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).