A sharp reverse Bonnesen-style inequality and generalization

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.

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. Let K be a domain of area A with simple boundary of perimeter P in $\mathbb{R}^{2}$, then

$$ P^{2}-4\pi A\ge 0, $$

(1.1)

the equality sign holds if and only if K is 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, 9,10,11, 15, 18, 22, 31,32,33,34,35]).

The quantity of the isoperimetric inequality (1.1)

$$ \Delta _{2}(K)=P^{2}-4\pi A $$

(1.2)

measures the deficit between K and a disc of radius $P/2\pi$, it is called the isoperimetric deficit of K.

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

$$ \Delta _{2} (K)=P^{2}-4\pi A\ge B_{K}, $$

(1.3)

where $B_{K}$ is a nonnegative invariant of geometric significance and $B_{K}=0$ if and only if K 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,

$$ \Delta _{2} (K)=P^{2}-4\pi A\le U_{K}, $$

(1.4)

where $U_{K}$ is a nonnegative invariant of geometric significance, it is called the reverse Bonnesen-style inequality.

For the oval domain K in $\mathbb{R}^{2}$, Bottema obtained the following reverse Bonnesen-style inequality (see [5]):

$$ P^{2}-4\pi A\le \pi ^{2}(\rho _{M}- \rho _{m})^{2}, $$

(1.5)

where $\rho _{m}$ and $\rho _{M}$ are the minimum and maximum of the continuous curvature radius ρ of the boundary ∂K, respectively. The equality holds if and only if $\rho _{m}=\rho _{M}$, that is, K is a disc. Howard, Gao, Pan, Zhang, and others (see [8, 17, 29]) obtained some reverse Bonnesen-style inequalities with the methods of analysis and curvature flow as follows:

$$ P^{2}-4\pi A\le c \vert \tilde{A} \vert , $$

(1.6)

where c is a constant and à is the area of K̃, the domain K̃ is bounded by the locus of the curvature centers of ∂K, where the equality sign holds if and only if K is a disc, that is, K̃ is a point. Some reverse Bonnesen-style inequalities for surface $X_{\epsilon }^{2}$ 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 inequality 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, 24,25,26, 30]). That is, let $K_{k}$ ($k=0,1$) be two domains of areas $A_{k}$ with simple boundaries of perimeters $P_{k}$ in $\mathbb{R}^{2}$. Then

$$ P_{0}^{2}P_{1}^{2}-16\pi ^{2} A_{0}A_{1}\ge 0, $$

(1.7)

where the equality sign holds if and only if both $K_{0}$ and $K_{1}$ are discs. When one of the domains is a disc, inequality (1.7) immediately reduces to (1.1). That is, the symmetric mixed isoperimetric inequality (1.7) is a generalization of the isoperimetric inequality (1.1).

The quantity

$$ \Delta _{2}(K_{0},K_{1})=P_{0}^{2}P_{1}^{2}-16 \pi ^{2} A_{0}A_{1} $$

(1.8)

is called the symmetric mixed isoperimetric deficit of $K_{0}$ and $K_{1}$.

We were motivated by Bonnesen’s works, we considered whether there is a nonnegative invariant $B_{K_{0},K_{1}}$ of geometric significance such that

$$ \Delta _{2}(K_{0},K_{1})=P_{0}^{2}P_{1}^{2}-16 \pi ^{2} A_{0}A_{1}\geq B _{K_{0},K_{1}}, $$

(1.9)

where $B_{K_{0},K_{1}}=0$ if and only if both $K_{0}$ and $K_{1}$ are 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, 24,25,26, 30]) obtained some Bonnesen-style symmetric mixed inequalities with the known kinematic formulae of Poincaré and Blaschke.

Conversely, we considered the upper bound of the symmetric mixed isoperimetric deficit of $K_{0}$ and $K_{1}$, that is,

$$ \Delta _{2}(K_{0},K_{1})=P_{0}^{2}P_{1}^{2}-16 \pi ^{2} A_{0}A_{1}\leq U _{K_{0},K_{1}}, $$

(1.10)

where $U_{K_{0},K_{1}}$ 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 inequality of the form (1.10) reduces to a reverse Bonnesen-style inequality. For any convex domain $K_{k}$ ($k=0,1$) of areas $A_{k}$ and perimeters $P_{k}$ in $\mathbb{R}^{2}$, Zhou, Xu, Zeng, and others obtained the following reverse Bonnesen-style symmetric mixed inequalities (see [21, 25, 30]):

$$\begin{aligned}& P_{0}^{2}P_{1}^{2}-16 \pi ^{2} A_{0}A_{1}\leq 4\pi ^{2}P_{0}P_{1} \bigl(R _{01}R_{1}^{2}-r_{01}r_{1}^{2} \bigr), \end{aligned}$$

(1.11)

$$\begin{aligned}& P_{0}^{2}P_{1}^{2}-16\pi ^{2} A_{0}A_{1}\le 16\pi ^{4} \bigl(R_{0} ^{2} R_{1}^{2} -r_{0}^{2} r_{1}^{2} \bigr), \end{aligned}$$

(1.12)

where $r_{01}=\max \{t: t(gK_{1})\subseteq K_{0}; g\in G_{2}\}$ and $R_{01}=\min \{t: t(gK_{1})\supseteq K_{0}; g\in G_{2}\}$ are the inradius of $K_{0}$ with respect to $K_{1}$ and the outradius of $K_{0}$ with respect to $K_{1}$, respectively. $G_{2}$ is a group of plane rigid motions. $R_{k}$ and $r_{k}$ are the radius of the minimum circumscribed disc and the radius of the maximum inscribed disc of $K_{k}$, respectively. Each equality sign holds if and only if both $K_{0}$ and $K_{1}$ are discs.

The purpose of this paper is to find some new reverse Bonnesen-style inequalities for the oval domain in $\mathbb{R}^{2}$, 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 Theorem 3.2 as follows:

$$ P^{2}-4\pi A\le (2\pi \rho _{M}-P) (P-2\pi \rho _{m}), $$

which improves Bottema’s result. Furthermore, we obtain two reverse Bonnesen-style symmetric mixed inequalities (4.10) and (4.11) in Theorem 4.2 as follows:

$$\begin{aligned}& P_{0}^{2}P_{1}^{2}-16\pi ^{2} A_{0}A_{1}\leq 4\pi ^{2} A_{1}^{2} \bigl(\rho _{01}^{M}-\rho _{01}^{m} \bigr)^{2}, \\& P_{0}^{2}P_{1}^{2}-16\pi ^{2} A_{0}A_{1}\leq 16\pi ^{2} A_{1}^{2} \biggl(\rho _{01}^{M}- \frac{P_{0}P_{1}}{4\pi A_{1}} \biggr) \biggl(\frac{P _{0}P_{1}}{4\pi A_{1}}-\rho _{01}^{m} \biggr). \end{aligned}$$

When $K_{1}$ 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).

Preliminaries

A set of points K in $\mathbb{R}^{n}$ is convex if the line segment $\lambda x+(1-\lambda )y\in K$ for all $x, y \in K$ and $0 \le \lambda \le 1$. A domain is a set with nonempty interior, and an oval domain is a convex domain of boundary at least $C^{2}$. A convex body is a compact convex domain. The Minkowski sum of convex sets K and L, the scalar product of convex set K with $\lambda \ge 0$ are, respectively, defined by

$$ K+L=\{x+y: x\in K, y\in L\}, $$

and

$$ \lambda K=\{\lambda x: x\in K\}. $$

A homothety of the convex set K is of the form $x+\lambda K$ for $x\in \mathbb{R}^{n}$, $\lambda > 0$.

For the proof of the main theorem, we cite Blaschke’s rolling theorem in $\mathbb{R}^{2}$ from [3, 7, 13, 25].

Lemma 2.1

(Blaschke’s rolling theorem)

Let K be an oval domain in $\mathbb{R} ^{2}$, $\rho _{m}$ and $\rho _{M}$ be the minimum and maximum of the curvature radius of ∂K, respectively, $B_{t}$ be a circle of radius t in $\mathbb{R}^{2}$.

If $t\in (0,\rho _{m}]$ and $B_{t}$ is tangent to ∂K inside, then $B_{t}$ has no other common point with ∂K.

If $t\in [\rho _{M}, +\infty )$ and $B_{t}$ is tangent to ∂K outside, then $B_{t}$ has no other common point with ∂K.

By Lemma 2.1, we obtain the following corollary.

Corollary 2.1

Let K be an oval domain in $\mathbb{R}^{2}$, $\rho _{m}$ and $\rho _{M}$ be the minimum and maximum of the curvature radius of ∂K, respectively, $B_{t}$ be a circle of radius t in $\mathbb{R}^{2}$. When $t\in (0,\rho _{m}]$ or $t\in [\rho _{M}, + \infty )$, and $\partial K\cap \partial (B_{t})\ne \emptyset$, then $B_{t}$ has two common points with ∂K or $B_{t}$ is tangent to ∂K.

Proof

Suppose that $B_{t}$ has more than two common points with ∂K when $t\in (0,\rho _{m}]$ or $t\in [\rho _{M}, +\infty )$, then we can move $B_{t}$ 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 $K_{k}$ ($k = 0, 1$) be two oval domains in $\mathbb{R}^{2}$, $\rho _{m}(\partial K_{k})$ and $\rho _{M}(\partial K_{k})$ be the minimum and maximum of the curvature radius of $\partial K_{k}$, respectively. When $\rho _{M}(\partial K_{1})\leq \rho _{m}(\partial K_{0})$ or $\rho _{m}(\partial K_{1})\geq \rho _{M}(\partial K_{0})$, and $\partial K_{0}\cap \partial K_{1}\ne \emptyset$, then $\partial K _{0}$ has two common points with $\partial K_{1}$ or $\partial K_{0}$ is tangent to $\partial K_{1}$.

Proof

Suppose that $\partial K_{0}$ has more than two common points with $\partial K_{1}$, we can draw a circle $B_{t}$ of radius t through three points among these common points. Therefore, we have $t\in (\rho _{m}(\partial K_{0}), \rho _{M}(\partial K_{0}))$ and $t\in (\rho _{m}( \partial K_{1}), \rho _{M}(\partial K_{1}))$; this is inconsistent with the conditions of Corollary 2.2. □

Reverse Bonnesen-style inequalities

Let K be an oval domain of area A and perimeter P in $\mathbb{R}^{2}$. Let $\rho (\partial K)$ be the curvature radius of boundary ∂K and $\rho _{m}=\min \{\rho (\partial K)\}$, $\rho _{M}=\max \{\rho (\partial K)\}$. Let dg denote the kinematic density of the group $G_{2}$ of plane rigid motions, and $B_{t}$ be a circle of radius t in $\mathbb{R}^{2}$. Let $n\{\partial K\cap \partial (gB_{t})\}$ denote the number of points of intersection $\partial K\cap \partial (gB_{t})$ and $\chi \{K\cap gB_{t}\}$ be the Euler–Poincaré characteristics of the intersection $K\cap gB_{t}$. Then we have the following kinematic formula of Poincaré (see [18]):

$$ \int _{\{g\in G_{2}: \partial K\cap \partial (gB_{t})\ne \emptyset \}} n\bigl\{ \partial K\cap \partial (gB_{t}) \bigr\} \,dg=8\pi Pt $$

(3.1)

and the kinematic formula of Blaschke

$$ \int _{\{g\in G_{2} : K\cap gB_{t}\ne \emptyset \}} \chi \{K\cap gB _{t}\}\,dg=2\pi ^{2}t^{2}+2\pi Pt+2\pi A. $$

(3.2)

If μ denotes a set of all positions of $B_{t}$ in which either $gB_{t}\subset K$ or $gB_{t}\supset K$, then the kinematic formula of Blaschke (3.2) can be rewritten as

$$ \int _{\mu }dg=2\pi ^{2}t^{2}+2\pi Pt+2\pi A- \int _{\{g\in G_{2} : \partial K\cap \partial (gB_{t})\neq \emptyset \}} \chi \{K\cap gB_{t}\}\,dg. $$

(3.3)

Since K is an oval domain in $\mathbb{R}^{2}$, then

$$ \int _{\{g\in G_{2} : \partial K\cap \partial (gB_{t})\neq \emptyset \}} \chi \{K\cap gB_{t}\}\,dg= \int _{\{g\in G_{2} : \partial K\cap \partial (gB_{t})\neq \emptyset \}}\,dg. $$

(3.4)

When $t\in (0,\rho _{m}]$ or $t\in [\rho _{M}, +\infty )$, by Corollary 2.1, we have $n\{\partial K\cap \partial (gB_{t})\}=2$ or $gB_{t}$ is tangent to ∂K. When $gB_{t}$ is tangent to ∂K, we have

$$ \int _{\{g\in G_{2}: \partial K\cap \partial (gB_{t})\ne \emptyset \}} n\bigl\{ \partial K\cap \partial (gB_{t}) \bigr\} \,dg=0, $$

(3.5)

therefore,

$$ \int _{\{g\in G_{2}: \partial K\cap \partial (gB_{t})\ne \emptyset \}} n\bigl\{ \partial K\cap \partial (gB_{t}) \bigr\} \,dg= \int _{\{g\in G_{2}: \partial K\cap \partial (gB_{t})\ne \emptyset \}} 2dg. $$

(3.6)

By (3.4) and (3.6), we have

$$ \int _{\{g\in G_{2} : \partial K\cap \partial (gB_{t})\neq \emptyset \}} \chi \{K\cap gB_{t}\}\,dg= \frac{1}{2} \int _{\{g\in G_{2}: \partial K\cap \partial (gB_{t})\ne \emptyset \}} n\bigl\{ \partial K\cap \partial (gB_{t}) \bigr\} \,dg. $$

(3.7)

Therefore, when $t\in (0,\rho _{m}]$ or $t\in [\rho _{M}, +\infty )$, by (3.3), (3.7), and (3.1), we obtain

$$\begin{aligned} \int _{\mu }dg =& 2\pi ^{2}t^{2}+2\pi Pt+2 \pi A- \int _{\{g\in G_{2} : \partial K\cap \partial (gB_{t})\neq \emptyset \}} \chi \{K\cap gB_{t}\}\,dg \\ =& 2\pi ^{2}t^{2}+2\pi Pt+2\pi A-\frac{1}{2} \int _{\{g\in G_{2}: \partial K\cap \partial (gB_{t})\ne \emptyset \}} n\bigl\{ \partial K\cap \partial (gB_{t}) \bigr\} \,dg \\ =& 2\pi ^{2}t^{2}+2\pi Pt+2\pi A-4\pi Pt \\ =& 2\pi ^{2}t^{2}-2\pi Pt+2\pi A \\ \ge & 0. \end{aligned}$$

(3.8)

Theorem 3.1

Let K be an oval domain of area A and perimeter P in $\mathbb{R}^{2}$, then

$$ \pi t^{2}-Pt+A\geq 0; \quad t\in (0,\rho _{m}] \textit{ or } t\in [\rho _{M}, +\infty ). $$

(3.9)

The inequality is strict whenever $t\in (0,\rho _{m})$ or $t\in (\rho _{M}, +\infty )$. When $t=\rho _{m}$ or $t=\rho _{M}$, the equality holds if and only if K is a disc.

Proof

We obtain inequality (3.9) directly from (3.8)

$$ \int _{\mu }dg = 2\pi ^{2}t^{2}-2\pi Pt+2\pi A \ge 0; \quad t\in (0,\rho _{m}] \mbox{ or } t\in [\rho _{M}, +\infty ). $$

By Blaschke’s rolling theorem (Lemma 2.1), we know $B_{t}$ has no other common point with ∂K when $B_{t}$ is tangent to ∂K inside with $t\in (0,\rho _{m}]$, or $B_{t}$ is tangent to ∂K outside with $t\in [\rho _{M}, +\infty )$. Therefore, we have $gB_{t}\subset K$ when $t\in (0,\rho _{m})$, $gB_{t}\supset K$ when $t\in (\rho _{M}, +\infty )$, and $\partial (gB_{t})$ has no common point with ∂K, then $\int _{\mu }dg > 0$ when $t\in (0,\rho _{m})$ or $t\in (\rho _{M}, +\infty )$. That is, inequality (3.9) is strict whenever $t\in (0,\rho _{m})$ or $t\in (\rho _{M}, +\infty )$.

When $t=\rho _{m}$ or $t=\rho _{M}$, the equality holds clearly in inequality (3.9) if K is a disc. Conversely, if K is not a disc, by Blaschke’s rolling theorem (Lemma 2.1), we know that $B_{\rho _{m}}$ has no other common point with ∂K when $B_{\rho _{m}}$ is tangent to ∂K inside, and $B_{\rho _{M}}$ has no other common point with ∂K when $B_{\rho _{M}}$ is tangent to ∂K outside. Therefore, if K is not a disc, we have $gB_{\rho _{m}}\subset K$ and $\partial (gB_{\rho _{m}})$ has no common point with ∂K, $gB_{\rho _{M}}\supset K$ and $\partial (gB_{\rho _{M}})$ has no common point with ∂K, then $\int _{\mu }dg > 0$ when K is not a disc. That is, K is a disc when $\int _{\mu }dg = 0$. Therefore, when $t=\rho _{m}$ or $t=\rho _{M}$, the equality holds in (3.9) if and only if K is a disc. □

Theorem 3.2

Let K be an oval domain of area A and perimeter P in $\mathbb{R}^{2}$, then

$$ P^{2}-4\pi A\le (2\pi \rho _{M}-P) (P-2\pi \rho _{m}), $$

(3.10)

where $\rho _{m}$ and $\rho _{M}$ are the minimum and maximum of the continuous curvature radius ρ of the boundary ∂K, respectively. The equality holds if and only if K is a disc.

Proof

By inequality (3.9),

$$ \pi t^{2}-Pt+A\geq 0; \quad t\in (0,\rho _{m}] \mbox{ or } t\in [\rho _{M}, +\infty ), $$

we have

$$\begin{aligned}& \pi {\rho _{m}}^{2}-P\rho _{m}+A\geq 0, \\& \pi {\rho _{M}}^{2}-P\rho _{M}+A\geq 0, \end{aligned}$$

that is,

$$\begin{aligned}& -4\pi A\leq 4\pi ^{2}{\rho _{m}}^{2}-4\pi P\rho _{m}, \\& -4\pi A\leq 4\pi ^{2}{\rho _{M}}^{2}-4\pi P\rho _{M}. \end{aligned}$$

Therefore, we have

$$\begin{aligned} P^{2}-4\pi A \leq & P^{2}-4\pi P\rho _{m}+4\pi ^{2}{\rho _{m}}^{2} \\ =&(P-2\pi \rho _{m})^{2}, \end{aligned}$$

and

$$\begin{aligned} P^{2}-4\pi A \leq & P^{2}-4\pi P\rho _{M}+4\pi ^{2}{\rho _{M}}^{2} \\ =&(2\pi \rho _{M}-P)^{2}. \end{aligned}$$

Since $B(t)=\pi t^{2}-Pt+A$ reaches the minimum when $t = \frac{P}{2 \pi }$ and inequality (3.9), we have $\rho _{m}\leq \frac{P}{2\pi }\leq \rho _{M}$, that is, $2\pi \rho _{m} \leq P\leq 2\pi \rho _{M}$. Therefore,

$$\begin{aligned}& \sqrt{P^{2}-4\pi A} \leq P-2\pi \rho _{m}, \\& \sqrt{P^{2}-4\pi A} \leq 2\pi \rho _{M}-P. \end{aligned}$$

By multiplying the last inequalities side by side, we have

$$ P^{2}-4\pi A\le (2\pi \rho _{M}-P) (P-2\pi \rho _{m}). $$

The equality holds in (3.10) if and only if the two equalities hold in (3.9) when $t =\rho _{m}$ and $t = \rho _{M}$, that is, K is a disc. □

For all $a\geq 0$, $b\geq 0$, we have $4ab\leq (a + b)^{2}$, that is,

$$ (2\pi \rho _{M}-P) (P-2\pi \rho _{m})\leq \pi ^{2}(\rho _{M}-\rho _{m})^{2}. $$

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.

Reverse Bonnesen-style symmetric mixed inequalities

Let $K_{k}$ ($k = 0, 1$) be two oval domains in $\mathbb{R}^{2}$. Let $\rho (\partial K_{k})$ be the curvature radii of boundaries $\partial K_{k}$, and let $\rho _{m}(\partial K_{k})=\min \{\rho ( \partial K_{k})\}$, $\rho _{M}(\partial K_{k})=\max \{\rho (\partial K _{k})\}$. Let

$$ \rho _{m}^{g}(K_{0}, K_{1})=\max \bigl\{ t: \rho _{M}\bigl(\partial \bigl(t(gK_{1})\bigr)\bigr) \leq \rho _{m}(\partial K_{0}); g\in G_{2} \bigr\} $$

and

$$ \rho _{M}^{g}(K_{0}, K_{1})=\min \bigl\{ t: \rho _{m}\bigl(\partial \bigl(t(gK_{1})\bigr)\bigr) \geq \rho _{M}(\partial K_{0}); g\in G_{2} \bigr\} $$

be the inradius and the outradius of curvature, $K_{0}$ with respect to $K_{1}$, where $G_{2}$ is a group of plane rigid motions. It is obvious that $\rho _{m}^{g}(K_{0}, K_{1})\leq \rho _{M}^{g}(K_{0}, K_{1})$. Since both $\rho _{m}^{g}(K_{0}, K_{1})$ and $\rho _{M}^{g}(K_{0}, K_{1})$ are rigid invariant, we simply denote them by $\rho _{01}^{m}$ and $\rho _{01}^{M}$, respectively. Note that, if $K_{1}$ is a unit disc, then $\rho _{01}^{m}$ and $\rho _{01}^{M}$ are the minimum $\rho _{m}( \partial K_{0})$ and the maximum $\rho _{M}(\partial K_{0})$ of the continuous curvature radius of the boundary $\partial K_{0}$, respectively.

Let $K_{k}$ ($k=0,1$) be two oval domains of areas $A_{k}$ and perimeters $P_{k}$ in $\mathbb{R}^{2}$. Let dg denote the kinematic density of the group $G_{2}$ of plane rigid motions. Let $n\{\partial K_{0} \cap \partial (t(gK_{1}))\}$ denote the number of points of intersection $\partial K_{0}\cap \partial (t(gK_{1}))$, and let $\chi \{K_{0} \cap t(gK_{1})\}$ be the Euler–Poincaré characteristics of the intersection $K_{0}\cap t(gK_{1})$. Then we have the following kinematic formula of Poincaré (see [14, 20, 21, 24,25,26, 30]):

$$ \int _{\{g\in G_{2}: \partial K_{0}\cap \partial (t(gK_{1}))\ne \emptyset \}} n\bigl\{ \partial K_{0}\cap \partial \bigl(t(gK_{1})\bigr)\bigr\} \,dg=4tP_{0}P_{1} $$

(4.1)

and the kinematic formula of Blaschke

$$ \int _{\{g\in G_{2} : K_{0}\cap t(gK_{1})\ne \emptyset \}} \chi \bigl\{ K _{0}\cap t(gK_{1})\bigr\} \,dg=2\pi \bigl(t^{2}A_{1}+A_{0} \bigr)+tP_{0}P_{1}. $$

(4.2)

Let μ denote a set of all positions of $K_{1}$ in which either $t(gK_{1})\subset K_{0}$ or $t(gK_{1})\supset K_{0}$, then (4.2) can be rewritten as

$$ \int _{\mu }dg=2\pi \bigl(t^{2}A_{1}+A_{0} \bigr)+tP_{0}P_{1}- \int _{\{g\in G_{2} : \partial K_{0}\cap \partial (t(gK_{1}))\neq \emptyset \}} \chi \bigl\{ K_{0}\cap t(gK_{1})\bigr\} \,dg. $$

(4.3)

Since $K_{k}$ ($k=0,1$) are two oval domains in $\mathbb{R}^{2}$, then

$$ \int _{\{g\in G_{2} : \partial K_{0}\cap \partial (t(gK_{1}))\neq \emptyset \}} \chi \bigl\{ K_{0}\cap t(gK_{1})\bigr\} \,dg= \int _{\{g\in G_{2} : \partial K_{0}\cap \partial (t(gK_{1}))\neq \emptyset \}}\,dg. $$

(4.4)

When $t\in (0, \rho _{01}^{m}]$ or $t\in [\rho _{01}^{M}, +\infty )$, we can obtain $\rho _{M}(\partial (t(gK_{1})))\leq \rho _{m}(\partial K _{0})$ or $\rho _{m}(\partial (t(gK_{1})))\geq \rho _{M}(\partial K_{0})$. By Corollary 2.2, we have $n\{\partial K_{0}\cap \partial (t(gK_{1}))\}= 2$ or $\partial (t(gK_{1}))$ is tangent to $\partial K _{0}$. When $\partial (t(gK_{1}))$ is tangent to $\partial K_{0}$, we have

$$ \int _{\{g\in G_{2}: \partial K_{0}\cap \partial (t(gK_{1}))\ne \emptyset \}} n\bigl\{ \partial K_{0}\cap \partial \bigl(t(gK_{1})\bigr)\bigr\} \,dg=0, $$

(4.5)

therefore,

$$ \int _{\{g\in G_{2}: \partial K_{0}\cap \partial (t(gK_{1}))\ne \emptyset \}} n\bigl\{ \partial K_{0}\cap \partial \bigl(t(gK_{1})\bigr)\bigr\} \,dg= \int _{\{g\in G_{2}: \partial K_{0}\cap \partial (t(gK_{1}))\ne \emptyset \}} 2 dg. $$

(4.6)

By (4.4) and (4.6), we have

$$\begin{aligned}& \int _{\{g\in G_{2} : \partial K_{0}\cap \partial (t(gK_{1}))\neq \emptyset \}} \chi \bigl\{ K_{0}\cap t(gK_{1})\bigr\} \,dg \\& \quad =\frac{1}{2} \int _{\{g\in G_{2} : \partial K_{0}\cap \partial (t(gK_{1}))\neq \emptyset \}} n\bigl\{ \partial K_{0}\cap \partial \bigl(t(gK_{1})\bigr)\bigr\} \,dg. \end{aligned}$$

(4.7)

Therefore, when $t\in (0, \rho _{01}^{m}]$ or $t\in [\rho _{01}^{M}, + \infty )$, by (4.3), (4.7), and (4.1), we obtain

$$\begin{aligned} \int _{\mu }dg =& 2\pi \bigl(t^{2}A_{1}+A_{0} \bigr)+tP_{0}P_{1}- \int _{\{g\in G_{2} : \partial K_{0}\cap \partial (t(gK_{1}))\neq \emptyset \}} \chi \bigl\{ K_{0}\cap t(gK_{1})\bigr\} \,dg \\ =& 2\pi \bigl(t^{2}A_{1}+A_{0} \bigr)+tP_{0}P_{1}-\frac{1}{2} \int _{\{g\in G_{2} : \partial K_{0}\cap \partial (t(gK_{1}))\neq \emptyset \}} n\bigl\{ \partial K_{0}\cap \partial \bigl(t(gK_{1})\bigr)\bigr\} \,dg \\ =& 2\pi A_{1}t^{2}-P_{0}P_{1}t+2\pi A_{0} \\ \ge & 0. \end{aligned}$$

(4.8)

Theorem 4.1

Let $K_{k}$ ($k = 0, 1$) be two oval domains of areas $A_{k}$ and perimeters $P_{k}$ in $\mathbb{R}^{2}$, then

$$ 2\pi A_{1}t^{2}-P_{0}P_{1}t+2 \pi A_{0}\geq 0; \quad t\in \bigl(0, \rho _{01}^{m}\big] \textit{ or } t\in \big[\rho _{01}^{M}, +\infty \bigr). $$

(4.9)

The inequality is strict whenever $t\in (0, \rho _{01}^{m})$ or $t\in (\rho _{01}^{M}, +\infty )$. When $t=\rho _{01}^{m}$ or $t=\rho _{01}^{M}$, the equality holds if and only if both $K_{0}$ and $K_{1}$ are discs.

Proof

We obtain inequality (4.9) directly from (4.8)

$$ \int _{\mu }dg = 2\pi A_{1}t^{2}-P_{0}P_{1}t+2 \pi A_{0} \ge 0; \quad t\in \bigl(0, \rho _{01}^{m}\big] \mbox{ or } t\in \big[\rho _{01}^{M}, +\infty \bigr). $$

When $t\in (0, \rho _{01}^{m})$, that is, $\rho _{M}(\partial (t(gK_{1})))< \rho _{m}(\partial K_{0})$, we have

$$ t(gK_{1})\subset B_{\rho _{M}(\partial (t(gK_{1})))}\subset B_{\rho _{m}(\partial K_{0})}\subset K_{0}; \quad t\in \bigl(0, \rho _{01}^{m}\bigr), $$

where $\partial (B_{\rho _{M}(\partial (t(gK_{1})))})$ has no common point with $\partial (B_{\rho _{m}(\partial K_{0})})$. Therefore, we have $t(gK_{1})\subset K_{0}$, and $\partial (t(gK_{1}))$ has no common point with $\partial (K_{0})$ when $t\in (0, \rho _{01}^{m})$. When $t\in (\rho _{01}^{M}, +\infty )$, that is, $\rho _{m}(\partial (t(gK _{1})))> \rho _{M}(\partial K_{0})$, we have

$$ t(gK_{1})\supset B_{\rho _{m}(\partial (t(gK_{1})))}\supset B_{\rho _{M}(\partial K_{0})}\supset K_{0}; \quad t\in \bigl(\rho _{01}^{M}, +\infty \bigr), $$

where $\partial (B_{\rho _{m}(\partial (t(gK_{1})))})$ has no common point with $\partial (B_{\rho _{M}(\partial K_{0})})$. Therefore, we have $t(gK_{1})\supset K_{0}$, and $\partial (t(gK_{1}))$ has no common point with $\partial (K_{0})$ when $t\in (\rho _{01}^{M}, +\infty )$. In summary, we have $\int _{\mu }dg > 0$ when $t\in (0, \rho _{01}^{m})$ or $t\in (\rho _{01}^{M}, +\infty )$. That is, inequality (4.9) is strict whenever $t\in (0, \rho _{01}^{m})$ or $t\in (\rho _{01}^{M}, + \infty )$.

When $t=\rho _{01}^{m}$ or $t=\rho _{01}^{M}$, the equality holds clearly in inequality (4.9) if both $K_{0}$ and $K_{1}$ are discs. Conversely, if $K_{0}$ and $K_{1}$ of which at least one is not a disc, it includes the following two types: Only one of them is not a disc; $K_{0}$ and $K_{1}$ are not discs. When only one of $K_{0}$ and $K_{1}$ is not a disc, we have $\rho _{01}^{m}(gK_{1}) \subset K_{0}$ and $\partial (\rho _{01}^{m}(gK_{1}))$ has no common point with $\partial K_{0}$, $\rho _{01}^{M}(gK_{1})\supset K_{0}$ and $\partial (\rho _{01}^{M}(gK_{1}))$ has no common point with $\partial K_{0}$, then $\int _{\mu }dg > 0$ when only one of $K_{0}$ and $K_{1}$ is not a disc. When $K_{0}$ and $K_{1}$ are not discs, we have

$$ \rho _{01}^{m}(gK_{1})\subset B_{\rho _{M}(\partial (\rho _{01}^{m}(gK _{1}))} \subset B_{\rho _{m}(\partial K_{0})}\subset K_{0}, $$

where $\partial (\rho _{01}^{m}(gK_{1}))$ has no common point with $\partial K_{0}$, and

$$ \rho _{01}^{M}(gK_{1})\supset B_{\rho _{m}(\partial (\rho _{01}^{M}(gK _{1})))} \supset B_{\rho _{M}(\partial K_{0})}\supset K_{0}, $$

where $\partial (\rho _{01}^{M}(gK_{1}))$ has no common point with $\partial K_{0}$, then $\int _{\mu }dg > 0$ when $K_{0}$ and $K_{1}$ are not discs. In summary, $\int _{\mu }dg > 0$ when $K_{0}$ and $K_{1}$ of which at least one is not a disc. That is, both $K_{0}$ and $K_{1}$ are discs when $\int _{\mu }dg = 0$. Therefore, when $t=\rho _{01}^{m}$ or $t=\rho _{01}^{M}$, the equality holds in inequality (4.9) if and only if both $K_{0}$ and $K_{1}$ are discs. □

When $K_{1}$ is 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 $K_{k}$ ($k = 0, 1$) be two oval domains of areas $A_{k}$ and perimeters $P_{k}$ in $\mathbb{R}^{2}$, then

$$\begin{aligned}& P_{0}^{2}P_{1}^{2}-16 \pi ^{2} A_{0}A_{1}\leq 4\pi ^{2} A_{1}^{2} \bigl(\rho _{01}^{M}-\rho _{01}^{m} \bigr)^{2}, \end{aligned}$$

(4.10)

$$\begin{aligned}& P_{0}^{2}P_{1}^{2}-16 \pi ^{2} A_{0}A_{1}\leq 16\pi ^{2} A_{1}^{2} \biggl(\rho _{01}^{M}- \frac{P_{0}P_{1}}{4\pi A_{1}} \biggr) \biggl(\frac{P_{0}P _{1}}{4\pi A_{1}}-\rho _{01}^{m} \biggr), \end{aligned}$$

(4.11)

where each equality holds if and only if both $K_{0}$ and $K_{1}$ are discs.

Proof

By inequality (4.9),

$$ 2\pi A_{1} t^{2}-P_{0}P_{1} t +2\pi A_{0}\geq 0; \quad t\in \bigl(0, \rho _{01}^{m}\big] \mbox{ or } t\in \big[\rho _{01}^{M}, +\infty \bigr), $$

we have

$$\begin{aligned}& 2\pi A_{1} {\bigl(\rho _{01}^{m} \bigr)}^{2}-P_{0}P_{1} \rho _{01}^{m}+2 \pi A _{0}\geq 0, \\& 2\pi A_{1} {\bigl(\rho _{01}^{M} \bigr)}^{2}-P_{0}P_{1} \rho _{01}^{M}+2 \pi A _{0}\geq 0, \end{aligned}$$

that is,

$$\begin{aligned}& -16\pi ^{2}A_{0}A_{1}\leq 16\pi ^{2}A_{1}^{2}{\bigl(\rho _{01}^{m} \bigr)}^{2}-8 \pi A_{1}P_{0}P_{1}\rho _{01}^{m}, \\& -16\pi ^{2}A_{0}A_{1}\leq 16\pi ^{2}A_{1}^{2}{\bigl(\rho _{01}^{M} \bigr)}^{2}-8 \pi A_{1}P_{0}P_{1}\rho _{01}^{M}. \end{aligned}$$

Therefore, we have

$$\begin{aligned} P_{0}^{2}P_{1}^{2}-16\pi ^{2} A_{0}A_{1} \leq & P_{0}^{2}P_{1}^{2}+16 \pi ^{2}A_{1}^{2}{\bigl(\rho _{01}^{m} \bigr)}^{2}-8\pi A_{1}P_{0}P_{1}\rho _{01} ^{m} \\ =&\bigl(P_{0}P_{1}-4\pi A_{1}\rho _{01}^{m}\bigr)^{2} \end{aligned}$$

and

$$\begin{aligned} P_{0}^{2}P_{1}^{2}-16\pi ^{2} A_{0}A_{1} \leq & P_{0}^{2}P_{1}^{2}+16 \pi ^{2}A_{1}^{2}{\bigl(\rho _{01}^{M} \bigr)}^{2}-8\pi A_{1}P_{0}P_{1}\rho _{01} ^{M} \\ =&\bigl(P_{0}P_{1}-4\pi A_{1}\rho _{01}^{M}\bigr)^{2}. \end{aligned}$$

Since $B_{K_{0},K_{1}}(t)=2\pi A_{1}t^{2}-P_{0}P_{1}t+2\pi A_{0}$ reaches the minimum at $t = \frac{P_{0}P_{1}}{4\pi A_{1}}$, and inequality (4.9), we have $\rho _{01}^{m} \leq \frac{P_{0}P_{1}}{4\pi A_{1}}\leq \rho _{01}^{M}$, that is, $4\pi A_{1}\rho _{01}^{m}\leq P_{0}P_{1}\leq 4\pi A_{1}\rho _{01}^{M}$. Therefore,

$$\begin{aligned}& \sqrt{P_{0}^{2}P_{1}^{2}-16 \pi ^{2} A_{0}A_{1}} \leq P_{0}P_{1}-4 \pi A_{1}\rho _{01}^{m}, \\& \sqrt{P_{0}^{2}P_{1}^{2}-16 \pi ^{2} A_{0}A_{1}} \leq 4\pi A_{1} \rho _{01}^{M}-P_{0}P_{1}. \end{aligned}$$

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

$$ P_{0}^{2}P_{1}^{2}-16\pi ^{2} A_{0}A_{1}\leq 4\pi ^{2} A_{1}^{2} \bigl(\rho _{01}^{M}-\rho _{01}^{m} \bigr)^{2} $$

and

$$ P_{0}^{2}P_{1}^{2}-16\pi ^{2} A_{0}A_{1}\leq 16\pi ^{2} A_{1}^{2} \biggl(\rho _{01}^{M}- \frac{P_{0}P_{1}}{4\pi A_{1}} \biggr) \biggl(\frac{P_{0}P _{1}}{4\pi A_{1}}-\rho _{01}^{m} \biggr). $$

Each equality holds in (4.10) and (4.11) if and only if the equalities hold in (4.9) when $t =\rho _{01}^{m}$ and $t = \rho _{01}^{M}$, that is, both $K_{0}$ and $K_{1}$ are discs. □

When $K_{1}$ is 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 all $a\geq 0$, $b\geq 0$, we have $4ab\leq (a + b)^{2}$, that is,

$$ 16\pi ^{2} A_{1}^{2} \biggl(\rho _{01}^{M}- \frac{P_{0}P_{1}}{4\pi A_{1}} \biggr) \biggl( \frac{P_{0}P_{1}}{4\pi A _{1}}-\rho _{01}^{m} \biggr)\leq 4\pi ^{2} A_{1}^{2} \bigl(\rho _{01} ^{M}-\rho _{01}^{m} \bigr)^{2}. $$

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).