A sharp reverse Bonnesen style inequality and generalization

R E S E A R C H

Open Access

A sharp reverse Bonnesen-style inequality

and generalization

Pengfu Wang

*_{Correspondence:}

[email protected] 1_{School of Mathematics and}

Statistics, Chongqing Technology and Business University, Chongqing, People’s Republic of China

Abstract

We investigate the isoperimetric deficit of the oval domain in the Euclidean plane. Via the kinematic formulae of Poincaré and Blaschke, and Blaschke’s rolling theorem, we obtain a sharp reverse Bonnesen-style inequality for a plane oval domain, which improves Bottema’s result. Furthermore, we extend the isoperimetric deficit to the symmetric mixed isoperimetric deficit for two plane oval domains, and we obtain two reverse Bonnesen-style symmetric mixed inequalities, which are generalizations of Bottema’s result and its strengthened form.

MSC: Primary 52A10; secondary 52A22

Keywords: Convex domain; Isoperimetric deficit; Isoperimetric inequality; Reverse Bonnesen-style inequality; Symmetric mixed isoperimetric deficit; Reverse

Bonnesen-style symmetric mixed inequality

1 Introduction and main results

Integral geometry originated from geometric probability. It is a very important branch of the global differential geometry, which investigates the global properties of manifolds and convex bodies. Geometric inequality is an important topic in integral geometry. Perhaps the classical isoperimetric inequality is the oldest geometric inequality, that is, the disc encloses the maximum area among all domains of fixed perimeter. LetKbe a domain of areaAwith simple boundary of perimeterPinR2_{, then}

P2– 4πA≥0, (1.1)

the equality sign holds if and only ifKis a disc.

The root of the classical isoperimetric problem can be traced back to ancient Greece. However, the rigorous mathematical proof of the isoperimetric inequality was obtained in the 19th century. Via the variational method, the first rigorous mathematical proof of the isoperimetric inequality was obtained by Weierstrass in 1870. By comparing a simple closed curve and a circle, Schmidt found a concise proof of the isoperimetric inequality in 1938. The isoperimetric inequality has been extended to the discrete case, the higher dimensions, and the surface of constant curvature (see [1,2,6,9–11,15,18,22,31–35]).

The quantity of the isoperimetric inequality (1.1)

2(K) =P2– 4πA (1.2)

measures the deficit betweenK and a disc of radiusP/2π, it is called the isoperimetric deficit ofK.

During the 1920s, Bonnesen proved some inequalities of the following form:

2(K) =P2– 4πA≥BK, (1.3)

whereBKis a nonnegative invariant of geometric significance andBK= 0 if and only ifK

is a disc. An inequality of the form (1.3) is called the Bonnesen-style inequality, and it is stronger than the isoperimetric inequality (1.1). Many Bonnesen-style inequalities have been found (see [1,4,12,16,19,33]).

Conversely, we considered the upper bound of the isoperimetric deficit, that is,

2(K) =P2– 4πA≤UK, (1.4)

where UK is a nonnegative invariant of geometric significance, it is called the reverse

Bonnesen-style inequality.

For the oval domainKinR2, Bottema obtained the following reverse Bonnesen-style inequality (see [5]):

P2– 4πA≤π2(ρM–ρm)2, (1.5)

where ρm andρM are the minimum and maximum of the continuous curvature radius

ρ of the boundary ∂K, respectively. The equality holds if and only ifρm=ρM, that is,

K is a disc. Howard, Gao, Pan, Zhang, and others (see [8,17, 29]) obtained some re-verse Bonnesen-style inequalities with the methods of analysis and curvature flow as fol-lows:

P2– 4πA≤c| ˜A|, (1.6)

wherecis a constant andA˜ is the area ofK˜, the domainK˜ is bounded by the locus of the curvature centers of∂K, where the equality sign holds if and only ifKis a disc, that is,K˜ is a point. Some reverse Bonnesen-style inequalities for surfaceX2

of constant curvature

have been obtained in [13,23,27,28]. Zhou et al. obtained some reverse Bonnesen-style inequalities for any convex domain in [33].

By comparing a simple closed curve and a circle, Schmidt proved the isoperimetric in-equality in 1938. We were motivated by Schmidt’s works, we compared the two simple closed curves directly and obtained the symmetric mixed isoperimetric inequality (see [14,20,21,24–26,30]). That is, letKk(k= 0, 1) be two domains of areasAkwith simple

boundaries of perimetersPkinR2. Then

P2₀P2₁– 16π2A0A1≥0, (1.7)

The quantity

2(K0,K1) =P02P12– 16π2A0A1 (1.8)

is called the symmetric mixed isoperimetric deficit ofK0andK1.

We were motivated by Bonnesen’s works, we considered whether there is a nonnegative invariantBK0,K1of geometric significance such that

2(K0,K1) =P20P12– 16π2A0A1≥BK0,K1, (1.9)

whereBK0,K1= 0 if and only if bothK0andK1are discs. An inequality of the form (1.9) is called the Bonnesen-style symmetric mixed inequality, it is stronger than the symmetric mixed isoperimetric inequality (1.7). Zhou, Xu, Zeng, and others (see [14,20,21,24–26, 30]) obtained some Bonnesen-style symmetric mixed inequalities with the known kine-matic formulae of Poincaré and Blaschke.

Conversely, we considered the upper bound of the symmetric mixed isoperimetric deficit ofK0andK1, that is,

2(K0,K1) =P20P12– 16π2A0A1≤UK0,K1, (1.10)

whereUK0,K1 is a nonnegative invariant of geometric significance, it is called the reverse

Bonnesen-style symmetric mixed inequality. When one of the domains is a disc, an in-equality of the form (1.10) reduces to a reverse Bonnesen-style inequality. For any con-vex domain Kk (k = 0, 1) of areas Ak and perimeters Pk in R2, Zhou, Xu, Zeng, and

others obtained the following reverse Bonnesen-style symmetric mixed inequalities (see [21,25,30]):

P2₀P2₁– 16π2A0A1≤4π2P0P1

R01R21–r01r21

, (1.11)

P2₀P2₁– 16π2A0A1≤16π4

R2₀R2₁–r2₀r2₁, (1.12)

wherer01=max{t:t(gK1)⊆K0;g∈G2}andR01=min{t:t(gK1)⊇K0;g∈G2}are the inra-dius ofK0with respect toK1and the outradius ofK0with respect toK1, respectively.G2 is a group of plane rigid motions.Rkandrkare the radius of the minimum circumscribed

disc and the radius of the maximum inscribed disc ofKk, respectively. Each equality sign

holds if and only if bothK0andK1are discs.

The purpose of this paper is to find some new reverse Bonnesen-style inequalities for the oval domain inR2_{, which generalize known reverse Bonnesen-style inequalities. Via the} kinematic formulae of Poincaré and Blaschke, and Blaschke’s rolling theorem, we obtain a sharp reverse Bonnesen-style inequality (3.10) in Theorem3.2as follows:

P2– 4πA≤(2πρM–P)(P– 2πρm),

which improves Bottema’s result. Furthermore, we obtain two reverse Bonnesen-style symmetric mixed inequalities (4.10) and (4.11) in Theorem4.2as follows:

P2₀P2₁– 16π2A0A1≤4π2A21

ρ₀₁M–ρ₀₁m2,

P2₀P2₁– 16π2A0A1≤16π2A21

ρ₀₁M– P0P1 4πA1

P0P1 4πA1

–ρ₀₁m

WhenK1 is a unit disc, (4.10) reduces to the known reverse Bonnesen-style inequality (1.5) of Bottema, inequality (4.11) reduces to (3.10).

2 Preliminaries

A set of pointsKinRn_{is convex if the line segmentλx+ (1 –λ)y∈Kfor allx,y∈Kand 0≤} λ≤1. A domain is a set with nonempty interior, and an oval domain is a convex domain of boundary at leastC2_{. A convex body is a compact convex domain. The Minkowski sum} of convex setsKandL, the scalar product of convex setKwithλ≥0 are, respectively, defined by

K+L={x+y:x∈K,y∈L}, and

λK={λx:x∈K}.

A homothety of the convex setKis of the formx+λKforx∈Rn_{,λ> 0.}

For the proof of the main theorem, we cite Blaschke’s rolling theorem inR2from [3,7, 13,25].

Lemma 2.1(Blaschke’s rolling theorem) Let K be an oval domain inR2_,ρ

mandρM be

the minimum and maximum of the curvature radius of∂K,respectively,Bt be a circle of

radius t inR2_.

If t∈(0,ρm]and Btis tangent to∂K inside,then Bthas no other common point with∂K.

If t∈[ρM, +∞)and Btis tangent to∂K outside,then Bthas no other common point with ∂K.

By Lemma2.1, we obtain the following corollary.

Corollary 2.1 Let K be an oval domain inR2,ρmandρMbe the minimum and maximum

of the curvature radius of∂K,respectively,Btbe a circle of radius t inR2.When t∈(0,ρm]

or t∈[ρM, +∞),and∂K∩∂(Bt)=∅, then Bt has two common points with∂K or Bt is

tangent to∂K.

Proof Suppose thatBt has more than two common points with∂K whent∈(0,ρm] or

t∈[ρM, +∞), then we can move Bt properly so that it is tangent to∂K and has other

common point with∂K; this is inconsistent with Blaschke’s rolling theorem.

Corollary 2.2 Let Kk(k= 0, 1)be two oval domains inR2,ρm(∂Kk)andρM(∂Kk)be the

minimum and maximum of the curvature radius of∂Kk, respectively.WhenρM(∂K1)≤

ρm(∂K0)orρm(∂K1)≥ρM(∂K0),and∂K0∩∂K1=∅,then∂K0has two common points with

∂K1or∂K0is tangent to∂K1.

Proof Suppose that∂K0has more than two common points with∂K1, we can draw a circle

Bt of radiustthrough three points among these common points. Therefore, we havet∈

(ρm(∂K0),ρM(∂K0)) andt∈(ρm(∂K1),ρM(∂K1)); this is inconsistent with the conditions of

3 Reverse Bonnesen-style inequalities

LetKbe an oval domain of areaAand perimeterPinR2_{. Letρ(∂K) be the curvature radius} of boundary∂K andρm=min{ρ(∂K)}, ρM=max{ρ(∂K)}. Let dgdenote the kinematic

density of the groupG2of plane rigid motions, andBt be a circle of radiustinR2. Let

n{∂K∩∂(gBt)}denote the number of points of intersection∂K∩∂(gBt) andχ{K∩gBt}be

the Euler–Poincaré characteristics of the intersectionK∩gBt. Then we have the following

kinematic formula of Poincaré (see [18]):

{g∈G2:∂K∩∂(gBt)=∅}

n∂K∩∂(gBt)

dg= 8πPt (3.1)

and the kinematic formula of Blaschke

{g∈G2:K∩gBt=∅}

χ{K∩gBt}dg= 2π2t2+ 2πPt+ 2πA. (3.2)

Ifμdenotes a set of all positions ofBtin which eithergBt⊂KorgBt⊃K, then the

kine-matic formula of Blaschke (3.2) can be rewritten as

μ

dg= 2π2t2+ 2πPt+ 2πA–

{g∈G2:∂K∩∂(gBt)=∅}

χ{K∩gBt}dg. (3.3)

SinceKis an oval domain inR2_{, then}

{g∈G2:∂K∩∂(gBt)=∅}

χ{K∩gBt}dg=

{g∈G2:∂K∩∂(gBt)=∅}

dg. (3.4)

Whent∈(0,ρm] ort∈[ρM, +∞), by Corollary2.1, we haven{∂K∩∂(gBt)}= 2 orgBt is

tangent to∂K. WhengBtis tangent to∂K, we have

{g∈G2:∂K∩∂(gBt)=∅}

n∂K∩∂(gBt)

dg= 0, (3.5)

therefore,

{g∈G2:∂K∩∂(gBt)=∅}

n∂K∩∂(gBt)

dg=

{g∈G2:∂K∩∂(gBt)=∅}

2dg. (3.6)

By (3.4) and (3.6), we have

{g∈G2:∂K∩∂(gBt)=∅}

χ{K∩gBt}dg=

1 2

{g∈G2:∂K∩∂(gBt)=∅}

n∂K∩∂(gBt)

dg. (3.7)

Therefore, whent∈(0,ρm] ort∈[ρM, +∞), by (3.3), (3.7), and (3.1), we obtain

μ

dg= 2π2t2+ 2πPt+ 2πA–

{g∈G2:∂K∩∂(gBt)=∅}

χ{K∩gBt}dg

= 2π2t2+ 2πPt+ 2πA–1 2

{g∈G2:∂K∩∂(gBt)=∅}

n∂K∩∂(gBt)

dg

= 2π2t2– 2πPt+ 2πA

≥0. (3.8)

Theorem 3.1 Let K be an oval domain of area A and perimeter P inR2,then

πt2–Pt+A≥0; t∈(0,ρm]or t∈[ρM, +∞). (3.9)

The inequality is strict whenever t∈(0,ρm)or t∈(ρM, +∞).When t=ρm or t=ρM,the

equality holds if and only if K is a disc.

Proof We obtain inequality (3.9) directly from (3.8)

μ

dg= 2π2t2– 2πPt+ 2πA≥0; t∈(0,ρm] ort∈[ρM, +∞).

By Blaschke’s rolling theorem (Lemma2.1), we knowBthas no other common point with ∂K whenBt is tangent to∂Kinside witht∈(0,ρm], orBt is tangent to∂K outside with

t∈[ρM, +∞). Therefore, we havegBt⊂Kwhent∈(0,ρm),gBt⊃Kwhent∈(ρM, +∞),

and∂(gBt) has no common point with∂K, then μdg> 0 whent∈(0,ρm) ort∈(ρM, +∞).

That is, inequality (3.9) is strict whenevert∈(0,ρm) ort∈(ρM, +∞).

Whent=ρmort=ρM, the equality holds clearly in inequality (3.9) ifKis a disc.

Con-versely, ifKis not a disc, by Blaschke’s rolling theorem (Lemma2.1), we know thatBρmhas no other common point with∂KwhenBρmis tangent to∂Kinside, andBρMhas no other common point with∂KwhenBρMis tangent to∂Koutside. Therefore, ifKis not a disc, we havegBρm⊂Kand∂(gBρm) has no common point with∂K,gBρM⊃Kand∂(gBρM) has no common point with∂K, then _μdg> 0 whenKis not a disc. That is,Kis a disc when

μdg= 0. Therefore, whent=ρmort=ρM, the equality holds in (3.9) if and only ifKis a

disc.

Theorem 3.2 Let K be an oval domain of area A and perimeter P inR2_,then

P2– 4πA≤(2πρM–P)(P– 2πρm), (3.10)

whereρmandρM are the minimum and maximum of the continuous curvature radiusρ

of the boundary∂K,respectively.The equality holds if and only if K is a disc.

Proof By inequality (3.9),

πt2–Pt+A≥0; t∈(0,ρm] ort∈[ρM, +∞),

we have

that is,

–4πA≤4π2ρm2– 4πPρm,

–4πA≤4π2ρM2– 4πPρM.

Therefore, we have

P2– 4πA≤P2– 4πPρm+ 4π2ρm2

= (P– 2πρm)2,

and

P2– 4πA≤P2– 4πPρM+ 4π2ρM2

= (2πρM–P)2.

SinceB(t) =πt2_{–Pt+Areaches the minimum whent=} P

2π and inequality (3.9), we have ρm≤₂Pπ ≤ρM, that is, 2πρm≤P≤2πρM. Therefore,

√

P2_{– 4πA}≤_{P– 2πρ}

m,

√

P2_{– 4πA≤2πρ}

M–P.

By multiplying the last inequalities side by side, we have

P2– 4πA≤(2πρM–P)(P– 2πρm).

The equality holds in (3.10) if and only if the two equalities hold in (3.9) whent=ρmand

t=ρM, that is,Kis a disc.

For alla≥0,b≥0, we have 4ab≤(a+b)2, that is, (2πρM–P)(P– 2πρm)≤π2(ρM–ρm)2.

Therefore, the upper bound of the isoperimetric deficit in inequality (3.10) is better than the upper bound in inequality (1.5), that is, the reverse Bonnesen-style inequality (3.10) strengthens Bottema’s result.

4 Reverse Bonnesen-style symmetric mixed inequalities

LetKk(k= 0, 1) be two oval domains inR2. Letρ(∂Kk) be the curvature radii of boundaries ∂Kk, and letρm(∂Kk) =min{ρ(∂Kk)},ρM(∂Kk) =max{ρ(∂Kk)}. Let

ρ_mg(K0,K1) =max

t:ρM∂t(gK1)

≤ρm(∂K0);g∈G2

and

ρ_Mg (K0,K1) =min

t:ρm

∂t(gK1)

≥ρM(∂K0);g∈G2

be the inradius and the outradius of curvature,K0with respect toK1, whereG2is a group of plane rigid motions. It is obvious thatρmg(K0,K1)≤ρMg (K0,K1). Since bothρmg(K0,K1) and

ρ_Mg (K0,K1) are rigid invariant, we simply denote them byρ01m andρM01, respectively. Note that, ifK1 is a unit disc, thenρ01m andρ01M are the minimumρm(∂K0) and the maximum

ρM(∂K0) of the continuous curvature radius of the boundary∂K0, respectively.

LetKk(k= 0, 1) be two oval domains of areasAkand perimetersPkinR2. Letdgdenote

the kinematic density of the groupG2of plane rigid motions. Letn{∂K0∩∂(t(gK1))} de-note the number of points of intersection∂K0∩∂(t(gK1)), and letχ{K0∩t(gK1)}be the Euler–Poincaré characteristics of the intersectionK0∩t(gK1). Then we have the following kinematic formula of Poincaré (see [14,20,21,24–26,30]):

{g∈G2:∂K0∩∂(t(gK1))=∅}

n∂K0∩∂

t(gK1)

dg= 4tP0P1 (4.1)

and the kinematic formula of Blaschke

{g∈G2:K0∩t(gK1)=∅}

χK0∩t(gK1)

dg= 2πt2A1+A0

+tP0P1. (4.2)

Letμdenote a set of all positions ofK1in which eithert(gK1)⊂K0ort(gK1)⊃K0, then (4.2) can be rewritten as

μ

dg= 2πt2A1+A0

+tP0P1–

{g∈G2:∂K0∩∂(t(gK1))=∅}

χK0∩t(gK1)

dg. (4.3)

SinceKk(k= 0, 1) are two oval domains inR2, then

{g∈G2:∂K0∩∂(t(gK1))=∅}χ

K0∩t(gK1)

dg=

{g∈G2:∂K0∩∂(t(gK1))=∅}dg. (4.4)

When t ∈ (0,ρ₀₁m] or t ∈ [ρ₀₁M, +∞), we can obtain ρM(∂(t(gK1))) ≤ ρm(∂K0) or

ρm(∂(t(gK1)))≥ρM(∂K0). By Corollary2.2, we haven{∂K0∩∂(t(gK1))}= 2 or ∂(t(gK1)) is tangent to∂K0. When∂(t(gK1)) is tangent to∂K0, we have

{g∈G2:∂K0∩∂(t(gK1))=∅}

n∂K0∩∂

t(gK1)

dg= 0, (4.5)

therefore,

{g∈G2:∂K0∩∂(t(gK1))=∅}

n∂K0∩∂

t(gK1)

dg=

{g∈G2:∂K0∩∂(t(gK1))=∅}

2dg. (4.6)

By (4.4) and (4.6), we have

{g∈G2:∂K0∩∂(t(gK1))=∅}χ

K0∩t(gK1)

dg

=1 2

{g∈G2:∂K0∩∂(t(gK1))=∅}n

∂K0∩∂

t(gK1)

Therefore, whent∈(0,ρ₀₁m] ort∈[ρM

01, +∞), by (4.3), (4.7), and (4.1), we obtain

μ

dg= 2πt2A1+A0

+tP0P1–

{g∈G2:∂K0∩∂(t(gK1))=∅}χ

K0∩t(gK1)

dg

= 2πt2A1+A0

+tP0P1– 1 2

{g∈G2:∂K0∩∂(t(gK1))=∅}n

∂K0∩∂

t(gK1)

dg

= 2πA1t2–P0P1t+ 2πA0

≥0. (4.8)

Theorem 4.1 Let Kk(k= 0, 1)be two oval domains of areas Akand perimeters PkinR2,

then

2πA1t2–P0P1t+ 2πA0≥0; t∈

0,ρm₀₁or t∈ρ₀₁M, +∞. (4.9)

The inequality is strict whenever t∈(0,ρ₀₁m)or t∈(ρ₀₁M, +∞).When t=ρ₀₁m or t=ρ₀₁M,the equality holds if and only if both K0and K1are discs.

Proof We obtain inequality (4.9) directly from (4.8)

μ

dg= 2πA1t2–P0P1t+ 2πA0≥0; t∈

0,ρ₀₁mort∈ρ₀₁M, +∞.

Whent∈(0,ρ₀₁m), that is,ρM(∂(t(gK1))) <ρm(∂K0), we have

t(gK1)⊂BρM(∂(t(gK1)))⊂Bρm(∂K0)⊂K0; t∈

0,ρ₀₁m,

where∂(BρM(∂(t(gK1)))) has no common point with∂(Bρm(∂K0)). Therefore, we havet(gK1)⊂

K0, and∂(t(gK1)) has no common point with∂(K0) whent∈(0,ρ01m). Whent∈(ρ01M, +∞), that is,ρm(∂(t(gK1))) >ρM(∂K0), we have

t(gK1)⊃Bρm(∂(t(gK1)))⊃BρM(∂K0)⊃K0; t∈

ρM₀₁, +∞,

where∂(Bρm(∂(t(gK1)))) has no common point with∂(BρM(∂K0)). Therefore, we havet(gK1)⊃ K0, and∂(t(gK1)) has no common point with∂(K0) whent∈(ρ01M, +∞). In summary, we have _μdg> 0 whent∈(0,ρ₀₁m) ort∈(ρ₀₁M, +∞). That is, inequality (4.9) is strict whenever t∈(0,ρ₀₁m) ort∈(ρ₀₁M, +∞).

Whent=ρ₀₁mort=ρ₀₁M, the equality holds clearly in inequality (4.9) if bothK0andK1are discs. Conversely, ifK0andK1of which at least one is not a disc, it includes the following two types: Only one of them is not a disc;K0andK1are not discs. When only one ofK0 andK1 is not a disc, we haveρ01m(gK1)⊂K0and∂(ρ01m(gK1)) has no common point with

∂K0,ρ01M(gK1)⊃K0and∂(ρ01M(gK1)) has no common point with∂K0, then μdg> 0 when

only one ofK0andK1is not a disc. WhenK0andK1are not discs, we have

ρ₀₁m(gK1)⊂BρM(∂(ρ01m(gK1))⊂Bρm(∂K0)⊂K0,

where∂(ρm

01(gK1)) has no common point with∂K0, and

where∂(ρM

01(gK1)) has no common point with∂K0, then μdg> 0 whenK0andK1are not

discs. In summary, _μdg> 0 whenK0andK1of which at least one is not a disc. That is, bothK0andK1are discs when μdg= 0. Therefore, whent=ρ

m

01ort=ρ01M, the equality

holds in inequality (4.9) if and only if bothK0andK1are discs.

WhenK1is a unit disc, inequality (4.9) immediately reduces to inequality (3.9). We now obtain the following reverse Bonnesen-style symmetric mixed inequalities.

Theorem 4.2 Let Kk(k= 0, 1)be two oval domains of areas Akand perimeters PkinR2,

then

P2₀P2₁– 16π2A0A1≤4π2A21

ρ₀₁M–ρ₀₁m2, (4.10)

P2₀P2₁– 16π2A0A1≤16π2A21

ρ₀₁M– P0P1 4πA1

P0P1 4πA1

–ρ₀₁m

, (4.11)

where each equality holds if and only if both K0and K1are discs.

Proof By inequality (4.9),

2πA1t2–P0P1t+ 2πA0≥0; t∈

0,ρm₀₁ort∈ρ₀₁M, +∞, we have

2πA1

ρm₀₁2–P0P1ρ01m+ 2πA0≥0, 2πA1

ρM₀₁2–P0P1ρM01+ 2πA0≥0, that is,

–16π2A0A1≤16π2A21

ρ₀₁m2– 8πA1P0P1ρ01m,

–16π2A0A1≤16π2A21

ρ₀₁M2– 8πA1P0P1ρ01M.

Therefore, we have

P2₀P2₁– 16π2A0A1≤P20P12+ 16π2A21

ρ₀₁m2– 8πA1P0P1ρm01

=P0P1– 4πA1ρ01m

2

and

P2₀P2₁– 16π2A0A1≤P20P12+ 16π2A21

ρ₀₁M2– 8πA1P0P1ρ01M

=P0P1– 4πA1ρ01M

2 .

SinceBK0,K1(t) = 2πA1t2–P0P1t+ 2πA0reaches the minimum att=₄P0P1_πA1, and inequality (4.9), we haveρ₀₁m≤₄P0P1_πA1 ≤ρ₀₁M, that is, 4πA1ρ01m≤P0P1≤4πA1ρ01M. Therefore,

P2

0P21– 16π2A0A1≤P0P1– 4πA1ρ01m,

P2₀P2₁– 16π2_A

By adding and multiplying the last inequalities side by side, we have

P2₀P2₁– 16π2A0A1≤4π2A21

ρ₀₁M–ρ₀₁m2

and

P2₀P2₁– 16π2A0A1≤16π2A21

ρ₀₁M– P0P1 4πA1

P0P1 4πA1

–ρ₀₁m

.

Each equality holds in (4.10) and (4.11) if and only if the equalities hold in (4.9) when

t=ρ₀₁m andt=ρ₀₁M, that is, bothK0andK1are discs.

WhenK1is a unit disc, the reverse Bonnesen-style symmetric mixed inequality (4.10) immediately reduces to the known reverse Bonnesen-style inequality (1.5) of Bottema, inequality (4.11) reduces to inequality (3.10). For alla≥0,b≥0, we have 4ab≤(a+b)2_, that is,

16π2A2₁

ρ₀₁M– P0P1 4πA1

P0P1 4πA1

–ρ₀₁m

≤4π2A2₁ρ₀₁M–ρ₀₁m2.

Therefore, the upper bound of the symmetric mixed isoperimetric deficit in inequality (4.11) is better than the upper bound in inequality (4.10), that is, the reverse Bonnesen-style symmetric mixed inequality (4.11) is stronger than inequality (4.10).

Acknowledgements

The author would like to express his deep gratitude to the anonymous referees for their valuable suggestions, which led to the improvement of the article.

Funding

The author is supported in part by the Chongqing Research Program of Basic Research and Frontier Technology (No. cstc2017jcyjAX0416), the Scientific and Technological Research Program of Chongqing Municipal Education Commission (No. KJ1706168), and the High Level Introduction of Talent Research Start-up Fund of Chongqing Technology and Business University (No. 1756005).

Availability of data and materials Not applicable.

Competing interests

The author declares that he has no competing interests.

Authors’ contributions

The author read and approved the final manuscript.

Publisher’s Note

Springer Nature remains neutral with regard to jurisdictional claims in published maps and institutional affiliations.

Received: 20 October 2018 Accepted: 26 March 2019

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