\mathrm{\Psi}\right)$ is
increasing on $\left[0,1\right]$ about the variable $t$, thus

$G\left(\mathrm{\Theta}\left(0\right),\mathrm{\Psi}\right)=G\left(\mathrm{\Phi},\mathrm{\Psi}\right)<G\left(\mathrm{\Psi},\mathrm{\Psi}\right)=G\left(\mathrm{\Theta}\left(1\right),\mathrm{\Psi}\right).$

Second, if $f^{\prime\prime}\left(\theta\right)>0$ and
$f^{\prime}\left(\theta\right)<0$ on $\left(0,l\right)$, then

$\sum_{i=1}^{n}\left(\psi_{i}-\phi_{i}\right)\left[f^{\prime}\left(\psi_{i}\right)-f^{\prime}\left(\theta_{i}\left(t\right)\right)\right]f^{\prime}\left(\theta_{i}\left(t\right)\right)<0.$

This shows that $G\left(\mathrm{\Theta}\left(t\right),\mathrm{\Psi}\right)$ is
decreasing on $\left[0,1\right]$ about the variable $t$, thus

$G\left(\mathrm{\Theta}\left(0\right),\mathrm{\Psi}\right)=G\left(\mathrm{\Phi},\mathrm{\Psi}\right)>G\left(\mathrm{\Psi},\mathrm{\Psi}\right)=G\left(\mathrm{\Theta}\left(1\right),\mathrm{\Psi}\right).$

∎

######  Theorem 8.

_([18])_ Assume that ${\Lambda}_{n}$ is an $n$-sided planar convex polygon
inscribed in a circle of radius $R$. Denote by $L_{n}$ and $A_{n}$ the
perimeter and area of ${\Lambda}_{n}$, then

$A_{n}-L_{n}R\cos{\sigma}+d_{n}\left(R\cos{\sigma}\right)^{2}\leq 0,{\ \ \ \ \
d}_{n}=n\tan{\frac{\pi}{n}}.$ (5.11)

The equality holds when and only when $\theta_{1}=\cdots=\theta_{n}=\sigma$.

###### Proof.

Let $f\left(\theta_{i}\right)=\sin{\theta_{i}}$,
$\theta_{i}\in(0,\frac{\pi}{2})$ and $f\left(\psi_{i}\right)=\sin{\sigma}$
where $\psi_{i}=\sigma=\frac{\pi}{n}$, $i=1,2,\cdots,n(n\geq 3)$. Then the
following inequality is the special case of (5.2), that is

$2\cos{\sigma\sum_{i=1}^{n}{\sin\theta_{i}}}\geq\sum_{i=1}^{n}{\sin\theta_{i}\cos{\theta_{i}}}+d_{n}\left(\cos{\sigma}\right)^{2}.$
(5.12)

Denote by $a_{i}$ the length of the $i$th side of ${\Lambda}_{n}$ and
$\theta_{i}$ the half of the central angle subtended by the $i$th side of
${\Lambda}_{n}$, $i=1,2,\cdots,n$, then

$L_{n}=\sum_{i=1}^{n}a_{i}=\sum_{i=1}^{n}{2R\sin{\theta_{i}}};\ \ \
A_{n}=\sum_{i=1}^{n}{\frac{1}{2}a_{i}R\cos{\theta_{i}}}=\sum_{i=1}^{n}R^{2}\sin{\theta_{i}}\cos{\theta_{i}};$
(5.13) $L_{n}^{\ast}=2Rn\sin{\frac{\pi}{n}};\ \ \
A_{n}^{\ast}=R^{2}n\sin{\sigma\cos{\sigma}}.$ (5.14)

Substituting (5.13) and (5.14) into (5.12), then (5.11) is obtained. ∎

######  Theorem 9.

_([6])_ Assume that ${\Lambda}_{n}$ is the planar convex $n$-sided polygon
with perimeter $L_{n}$ and area $A_{n}$. And ${\Lambda}_{n}$ is inscribed in a
circle with $R$ as its radius. Meanwhile, there exist a regular polygon with
perimeter $L_{n}^{\ast}$ and area $A_{n}^{\ast}$ inscribed in the same circle
with ${\Lambda}_{n}$, then

$L_{n}^{2}-4d_{n}A_{n}\geq\frac{1}{R^{2}}\left(A_{n}^{\ast}-A_{n}\right)^{2},\
\ \ \ d_{n}=n\tan{\frac{\pi}{n}}.$ (5.15)

The equality holds when and only when ${\Lambda}_{n}$ is regular polygon.

###### Proof.

From Theorem 8, we have

$\displaystyle L_{n}^{2}$
$\displaystyle\geq\left[\frac{A_{n}}{R\cos{\sigma}}+d_{n}\left(R\cos{\sigma}\right)\right]^{2}$
(5.16)
$\displaystyle=\frac{A_{n}^{2}}{\left(R\cos{\sigma}\right)^{2}}+2d_{n}A_{n}+d_{n}^{2}\left(R\cos{\sigma}\right)^{2}.$

(5.16) is equivalent to

$\displaystyle L_{n}^{2}-4d_{n}A_{n}$
$\displaystyle\geq\frac{A_{n}^{2}}{\left(R\cos{\sigma}\right)^{2}}-2d_{n}A_{n}+d_{n}^{2}\left(R\cos{\sigma}\right)^{2}$
$\displaystyle=\left[\frac{A_{n}}{R\cos{\sigma}}-d_{n}\left(R\cos{\sigma}\right)\right]^{2}$
$\displaystyle=\frac{1}{\left(R\cos{\sigma}\right)^{2}}\left(A_{n}-R^{2}n\sin{\sigma\cos{\sigma}}\right)^{2}$
$\displaystyle=\frac{1}{\left(R\cos{\sigma}\right)^{2}}\left(A_{n}^{\ast}-A_{n}\right)^{2}.$

Notice that $A_{n}^{\ast}=R^{2}n\sin{\sigma\cos{\sigma}}$,
$d_{n}=n\tan{\sigma}$, and $\sigma=\frac{\pi}{n}$. It follows that

$\displaystyle L_{n}^{2}-4d_{n}A_{n}$
$\displaystyle\geq\frac{\cos^{2}{\sigma}+\sin^{2}{\sigma}}{\left(R\cos{\sigma}\right)^{2}}\left(A_{n}^{\ast}-A_{n}\right)^{2}$
$\displaystyle=\frac{1}{R^{2}}\left(1+{tan}^{2}{\sigma}\right)\left(A_{n}^{\ast}-A_{n}\right)^{2}$
$\displaystyle\geq\frac{1}{R^{2}}\left(A_{n}^{\ast}-A_{n}\right)^{2}.$

Therefore

$L_{n}^{2}-4d_{n}A_{n}\geq\frac{1}{R^{2}}\left(A_{n}^{\ast}-A_{n}\right)^{2}.$

∎

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