We begin this section with three lemmas, which will be used in the sequel.
From Yang [13] we get the following lemma.
Lemma 2.1
([13])
If A and B are positive numbers, then
$$ (\ln A-\ln B ) \bigl(A^{t}-B^{t} \bigr) \textstyle\begin{cases} \geq 0,& t\geq 0,\\ \leq 0,& t\leq 0, \end{cases} $$
and the equality holds if and only if
\((A-B)t=0\).
Lemma 2.2
([14])
Let
\(rp_{k}>1\), \(\sum_{k=1}^{n}\frac{1}{p_{k}}=r\), \(a_{ij}>0\), \(1\leq i\leq m\), \(1\leq j\leq n\), \(b_{i}= (\prod_{j=1} ^{n} a_{ij} )^{\frac{1}{r}}\), \(d_{ik}=a_{ik}^{rp_{k}}/\prod_{j=1} ^{n} a_{ij}\), \(1\leq i\leq m\), \(1\leq k\leq n\). Then
$$ \sum_{k=1}^{n}\frac{1}{p_{k}} \Biggl(\sum _{i=1}^{m} b_{i}\ln d_{ik} \Biggr)=0. $$
Proof
By the definition of \(b_{i}\) and \(d_{ik}\) we have for \(i=1,2,\ldots ,m\):
$$ \begin{aligned} \sum_{k=1}^{n} \frac{1}{p_{k}} \ln d_{ik} &=\sum_{k=1}^{n} \frac{1}{p _{k}} \ln \biggl(\frac{a_{ik}^{rp_{k}}}{b_{i}^{r}} \biggr) \\ &=\sum_{k=1} ^{n}r \ln a_{ik}- \sum_{k=1}^{n}\frac{r}{p_{k}} \ln b_{i} \\ &=r \Biggl( \ln \Biggl(\prod_{k=1}^{n} \frac{a_{ik}}{b_{i}^{r}} \Biggr) \Biggr) \\ &=0. \end{aligned} $$
Therefore
$$ \sum_{k=1}^{n}\frac{1}{p_{k}} \Biggl(\sum _{i=1}^{m} b_{i}^{r}\ln d_{ik} \Biggr)= \sum_{i=1}^{m} b_{i}^{r} \Biggl(\sum_{k=1}^{n} \frac{1}{p_{k}} \ln d_{ik} \Biggr)=0. $$
Thus the proof of the lemma is completed. □
Lemma 2.3
([14])
Let
\(rp_{k}>1\), \(\sum_{k=1}^{n}\frac{1}{p_{k}}=r\), \(f_{k}(x)>0\), \(F(x)= (\prod_{j=1}^{n}f_{j}(x) )^{\frac{1}{r}}\), \(g_{k}(x)=f_{k}^{rp_{k}}/F^{r}(x)\), \(x\in [a, b]\), \(k=1,2,\ldots ,n\). Then
$$ \sum_{k=1}^{n} p_{k}^{-1} \ln g_{k}(x)=0,\quad x\in [a,b]. $$
Proof
After a simple operation, we have
$$ \begin{aligned} \sum_{k=1}^{n} p_{k}^{-1}\ln g_{k}(x) &=\sum _{k=1}^{n}\frac{1}{p_{k}} \ln \frac{f_{k}^{rp_{k}}(x)}{F^{r}(x)} \\ &=\sum_{k=1}^{n}\frac{1}{p _{k}} \bigl[rp_{k}\ln f_{k}(x)-r\ln F(x) \bigr] \\ &=\sum_{k=1}^{n} \ln f_{k}(x)- \sum_{k=1}^{n}\frac{r}{p_{k}}\ln F(x) \\ &=r\ln \prod_{k=1} ^{n}f_{k}(x)-r^{2} \ln F(x) \\ &=r\ln \frac{\prod_{k=1}^{n}f_{k}(x)}{F ^{r}(x)} \\ &=0. \end{aligned} $$
Thus the proof of the lemma is completed. □
The main result of this paper is the following theorem.
As can be seen from Kwon and Bae [25], we can get the continuous function \(h_{r}(t)\) proposed by Theorem 2.4. In this paper, we consider continuous variables \(f(x,y)\) instead of discrete variables \(a_{ij}\). At the same time, we will introduce the discrete form of \(h_{r}(t)\) used in this paper by the continuous function.
Theorem 2.4
Let
\(X= (X,\mu )\)
and
\((Y,\nu )\)
be positive measure spaces with
\(\mu (X)=1\)
and
\(0<\nu (Y)<\infty \). Let
\(f(x,y)\)
be a real-valued bounded measurable function on
\(X\times Y\). Define the function
\(h_{r}(t):\mathbb{R}\to \mathbb{R}^{+}\)
by
$$ h_{r}(t)=\exp { \biggl[ \int _{X} \ln \biggl( \int _{Y} G(y)e^{tH(x,y)}\,d \nu (y) \biggr)\,d \mu (x) \biggr]}, $$
(6)
where
\(G(y)=e^{\int _{X} f(x,y) \,d\mu (x)}\)
and
\(H(x,y)=rf(x,y)-\int _{X} f(x,y) \,d\mu (x)\).
Then the discrete form of
\(h_{r}(t)\)
can be further pushed out:
$$ h_{r}(t)=\prod_{k=1}^{n} \Biggl[ \sum_{i=1}^{n} \Biggl(\prod _{j=1}^{m} X _{ij} \Biggr)^{1-t} \bigl(X_{ik}^{rp_{k}} \bigr)^{t} \Biggr]^{\frac{1}{p _{k}}}, $$
(7)
where
\(rp_{k}>1\), \(\sum_{k=1}^{n}\frac{1}{p_{k}}=r\), \(X_{ij}>0\)
\((1\leq i\leq m,1\leq j\leq n)\), and
\(X_{ij}\)
are measurable.
Proof
Take X, Y as two discrete spaces \(X={1,2,3,\ldots ,n}\), \(Y={1,2,3, \ldots ,m}\), and take a positive sequence \(p_{j}\) satisfying \(\sum_{j=1}^{n}\frac{1}{p_{j}}=r\). Set μ, ν as
$$\begin{aligned} &\mu =\sum_{j=1}^{n}\frac{1}{p_{j}}\delta _{j},\\ & \nu =\sum_{i=1}^{m} \delta _{i}, \end{aligned}$$
where \(\delta _{j}\) and \(\delta _{i}\) denote the unit mass concentrated at j and i, respectively. Then \((X,2^{X},\mu )\) and \((Y,2^{Y}, \nu )\) are positive measure spaces with \(\mu (X)=1\) and \(0<\nu (Y)=m< \infty \). Here \(2^{\lbrace \cdot \rbrace }\) means the set of all subsets of \(\lbrace \cdot \rbrace \). Let
$$ e^{f(j,i)}=X_{ij}^{p_{j}}. $$
Then
$$ \begin{aligned} \exp { \biggl( \int _{X} f(x,y) \,d\mu (x) \biggr)}&= \exp { \Biggl(\sum _{j=1} ^{n}\frac{1}{p_{j}}\ln X_{ij}^{p_{j}} \Biggr)}\\ &=\exp { \Biggl(\sum _{j=1} ^{n} \ln X_{ij} \Biggr)} \\ &=\exp { \Biggl(\ln \prod_{j=1}^{n} X_{ij} \Biggr)}=\prod_{j=1}^{n} X_{ij}. \end{aligned} $$
Therefore \(h_{r}(t)\) has the expression
$$ \begin{aligned} h_{r}(t) &=\exp { \biggl[ \int _{X} \ln \biggl( \int _{Y} G(y)e^{tH(x,y)} \,d \nu (y) \biggr)\,d\mu (x) \biggr]} \\ &=\exp { \biggl[ \int _{X} \ln \biggl( \int _{Y} e^{\int _{X} f(x,y) \,d\mu (x)}e^{rtf(x,y)-t\int _{X} f(x,y) \,d\mu (x)} \,d\nu (y) \biggr)\,d\mu (x) \biggr]} \\ &=\exp { \biggl[ \int _{X} \ln \biggl( \int _{Y} \bigl(e^{\int _{X} f(x,y) \,d\mu (x)} \bigr)^{1-t} \bigl(e^{f(x,y)} \bigr)^{rt} \,d\nu (y) \biggr) \,d\mu (x) \biggr]} \\ &=\exp { \Biggl[ \int _{X} \ln \Biggl( \int _{Y} \Biggl(\prod_{j=1}^{n} X_{ij} \Biggr)^{1-t} \bigl(X_{ik} ^{rp_{k}} \bigr)^{t} \,d\nu (y) \Biggr) \,d\mu (x) \Biggr]} \\ &=\exp { \Biggl[ \int _{X} \ln \Biggl(\sum_{i=1}^{m} \Biggl(\prod_{j=1}^{n} X_{ij} \Biggr)^{1-t} \bigl(X_{ik}^{rp_{k}} \bigr)^{t} \Biggr) \,d\mu (x) \Biggr]} \\ &=\exp { \Biggl[ \sum_{k=1}^{n} \frac{1}{p_{k}} \ln \Biggl(\sum_{i=1}^{m} \Biggl(\prod_{j=1} ^{n}X_{ij} \Biggr)^{1-t} \bigl(X_{ik}^{rp_{k}} \bigr)^{t} \Biggr) \Biggr]} \\ &= \prod_{k=1}^{n} \Biggl[\sum _{i=1}^{m} \Biggl(\prod_{j=1}^{n}X_{ij} \Biggr)^{1-t} \bigl(X_{ik}^{rp_{k}} \bigr)^{t} \Biggr]^{\frac{1}{p_{k}}}. \end{aligned} $$
Thus the proof of the theorem is completed. □
Theorem 2.5
Let
\(a_{ij}>0\)
\((1\leq i\leq m, 1\leq j\leq n)\), \(rp_{k}>1\), \(\sum_{k=1}^{n} 1/p_{k}=r\). Define the positive function
$$ h_{r}(t)= \prod_{k=1}^{n} \Biggl[\sum_{i=1}^{m} \Biggl(\prod _{j=1}^{n} a _{ij} \Biggr)^{1-t} \bigl(a_{ik}^{rp_{k}} \bigr)^{t} \Biggr]^{1/p_{k}}, \quad t \in \mathbb{R}. $$
(8)
Then the function
\(h_{r}(t)\)
defined by (8) is concave, that is, \(h_{r}^{\prime \prime }(t)\geq 0\)
for
\(t\in \mathbb{R}\), and the equality holds if and only if
$$ \frac{a_{ik}^{rp_{k}}}{\prod_{j=1}^{n} a_{ij}}=\frac{a_{jk}^{rp_{k}}}{ \prod_{l=1}^{n} a_{jl}},\quad 1\leq i,j\leq m, k=1,2, \ldots ,n. $$
(9)
In this case, \(h_{r}(t)\)
is a constant.
Proof
It is easy to see from (2) that \(h_{r}(t)\) can be expressed in the following equivalent form:
Let \(b_{i}= (\prod_{j=1}^{n} a_{ij} )^{\frac{1}{r}}\), \(d_{ik}=a_{ik}^{rp_{k}}/b_{i}^{r}\),
$$ h_{r}(t)= \prod_{k=1}^{n} \Biggl[\sum_{i=1}^{m} b_{i}^{r} \biggl(\frac{a _{ik}^{rp_{k}}}{b_{i}^{r}} \biggr)^{t} \Biggr]^{\frac{1}{p_{k}}}. $$
Let \(H_{r}(t)=\ln h_{r}(t)\),
$$ H_{r}(t)= \sum_{k=1}^{n} \frac{1}{p_{k}}\ln \Biggl(\sum_{i=1}^{m} b_{i} ^{r} d_{ik}^{t} \Biggr). $$
Then by Lemmas 2.1 and 2.2 we obtain
$$ \begin{aligned} H_{r}^{\prime }(t) = \frac{h_{r}^{\prime }(t)}{h_{r}(t)} &=\sum_{k=1}^{n} \frac{1}{p _{k}}\frac{\sum_{i=1}^{m} b_{i}^{r}d_{ik}^{t}\ln d_{ik}}{\sum_{i=1} ^{m} b_{i}^{r}d_{ik}^{t}} \\ &=\sum_{k=1}^{n}\frac{1}{p_{k}} \frac{ \sum_{i=1}^{m} b_{i}^{r}d_{ik}^{t}\ln d_{ik}}{\sum_{i=1}^{m} b_{i} ^{r}d_{ik}^{t}}-\sum_{k=1}^{n} \frac{1}{p_{k}}\frac{\sum_{i=1}^{m} b _{i}^{r}\ln d_{ik}}{\sum_{i=1}^{m} b_{i}^{r}} \\ &=\sum_{k=1}^{n}\frac{1}{p _{k}} \frac{\sum_{1\leq i< j\leq m}b_{i}^{r}b_{j}^{r} (\ln d_{ik}- \ln d_{jk} ) (d_{ik}^{t}-d_{jk}^{t} )}{ (\sum_{i=1}^{m} b _{i}^{r} ) (\sum_{i=1}^{m} b_{i}^{r}d_{ik}^{t} )} & \textstyle\begin{cases} \geq 0,& t\geq 0,\\ \leq 0,& t\leq 0. \end{cases}\displaystyle \end{aligned} $$
Therefore \(H_{r}^{\prime }(t)=0\) if and only if \(t=0\) or
$$ d_{ik}=d_{jk}, \quad 1\leq i< j\leq m, k=1,2,\ldots ,n. $$
Similarly, by Lemmas 2.1 and 2.2 we obtain
$$ H_{r}^{\prime \prime }(t)=\sum_{k=1}^{n} \frac{1}{p_{k}}\frac{ (\sum_{i=1}^{n} b _{i}^{r}d_{ik}^{t} ) (\sum_{i=1}^{m}b_{i}^{r}d_{ik}^{t} ( \ln d_{ik} )^{2} )- (\sum_{i=1}^{m}b_{i}^{r}d_{ik}^{t}\ln d _{ik} )^{2}}{ (\sum_{i=1}^{m} b_{i}^{r}d_{ik}^{t} )^{2}}. $$
It is easy to see that \(H_{r}^{\prime \prime }\geq 0\) and the equality holds at some \(t_{0}\) if and only if
$$ d_{ik}=d_{jk},\quad 1\leq i< j\leq m,k=1,\ldots ,n. $$
Since \(h_{r}^{\prime }(t)=h_{r}(t)H_{r}^{\prime }(t)\), we have \(h_{r}^{\prime \prime }(t)=h _{r}(t) [ (H_{r}^{\prime }(t) )^{2}+H_{r}^{\prime \prime }(t) ]\geq 0\), and \(h_{r}^{\prime \prime }(t)=0\) if and only if \(H_{r}^{\prime }(t)=0\) and \(H_{r}^{\prime \prime }(t)=0\).
The proof of the theorem is completed. □
From our discussion we see that \(h_{r}^{\prime \prime }(t)\geq 0\) and the equality holds at some \(t_{0}\in \mathbb{R}\) if and only if (9) holds. In this case, \(h_{r}^{\prime \prime }(t)\equiv 0\) and \(h_{r}^{\prime }(t)=h_{r}^{\prime }(0)=\mathrm{const}\). Since, by by the proof, \(h_{r}^{\prime }(t)=h _{r}(0)H_{r}^{\prime }(0)=0\), which yields \(h_{r}(t)=h_{r}(0)=\mathrm{const}\), \(t\in \mathbb{R}\).
Corollary 2.6
If (9) holds, then
\(h_{r}(t)=\mathrm{const}\). Otherwise, \(h_{r}^{\prime \prime }(t)>0\), \(t\in \mathbb{R}\), and
\(h_{r}^{\prime }(t)t>0\)
for
\(t\neq 0\); in particular, for
\(0=t_{1}< t_{2}<\cdots <t_{N}=1\), we have
$$\begin{aligned} & h_{r}(0)=h_{r}(t_{1})< h_{r}(t_{2})< \cdots < h_{r}(t_{N})=h_{r}(1), \\ &\begin{aligned} h _{r}(0)&=\min_{t\in R}h_{r}(t)= \prod_{k=1}^{n} \Biggl[\sum _{i=1}^{m} b _{i}^{r} \Biggr]^{\frac{1}{p_{k}}}= \Biggl(\sum_{i=1}^{m} b_{i}^{r} \Biggr)^{r} \\ &= \Biggl(\sum_{i=1}^{m}\prod _{j=1}^{n} a_{ij} \Biggr)^{r}\leq h_{r}(1)= \prod_{k=1}^{n} \Biggl[ \sum_{i=1}^{m} a_{ik}^{rp_{k}} \Biggr]^{\frac{1}{p _{k}}}. \end{aligned} \end{aligned}$$
If \(r=1\), then we get a refinement of (1),
$$ \sum_{i=1}^{m}\prod _{j=1}^{n} a_{ij}\leq \prod _{j=1}^{n} \Biggl(\sum_{i=1}^{m} a_{ij}^{p_{j}} \Biggr)^{\frac{1}{p_{j}}}. $$
Corollary 2.7
If (9) holds, then
\(h_{r}(t)=\mathrm{const}\). Otherwise,
$$ h_{r}(0)=h_{r}(1)-h_{r}^{\prime }(\tau ), \quad \tau \in (0,1), $$
and
\(0< h_{r}^{\prime }(\tau )< h_{r}^{\prime }(1)\), where
$$ h_{r}^{\prime }(1)=H_{r}^{\prime }(1)h_{r}(1) = \Biggl(\sum_{k=1}^{n} \frac{1}{p_{k}} \frac{\sum_{1\leq i< j\leq m} b_{i}^{r}b_{j}^{r} ( \ln d_{ik}-\ln d_{jk} ) (d_{ik}-d_{jk} )}{ (\sum_{i=1} ^{m} b_{i}^{r} ) (\sum_{i=1}^{m} b_{i}^{r}d_{ik} )} \Biggr) \prod_{j=1}^{n} \Biggl(\sum_{j=1}^{n} a_{ij}^{p_{j}} \Biggr)^{\frac{1}{p _{j}}}. $$
Theorem 2.8
Let
\(f_{k}(x)>0\), \(x\in [a,b]\), \(k=1,2,\ldots ,n\), and
\(f_{k}\in L ^{p_{k}}[a,b]\). Define the positive function
$$ g_{r}(t)= \prod_{k=1}^{n} \Biggl[ \int _{a}^{b} \Biggl(\prod _{j=1}^{n} f_{j}(x) \Biggr)^{1-t}f_{k}^{rp_{k}t}(x)\,dx \Biggr]^{\frac{1}{p_{k}}},\quad t\in \mathbb{R}. $$
(10)
The function
\(g_{r}(t)\)
defined by (10) is concave, that is, \(g_{r}^{\prime \prime }(t)\geq 0\)
for
\(t\in \mathbb{R}\), and the equality holds if and only if
$$ \frac{f_{k}^{rp_{k}}(x)}{\prod_{j=1}^{n} f_{j}(x)}=c_{k}=\mathrm{const}. $$
(11)
Proof
It is known from Lemma 2.3 that if \(F(x)= (\prod_{j=1} ^{n} f_{j}(x) )^{\frac{1}{r}}\) and \(g_{k}(x)=f_{k}^{rp_{k}}(x)/F ^{r}(x)\), then
$$ g_{r}(t)=\prod_{k=1}^{n} \biggl[ \int _{a}^{b} F^{r}(x)g_{k}^{t}(x)\,dx \biggr]^{\frac{1}{p_{k}}}. $$
Let \(G_{r}(t)=\ln g_{r}(t)\). Then
$$ G_{r}(t)=\sum_{k=1}^{n} \frac{1}{p_{k}} \ln \biggl( \int _{a}^{b} F^{r}(x)g _{k}^{t}(x)\,dx \biggr). $$
Then by Lemmas 2.1 and 2.3 we obtain
$$ \begin{aligned} &G_{r}^{\prime }(t)\\ &\quad = \sum _{k=1}^{n}\frac{1}{p_{k}}\frac{\int _{a}^{b} F ^{r}(x)g_{k}^{t}(x)\ln g_{k}(x)\,dx}{\int _{a}^{b} F^{r}(x)g_{k}^{t}(x)\,dx} \\ &\quad =\sum_{k=1}^{n}\frac{1}{p_{k}} \frac{\int _{a}^{b} F^{r}(x)g_{k} ^{t}(x)\ln g_{k}(x)\,dx}{\int _{a}^{b} F^{r}(x)g_{k}^{t}(x)\,dx} -\frac{ \int _{a}^{b} F^{r}(x) (\sum_{k=1}^{n}\frac{1}{p_{k}}\ln g_{k}(x) )\,dx}{\int _{a}^{b} F^{r}(x)\,dx} \\ &\quad =\sum_{k=1}^{n} \frac{1}{p_{k}} \biggl[ \frac{\int _{a}^{b} F^{r}(x)g_{k}^{t}(x)\ln g _{k}(x)\,dx}{\int _{a}^{b} F^{r}(x)g_{k}^{t}(x)\,dx}- \frac{\int _{a} ^{b} F^{r}(x)\ln g_{k}(x)\,dx}{\int _{a}^{b} F^{r}(x)\,dx} \biggr] \\ &\quad = \sum_{k=1}^{n}\frac{1}{2p_{k}} \frac{\int _{a}^{b}\int _{a}^{b} F^{r}(x)F ^{r}(y) (g_{k}^{t}(x)-g_{k}^{t}(y) ) (\ln g_{k}(x)-\ln g _{k}(y) )\,dx\,dy}{\int _{a}^{b} F^{r}(x)g_{k}^{t}(x)\,dx\int _{a}^{b} F ^{r}(x)\,dx} & \textstyle\begin{cases} \geq 0,& t\geq 0,\\ \leq 0,& t\leq 0. \end{cases}\displaystyle \end{aligned} $$
Therefore \(G_{r}^{\prime }(t)=0\) if and only if
$$ \frac{f_{k}^{rp_{k}}(x)}{\prod_{j=1}^{n} f_{j}(x)}=c_{k}=\mathrm{const}. $$
Similarly, by Lemmas 2.1 and 2.3 we obtain
$$ G_{r}^{\prime \prime }(t)=\sum_{k=1}^{n} \frac{1}{p_{k}}\frac{\int _{a}^{b}\int _{a} ^{b} F^{r}(x)F^{r}(y)g_{k}^{t}(x)g_{k}^{t}(y) (\ln g_{k}(x)- \ln g_{k}(y) )^{2}\,dx\,dy}{2 (\int _{a}^{b} F^{r}(x)g_{k}^{t}(x)\,dx )^{2}}\geq 0. $$
It is easy to see \(G_{r}^{\prime \prime }(t)\geq 0\) and the equality holds at some \(t_{0}\) if and only if
$$ \frac{f_{k}^{rp_{k}}(x)}{\prod_{j=1}^{n} f_{j}(x)}=c_{k}=\mathrm{const}. $$
Since \(g_{r}^{\prime }(t)=g_{r}(t)G_{r}^{\prime }(t)\), we have \(g_{r}^{\prime \prime }(t)=g _{r}(t) [ (G_{r}^{\prime }(t) )^{2}+G_{r}^{\prime \prime }(t)]\geq 0\), and \(g_{r}^{\prime \prime }(t)=0\) if and only if \(G_{r}^{\prime }(t)=0\) and \(G_{r}^{\prime \prime }(t)=0\).
Thus the proof of the theorem is completed. □
Corollary 2.9
If
\(g_{k}(x)=\mathrm{const}\), then
\(g_{r}(t)=\mathrm{const}\). Otherwise, \(g_{r}^{\prime \prime }(t)>0\), \(t\in \mathbb{R}\), and
\(g_{r}^{\prime }(t)t>0\)
for
\(t\neq 0\); in particular, for
\(0=t_{1}< t_{2}<\cdots <t_{N}=1\), we have
$$\begin{aligned} & g_{r}(0)=g_{r}(t_{1})< g_{r}(t_{2})< \cdots < g_{r}(t_{N})=g_{r}(1), \\ & g_{r}(0)=\prod_{k=1}^{n} \biggl[ \int _{a}^{b} F^{r}(x)\,dx \biggr]^{\frac{1}{p _{k}}}\leq g_{r}(1)=\prod_{k=1}^{n} \biggl[ \int _{a}^{b} F^{r}(x)g _{k}(x)\,dx \biggr]^{\frac{1}{p_{k}}}. \end{aligned}$$
Corollary 2.10
If (11) holds, then
\(g_{r}(t)=\mathrm{const}\). Otherwise,
$$ g_{r}(0)=g_{r}(1)-g_{r}^{\prime }(s),\quad s\in (0,1), $$
and
\(0< g_{r}^{\prime }(s)< g_{r}^{\prime }(1)\), where
$$ \begin{aligned} g_{r}^{\prime }(1) ={}&G_{r}^{\prime }(1)g_{r}(1) \\ ={}&\sum_{k=1}^{n}\frac{1}{2p_{k}} \frac{ \int _{a}^{b}\int _{a}^{b} F^{r}(x)F^{r}(y) (g_{k} (x)-g_{k} (y) ) (\ln g_{k}(x)-\ln g_{k}(y) )\,dx\,dy}{\int _{a}^{b} F ^{r}(x)g_{k} (x)\,dx\int _{a}^{b} F^{r}(x)\,dx} \\ &{}\times \prod_{k=1}^{n} \biggl( \int _{a}^{b} f_{k}^{rp_{k}}(x)\,dx \biggr)^{\frac{1}{p _{k}}}. \end{aligned} $$
Next, we use the existing inequalities to derive the inequalities obtained in Kwon and Bae [25].
Theorem 2.11
Let
\(a_{ij}>0\), \(p_{k}>1\), \(\alpha _{kj}\in \mathbb{R}\), \(1\leq i\leq m\), \(1 \leq j\), \(k\leq n\), \(\sum_{k=1}^{n}\frac{1}{p_{k}}=1\), and
\(\sum_{k=1} ^{n}\alpha _{kj}=0\). Then
$$ \sum_{i=1}^{m}\prod _{j=1}^{n} a_{ij}\leq \prod _{k=1}^{n} \Biggl(\sum_{i=1}^{m} \Biggl(\prod_{j=1}^{n} a_{ij}^{1+p_{k}\alpha _{kj}} \Biggr) \Biggr)^{\frac{1}{p _{k}}}. $$
(12)
Moreover, for the integral form of this inequality, if
\(f_{j}(x)>0\)
\((j=1,2,\ldots ,n)\), \(x\in [a,b]\), \(-\infty < a< b<+\infty \), and
\(f_{j}(x)\in C[a,b]\), then
$$ \int _{a}^{b} \Biggl(\prod _{j=1}^{n} f_{j}(x) \Biggr)\,dx\leq \prod _{k=1} ^{n} \Biggl( \int _{a}^{b}\prod_{j=1}^{n} f_{j}^{1+p_{k}\alpha _{kj}}(x)\,dx \Biggr)^{\frac{1}{p_{k}}}. $$
(13)
Proof
From (2) we can see that \(h(0)\) is the minimum point of \(h(t)\).
Thus \(h(0)\leq h(t)\), that is,
$$ \sum_{i=1}^{m}\prod _{j=1}^{n} a_{ij}\leq \prod _{k=1}^{n} \Biggl[\sum_{i=1}^{m} \Biggl(\prod_{j=1}^{n} a_{ij} \Biggr)^{1-t} \bigl(a_{ik}^{p_{k}} \bigr)^{t} \Biggr]^{\frac{1}{p_{k}}}. $$
Under the assumptions of (2), taking \(t=-p_{k}\alpha _{kj}\) for \(j\neq k\) and \(t=\alpha _{kk}/(1-\frac{1}{p_{k}})\) for \(j=k\) with \(\sum_{k=1}^{n}\alpha _{kj}=0\), we have
$$ \sum_{i=1}^{m}\prod _{j=1}^{n} a_{ij}\leq \prod _{k=1}^{n} \Biggl(\sum_{i=1}^{m} \Biggl(\prod_{j=1}^{n} a_{ij}^{1+p_{k}\alpha _{kj}} \Biggr) \Biggr)^{\frac{1}{p_{k}}}. $$
Similarly, the integral expression of Hölder’s inequality now becomes \(g(0)\leq g(t)\), that is,
$$ \int _{a}^{b} \Biggl(\prod _{k=1}^{n}f_{k}(x) \Biggr)\,dx\leq \prod _{k=1}^{n} \Biggl( \int _{a}^{b} \Biggl(\prod _{j=1}^{n}f_{j}(x) \Biggr)^{1-t} f_{k}^{p _{k}t}(x)\,dx \Biggr)^{\frac{1}{p_{k}}},\quad t\in \mathbb{R}. $$
Next, we make some changes to \(g(t)\). Taking \(t=-p_{k}\alpha _{kj}\) for \(j\neq k\) and \(t=\alpha _{kk}/(1-\frac{1}{p_{k}})\) for \(j=k\) with \(\sum_{k=1}^{n}\alpha _{kj}=0\), we have
$$ \int _{a}^{b} \Biggl(\prod _{j=1}^{n} f_{j}(x) \Biggr)\,dx\leq \prod _{k=1} ^{n} \Biggl( \int _{a}^{b}\prod_{j=1}^{n} f_{j}^{1+p_{k}\alpha _{kj}}(x)\,dx \Biggr)^{\frac{1}{p_{k}}}. $$
Thus, the proof of the theorem is completed. □
Similarly, we also consider the case \(p_{j}>1\), \(\sum_{k=1}^{n}\frac{1}{p _{k}}=r\).
Theorem 2.12
Let
\(a_{ij}>0\), \(rp_{k}>1\), \(\alpha _{kj}\in \mathbb{R}\), \(1\leq i\leq m\), \(1 \leq j\), \(k\leq n\), \(\sum_{k=1}^{n}\frac{1}{p_{k}}=r\), and
\(\sum_{k=1} ^{n}\alpha _{kj}=0\). Then
$$ \sum_{i=1}^{m}\prod _{j=1}^{n} a_{ij}\leq \prod _{k=1}^{n} \Biggl(\sum_{i=1}^{m} \Biggl(\prod_{j=1}^{n} a_{ij}^{1+rp_{k}\alpha _{kj}} \Biggr) \Biggr)^{\frac{1}{p_{k}}}. $$
(14)
Moreover, for the integral form of this inequality, if
\(f_{j}(x)>0\)
\((j=1,2,\ldots ,n)\), \(x\in [a,b]\), \(-\infty < a< b<+\infty \), and
\(f_{j}(x)\in C[a,b]\), then
$$ \int _{a}^{b} \Biggl(\prod _{j=1}^{n} f_{j}(x) \Biggr)\,dx\leq \prod _{k=1} ^{n} \Biggl( \int _{a}^{b}\prod_{j=1}^{n} f_{j}^{1+rp_{k}\alpha _{kj}}(x)\,dx \Biggr)^{\frac{1}{p_{k}}}. $$
(15)
Proof
From (8) we get that \(h_{r}(0)\) is the minimum point of \(h_{r}(t)\). Thus, \(h_{r}(0)\leq h_{r}(t)\), that is,
$$ \sum_{i=1}^{m}\prod _{j=1}^{n} a_{ij}\leq \prod _{k=1}^{n} \Biggl[\sum_{i=1}^{m} \Biggl(\prod_{j=1}^{n} a_{ij} \Biggr)^{1-t} \bigl(a_{ik}^{rp _{k}} \bigr)^{t} \Biggr]^{\frac{1}{p_{k}}}. $$
Under the assumptions of (8), taking \(t=-rp_{k}\alpha _{kj}\) for \(j\neq k\) and \(t=\alpha _{kk}/(1-\frac{1}{rp_{k}})\) for \(j=k\) with \(\sum_{k=1}^{n}\alpha _{kj}=0\), we have
$$ \sum_{i=1}^{m}\prod _{j=1}^{n} a_{ij}\leq \prod _{k=1}^{n} \Biggl(\sum_{i=1}^{m} \Biggl(\prod_{j=1}^{n} a_{ij}^{1+rp_{k}\alpha _{kj}} \Biggr) \Biggr)^{\frac{1}{p_{k}}}. $$
Similarly, the integral expression of Hölder’s inequality now becomes \(g_{r}(0)\leq g_{r}(t)\), that is,
$$ \int _{a}^{b} \Biggl(\prod _{k=1}^{n} f_{k}(x) \Biggr)\,dx\leq \prod _{k=1} ^{n} \Biggl( \int _{a}^{b} \Biggl(\prod _{j=1}^{n} f_{j}(x) \Biggr)^{1-t}f_{k} ^{rp_{k}t}(x)\,dx \Biggr)^{\frac{1}{p_{k}}},\quad t\in \mathbb{R}. $$
Similarly to discrete \(h_{r}(t)\), we also make some changes to the continuous \(g_{r}(t)\). Taking \(t=-rp_{k}\alpha _{kj}\) for \(j\neq k\) and \(t=\alpha _{kk}/(1-\frac{1}{p_{k}})\) for \(j=k\) with \(\sum_{k=1}^{n} \alpha _{kj}=0\), we have
$$ \int _{a}^{b} \Biggl(\prod _{j=1}^{n} f_{j}(x) \Biggr)\,dx\leq \prod _{k=1} ^{n} \Biggl( \int _{a}^{b}\prod_{j=1}^{n} f_{j}^{1+rp_{k}\alpha _{kj}}(x)\,dx \Biggr)^{\frac{1}{p_{k}}}. $$
Thus the proof of the theorem is completed. □