This paper was converted on www.awesomepapers.org from LaTeX by an anonymous user.
Want to know more? Visit the Converter page.

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 protected] School of Mathematics and Statistics, Chongqing Normal University, Chongqing 401332, People’s Republic of China [email protected]
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 2\mathbb{R}^{2} is: assume that KK is a domain with length LL and area AA, then

L24πA0,L^{2}-4\pi A\geq 0,

where the equality holds if and only if KK is a disc.

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

Δ(K)=L24πABK,\Delta\left(K\right)=L^{2}-4\pi A\geq B_{K}, (1.1)

where BkB_{k} is non-negative with geometric significance and vanishes only when KK is a disc.

Denote by Δ(K)=L24πA\Delta\left(K\right)=L^{2}-4\pi A the isoperimetric deficit of K,K, and BKB_{K} measures the “deviation” between KK and a disc. Many BKB_{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 KK is the domain of length LL and area A.A. Let rKr_{K} and RKR_{K} be the maximum inscribed radius and minimum circumscribed radius, respectively. Then

πr2Lr+A0;\pi r^{2}-Lr+A\leq 0;
LL24πA2πrKL2πRkLL2+4πA2π;\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

L24πAA2(1RK1rK)2;L24πAL2(RKrKRK+rK)2;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};
L24πAA2(1RK1r)2;L24πAL2(rRKr+RK)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};
L24πAA2(1r1rK)2;L24πAL2(rKrrK+r)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 KK is a disc.

Comparing these BKB_{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 2\mathbb{R}^{2}.

Assume that HnH_{n} is an nn-sided planar convex polygon. Denote by LnL_{n} and AnA_{n} length and area of HnH_{n}, respectively. Then

Ln24dnAn0,dn=ntanπn.L_{n}^{2}-4d_{n}A_{n}\geq 0,\ \ \ \ d_{n}=n\tan{\frac{\pi}{n}}. (1.2)

The quantity Ln24dnAnL_{n}^{2}-4d_{n}A_{n} is called the isoperimetric deficit of HnH_{n}. The sharp discrete Bonnesen-style isoperimetric inequality has the following form

Ln24dnAnUn,L_{n}^{2}-4d_{n}A_{n}\geq U_{n}, (1.3)

where UnU_{n} is non-negative with geometric significance and vanishes only when HnH_{n} is a regular polygon. UnU_{n} measures the “deviation” of HnH_{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 2\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 Λn{\Lambda}_{n} is an nn-sided planar convex polygon with perimeter LnL_{n} and area AnA_{n}, which is inscribed in a circle with radius RR. Meanwhile, there exist a regular polygon with perimeter LnL_{n}^{\ast} and area AnA_{n}^{\ast} inscribed in the same circle Λn{\Lambda}_{n}. Zhang [18] achieved the following geometric inequality for planar convex polygon in 2\mathbb{R}^{2}

Ln24dnAn(LnLn)2,dn=ntanπn,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 Λn\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×nn\times n matrix P=[pij]P=\left[p_{ij}\right] is called a doubly stochastic matrix if pij0p_{ij}\geq 0 for 1i1\leq ijnj\leq n, and

j=1npij=1,i=1,2,,n;i=1npij=1,j=1,2,,n.\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=[pij]P=\left[p_{ij}\right] with pij=1np_{ij}=\frac{1}{n}, 1i1\leq ijnj\leq n is a doubly stochastic matrix.

Let II be an open interval of the real number line R, and In=I×I××II^{n}=I\times I\times\cdots\times I. A real function f:InRf:I^{n}\rightarrow\textbf{R} (n2n\geq 2) is Schur-convex if for any doubly stochastic matrix PP and xInx\in I^{n}, then

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

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

A real function f:InRf:I^{n}\rightarrow\textbf{R} (n2n\geq 2) is symmetric if for any permutation matrix TT and xInx\in I^{n}

f(Tx)=f(x).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(x)=f(x1,x2,,xn)f\left(x\right)=f\left(x_{1},x_{2},\cdots,x_{n}\right) with symmetry has continuous partial derivatives on InI^{n}. Then f:InRf:I^{n}\rightarrow\textbf{R} is Schur-convex (concave) when and only when

(xixj)(fxifxj)0(0).\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 ff is strictly Schur-convex if inequality (2.3) is strict for xixjx_{i}\neq x_{j}1i1\leq ijnj\leq n.

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

(x1x2)(fx1fx2)0(0),\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 ff is strictly Schur-convex if (2.4) is strict for x1x2x_{1}\neq x_{2}.

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

I=(0,l);Hn={Θ=(θ1,,θn)n,i=1nθi=ml}(0<m<n);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);
Dn=InHn;Ω=(σ,,σ)whereσ=1ni=1nθi=mln.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 ff: InRI^{n}\rightarrow\textbf{R} is a Schur-concave function on InI^{n}, then f(Ω)f\left(\mathrm{\Omega}\right) is a global maximum in DnD_{n}. If ff is strictly Schur-concave on InI^{n}, then f(Ω)f\left(\mathrm{\Omega}\right) is the unique global maximum in DnD_{n}.

Proof.

Because that ff is Schur-concave, then

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

for any doubly stochastic matrix PP. Then for Θ=(θ1,,θn)\mathrm{\Theta}=\left(\theta_{1},\cdots,\theta_{n}\right) in DnD_{n}, and

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

It yields

f(Ω)=f(PΘ)f(Θ).f\left(\mathrm{\Omega}\right)=\ f\left(P\mathrm{\Theta}\right)\geq f\left(\mathrm{\Theta}\right).

If ff is strictly Schur-concave on InI^{n}, then f(Ω)>f(Θ)f\left(\mathrm{\Omega}\right)>f\left(\mathrm{\Theta}\right) for all Θ\mathrm{\Theta} in DnD_{n}, ΘΩ\mathrm{\Theta}\neq\mathrm{\Omega}. ∎

Lemma 3.

([19]) If ff: InRI^{n}\rightarrow\textbf{R} is a Schur-convex function, then f(Ω)f\left(\mathrm{\Omega}\right) is a global minimum in DnD_{n}. If ff is strictly Schur-convex on InI^{n}, then f(Ω)f\left(\mathrm{\Omega}\right) is unique global minimum in DnD_{n}.

We obtain the following analytic inequality.

Theorem 1.

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

(i=1nf(θi))2α(nf(σ))α(i=1nf(θi))α(f(σ))α[(i=1nf(θi))α(nf(σ))α],\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 α+.\alpha\in\mathbb{N}^{+}. The equality holds when and only when θ1=θ2==θn=σ\theta_{1}=\theta_{2}=\cdots=\theta_{n}=\sigma.

Proof.

Consider the following function

F(Θ)=(i=1nf(θi))2α(nf(σ))α(i=1nf(θi))α(f(σ))α[(i=1nf(θi))α(nf(σ))α].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(Θ)F\left(\Theta\right) is symmetric on InI^{n} and F(Ω)=0F\left(\mathrm{\Omega}\right)=0. In order to prove Theorem 1, we need to prove that F(Θ)0F\left(\Theta\right)\geq 0 and F(Θ)F\left(\Theta\right) is strictly Schur-convex on InI^{n}. Then by Lemma 3 it implies that F(Θ)F\left(\mathrm{\Theta}\right) has the unique global minimum F(Ω)F\left(\mathrm{\Omega}\right) in DnD_{n}. So the key of this proof is to determine whether F(Θ)F\left(\Theta\right) satisfies strict Schur-convexity on InI^{n}. By Lemma 1 and (2.4), in order to demonstrate that, it is necessary to verify the following inequality

(θ1θ2)(Fθ1Fθ2)>0,ifθ1θ2.\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 Pn=i=1nf(θi)P_{n}=\sum_{i=1}^{n}f\left(\theta_{i}\right), then

F(Θ)=(Pn)2α(nα+1)(f(σ))α(Pn)α+nα(f(σ))2α.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

Fθi=2α(Pn)2α1f(θi)α(nα+1)(f(σ))α(Pn)α1f(θi)\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

(θ1θ2)(Fθ1Fθ2)=\displaystyle\left(\theta_{1}-\theta_{2}\right)\left(\frac{\partial F}{\partial\theta_{1}}-\frac{\partial F}{\partial\theta_{2}}\right)= (θ1θ2)(f(θ1)f(θ2))\displaystyle\left(\theta_{1}-\theta_{2}\right)\cdot\left(f^{\prime}\left(\theta_{1}\right)-f^{\prime}\left(\theta_{2}\right)\right)
[2α(Pn)2α1α(nα+1)(f(σ))α(Pn)α1].\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 ff has strict convexity, then f′′>0f^{\prime\prime}>0 and ff^{\prime} is increasing on (0,l).\left(0,l\right). This leads to

(θ1θ2)(f(θ1)f(θ2))>0.\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(Θ)=2α(Pn)2α1α(nα+1)(f(σ))α(Pn)α1.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 ff is strictly convex and by Jensen’s inequality it follows that

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

(2.9) is strict and it implies that

(Pn)α+(Pn)α>(nf(σ))α+(f(σ))α=(nα+1)(f(σ))α,\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(Pn)α>(nα+1)(f(σ))α.{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

2α(Pn)2α1>α(nα+1)(f(σ))α(Pn)α1.2\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(Θ)=2α(Pn)2α1α(nα+1)(f(σ))α(Pn)α1>0.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(θ)f\left(\theta\right) is a positive convex (resp. concave) function, and taking α=1,\alpha=1, then (2.9) becomes the classical Jensen’s inequality:

i=1nf(θi)()nf(σ),\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 θ1=θ2==θn=σ\theta_{1}=\theta_{2}=\cdots=\theta_{n}=\sigma. For example

(1)i=1ntanθintanσ,(2)i=1nsinθinsinσ,\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 θi(0,π2)\theta_{i}\in\left(0,\frac{\pi}{2}\right), i=1,2,,ni=1,2,\cdots,n; and   i=1nθi=π\sum_{i=1}^{n}\theta_{i}=\pi,  σ=1ni=1nθi=πn\sigma=\frac{1}{n}\sum_{i=1}^{n}\theta_{i}=\frac{\pi}{n}.

In particular, if f(θ)=tanθf\left(\theta\right)=\tan{\theta} and f(θ)=secθf\left(\theta\right)=\sec{\theta} for θ(0,π2)\theta\in\left(0,\frac{\pi}{2}\right), then

Corollary 1.

Let θi(0,π2)\theta_{i}\in\left(0,\frac{\pi}{2}\right), i=1,2,,ni=1,2,\cdots,n; and   i=1nθi=π\sum_{i=1}^{n}\theta_{i}=\pi, then for α+\alpha\in\mathbb{N}^{+}

(i=1ntanθi)2α(ntanπn)α(i=1ntanθi)α(tanπn)α[(i=1ntanθi)α(ntanπn)α];\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 α=1\alpha=1, then

(i=1ntanθi)2(ntanπn)(i=1ntanθi)tanπn(i=1ntanθintanπn),\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 θ1=θ2==θn=πn\theta_{1}=\theta_{2}=\cdots=\theta_{n}=\frac{\pi}{n}.

Corollary 2.

Let θi(0,π2)\theta_{i}\in\left(0,\frac{\pi}{2}\right), i=1,2,,ni=1,2,\cdots,n; and  i=1nθi=π\sum_{i=1}^{n}\theta_{i}=\pi, then

(i=1nsecθi)2α(nsecπn)α(i=1nsecθi)α(secπn)α[(i=1nsecθi)α(nsecπn)α],\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 α+.\alpha\in\mathbb{N}^{+}. The equality holds when and only when θ1=θ2==θn=πn\theta_{1}=\theta_{2}=\cdots=\theta_{n}=\frac{\pi}{n}.

Notice that f(x)=x2f\left(x\right)=x^{2} on (0,1)\left(0,1\right) has strict convexity, then

Corollary 3.

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

(i=1nxi2)2α(m2n)α(i=1nxi2)α(m2n2)α[(i=1nxi2)α(m2n)α].\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 α+.\alpha\in\mathbb{N}^{+}. The equality holds when and only when x1=x2==xn=mnx_{1}=x_{2}=\cdots=x_{n}=\frac{m}{n} (0<m<n)\left(0<m<n\right).

Theorem 2.

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

(i=1nf(θi))2α(nf(σ))α(i=1nf(θi))α(i=1nf(θi))kα(nf(σ))kα,\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 α,k+\alpha,k\in\mathbb{N}^{+} and k2k\geq 2. The equality holds when and only when θ1=θ2==θn=σ\theta_{1}=\theta_{2}=\cdots=\theta_{n}=\sigma.

Proof.

Consider the following function

F(Θ)=(i=1nf(θi))2α(nf(σ))α(i=1nf(θi))α(i=1nf(θi))kα+(nf(σ))kα.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(Θ)F\left(\Theta\right) is symmetrical on InI^{n} and F(Ω)=0F\left(\mathrm{\Omega}\right)=0. Lemma 2 implies that F(Ω)F\left(\mathrm{\Omega}\right) is the unique global maximum in DnD_{n} when F(Θ)F\left(\Theta\right) is strictly Schur-concave. So the key is to determine whether F(Θ)F\left(\Theta\right) is strictly Schur-concave on InI^{n}. By Lemma 1 and (2.4), in order to demonstrate that, it is necessary to verify the following inequality

(θ1θ2)(Fθ1Fθ2)<0,ifθ1θ2.\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 Pn=i=1nf(θi)P_{n}=\sum_{i=1}^{n}f\left(\theta_{i}\right), then

F(Θ)=(Pn)2α(nf(σ))α(Pn)α(Pn)kα+(nf(σ))kα.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

Fθi\displaystyle\frac{\partial F}{\partial\theta_{i}} =2α(Pn)2α1f(θi)n(f(σ))aα(Pn)α1f(θi)kα(Pn)kα1f(θi)\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)
=f(θi)[2α(Pn)2α1n(f(σ))aα(Pn)α1kα(Pn)kα1]\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

(θ1θ2)(fθ1fθ2)=\displaystyle\left(\theta_{1}-\theta_{2}\right)\left(\frac{\partial f}{\partial\theta_{1}}-\frac{\partial f}{\partial\theta_{2}}\right)= (θ1θ2)(f(θ1)f(θ2))\displaystyle\left(\theta_{1}-\theta_{2}\right)\cdot\left(f^{\prime}\left(\theta_{1}\right)-f^{\prime}\left(\theta_{2}\right)\right) (2.20)
[2α(Pn)2α1n(f(σ))aα(Pn)α1kα(Pn)kα1].\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 ff is strictly convex, then f′′>0f^{\prime\prime}>0 and ff^{{}^{\prime}} is increasing. It leads to

(θ1θ2)(f(θ1)f(θ2))>0,\left(\theta_{1}-\theta_{2}\right)\left(f^{\prime}\left(\theta_{1}\right)-f^{\prime}\left(\theta_{2}\right)\right)>0,

and

2α(Pn)2α1n(f(σ))aα(Pn)α1kα(Pn)kα1<0.2\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(θ)=tanθf\left(\theta\right)=\tan{\theta} and f(θ)=cscθf\left(\theta\right)=\csc{\theta} for θ(0,π2)\theta\in\left(0,\frac{\pi}{2}\right), then we have

Corollary 4.

Let θi(0,π2)\theta_{i}\in\left(0,\frac{\pi}{2}\right), i=1,2,,n;i=1,2,\cdots,n; and i=1nθi=π\sum_{i=1}^{n}\theta_{i}=\pi, then

(i=1ntanθi)2α(ntanπn)α(i=1ntanθi)α(i=1ntanθi)kα(ntanπn)kα,\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 α,k+\alpha,k\in\mathbb{N}^{+} and k2k\geq 2. Especially, if α=1\alpha=1 and k=3k=3, then

(i=1ntanθi)2(ntanπn)(i=1ntanθi)(i=1ntanθi)3(ntanπn)3,\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 θ1=θ2==θn=πn\theta_{1}=\theta_{2}=\cdots=\theta_{n}=\frac{\pi}{n}.

Corollary 5.

Let f(θ)=cscθf\left(\theta\right)=\csc{\theta}θi(0,π2)\theta_{i}\in\left(0,\frac{\pi}{2}\right), i=1,2,,ni=1,2,\cdots,n; and i=1nθi=π\sum_{i=1}^{n}\theta_{i}=\pi, then

(i=1ncscθi)2α(ncscπn)α(i=1ncscθi)α(i=1ncscθi)kα(ncscπn)kα,\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 α,k+\alpha,k\in\mathbb{N}^{+} and k2k\geq 2. Especially, when α=1\alpha=1 and k=3k=3, then

(i=1ncscθi)2(ncscπn)(i=1ncscθi)(i=1ncscθi)3(ncscπn)3,\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 θ1=θ2==θn=πn\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 Γn{\Gamma}_{n} is an nn-sided planar convex polygon circumscribed in a circle of radius RR. Denote by LnL_{n} and AnA_{n} the perimeter and area of Γn{\Gamma}_{n}, then

Ln2α4α(dnAn)α2αRα(tanπn)α[Lnα(Ln)α],dn=ntanπn,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)
(AnR2)2αdnα(Ln2R)α(tanπn)α[(Ln2R)α(Ln2R)α],dn=ntanπn,\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 α+\alpha\in\mathbb{N}^{+}, LnL_{n}^{\ast} is the perimeter of the regular nn-sided convex polygon circumscribed in the same circle with Γn{\Gamma}_{n}. Both equalities hold when and only when Γn{\Gamma}_{n} is a regular polygon.

Proof.

Denote by aia_{i} and θi\theta_{i} the length of the iith side of Γn{\Gamma}_{n}, and the half of the central angle subtended by the iith vertex AiA_{i} of Γn{\Gamma}_{n}, i=1,2,,ni=1,2,\cdots,n, respectively. Then

[Uncaptioned image]
Ln=i=1nai=2Ri=1ntanθi;Ln=2nRtanπn.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)
i=1nθi=π;An=12i=1naiR=R2i=1ntanθi;An=nR2tanπn.\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 α=1\alpha=1, we have the following special case.

Corollary 6.

Assume that Γn{\Gamma}_{n} is an nn-sided planar convex polygon circumscribed in a circle of radius RR. Denote by LnL_{n} and AnA_{n} the perimeter and area of Γn{\Gamma}_{n}, then

Ln24dnAn2R(tanπn)[LnLn],L_{n}^{2}-4d_{n}A_{n}\geq 2R\left(\tan{\frac{\pi}{n}}\right)\left[L_{n}-L_{n}^{\ast}\right], (3.5)
(AnR2)2dn(Ln2R)(tanπn)(Ln2RLn2R),\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 dn=ntanπnd_{n}=n\tan{\frac{\pi}{n}}. Both equalities hold when and only when Γn{\Gamma}_{n} is a regular polygon.

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

Theorem 4.

Assume that Γn{\Gamma}_{n} is an nn-sided planar convex polygon circumscribed in a circle of radius RR. Denote by LnL_{n} and AnA_{n} the perimeter and area of Γn{\Gamma}_{n}, then

Ln2α4α(dnAn)α4α(tanπn)α[Anα(An)α],dn=ntanπn,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)
(AnR2)2αdnα(Ln2R)α(tanπn)α[(AnR2)α(AnR2)α],dn=ntanπn,\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 α+\alpha\in\mathbb{N}^{+}, and AnA_{n}^{\ast} denotes the area of the regular nn-sided convex polygon circumscribed in the same circle with Γn{\Gamma}_{n}. Both equalities hold when and only when Γn{\Gamma}_{n} is a regular polygon.

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

Corollary 7.

Assume that Γn{\Gamma}_{n} is an nn-sided planar convex polygon circumscribed in a circle of radius RR. Denote by LnL_{n} and AnA_{n} the perimeter and area of Γn{\Gamma}_{n}, then

Ln24dnAn4(tanπn)[AnAn],L_{n}^{2}-4d_{n}A_{n}\geq 4\left(\tan{\frac{\pi}{n}}\right)\left[A_{n}-A_{n}^{\ast}\right], (3.9)
(AnR2)2dn(Ln2R)(tanπn)[(AnR2)(AnR2)],\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 dn=ntanπnd_{n}=n\tan{\frac{\pi}{n}}. Both equalities hold when and only when Γn{\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 2\mathbb{R}^{2}([10]). The discrete cases are more complex and difficult.

Assume that KK is a convex domain with C2C^{2} smooth boundary in 2\mathbb{R}^{2}, then ([10])

Δ(K)=L24πAπ2(ρMρm)2,\mathrm{\Delta}\left(K\right)=L^{2}-4\pi A\leq\pi^{2}\left(\rho_{M}-\rho_{m}\right)^{2},

where ρM\rho_{M} and ρm\rho_{m} are the maximum and minimum value of ρ(K)\rho\left(\partial K\right), respectively. The equality holds when and only when K\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 nn-sided planar convex polygon if there is a geometric invariant DnD_{n} such that

Ln24dnAnDn,dn=ntanπn,L_{n}^{2}-4d_{n}A_{n}\leq D_{n},\ \ \ d_{n}=n\tan{\frac{\pi}{n}},

where DnD_{n} satisfies

  1. (1)

    Dn0D_{n}\geq 0;

  2. (2)

    Dn=0D_{n}=0 only when HnH_{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 Γn{\Gamma}_{n} is an nn-sided planar convex polygon circumscribed in a circle of radius RR. Denote by LnL_{n} and AnA_{n} the perimeter and area of Γn{\Gamma}_{n}, then

Ln2α(4dnAn)αLnkα(Ln)kα,dn=ntanπn,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)
(AnR2)2αdnα(Ln2R)α(Ln2R)kα(Ln2R)kα,\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 α,k+\alpha,k\in\mathbb{N}^{+} and k2k\geq 2. Both equalities hold when and only when Γn{\Gamma}_{n} is a regular polygon.

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

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

Corollary 8.

Assume that Γn{\Gamma}_{n} is an nn-sided planar convex polygon circumscribed in a circle of radius RR. Denote by LnL_{n} and AnA_{n} the perimeter and area of Γn{\Gamma}_{n}, then

Ln24dnAnLn3(Ln)3,L_{n}^{2}-4d_{n}A_{n}\leq L_{n}^{3}-(L_{n}^{\ast})^{3}, (4.3)
(AnR2)2dn(Ln2R)(Ln2R)2(Ln2R)2,\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 dn=ntanπnd_{n}=n\tan{\frac{\pi}{n}}. Both equalities hold when and only when Γn{\Gamma}_{n} is a regular polygon.

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

Theorem 6.

Assume that Γn{\Gamma}_{n} is an nn-sided planar convex polygon circumscribed in a circle of radius RR. Denote by LnL_{n} and AnA_{n} the perimeter and area of Γn{\Gamma}_{n}, then

Ln2α4α(dnAn)α4αR2(k1)α[Ankα(An)kα],dn=ntanπn,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)
(AnR2)2αdnα(Ln2R)α(AnR2)kα(AnR2)kα,\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 α,k\alpha,k and k2k\geq 2. Both equalities hold when and only when Γn{\Gamma}_{n} is a regular polygon.

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

Corollary 9.

Assume that Γn{\Gamma}_{n} is an nn-sided planar convex polygon circumscribed in a circle of radius RR. Denote by LnL_{n} and AnA_{n} the perimeter and area of Γn{\Gamma}_{n}, then

Ln24dnAn4R2[An2(An)2],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)
(AnR2)2dn(Ln2R)(AnR2)3(AnR2)3,\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 dn=ntanπnd_{n}=n\tan{\frac{\pi}{n}}. Both equalities hold when and only when Γn{\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 θi\theta_{i} and ψi\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 θi\theta_{i} and ψi\psi_{i} are real numbers in (0,l)\left(0,l\right), i=1,2,,ni=1,2,\cdots,n, i=1nθi=ml=i=1nψi\sum_{i=1}^{n}\theta_{i}=ml=\sum_{i=1}^{n}\psi_{i}, where mm is a positive constant less then nn. Then

Theorem 7.

Assume that f(θ)f\left(\theta\right) is a positive C2C^{2}-function on (0,l)\left(0,l\right), and it satisfies

f2(θ)f(θ)f′′(θ)=μ,θ(0,l),{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 μ\mu is a constant and f(θ)f′′(θ)0f^{\prime}\left(\theta\right)f^{\prime\prime}\left(\theta\right)\neq 0.

  1. (i)
    1. (1)

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

      2i=1nf(θi)f(ψi)i=1nf(θi)f(θi)+i=1nf(ψi)f(ψi);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′′(θ)<0f^{\prime\prime}\left(\theta\right)<0 and f(θ)<0f^{\prime}\left(\theta\right)<0 on (0,l)\left(0,l\right), then

      2i=1nf(θi)f(ψi)i=1nf(θi)f(θi)+i=1nf(ψi)f(ψi);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′′(θ)>0f^{\prime\prime}\left(\theta\right)>0 and f(θ)>0f^{\prime}\left(\theta\right)>0 on (0,l)\left(0,l\right), then

      2i=1nf(θi)f(ψi)i=1nf(θi)f(θi)+i=1nf(ψi)f(ψi);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′′(θ)>0f^{\prime\prime}\left(\theta\right)>0 and f(θ)<0f^{\prime}\left(\theta\right)<0 on (0,l)\left(0,l\right), then

      2i=1nf(θi)f(ψi)i=1nf(θi)f(θi)+i=1nf(ψi)f(ψi).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 θi=ψi\theta_{i}=\psi_{i}.

Proof.

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

Θ=(θ1,θ2,,θn)n;Ψ=(ψ1,ψ2,,ψn)n;\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};
Hn={Θn,i=1nθi=ml}(0<m<n);H_{n}=\Big{\{}\mathrm{\Theta}\in\mathbb{R}^{n},\sum_{i=1}^{n}\theta_{i}=ml\Big{\}}\quad\left(0<m<n\right);
In=(0,l)nn;Dn=HnIn.I_{n}=\left(0,l\right)^{n}\subset\mathbb{R}^{n};\ \ \ D_{n}=H_{n}\cap I_{n}.

We set

G(Θ,Ψ)\displaystyle G\left(\mathrm{\Theta},\mathrm{\Psi}\right) =2i=1nf(θi)f(ψi)i=1nf(θi)f(θi)i=1nf(ψi)f(ψi).\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′′(θ)<0f^{\prime\prime}\left(\theta\right)<0 and f(θ)>0f^{\prime}\left(\theta\right)>0 on (0,l)\left(0,l\right), then f(θ)f^{\prime}\left(\theta\right) is a strict decreasing function on (0,l)\left(0,l\right). Since G(Ψ,Ψ)=0G\left(\mathrm{\Psi},\mathrm{\Psi}\right)=0, we should demonstrate that G(Θ,Ψ)0G\left(\mathrm{\Theta},\mathrm{\Psi}\right)\geq 0 with equality only when Θ=Ψ\Theta=\Psi. It suffices for us to verify the following inequality

G(Φ,Ψ)>G(Ψ,Ψ)foranyΦ=(ϕ1,ϕ2,,ϕn)ΨinDn.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 Φ\Phi and Ψ\Psi can be defined as the following form

Θ(t)=(θ1(t),θ2(t),,θn(t)),\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

θi(t)=tψi+(1t)ϕi,i=1,2,,n,t[0,1].\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

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

Observe that

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

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

(ψiϕi)[f(ψi)f(θi(t))]<0,i=1,2,,n,t[0,1].\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(θ)f^{\prime}\left(\theta\right) is decreasing on (0,l)\left(0,l\right). If ψi>ϕi\psi_{i}>\phi_{i}, then

θi(t)=tψi+(1t)ϕi<tψi+(1t)ψi=ψi,\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(θi(t))>f(ψi)f^{\prime}\left(\theta_{i}\left(t\right)\right)>f^{\prime}\left(\psi_{i}\right). If ψi<ϕi\psi_{i}<\phi_{i}, then

θi(t)=tψi+(1t)ϕi>tψi+(1t)ψi=ψi,\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(θi(t))<f(ψi)f^{\prime}\left(\theta_{i}\left(t\right)\right)<f^{\prime}\left(\psi_{i}\right).

Hence, the inequality (5.8) also holds for 0t10\leq t\leq 1

i=1n(ψiϕi)[f(ψi)f(θi(t))]f(θi(t))<0.\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

μi=1n(ψiϕi)\displaystyle\mu\sum_{i=1}^{n}\left(\psi_{i}-\phi_{i}\right) =i=1n[f2(θi(t))f(θi(t))f′′(θi(t))](ψiϕi)=0.\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

i=1n(ψiϕi)f(θi(t))f′′(θi(t))\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)} =i=1n(ψiϕi)[f(θi(t))]2.\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(Θ(t),Ψ)G\left(\mathrm{\Theta}\left(t\right),\mathrm{\Psi}\right) with respect to tt and by (5.9), we have

G(Θ(t),Ψ)\displaystyle G^{\prime}\left(\mathrm{\Theta}\left(t\right),\mathrm{\Psi}\right) =2i=1nf(ψi)f(θi(t))(ψiϕi)i=1n[2f2(θi(t))1](ψiϕi)\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)
=2i=1nf(ψi)f(θi(t))(ψiϕi)2i=1nf2(θi(t))(ψiϕi)\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)
=2i=1n(ψiϕi)[f(ψi)f(θi(t))]f(θi(t))<0.\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(Θ(t),Ψ)G\left(\mathrm{\Theta}\left(t\right),\mathrm{\Psi}\right) is decreasing on [0,1]\left[0,1\right] about the variable tt, thus

G(Θ(0),Ψ)=G(Φ,Ψ)>G(Ψ,Ψ)=G(Θ(1),Ψ).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 Φ\Phi is arbitrarily chosen, the proof of case (i)(1) is complete.

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

G(Θ(0),Ψ)=G(Φ,Ψ)<G(Ψ,Ψ)=G(Θ(1),Ψ).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′′(θ)>0f^{\prime\prime}\left(\theta\right)>0 and f(θ)>0f^{\prime}\left(\theta\right)>0 on (0,l)\left(0,l\right). The proof is similar to case (i)(1), then

i=1n(ψiϕi)[f(ψi)f(θi(t))]f(θi(t))>0.\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(Θ(t),Ψ)G\left(\mathrm{\Theta}\left(t\right),\mathrm{\Psi}\right) is increasing on [0,1]\left[0,1\right] about the variable tt, thus

G(Θ(0),Ψ)=G(Φ,Ψ)<G(Ψ,Ψ)=G(Θ(1),Ψ).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′′(θ)>0f^{\prime\prime}\left(\theta\right)>0 and f(θ)<0f^{\prime}\left(\theta\right)<0 on (0,l)\left(0,l\right), then

i=1n(ψiϕi)[f(ψi)f(θi(t))]f(θi(t))<0.\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(Θ(t),Ψ)G\left(\mathrm{\Theta}\left(t\right),\mathrm{\Psi}\right) is decreasing on [0,1]\left[0,1\right] about the variable tt, thus

G(Θ(0),Ψ)=G(Φ,Ψ)>G(Ψ,Ψ)=G(Θ(1),Ψ).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 Λn{\Lambda}_{n} is an nn-sided planar convex polygon inscribed in a circle of radius RR. Denote by LnL_{n} and AnA_{n} the perimeter and area of Λn{\Lambda}_{n}, then

AnLnRcosσ+dn(Rcosσ)20,dn=ntanπn.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 θ1==θn=σ\theta_{1}=\cdots=\theta_{n}=\sigma.

Proof.

Let f(θi)=sinθif\left(\theta_{i}\right)=\sin{\theta_{i}}, θi(0,π2)\theta_{i}\in(0,\frac{\pi}{2}) and f(ψi)=sinσf\left(\psi_{i}\right)=\sin{\sigma} where ψi=σ=πn\psi_{i}=\sigma=\frac{\pi}{n}, i=1,2,,n(n3)i=1,2,\cdots,n(n\geq 3). Then the following inequality is the special case of (5.2), that is

2cosσi=1nsinθii=1nsinθicosθi+dn(cosσ)2.2\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 aia_{i} the length of the iith side of Λn{\Lambda}_{n} and θi\theta_{i} the half of the central angle subtended by the iith side of Λn{\Lambda}_{n}, i=1,2,,ni=1,2,\cdots,n, then

[Uncaptioned image]
Ln=i=1nai=i=1n2Rsinθi;An=i=1n12aiRcosθi=i=1nR2sinθicosθi;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)
Ln=2Rnsinπn;An=R2nsinσcosσ.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 Λn{\Lambda}_{n} is the planar convex nn-sided polygon with perimeter LnL_{n} and area AnA_{n}. And Λn{\Lambda}_{n} is inscribed in a circle with RR as its radius. Meanwhile, there exist a regular polygon with perimeter LnL_{n}^{\ast} and area AnA_{n}^{\ast} inscribed in the same circle with Λn{\Lambda}_{n}, then

Ln24dnAn1R2(AnAn)2,dn=ntanπn.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 Λn{\Lambda}_{n} is regular polygon.

Proof.

From Theorem 8, we have

Ln2\displaystyle L_{n}^{2} [AnRcosσ+dn(Rcosσ)]2\displaystyle\geq\left[\frac{A_{n}}{R\cos{\sigma}}+d_{n}\left(R\cos{\sigma}\right)\right]^{2} (5.16)
=An2(Rcosσ)2+2dnAn+dn2(Rcosσ)2.\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

Ln24dnAn\displaystyle L_{n}^{2}-4d_{n}A_{n} An2(Rcosσ)22dnAn+dn2(Rcosσ)2\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}
=[AnRcosσdn(Rcosσ)]2\displaystyle=\left[\frac{A_{n}}{R\cos{\sigma}}-d_{n}\left(R\cos{\sigma}\right)\right]^{2}
=1(Rcosσ)2(AnR2nsinσcosσ)2\displaystyle=\frac{1}{\left(R\cos{\sigma}\right)^{2}}\left(A_{n}-R^{2}n\sin{\sigma\cos{\sigma}}\right)^{2}
=1(Rcosσ)2(AnAn)2.\displaystyle=\frac{1}{\left(R\cos{\sigma}\right)^{2}}\left(A_{n}^{\ast}-A_{n}\right)^{2}.

Notice that An=R2nsinσcosσA_{n}^{\ast}=R^{2}n\sin{\sigma\cos{\sigma}}, dn=ntanσd_{n}=n\tan{\sigma}, and σ=πn\sigma=\frac{\pi}{n}. It follows that

Ln24dnAn\displaystyle L_{n}^{2}-4d_{n}A_{n} cos2σ+sin2σ(Rcosσ)2(AnAn)2\displaystyle\geq\frac{\cos^{2}{\sigma}+\sin^{2}{\sigma}}{\left(R\cos{\sigma}\right)^{2}}\left(A_{n}^{\ast}-A_{n}\right)^{2}
=1R2(1+tan2σ)(AnAn)2\displaystyle=\frac{1}{R^{2}}\left(1+{tan}^{2}{\sigma}\right)\left(A_{n}^{\ast}-A_{n}\right)^{2}
1R2(AnAn)2.\displaystyle\geq\frac{1}{R^{2}}\left(A_{n}^{\ast}-A_{n}\right)^{2}.

Therefore

Ln24dnAn1R2(AnAn)2.L_{n}^{2}-4d_{n}A_{n}\geq\frac{1}{R^{2}}\left(A_{n}^{\ast}-A_{n}\right)^{2}.

References

  • [1] W. Blaschke, Kreis und Kugel.(German) Chelsea Publishing Co., New York, 1949.
  • [2] X. Gao, A new reverse isoperimetric inequality and its stability, Math. Inequal. Appl. 15(2012), 733-743.
  • [3] H. Groemer, Geometric applications of Fourier series and spherical harmonics, Encyclopedia Math. Appl., vol. 61, Cambridge Univ. Press, Cambridge, 1996.
  • [4] G. Hardy, J. Littlewood, G. Polya, Inequalities, Cambridge Math. Library, Cambridge Unive. Press, Cambridge, 1988.
  • [5] A. Hurwitz, Sur quelques applications géométriques des séries de Fourier (French), Ann. Sci. École Norm. Sup. 19(1902), 357-408.
  • [6] L. Ma, A Bonnesen-style inequality for the planar convex polygon, J. of Math.(PRC) 35(2015), 154-158.
  • [7] A. W. Marshall, I. Olkin, Inequalities: Theory of Majorization and Its Applications, Academic Press, 1979.
  • [8] R. Osserman, The isoperimetric inequality, Bull. Amer. Math. Soc. (84)1978, 1182-1238.
  • [9] R. Osserman, Bonnesen-style isoperimetric inequalities, Amer. Math. Monthly 86(1979), 1-29.
  • [10] D. Ren, Topics in integral geometry, World Scientific Publishing Co., Inc., River Edge, NJ, 1994.
  • [11] I. Schur, Uber eine Klasse von Mittelbildungen mit Anwendungen auf die Determinantentheorie, Sitzunsber. Berlin. Math. Ges. 22(1923) 9-20.
  • [12] D. Tang, Discrete wirtinger and isoperimetric type inequalities, Bull. Austral. Math. Soc. 43(1991) 467-474.
  • [13] J. Qi, W. Wang, Schur convex functions and the Bonnesen-style isoperimetric inequalities for planar convex polygons, J. Math. Inequal. 12(2018), 23-29.
  • [14] A. Wayne, E. Dale, Convex functions, Pure and Applied Mathematics, vol. 57, Academic Press, New York-London, 1973.
  • [15] C. Zeng, L. Ma, J. Zhou, The Bonnesen isoperimetric inequality in a surface of constant curvature, Sci. China. Math. 55(2012), 1913-1919.
  • [16] C. Zeng, J. Zhou, S. Yue, The symmetric mixed isoperimetric inequality of two planar convex domains, Acta Math. Sinica (Chinese Series) 55(2012), 355-362.
  • [17] X. Zhang, A Refinement of the Discrete Wirtinger Ineqality, J. Math. Anal. Appl. 200(1996), 687-697.
  • [18] X. Zhang, Bonnesen-style inequalities and Pseudo-perimeters for polygons, J. Geom. 60(1997), 188-201.
  • [19] X. Zhang, Schur-convex functions and isoperimetric inequalities, Proc. Amer. Math. Soc. 126(1998), 461-470.
  • [20] G. Zhang, The affine Sobolev inequality, J. Differential Geom. 53(1999), 183-202.
  • [21] J. Zhou, Kinematic formulas for mean curvature powers of hypersurfaces and Hadwiger’s theorem in 2n\mathbb{R}^{2n}, Trans. Amer. Math. Soc. 345(1994), 243-262.
  • [22] J. Zhou, Kinematic formula for square mean curvature of hypersurfaces, Bull. Inst. Math. Acad. Sinica 22(1994), 31-47.