1 Introduction and main results
The setting for this paper is the Euclidean n-space \(\mathbb{R}^{n}\). We use \(S^{n-1}\) and \(V(K)\) to denote the unit sphere and the n-dimensional volume of a body K, respectively. For the standard unit ball B, we write \(V(B)=\omega _{n}\).
If K is a nonempty compact convex set in \(\mathbb{R}^{n}\), then the support function of K, \(h_{K}=h(K,\cdot ): \mathbb{R}^{n}\rightarrow \mathbb{R}\), is defined by (see [1])
for \(x\in \mathbb{R}^{n}\), where \(x\cdot y\) is the standard inner product of x and y. If K is a compact convex set with nonempty interiors in \(\mathbb{R}^{n}\), then K is called a convex body. Let \(\mathcal{K}^{n}\) denote the set of convex bodies in \(\mathbb{R}^{n}\).
$$ h(K,x)= \max \{x\cdot y: y\in K\} $$
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The radial function \(\rho _{K}=\rho (K,\cdot ): \mathbb{R}^{n}\setminus \{0\}\rightarrow [0,\infty )\) of a compact star-shaped (about the origin) set \(K\subset \mathbb{R}^{n}\) is defined by (see [1])
If \(\rho _{K}\) is continuous, then K will be called a star body (about the origin). Let \(\mathcal{S}^{n}_{o}\) denote the subset of star bodies containing the origin in \(\mathbb{R}^{n}\). Two star bodies K and L are dilates (of one another) if \(\rho _{K}(u)/\rho _{L}(u)\) is independent of \(u\in S^{n-1}\).
$$ \rho (K,x)=\max \{\lambda \geq 0: \lambda x\in K\},\quad x\in \mathbb{R}^{n}\setminus \{0\}. $$
The research of width-integrals has a long history. The width-integrals were first considered by Blaschke (see [2]) and were further researched by Hadwiger (see [3]). In 1975, Lutwak [4] gave the ith width-integrals as follows:
For \(K\in \mathcal{K}^{n}\) and any real i, the ith width-integrals \(B_{i}(K)\) of K are defined by
Here \(b(K,u)\) denotes the half width of K in the direction \(u\in S^{n-1}\) which is defined by \(b(K,u)=\frac{1}{2}h(K,u)+ \frac{1}{2}h(K,-u)\). If there exists a constant \(\lambda >0\) such that \(b(K,u)=\lambda b(L,u)\) for all \(u\in S^{n-1}\), then K and L are said to have similar width. Further, Lutwak [4] established the following Brunn–Minkowski inequality and cyclic inequality for the ith width-integrals, respectively.
$$ B_{i}(K)=\frac{1}{n} \int _{S^{n-1}}b(K,u)^{n-i}\,dS(u). $$
(1.1)
Theorem 1.A
If
\(K,L\in \mathcal{K}^{n}\)
and real
\(i< n-1\), then
with equality if and only if
K
and
L
have similar width. Here
\(K+L\)
denotes the Minkowski sum of
K
and
L.
$$ B_{i}(K+L)^{\frac{1}{n-i}}\leq B_{i}(K)^{\frac{1}{n-i}}+B_{i}(L)^{ \frac{1}{n-i}} $$
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Theorem 1.B
If
\(K\in \mathcal{K}^{n}\)
and reals
i, j, k
satisfy
\(i< j< k\), then
with equality if and only if
K
is of constant width.
$$ B_{j}(K)^{k-i}\leq B_{i}(K)^{k-j}B_{k}(K)^{j-i} $$
Whereafter, Lutwak [5] showed that the mixed width-integral \(B(K_{1},\ldots,K_{n})\) of \(K_{1},\ldots, K_{n}\in \mathcal{K}^{n}\) was defined by
$$ B(K_{1},\ldots,K_{n})=\frac{1}{n} \int _{S^{n-1}}b(K_{1},u)\cdots b(K_{n},u) \,dS(u). $$
(1.2)
In 2016, based on (1.2), Feng [6] introduced the general mixed width-integrals as follows: For \(K_{1},\ldots,K_{n}\in \mathcal{K} ^{n}\) and \(\tau \in (-1,1)\), the general mixed width-integral \(B^{(\tau )}(K_{1},\ldots,K_{n})\) of \(K_{1}, \ldots, K_{n}\) is given by
where \(b^{\tau }(K,u)=f_{1}(\tau )h(K,u)+f_{2}(\tau )h(K,-u)\) and
Obviously,
Combined with (1.4), the case of \(\tau =0\) in (1.3) is just (1.2). If there exists a constant \(\lambda >0\) such that \(b^{\tau }(K,u)=\lambda b^{\tau }(L,u)\) for all \(u\in S^{n-1}\), then we say convex bodies K and L have similar general width. K and L have joint constant general width means that \(b^{(\tau )}(K,u)b^{(\tau )}(L,u)\) is a constant for all \(u\in S^{n-1}\).
$$ B^{(\tau )}(K_{1},\ldots,K_{n})= \frac{1}{n} \int _{S^{n-1}}b^{(\tau )}(K _{1},u)\cdots b^{(\tau )}(K_{n},u)\,dS(u), $$
(1.3)
$$ f_{1}(\tau )=\frac{(1+\tau )^{2}}{2(1+\tau ^{2})},\qquad f_{2}(\tau )=\frac{(1-\tau )^{2}}{2(1+\tau ^{2})}. $$
(1.4)
$$\begin{aligned}& f_{1}(\tau )+f_{2}(\tau )=1; \\& f_{1}(-\tau )=f_{2}(\tau ),\qquad f_{2}(- \tau )=f_{1}(\tau ). \end{aligned}$$
Taking \(K_{1}=\cdots=K_{n-i}=K\), \(K_{n-i+1}=\cdots=K_{n}=B\) in (1.3) and allowing i to be any real, the general ith width-integrals \(B_{i}^{(\tau )}(K)\) of \(K\in \mathcal{K}^{n}\) were given by (see [6])
From (1.1), (1.4), and (1.5), we easily see that if \(\tau =0\), then \(B_{i}^{(0)}(K)=B_{i}(K)\).
$$ B_{i}^{(\tau )}(K)=\frac{1}{n} \int _{S^{n-1}}b^{(\tau )}(K,u)^{n-i}\,dS(u). $$
(1.5)
In 2006, motivated by Lutwak’s ith width-integrals and together with the notion of radial function, Li, Yuan, and Leng [7] gave the ith chord-integrals as follows: For \(K\in \mathcal{S}_{o}^{n}\) and i is any real, the ith chord-integrals \(C_{i}(K)\) of K are defined by
Here \(c(K,u)\) denotes the half chord of K in the direction u and \(c(K,u)=\frac{1}{2}\rho (K,u)+\frac{1}{2}\rho (K,-u)\). If there exists a constant \(\lambda >0\) such that \(c(K,u)=\lambda c(L,u)\) for all \(u\in \mathcal{S}^{n-1}\), then we say that K and L have similar chord.
$$ C_{i}(K)=\frac{1}{n} \int _{S^{n-1}}c(K,u)^{n-i}\,dS(u). $$
(1.6)
For the ith chord-integrals, the authors [7] proved the following Brunn–Minkowski inequality and cyclic inequality.
Theorem 1.C
For
\(K,L\in \mathcal{S}_{o}^{n}\)
and real
\(i\neq n\). If
\(i< n-1\), then
if
\(i>n-1\), then
In each inequality, equality holds if and only if
K
and
L
are dilates. Here
\(K\mathbin{\tilde{+}}L\)
denotes the radial sum of
K
and
L.
$$ C_{i}(K\mathbin{\tilde{+}}L)^{\frac{1}{n-i}}\leq C_{i}(K)^{\frac{1}{n-i}}+C_{i}(L)^{ \frac{1}{n-i}}; $$
$$ C_{i}(K\mathbin{\tilde{+}}L)^{\frac{1}{n-i}}\geq C_{i}(K)^{\frac{1}{n-i}}+C_{i}(L)^{ \frac{1}{n-i}}. $$
Theorem 1.D
If
\(K\in \mathcal{S}_{o}^{n}\)
and reals
i, j, k
satisfy
\(i< j< k\), then
with equality if and only if
K
is of constant chord.
$$ C_{j}(K)^{k-i}\leq C_{i}(K)^{k-j}C_{k}(K)^{j-i}, $$
The mixed chord-integrals of star bodies were defined by Lu (see [8]): For \(K_{1},\ldots,K_{n}\in \mathcal{S}_{o}^{n}\), the mixed chord-integrals \(C(K_{1},\ldots,K_{n})\) of \(K_{1}, \ldots, K_{n}\) are defined by
$$ C(K_{1},\ldots,K_{n})=\frac{1}{n} \int _{S^{n-1}}c(K_{1},u)\cdots c(K_{n},u) \,dS(u). $$
Recently, Feng and Wang [9] gave the general mixed chord-integrals \(C^{(\tau )}(K_{1},\ldots,K_{n})\) of \(K_{1},\ldots,K_{n} \in \mathcal{S}_{o}^{n}\) defined by
where \(c^{(\tau )}(K,u)=f_{1}(\tau )\rho (K,u)+f_{2}(\tau )\rho (K,-u)\) and functions \(f_{1}(\tau )\), \(f_{2}(\tau )\) satisfy (1.4). By (1.4), let \(\tau =0\) in (1.7), this is just Lu’s mixed chord-integrals \(C(K_{1},\ldots,K_{n})\). Star bodies K and L are said to have similar general chord mean that there exist constants \(\lambda , \mu >0\) such that \(\lambda c^{(\tau )}(K,u)=\mu c^{(\tau )}(L,u)\) for all \(u\in S^{n-1}\). If the product \(c^{(\tau )}(K,u)c^{(\tau )}(L,u)\) is constant for all \(u\in S^{n-1}\), then they are said to have joint constant general chord.
$$ C^{(\tau )}(K_{1},\ldots,K_{n})= \frac{1}{n} \int _{S^{n-1}}c^{(\tau )}(K _{1},u)\cdots c^{(\tau )}(K_{n},u)\,dS(u), $$
(1.7)
Taking \(K_{1}=\cdots=K_{n-i}=K\) and \(K_{n-i+1}=\cdots=K_{n}=B\) in (1.7) and allowing i to be any real, the general ith chord-integral \(C_{i}^{(\tau )}(K)\) of \(K\in \mathcal{S}_{o}^{n}\) was given by (see [9])
Obviously, (1.4), (1.6), and (1.8) give \(C_{i}^{(0)}(K)=C_{i}(K)\).
$$ C_{i}^{(\tau )}(K)=\frac{1}{n} \int _{S^{n-1}}c^{(\tau )}(K,u)^{n-i}\,dS(u). $$
(1.8)
In this paper, based on Theorems 1.A–1.B and Theorems 1.C–1.D, we respectively establish two cyclic Brunn–Minkowski inequalities for general ith width-integrals and general ith chord-integrals by using Zhao’s ideas (see [10] and [11]). Our works bring the cyclic inequality and Brunn–Minkowski inequality together. Our main results can be stated as follows.
Theorem 1.1
Let
\(K,L\in \mathcal{K}^{n}\), \(\tau \in (-1,1)\), and
i, j, k
all be reals. If
\(j< n-1\)
and
\(i\leq j< k\), then
with equality if and only if
K
and
L
have similar general width. If
\(n-1< j< n\)
and
\(j\leq i< k\)
or
\(j>n\)
and
\(i\leq j< k\), then inequality (1.9) is reversed.
$$ B_{j}^{(\tau )}(K+L)^{\frac{k-i}{n-j}} \leq B_{i}^{(\tau )}(K)^{ \frac{k-j}{n-j}}B_{k}^{(\tau )}(K)^{\frac{j-i}{n-j}}+B_{i}^{(\tau )}(L)^{ \frac{k-j}{n-j}}B_{k}^{(\tau )}(L) ^{\frac{j-i}{n-j}} $$
(1.9)
Theorem 1.2
Let
\(K,L\in \mathcal{S}_{o}^{n}\), \(\tau \in (-1,1)\), and
i, j, k
all be reals. If
\(j< n-1\)
and
\(i\leq j< k\), then
with equality if and only if
K
and
L
have similar general chord. If
\(n-1< j< n\)
and
\(j\leq i< k\)
or
\(j>n\)
and
\(i\leq j< k\), then inequality (1.10) is reversed.
$$ C_{j}^{(\tau )}(K\mathbin{\tilde{+}}L)^{\frac{k-i}{n-j}}\leq C_{i}^{(\tau )}(K)^{ \frac{k-j}{n-j}}C_{k}^{(\tau )}(K)^{\frac{j-i}{n-j}}+C_{i}^{(\tau )}(L)^{ \frac{k-j}{n-j}}C_{k}^{(\tau )}(L)^{\frac{j-i}{n-j}} $$
(1.10)
Remark 1.1
Corollary 1.1
Let
\(K,L\in \mathcal{K}^{n}\), real
\(i\neq n\), and
\(\tau \in (-1,1)\). If
\(i< n-1\), then
with equality if and only if
K
and
L
have similar general width. If
\(i>n-1\), then inequality (1.11) is reversed.
$$ B_{i}^{(\tau )}(K+L)^{\frac{1}{n-i}}\leq B_{i}^{(\tau )}(K)^{ \frac{1}{n-i}}+B_{i}^{(\tau )}(L)^{\frac{1}{n-i}} $$
(1.11)
Corollary 1.2
Let
\(K,L\in \mathcal{S}_{o}^{n}\), real
\(i\neq n\), and
\(\tau \in (-1,1)\). If
\(i< n-1\), then
with equality holds if and only if
K
and
L
have similar general chord. If
\(i>n-1\), then the above inequality is reversed.
$$ C_{i}^{(\tau )}(K\mathbin{\tilde{+}}L)^{\frac{1}{n-i}}\leq C_{i}^{(\tau )}(K)^{ \frac{1}{n-i}}+C_{i}^{(\tau )}(L)^{\frac{1}{n-i}} $$
Obviously, if \(\tau =0\), then inequality (1.11) yields Theorem 1.A, Corollary 1.2 gives Theorem 1.C, respectively.
Remark 1.2
Corollary 1.3
If
\(K\in \mathcal{K}^{n}\), \(\tau \in (-1,1)\), and
\(i< j< k\), then
with equality if and only if
K
is of constant width.
$$ B^{(\tau )}_{j}(K)^{k-i}\leq B^{(\tau )}_{i}(K)^{k-j}B^{(\tau )}_{k}(K)^{j-i} $$
Because of \(\{o\}\in \mathcal{S}_{o}^{n}\), hence let \(L=\{o\}\) in Theorem 1.2, we may obtain the following.
Corollary 1.4
If
\(K\in \mathcal{S}_{o}^{n}\), \(\tau \in (-1,1)\), and
\(i< j< k\), then
with equality if and only if
K
is of constant chord.
$$ C^{(\tau )}_{j}(K)^{k-i}\leq C^{(\tau )}_{i}(K)^{k-j}C^{(\tau )}_{k}(K)^{j-i} $$
Our works belong to the asymmetric Brunn–Minkowski theory, which has its starting point in the theory of valuations in connection with isoperimetric and analytic inequalities. As an important research object in convex geometry, asymmetric Brunn–Minkowski theory has gotten rich development, readers can refer to [12‐19].
2 Preliminaries
For nonempty compact convex bodies K and L, \(\lambda , \mu \geq 0\) (not both zero), the Minkowski combination \(\lambda K+ \mu L\) of K and L is defined by (see [1])
where “+” and “λK” are called Minkowski addition and Minkowski scalar multiplication, respectively. When \(\lambda =\mu =1\), the \(K+L\) is called Minkowski sum.
$$ h(\lambda K+\mu L,\cdot )=\lambda h(K,\cdot )+\mu h(L,\cdot ), $$
(2.1)
For \(K,L\in \mathcal{S}_{o}^{n}\), \(\lambda ,\mu \geq 0\) (not both zero), the radial combination \(\lambda \circ K\mathbin{\tilde{+}}\mu \circ L\) of K, L is given by (see [1, 20])
where “\(\mathbin{\tilde{+}}\)” and “\(\lambda \circ K\)” are radial addition and radial scalar multiplication, respectively. When \(\lambda =\mu =1\), the \(K\mathbin{\tilde{+}}L\) is radial sum of K and L.
$$ \rho (\lambda \circ K\mathbin{\tilde{+}}\mu \circ L,\cdot )=\lambda \rho (K, \cdot )+\mu \rho (L,\cdot ), $$
(2.2)
3 Proofs of theorems
In this part, we give the proofs of Theorem 1.1 and Theorem 1.2. First, we give the following lemmas.
Lemma 3.1
Let
\(f\in L^{p}(E)\), \(g\in L^{q}(E)\), real number
\(p,q\neq 0\), and
\(\frac{1}{p}+\frac{1}{q}=1\), if
\(p>1\), then
with equality if and only if there exists constants
\(c_{1}\)
and
\(c_{2}\)
such that
\(c_{1}|f(x)|^{p}=c_{2}|g(x)|^{q}\). The inequality is reversed if
\(p<0\)
or
\(0< p<1\). Here
\(L^{p}(E)\)
denotes all function sets defined on a measurable set
E
in
\(L_{p}\)
spaces.
$$ \biggl( \int _{E} \bigl\vert f(x) \bigr\vert ^{p} \,dx \biggr)^{\frac{1}{p}} \biggl( \int _{E} \bigl\vert g(x) \bigr\vert ^{q} \,dx \biggr)^{\frac{1}{q}}\geq \int _{E} \bigl\vert f(x)g(x) \bigr\vert \,dx $$
(3.1)
Lemma 3.2
Let
\(f,g\in L^{p}(E)\), if real number
\(p\neq 0\)
and
\(p>1\), then
with equality if and only if there exists constants
\(c_{1}\)
and
\(c_{2}\)
such that
\(c_{1}|f(x)|^{p}=c_{2}|g(x)|^{q}\). The inequality is reversed if
\(p<0\)
or
\(0< p<1\).
$$ \biggl( \int _{E} \bigl\vert f(x) \bigr\vert ^{p} \,dx \biggr)^{\frac{1}{p}}+ \biggl( \int _{E} \bigl\vert g(x) \bigr\vert ^{p} \,dx \biggr)^{\frac{1}{p}}\geq \biggl( \int _{E} \bigl\vert f(x)+g(x) \bigr\vert ^{p} \biggr)^{ \frac{1}{p}}\,dx $$
(3.2)
Proof of Theorem 1.1
If \(j< n-1\), i.e., \(n-j>1\), then from (1.5), (2.1), and (3.2), we have
In the inequality on the right above, notice that \(i< j< k\) means \(\frac{k-i}{k-j}>1\), thus by (3.1) we obtain
From this, for \(j< n-1\), we can get the following inequality by (3.4):
$$\begin{aligned} B_{j}^{(\tau )}(K+L)^{\frac{1}{n-j}} =& \biggl[ \frac{1}{n} \int _{S^{n-1}}b ^{(\tau )}(K+L,u)^{n-j}\,dS(u) \biggr]^{\frac{1}{n-j}} \\ =& \biggl[\frac{1}{n} \int _{S^{n-1}} \bigl(f_{1}(\tau )h(K+L,u)+f_{2}( \tau )h(K+L,-u) \bigr)^{n-j}\,dS(u) \biggr]^{\frac{1}{n-j}} \\ =& \biggl[\frac{1}{n} \int _{S^{n-1}} \bigl(f_{1}(\tau )h(K,u)+f_{1}( \tau )h(L,u) \\ &{}+f_{2}(\tau )h(K,-u)+f_{2}(\tau )h(L,-u) \bigr)^{n-j}\,dS(u) \biggr]^{{ \frac{1}{n-j}}} \\ =& \biggl[\frac{1}{n} \int _{S^{n-1}} \bigl(b^{(\tau )}(K,u)+b^{(\tau )}(L,u) \bigr)^{n-j}\,dS(u) \biggr]^{\frac{1}{n-j}} \\ =& \biggl[\frac{1}{n} \int _{S^{n-1}} \bigl(b^{(\tau )}(K,u)^{ \frac{(k-j)(n-i)}{(k-i)(n-j)}}b^{(\tau )}(K,u)^{ \frac{(j-i)(n-k)}{(k-i)(n-j)}} \\ &{}+b^{(\tau )}(L,u)^{\frac{(k-j)(n-i)}{(k-i)(n-j)}}b^{(\tau )}(L,u)^{ \frac{(j-i)(n-k)}{(k-i)(n-j)}} \bigr)^{n-j}\,dS(u) \biggr]^{\frac{1}{n-j}} \\ \leq& \biggl[\frac{1}{n} \int _{S^{n-1}}b^{(\tau )}(K,u)^{ \frac{(k-j)(n-i)}{k-i}}b^{(\tau )}(K,u)^{\frac{(j-i)(n-k)}{k-i}} \,dS(u) \biggr]^{\frac{1}{n-j}} \\ &{}+ \biggl[\frac{1}{n} \int _{S^{n-1}}b^{(\tau )}(L,u)^{ \frac{(k-j)(n-i)}{k-i}}b^{(\tau )}(L,u)^{\frac{(j-i)(n-k)}{k-i}} \,dS(u) \biggr]^{\frac{1}{n-j}}. \end{aligned}$$
(3.3)
$$\begin{aligned}& \frac{1}{n} \int _{S^{n-1}}b^{(\tau )}(K,u)^{\frac{(k-j)(n-i)}{k-i}}b ^{(\tau )}(K,u)^{\frac{(j-i)(n-k)}{k-i}}\,dS(u) \\ & \quad \leq \biggl[\frac{1}{n} \int _{S^{n-1}} \bigl(b^{(\tau )}(K,u)^{ \frac{(k-j)(n-i)}{k-i}} \bigr)^{\frac{k-i}{k-j}}\,dS(u) \biggr]^{ \frac{k-j}{k-i}} \\ & \qquad {}\times \biggl[\frac{1}{n} \int _{S^{n-1}} \bigl(b^{(\tau )}(K,u)^{ \frac{(j-i)(n-k)}{k-i}} \bigr)^{\frac{k-i}{j-i}}\,dS(u) \biggr]^{ \frac{j-i}{k-i}} \\ & \quad =B_{i}^{(\tau )}(K)^{\frac{k-j}{k-i}}B_{k}^{(\tau )}(K)^{ \frac{j-i}{k-i}}. \end{aligned}$$
(3.4)
$$\begin{aligned}& \biggl[\frac{1}{n} \int _{S^{n-1}}b^{(\tau )}(K,u)^{ \frac{(k-j)(n-i)}{k-i}}b^{(\tau )}(K,u)^{\frac{(j-i)(n-k)}{k-i}} \,dS(u) \biggr]^{\frac{1}{n-j}} \\ & \quad \leq B_{i}^{(\tau )}(K)^{\frac{k-j}{(k-i)(n-j)}}B_{k}^{(\tau )}(K)^{ \frac{j-i}{(k-i)(n-j)}}. \end{aligned}$$
(3.5)
Similarly,
Hence, by (3.3), (3.5), and (3.6) we have
This yields inequality (1.9).
$$\begin{aligned}& \biggl[\frac{1}{n} \int _{S^{n-1}}b^{(\tau )}(L,u)^{ \frac{(k-j)(n-i)}{k-i}}b^{(\tau )}(L,u)^{\frac{(j-i)(n-k)}{k-i}} \,dS(u) \biggr]^{\frac{1}{n-j}} \\ & \quad \leq B_{i}^{(\tau )}(L)^{\frac{k-j}{(k-i)(n-j)}}B_{k}^{(\tau )}(L)^{ \frac{j-i}{(k-i)(n-j)}}. \end{aligned}$$
(3.6)
$$ B_{j}^{(\tau )}(K+L)^{\frac{1}{n-j}} \leq B_{i}^{(\tau )}(K)^{\frac{k-j}{(k-i)(n-j)}}B_{k}^{(\tau )}(K)^{ \frac{j-i}{(k-i)(n-j)}}+B_{i}^{(\tau )}(L)^{\frac{k-j}{(k-i)(n-j)}}B _{k}^{(\tau )}(L) ^{\frac{j-i}{(k-i)(n-j)}}. $$
If \(n-1< j< n\), then \(0< n-j<1\), this gives (3.3) is reversed. But \(j< i< k\) means \(0<\frac{k-i}{k-j}<1\), thus (3.4) is reversed. Hence inequality (3.5) is reversed. Similarly, inequality (3.6) is also reversed. From this, we know that inequality (1.9) is reversed.
Proof of Theorem 1.2
For \(j< n-1\) and \(i\leq j< k\), since \(n-j>1\), thus by (1.8), (2.2), and (3.2) we get
$$\begin{aligned} C_{j}^{(\tau )}(K\mathbin{\tilde{+}}L)^{\frac{1}{n-j}} =& \biggl[\frac{1}{n} \int _{S^{n-1}}c^{(\tau )}(K\mathbin{\tilde{+}}L,u)^{n-j} \,dS(u) \biggr]^{ \frac{1}{n-j}} \\ =& \biggl[\frac{1}{n} \int _{S^{n-1}} \bigl(f_{1}(\tau )\rho (K\mathbin{ \tilde{+}}L,u)+f _{2}(\tau )\rho (K\mathbin{\tilde{+}}L,-u) \bigr)^{n-j}\,dS(u) \biggr]^{ \frac{1}{n-j}} \\ =& \biggl[\frac{1}{n} \int _{S^{n-1}} \bigl(f_{1}(\tau )\rho (K,u)+f_{1}( \tau )\rho (L,u) \\ &{}+f_{2}(\tau )\rho (K,-u)+f_{2}(\tau )\rho (L,-u) \bigr)^{n-j}\,dS(u) \biggr]^{{\frac{1}{n-j}}} \\ =& \biggl[\frac{1}{n} \int _{S^{n-1}} \bigl(c^{(\tau )}(K,u)+c^{(\tau )}(L,u) \bigr)^{n-j}\,dS(u) \biggr]^{\frac{1}{n-j}} \\ =& \biggl[\frac{1}{n} \int _{S^{n-1}} \bigl(c^{(\tau )}(K,u)^{ \frac{(k-j)(n-i)}{(k-i)(n-j)}}c^{(\tau )}(K,u)^{ \frac{(j-i)(n-k)}{(k-i)(n-j)}} \\ &{}+c^{(\tau )}(L,u)^{\frac{(k-j)(n-i)}{(k-i)(n-j)}}c^{(\tau )}(L,u)^{ \frac{(j-i)(n-k)}{(k-i)(n-j)}} \bigr)^{n-j}\,dS(u) \biggr]^{\frac{1}{n-j}} \\ \leq& \biggl[\frac{1}{n} \int _{S^{n-1}}c^{(\tau )}(K,u)^{ \frac{(k-j)(n-i)}{k-i}}c^{(\tau )}(K,u)^{\frac{(j-i)(n-k)}{k-i}} \,dS(u) \biggr]^{\frac{1}{n-j}} \\ &{}+ \biggl[\frac{1}{n} \int _{S^{n-1}}c^{(\tau )}(L,u)^{ \frac{(k-j)(n-i)}{k-i}}c^{(\tau )}(L,u)^{\frac{(j-i)(n-k)}{k-i}} \,dS(u) \biggr]^{\frac{1}{n-j}}. \end{aligned}$$
(3.7)
On the right-hand side of the above inequality, when \(i< j< k\), i.e., \(\frac{k-i}{k-j}>1\), by (3.1) we have
Hence, for \(j< n-1\),
Similarly,
According to (3.7), (3.9), and (3.10), we see that
This gives inequality (1.10).
$$\begin{aligned}& \frac{1}{n} \int _{S^{n-1}}c^{(\tau )}(K,u)^{\frac{(k-j)(n-i)}{k-i}}c ^{(\tau )}(K,u)^{\frac{(j-i)(n-k)}{k-i}}\,dS(u) \\& \quad \leq \biggl[\frac{1}{n} \int _{S^{n-1}} \bigl(c^{(\tau )}(K,u)^{ \frac{(k-j)(n-i)}{k-i}} \bigr)^{\frac{k-i}{k-j}}\,dS(u) \biggr]^{ \frac{k-j}{k-i}} \\& \qquad {}\times \biggl[\frac{1}{n} \int _{S^{n-1}} \bigl(c^{(\tau )}(K,u)^{ \frac{(j-i)(n-k)}{k-i}} \bigr)^{\frac{k-i}{j-i}}\,dS(u) \biggr]^{ \frac{j-i}{k-i}} \\& \quad =C_{i}^{(\tau )}(K)^{\frac{k-j}{k-i}}C_{k}^{(\tau )}(K)^{ \frac{j-i}{k-i}}. \end{aligned}$$
(3.8)
$$\begin{aligned}& \biggl[\frac{1}{n} \int _{S^{n-1}}c^{(\tau )}(K,u)^{ \frac{(k-j)(n-i)}{k-i}}c^{(\tau )}(K,u)^{\frac{(j-i)(n-k)}{k-i}} \,dS(u) \biggr]^{\frac{1}{n-j}} \\& \quad \leq C_{i}^{(\tau )}(K)^{\frac{k-j}{(k-i)(n-j)}}C_{k}^{(\tau )}(K)^{ \frac{j-i}{(k-i)(n-j)}}. \end{aligned}$$
(3.9)
$$\begin{aligned}& \biggl[\frac{1}{n} \int _{S^{n-1}}c^{(\tau )}(L,u)^{ \frac{(k-j)(n-i)}{k-i}}c^{(\tau )}(L,u)^{\frac{(j-i)(n-k)}{k-i}} \,dS(u) \biggr]^{\frac{1}{n-j}} \\& \quad \leq C_{i}^{(\tau )}(L)^{\frac{k-j}{(k-i)(n-j)}}C_{k}^{(\tau )}(L)^{ \frac{j-i}{(k-i)(n-j)}}. \end{aligned}$$
(3.10)
$$ C_{j}^{(\tau )}(K\mathbin{\tilde{+}}L)^{\frac{1}{n-j}} \leq C_{i}^{(\tau )}(K)^{\frac{k-j}{(k-i)(n-j)}}C_{k}^{(\tau )}(K)^{ \frac{j-i}{(k-i)(n-j)}}+C_{i}^{(\tau )}(L)^{\frac{k-j}{(k-i)(n-j)}}C _{k}^{(\tau )}(L) ^{\frac{j-i}{(k-i)(n-j)}}. $$
For \(n-1< j< n\) and \(j\leq i< k\), we easily obtain that (3.7) and (3.8) are reversed, thus inequalities (3.9) and (3.10) both are reversed. So, we can get that inequality (1.10) is reversed.
Acknowledgements
The authors want to express earnest thankfulness for the referees who provided extremely precious and helpful comments and suggestions. This made the article more accurate and readable.
Competing interests
The authors state that they have no competing interests.
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