Skip to content

Advertisement

  • Research
  • Open Access

On a generalized Hölder inequality

Journal of Inequalities and Applications20152015:88

https://doi.org/10.1186/s13660-015-0612-9

  • Received: 28 October 2014
  • Accepted: 25 February 2015
  • Published:

Abstract

Refining the Hölder inequality, a result of S Wu is extended to the case of multiple sequences.

Keywords

  • Hölder inequality
  • reversed Hölder inequality

MSC

  • 26D15

1 Introduction

There are a lot of generalizations and extensions of the classical Hölder inequality. Among them is the existence of a function
$$g(a, b) $$
for which
$$\sum_{i=1} ^{n} a_{i} b_{i} \leq \Biggl( \sum_{i=1}^{n} a_{i} ^{p} \Biggr) ^{ \frac{1}{p}} \Biggl( \sum _{i=1} ^{n} b_{i}^{q} \Biggr) ^{ \frac{1}{q}} g (a, b) \leq \Biggl( \sum_{i=1}^{n} a_{i} ^{p} \Biggr) ^{ \frac{1}{p}} \Biggl( \sum _{i=1}^{n} b_{i}^{q} \Biggr) ^{ \frac{1}{q}}. $$
Here \(a= \{ a _{i} \} \), \(b= \{ b _{i} \} \) are positive sequences and \(1 \leq p, q < \infty\), \(\frac{1}{p} + \frac{1}{q} = 1\). Hu and Wu published several results [13] on this topic. Among them is the following theorem.

Theorem A

([3], Theorem 1)

Let \(a_{i} >0\), \(b_{i} >0\) (\(i=1,2, \ldots,n\)), \(1-e_{i} +e_{j} \geq0\) (\(i, j=1, 2, \ldots,n\)), and let \(p \geq q > 0\). Then
$$\sum_{i=1} ^{n} a _{i} b _{i} \leq n ^{1-\min\{ \frac{1}{p} + \frac {1}{q} ,1 \} } \Biggl( \sum_{i=1} ^{n} a _{i} ^{p} \Biggr) ^{\frac{1}{p}} \Biggl( \sum_{i=1} ^{n} b _{i} ^{q} \Biggr) ^{\frac{1}{q}} \biggl[ 1- \biggl\{ \frac{\sum_{i=1} ^{n} a _{i} ^{p} e _{i}}{ \sum_{i=1} ^{n} a _{i} ^{p}} - \frac{\sum_{i=1} ^{n} b _{i} ^{q} e _{i}}{\sum_{i=1} ^{n} b _{i} ^{q}} \biggr\} ^{2} \biggr] ^{\frac{1}{2p}} . $$

Our purpose of this paper is to extend Theorem A to the case of multiple sequences. The following are the main results of this paper.

Theorem 1.1

Let \(a_{j i} >0\), \(p_{j}>0\) (\(i=1,2, \ldots,n \); \(j=1,2, \ldots ,m\)), \(1-e_{i} +e_{j} \geq0\) (\(i, j=1,2, \ldots,n \)), \(\Lambda:=\sum_{j=1} ^{m} \frac{1}{p_{j}}\) and \(\{p_{j_{1}}, p_{j_{2}} \} \subset \{p_{1}, p_{2},\ldots, p_{m} \}\). Then
$$\sum_{i=1} ^{n} \prod _{j=1}^{m} a _{j i} \leq n ^{1-\min \{ \Lambda,1 \} } \biggl\{ 1- \biggl( \frac{\sum_{i=1} ^{n} a _{j_{1} i} ^{p_{j_{1}}} e _{i}}{\sum_{i=1} ^{n} a _{j_{1} i} ^{p_{j_{1}}} } - \frac{\sum_{i=1} ^{n} a _{j_{2} i} ^{p_{j_{2}}} e _{i}}{\sum_{i=1} ^{n} a _{j_{2} i} ^{p_{j_{2}}} } \biggr) ^{2} \biggr\} ^{\frac{1}{2 \max \{ p_{j_{1}}, p_{j_{2}} \} }} \prod_{j=1}^{m} \Biggl( \sum_{i=1} ^{n} a_{j i}^{p_{j}} \Biggr) ^{\frac{1}{p_{j}}} . $$

Theorem 1.2

Let \(a_{j i} >0\) (\(i=1,2, \ldots,n \); \(j=1,2, \ldots,m\)), \(1-e_{i} +e_{j} \geq0\) (\(i, j=1,2, \ldots,n\)), \(p_{j}<0\) (\(j=1,2, \ldots,m-1\)), \(0<\sum_{j=1}^{m} \frac{1}{p_{j}}\leq 1\) and \(\{p_{j_{1}}, p_{j_{2}} \} \subset \{p_{1}, p_{2},\ldots, p_{m} \}\). Then
$$\sum_{i=1} ^{n} \prod _{j=1}^{m} a _{j i} \geq \biggl\{ 1- \biggl( \frac {\sum_{i=1} ^{n} a _{j_{1} i} ^{p_{j_{1}}} e _{i}}{\sum_{i=1} ^{n} a _{j_{1} i} ^{p_{j_{1}}} } - \frac{\sum_{i=1} ^{n} a _{j_{2} i} ^{p_{j_{2}}} e _{i}}{\sum_{i=1} ^{n} a _{j_{2} i} ^{p_{j_{2}}} } \biggr) ^{2} \biggr\} ^{\frac{1}{2 \min \{ p_{j_{1}}, p_{j_{2}} \} }} \prod_{j=1}^{m} \Biggl( \sum_{i=1} ^{n} a_{j i}^{p_{j}} \Biggr) ^{\frac{1}{p_{j}}} . $$

If \(m=2\), then Theorem 1.1 reduces to Theorem A. The case \(\Lambda=1\) and \(\frac{\sum_{i=1} ^{n} a _{j_{1} i} ^{p_{j_{1}}} e _{i}}{\sum_{i=1} ^{n} a _{j_{1} i} ^{p_{j_{1}}} } = \frac{\sum_{i=1} ^{n} a _{j_{2} i} ^{p_{j_{2}}} e _{i}}{\sum_{i=1} ^{n} a _{j_{2} i} ^{p_{j_{2}}} }\) of Theorem 1.1 is the well-known Hölder inequality.

Theorem 1.2 is a reversed version of Theorem 1.1. The case \(m=2\) with \(\Lambda=1\) and \(\frac{\sum_{i=1} ^{n} a _{j_{1} i} ^{p_{j_{1}}} e _{i}}{\sum_{i=1} ^{n} a _{j_{1} i} ^{p_{j_{1}}} } = \frac{\sum_{i=1} ^{n} a _{j_{2} i} ^{p_{j_{2}}} e _{i}}{\sum_{i=1} ^{n} a _{j_{2} i} ^{p_{j_{2}}} }\) of Theorem 1.2 reduces to the well-known reversed Hölder inequality.

The idea of proofs of Theorem 1.1 and Theorem 1.2 are essentially the same as that of Theorem A. The point is that a variant of the methods of Wu in [3] works for the case of multiple sequences. The process will be done in Section 3 and Section 4 after the preparing lemmas in Section 2. In Section 5, as an application of Theorem 1.1, we refine a reversed Hölder inequality different from Theorem 1.2. We refer to [4, 5] for the general theory of inequalities.

2 Lemmas

In order to prove Theorem 1.1 and Theorem 1.2, we need the following lemmas.

Lemma 2.1

([2], Lemma 1(1))

Let \(a _{j i} >0\), \(\lambda _{j} > 0\) (\(i=1, 2, \ldots, n \); \(j=1, 2, \ldots, m\)) and \(\Lambda:= \sum_{j=1} ^{m} \lambda _{j} \). Then
$$\sum_{i=1} ^{n} \prod _{j=1} ^{m} a _{j i} ^{\lambda _{j}} \leq n ^{1- \min \{ \Lambda, 1 \}} \prod_{j=1} ^{m} \Biggl( \sum_{i=1} ^{n} a _{j i} \Biggr) ^{\lambda _{j}} $$
with equality holding if and only if \(a _{j 1} =a _{j2} = \cdots=a _{j n}\) (\(j=1, 2, \ldots, m\)) for \(\Lambda <1\), or \(\frac{a _{1 i}}{\sum_{i=1} ^{n} a _{1 i}} = \frac{a _{2i}}{\sum_{i=1} ^{n} a _{2 i}} = \cdots= \frac{a _{m i}}{\sum_{i=1} ^{n} a _{m i}}\) (\(i=1, 2, \ldots, n\)) for \(\Lambda=1\).

Lemma 2.2

([2], Lemma 1(2))

Let \(a _{j i} >0\) (\(i=1, 2, \ldots, n \); \(j=1, 2, \ldots, m \)), \(\lambda _{j} <0\) (\(j=1, 2, \ldots, m-1\)), \(\lambda _{m}\) be a real number, and \(\Lambda:= \sum_{j=1} ^{m} \lambda _{j}\). Then
$$\sum_{i=1} ^{n} \prod _{j=1} ^{m} a _{j i} ^{\lambda _{j}} \geq n ^{1- \max \{ \Lambda, 1 \}} \prod_{j=1} ^{m} \Biggl( \sum_{i=1} ^{n} a _{j i} \Biggr) ^{\lambda _{j}} $$
with equality holding if and only if \(a _{j 1} =a _{j2} = \cdots=a _{j n}\) (\(j=1, 2, \ldots, m\)) for \(\Lambda>1\), or \(\frac{a _{1 i}}{\sum_{i=1} ^{n} a _{1 i}} = \frac{a _{2i}}{\sum_{i=1} ^{n} a _{2 i}} = \cdots= \frac{a _{m i}}{\sum_{i=1} ^{n} a _{m i}}\) (\(i=1, 2, \ldots, n\)) for \(\Lambda=1\).

3 Proof of Theorem 1.1

We may suppose \(p _{j _{1}} \leq p _{j _{2}}\).

Case (1): \(\Lambda> 1\). We show
$$ \sum_{i=1} ^{n} \prod _{j=1} ^{m} a _{j i} \leq \biggl\{ 1- \biggl( \frac {\sum_{i=1} ^{n} a _{j _{1} i} ^{p _{j _{1}}} e _{i}}{\sum_{i=1} ^{n} a _{j _{1} i} ^{p _{j _{1}}}} - \frac{\sum_{k=1} ^{n} a _{j _{2} i} ^{p _{j _{2}}} e _{i}}{\sum_{k=1} ^{n} a _{j _{2} i} ^{p _{j _{2}}}} \biggr) ^{2} \biggr\} ^{\frac{1}{2 p _{j _{2}} }} \prod_{j=1} ^{m} \Biggl( \sum_{i=1} ^{n} a _{ji} ^{p _{j}} \Biggr) ^{\frac{1}{p _{j}}} . $$
(3.1)
It is simple to see that
$$\begin{aligned} \Biggl( \sum_{i=1} ^{n} \prod _{j=1} ^{m} a _{ji} \Biggr) ^{2} &=\sum_{i=1} ^{n} \sum _{k=1} ^{n} \prod _{l=1} ^{m} a _{lk} \prod _{j=1} ^{m} a _{ji} \\ &= \Biggl\{ \sum_{i=1} ^{n} \sum _{k=1} ^{n} \prod_{l=1} ^{m} a _{lk} \prod_{j=1} ^{m} a _{ji} \Biggr\} ^{1- \Lambda} \Biggl\{ \sum _{i=1} ^{n} \sum _{k=1} ^{n} ( 1-e _{i} +e _{k} ) \prod_{l=1} ^{m} a _{lk} \prod_{j=1} ^{m} a _{ji} \Biggr\} ^{\Lambda} \\ &\leq\sum_{i=1} ^{n} \sum _{k=1} ^{n} \Biggl[ \Biggl\{ \prod _{l=1} ^{m} a _{lk} \prod _{j=1} ^{m} a _{ji} \Biggr\} ^{1- \Lambda} \Biggl\{ ( 1-e _{i} +e _{k} ) \prod _{l=1} ^{m} a _{lk} \prod _{j=1} ^{m} a _{ji} \Biggr\} ^{\Lambda} \Biggr] \\ &=\sum_{i=1} ^{n} \sum _{k=1} ^{n} \Biggl\{ ( 1-e _{i} +e _{k} )^{\Lambda} \prod_{l=1} ^{m} a _{lk} \prod_{j=1} ^{m} a _{ji} \Biggr\} \\ &=\sum_{i=1} ^{n} \sum _{k=1} ^{n} \prod_{l=1} ^{m} \bigl\{ a _{lk} ( 1-e _{i} +e _{k} )^{\frac{1}{p _{l}}} \bigr\} \prod_{j=1} ^{m} a _{j i} \\ &\leq\sum_{i=1} ^{n} \prod _{l=1} ^{m} \Biggl[ \sum _{k=1} ^{n} \bigl\{ a _{l k} ^{p _{l}} ( 1-e _{i} +e _{k} ) \bigr\} \Biggr] ^{\frac{1}{p _{l}}}\prod_{j=1} ^{m} a _{j i} \end{aligned}$$
by applying Lemma 2.2 and Lemma 2.1. On the other hand, it is straightforward to see that
$$\begin{aligned}& \Biggl[ \sum_{k=1} ^{n} \bigl\{ a _{j _{1} k} ^{p _{j _{1}}} ( 1-e _{i} +e _{k} ) \bigr\} \Biggr] ^{\frac{1}{p _{j _{1}}}} \Biggl[ \sum_{k=1} ^{n} \bigl\{ a _{j _{2} k} ^{p _{j _{2}}} ( 1-e _{i} +e _{k} ) \bigr\} \Biggr] ^{\frac{1}{p _{j _{2}}}}a _{j _{1} i} a _{j _{2} i} \\ & \quad = \Biggl[ \sum_{k=1} ^{n} \bigl\{ a _{j _{1} k} ^{p _{j _{1}}} ( 1-e _{i} +e _{k} ) \bigr\} \Biggr] ^{\frac{1}{p _{j _{1}}}- \frac {1}{p _{j _{2}} } } \Biggl[ \sum_{k=1} ^{n} \bigl\{ a _{j _{1} k} ^{p _{j _{1}}} ( 1-e _{i} +e _{k} ) \bigr\} \Biggr] ^{\frac{1}{p _{j _{2}}}} \\ & \qquad {}\times\Biggl[ \sum _{k=1} ^{n} \bigl\{ a _{j _{2} k} ^{p _{j _{2}}} ( 1-e _{i} +e _{k} ) \bigr\} \Biggr] ^{\frac{1}{p _{j _{2}}}}a _{j _{1} i} a _{j _{2} i} \\ & \quad = \Biggl( \sum_{k=1} ^{n} A _{j _{1} j _{1}} \Biggr) ^{\frac{1}{p _{j _{1}}} - \frac{1}{p _{j _{2}}} } \Biggl( \sum _{k=1} ^{n} A _{j _{2} j _{1}} \Biggr) ^{\frac{1}{p _{j _{2}}}} \Biggl( \sum_{k=1} ^{n} A _{j _{1} j _{2}} \Biggr) ^{\frac{1}{p _{j _{2}}}} \end{aligned}$$
and
$$ \Biggl[ \sum_{k=1} ^{n} \bigl\{ a _{j k} ^{p _{j}} ( 1-e _{i} +e _{k} ) \bigr\} \Biggr] ^{\frac{1}{p _{j}}}a _{ji}= \Biggl( \sum _{k=1} ^{n} A _{jj} \Biggr) ^{\frac{1}{p _{j}}} , $$
where \(A _{j l} =A _{j l} ( i, k ) =a _{j k} ^{p _{j}} a _{l i} ^{p _{l}} ( 1-e _{i} +e _{k} )\) (\(j,l=1,2,3,\ldots,m\)). Hence
$$\begin{aligned} \begin{aligned} &\sum_{i=1} ^{n} \prod _{l=1} ^{m} \Biggl[ \sum _{k=1} ^{n} \bigl\{ a _{l k} ^{p _{l}} ( 1-e _{i} +e _{k} ) \bigr\} \Biggr] ^{\frac{1}{p _{l}}}\prod_{j=1} ^{m} a _{j i} \\ &\quad = \sum_{i=1} ^{n} \Biggl\{ \Biggl( \sum_{k=1} ^{n} A _{j _{1} j _{1}} \Biggr) ^{ \frac{1}{p _{j _{1}}} - \frac{1}{p _{j _{2}}} } \Biggl( \sum_{k=1} ^{n} A _{j _{2} j _{1}} \Biggr) ^{\frac{1}{p _{j _{2}}}} \Biggl( \sum _{k=1} ^{n} A _{j _{1} j _{2}} \Biggr) ^{\frac{1}{p _{j _{2}}}} \prod_{ j \ne j _{1}, j _{2}} \Biggl(\sum _{k=1} ^{n} A _{jj} \Biggr) ^{\frac{1}{p _{j}}} \Biggr\} \\ &\quad \leq \Biggl( \sum_{i=1} ^{n} \sum _{k=1} ^{n} A _{j _{1} j _{1}} \Biggr) ^{ \frac{1}{p _{j _{1}}} - \frac{1}{p _{j _{2}}} } \Biggl( \sum_{i=1} ^{n} \sum_{k=1} ^{n} A _{j _{2} j _{1}} \Biggr) ^{\frac{1}{p _{j _{2}}}} \Biggl( \sum_{i=1} ^{n} \sum_{k=1} ^{n} A _{j _{1} j _{2}} \Biggr) ^{\frac{1}{p _{j _{2}}}} \prod_{ j \ne j _{1} ,j _{2}} \Biggl( \sum _{i=1} ^{n} \sum_{k=1} ^{n} A _{j j} \Biggr) ^{\frac{1}{p _{j}}} \end{aligned} \end{aligned}$$
by applying Lemma 2.1 again. On the other hand, it is also straightforward to see that
$$\begin{aligned}& \sum_{i=1} ^{n} \sum _{k=1} ^{n} A _{j _{2} j _{1}} \sum _{i=1} ^{n} \sum_{k=1} ^{n} A _{j _{1} j _{2}} \\& \quad = \Biggl( \sum_{i=1} ^{n} \sum_{k=1} ^{n} a _{j _{2} i} ^{p _{j _{2}}} a _{j _{1} k} ^{p _{j _{1}}} - \sum _{i=1} ^{n} \sum _{k=1} ^{n} a _{j _{2} i } ^{p _{j _{2}}} a _{j _{1} k} ^{p _{j _{1}}} e _{i} + \sum _{i=1} ^{n} \sum_{k=1} ^{n} a _{j _{2} i} ^{p _{j _{2}}} a _{j _{1} k} ^{p _{j _{1}}} e _{k} \Biggr) \\& \qquad {}\times \Biggl( \sum_{i=1} ^{n} \sum _{k=1} ^{n} a _{j _{1} i} ^{p _{j _{1}}} a _{j _{2} k} ^{p _{j _{2}}} - \sum _{i=1} ^{n} \sum_{k=1} ^{n} a _{j _{1} i} ^{p _{j _{1}}} a _{j _{2} k} ^{p _{j _{2}}} e _{i} + \sum_{i=1} ^{n} \sum_{k=1} ^{n} a _{j _{1} i} ^{p _{j _{1}}} a _{j _{2} k} ^{p _{j _{2}}} e _{k} \Biggr) \\& \quad = \Biggl( \sum_{i=1} ^{n} \sum _{k=1} ^{n} a _{j _{1} i} ^{p _{j _{1}}} a _{j _{2} k} ^{p _{j _{2}}} \Biggr) ^{2} - \Biggl( \sum _{i=1} ^{n} \sum_{k=1} ^{n} a _{j _{1} i} ^{p _{j _{1}}} a _{j _{2} k} ^{p _{j _{2}}} e _{i} - \sum_{i=1} ^{n} \sum_{k=1} ^{n} a _{j _{1} i} ^{p _{j _{1}}} a _{j _{2} k} ^{p _{j _{2}}} e _{k} \Biggr) ^{2} \\& \quad = \Biggl( \sum_{i=1} ^{n} a _{j _{1} i} ^{p _{j _{1}}} \Biggr) ^{2} \Biggl( \sum _{k=1} ^{n} a _{j _{2} k} ^{p _{j _{2}}} \Biggr) ^{2} \biggl\{ 1- \biggl( \frac{\sum_{i=1} ^{n} a _{j _{1} i} ^{p _{j _{1}}} e _{i}}{\sum_{i=1} ^{n} a _{j _{1} i} ^{p _{j _{1}}}} - \frac{\sum_{k=1} ^{n} a _{j _{2} } ^{p _{j _{2}}} e _{k}}{\sum_{k=1} ^{n} a _{j _{2} k} ^{p _{j _{2}}}} \biggr) ^{2} \biggr\} \end{aligned}$$
and, for \(j=1, 2, 3, \ldots, m\), that
$$ \sum_{i=1} ^{n} \sum _{k=1} ^{n} A _{j j} = \sum _{i=1} ^{n} \sum_{k=1} ^{n} a _{j i} ^{p _{j}} a _{j k} ^{p _{j}} - \sum_{i=1} ^{n} \sum _{k=1} ^{n} a _{j i} ^{p _{j}} a _{j k} ^{p _{j}} e _{i} + \sum _{i=1} ^{n} \sum _{k=1} ^{n} a _{j i} ^{p _{j}} a _{j k} ^{p _{j}} e _{k} = \Biggl( \sum _{i=1} ^{n} a _{j i} ^{p _{j}} \Biggr) ^{2}. $$
Hence,
$$\begin{aligned}& \Biggl( \sum_{i=1} ^{n} \sum _{k=1} ^{n} A _{j _{1} j _{1}} \Biggr) ^{ \frac{1}{p _{j _{1}}} - \frac{1}{p _{j _{2}}} } \Biggl( \sum_{i=1} ^{n} \sum_{k=1} ^{n} A _{j _{2} j _{1}} \Biggr) ^{\frac{1}{p _{j _{2}}}} \Biggl( \sum_{i=1} ^{n} \sum_{k=1} ^{n} A _{j _{1} j _{2}} \Biggr) ^{\frac{1}{p _{j _{2}}}} \\& \quad = \Biggl( \sum_{i=1} ^{n} a _{j _{1} i} ^{p _{j _{1}}} \Biggr) ^{ \frac {2}{p _{j _{1}}} - \frac{2}{p _{j _{2}}} } \Biggl( \sum _{i=1} ^{n} a _{j _{1} i} ^{p _{j _{1}}} \Biggr) ^{\frac{2}{p _{j _{2}}}} \Biggl( \sum_{i=1} ^{n} a _{j _{2} i} ^{p _{j _{2}}} \Biggr) ^{\frac{2}{p _{j _{2}}}} \\& \qquad {}\times\biggl\{ 1- \biggl( \frac{\sum_{i=1} ^{n} a _{j _{1} i} ^{p _{j _{1}}} e _{i}}{\sum_{i=1} ^{n} a _{j _{1} i} ^{p _{j _{1}}}} - \frac {\sum_{k=1} ^{n} a _{j _{2} i} ^{p _{j _{2}}} e _{i}}{\sum_{k=1} ^{n} a _{j _{2} i} ^{p _{j _{2}}}} \biggr) ^{2} \biggr\} ^{\frac{1}{p _{j _{2}}}} \\& \quad = \Biggl( \sum_{i=1} ^{n} a _{j _{1} i} ^{p _{j _{1}}} \Biggr) ^{\frac {2}{p _{j _{1}}}} \Biggl( \sum _{i=1} ^{n} a _{j _{2} i} ^{p _{j _{2}}} \Biggr) ^{\frac{2}{p _{j _{2}}}} \biggl\{ 1- \biggl( \frac{\sum_{i=1} ^{n} a _{j _{1} i} ^{p _{j _{1}}} e _{i}}{\sum_{i=1} ^{n} a _{j _{1} i} ^{p _{j _{1}}}} - \frac{\sum_{k=1} ^{n} a _{j _{2} i} ^{p _{j _{2}}} e _{i}}{\sum_{k=1} ^{n} a _{j _{2} i} ^{p _{j _{2}}}} \biggr) ^{2} \biggr\} ^{\frac{1}{p _{j _{2}}}} \end{aligned}$$
and
$$ \prod_{j \ne j _{1} , j _{2}} \Biggl( \sum _{i=1} ^{n} \sum_{k=1} ^{n} A _{j j} \Biggr) ^{\frac{1}{p _{j}}}=\prod _{ j\ne j _{1} ,j _{2}} \Biggl( \sum_{i=1} ^{n} a _{j i} ^{p _{j}} \Biggr) ^{\frac{2}{p _{j}}}. $$
Therefore (3.1) follows.
Case (2): \(0<\Lambda \leq1\). We show
$$ \sum_{i=1} ^{n} \prod _{j=1} ^{m} a _{j i} \leq n ^{ 1- \Lambda} \biggl\{ 1- \biggl( \frac{\sum_{i=1} ^{n} a _{j _{1} i} ^{p _{j _{1}}} e _{i}}{\sum_{i=1} ^{n} a _{j _{1} i} ^{p _{j _{1}}}} - \frac{\sum_{k=1} ^{n} a _{j _{2} i} ^{p _{j _{2}}} e _{i}}{\sum_{k=1} ^{n} a _{j _{2} i} ^{p _{j _{2}}}} \biggr) ^{2} \biggr\} ^{\frac{1}{2 p _{j _{2}} }} \prod _{j=1} ^{m} \Biggl( \sum _{i=1} ^{n} a _{ji} ^{p _{j}} \Biggr) ^{\frac{1}{p _{j}}} . $$
(3.2)
Since \(\sum_{j=1} ^{m} \frac{1}{p _{j} \Lambda} =1\), we have
$$\begin{aligned} \Biggl( \sum_{i=1} ^{n} \prod _{j=1} ^{m} a _{j i} \Biggr) ^{2} &=\sum_{i=1} ^{n}\sum _{k=1} ^{n} \Biggl\{ ( 1-e _{i} +e _{k} )^{\sum_{l=1}^{m}\frac{1}{p_{l}\Lambda}} \prod_{l=1} ^{m} a _{l k} \prod_{j=1}^{m} a _{j i} \Biggr\} \\ &=\sum_{i=1} ^{n}\sum _{k=1} ^{n}\prod_{l=1} ^{m} \bigl\{ a _{l k} ( 1-e _{i} +e _{k} )^{\frac{1}{p _{l}\Lambda}} \bigr\} \prod_{j=1}^{m} a _{j i} \\ &\leq\sum_{i=1} ^{n}\prod _{l=1} ^{m} \Biggl[ \sum _{k=1} ^{n} \bigl\{ a _{l k} ^{p _{l} \Lambda} ( 1-e _{i} +e _{k} ) \bigr\} \Biggr] ^{\frac{1}{p _{l} \Lambda}} \prod_{j=1} ^{m} a _{j i} \end{aligned}$$
by applying Lemma 2.1. On the other hand, it is straightforward to see that
$$\begin{aligned}& \Biggl[ \sum_{k=1} ^{n} \bigl\{ a _{j _{1} k} ^{p _{j _{1}} \Lambda} ( 1-e _{i} +e _{k} ) \bigr\} \Biggr] ^{\frac{1}{p _{j _{1}} \Lambda}} \Biggl[ \sum_{k=1} ^{n} \bigl\{ a _{j _{2} k} ^{p _{j _{2}} \Lambda} ( 1-e _{i} +e _{k} ) \bigr\} \Biggr] ^{\frac{1}{p _{j _{2}} \Lambda}}a _{j _{1} i} a _{j _{2} i} \\ & \quad = \Biggl( \sum_{k=1} ^{n} B _{j _{1} j _{1}} \Biggr) ^{ \frac{1}{p _{j _{1}}\Lambda} - \frac{1}{p _{j _{2}} \Lambda} } \Biggl( \sum _{k=1} ^{n} B _{j _{2} j _{1}} \Biggr) ^{\frac{1}{p _{j _{2}} \Lambda}} \Biggl(\sum_{k=1} ^{n} B _{j _{1} j _{2}} \Biggr) ^{\frac{1}{p _{j _{2}} \Lambda}} \end{aligned}$$
and
$$ \Biggl[\sum_{k=1} ^{n} \bigl\{ a _{jk} ^{p _{j} \Lambda} ( 1-e _{i} +e _{k} ) \bigr\} \Biggr] ^{\frac{1}{p _{j} \Lambda}}a _{j i} = \Biggl( \sum _{k=1} ^{n} B _{jj} \Biggr) ^{\frac{1}{p _{j}\Lambda}}, $$
where \(B _{jl} =B _{jl} ( i,k ) =a _{jk} ^{p _{j} \Lambda} a _{li} ^{p _{l} \Lambda} ( 1-e _{i} +e _{k} )\) (\(j,l=1,2,3,\ldots,m\)). Hence
$$\begin{aligned}& \sum_{i=1} ^{n}\prod _{l=1} ^{m} \Biggl[ \sum _{k=1} ^{n} \bigl\{ a _{l k} ^{p _{l} \Lambda} ( 1-e _{i} +e _{k} ) \bigr\} \Biggr] ^{\frac{1}{p _{l} \Lambda}} \prod_{j=1} ^{m} a _{j i} \\ & \quad = \sum_{i=1} ^{n} \Biggl\{ \Biggl( \sum_{k=1} ^{n} B _{j _{1} j _{1}} \Biggr) ^{ \frac{1}{p _{j _{1}} \Lambda} - \frac{1}{p _{j _{2}} \Lambda} } \Biggl( \sum_{k=1} ^{n} B _{j _{2} j _{1}} \Biggr) ^{\frac {1}{p _{j _{2}}\Lambda}} \Biggl( \sum _{k=1} ^{n} B _{j _{1} j _{2}} \Biggr) ^{\frac{1}{p _{j _{2}}\Lambda}} \prod_{j \ne j _{1} ,j _{2}} \Biggl(\sum _{k=1} ^{n} B _{jj} \Biggr) ^{\frac{1}{p _{j} \Lambda}} \Biggr\} \\ & \quad \leq \Biggl( \sum_{i=1} ^{n} \sum _{k=1} ^{n} B _{j _{1} j _{1}} \Biggr) ^{ \frac{1}{p _{j _{1}} \Lambda} - \frac{1}{p _{j _{2}}\Lambda }} \Biggl(\sum_{i=1} ^{n} \sum _{k=1} ^{n} B _{j _{2} j _{1}} \Biggr) ^{\frac{1}{p _{j _{2}} \Lambda}} \\ & \qquad {}\times \Biggl( \sum_{i=1} ^{n} \sum_{k=1} ^{n} B _{j _{1} j _{2}} \Biggr) ^{\frac{1}{p _{j _{2}} \Lambda}} \prod_{j \ne j _{1} ,j _{2}} \Biggl( \sum _{i=1} ^{n} \sum_{k=1} ^{n} B _{jj} \Biggr) ^{\frac{1}{p _{j} \Lambda}} \end{aligned}$$
by applying Lemma 2.1. Applying Lemma 2.1 again, we have
$$\begin{aligned} \begin{aligned} \Biggl(\sum_{i=1} ^{n} \sum _{k=1} ^{n} B _{ts} \Biggr) ^{\frac{1}{p _{h}\Lambda}} &= \Biggl[ \sum_{i=1} ^{n} \sum_{k=1} ^{n} \bigl\{ a _{ti} ^{p _{t}\Lambda} a _{sk} ^{p _{s} \Lambda} ( 1-e _{i} +e _{k} ) \bigr\} \Biggr] ^{\frac{1}{p _{h} \Lambda}} \\ &\leq \Biggl[ \sum_{i=1} ^{n} \sum _{k=1} ^{n} \bigl\{ a _{ti} ^{p _{t}} a _{sk} ^{p _{s}} ( 1-e _{i} +e _{k} ) \bigr\} \Biggr] ^{\frac{\Lambda}{p _{h} \Lambda}} \Biggl\{ \sum _{i=1} ^{n} \sum_{k=1} ^{n} ( 1-e _{i} +e _{k} ) \Biggr\} ^{\frac{1-\Lambda}{p _{h} \Lambda}} \\ &= \Biggl( \sum_{i=1} ^{n} \sum _{k=1} ^{n} A _{ts} \Biggr) ^{\frac{1}{p _{h}}} \Biggl\{ \sum_{i=1} ^{n} \sum_{k=1} ^{n} ( 1-e _{i} +e _{k} ) \Biggr\} ^{\frac{1-\Lambda}{p _{h}\Lambda}} \end{aligned} \end{aligned}$$
and
$$ \Biggl( \sum_{i=1} ^{n} \sum _{k=1} ^{n} B _{t t} \Biggr) ^{\frac{1}{p _{t} \Lambda} - \frac{1}{p _{s} \Lambda} } \leq \Biggl( \sum_{i=1} ^{n} \sum_{k=1} ^{n} A _{t t} \Biggr) ^{ \frac {1}{p _{t}} - \frac{1}{p _{s}} } \Biggl\{ \sum _{i=1} ^{n} \sum_{k=1} ^{n} ( 1-e _{i} +e _{k} ) \Biggr\} ^{ \frac{1-\Lambda}{p _{t} \Lambda} - \frac{1-\Lambda}{p _{s} \Lambda} }. $$
Hence
$$\begin{aligned}& \Biggl( \sum_{i=1} ^{n} \sum _{k=1} ^{n} B _{j _{1} j _{1}} \Biggr) ^{\frac{1}{p _{j _{1}} \Lambda} - \frac{1}{p _{j _{2}}\Lambda} } \Biggl( \sum_{i=1} ^{n} \sum_{k=1} ^{n} B _{j _{2} j _{1}} \Biggr) ^{\frac {1}{p _{j _{2}} \Lambda}} \\& \qquad {}\times\Biggl( \sum_{i=1} ^{n} \sum_{k=1} ^{n} B _{j _{1} j _{2}} \Biggr) ^{\frac{1}{p _{j _{2}} \Lambda}} \prod_{j \ne j _{1} ,j _{2}} \Biggl( \sum _{i=1} ^{n} \sum_{k=1} ^{n} B _{jj} \Biggr) ^{\frac{1}{p _{j} \Lambda}} \\& \quad \leq n ^{2(1-\Lambda)} \Biggl( \sum_{i=1} ^{n} \sum_{k=1} ^{n} A _{j _{1} j _{1}} \Biggr) ^{\frac{1}{p _{j _{1}}} - \frac{1}{p _{j _{2}}} } \Biggl( \sum _{i=1} ^{n} \sum_{k=1} ^{n} A _{j _{2} j _{1}} \Biggr) ^{\frac{1}{p _{j _{2}}}} \\& \qquad {}\times\Biggl( \sum _{i=1} ^{n} \sum_{k=1} ^{n} A _{j _{1} j _{2}} \Biggr) ^{\frac{1}{p _{j _{2}}}} \prod _{j \ne j _{1} ,j _{2}} \Biggl( \sum_{i=1} ^{n} \sum_{k=1} ^{n} A _{jj} \Biggr) ^{\frac{1}{p _{j}}}. \end{aligned}$$
By the same process as in the proof of Case (1),
$$\begin{aligned}& \Biggl( \sum_{i=1} ^{n} \sum _{k=1} ^{n} A _{j _{1} j _{1}} \Biggr) ^{\frac{1}{p _{j _{1}}} - \frac{1}{p _{j _{2}}} } \Biggl( \sum_{i=1} ^{n} \sum_{k=1} ^{n} A _{j _{2} j _{1}} \Biggr) ^{\frac{1}{p _{j _{2}}}} \Biggl( \sum_{i=1} ^{n} \sum_{k=1} ^{n} A _{j _{1} j _{2}} \Biggr) ^{\frac{1}{p _{j _{2}}}} \prod_{j \ne j _{1} ,j _{2}} \Biggl( \sum _{i=1} ^{n} \sum_{k=1} ^{n} A _{jj} \Biggr) ^{\frac{1}{p _{j}}} \\& \quad = \biggl\{ 1- \biggl( \frac{\sum_{i=1} ^{n} a _{j _{1} i} ^{p _{j _{1}}} e _{i}}{\sum_{i=1} ^{n} a _{j _{1} i} ^{p _{j _{1}}}} - \frac{\sum_{k=1} ^{n} a _{j _{2} i} ^{p _{j _{2}}} e _{i}}{\sum_{k=1} ^{n} a _{j _{2} i} ^{p _{j _{2}}}} \biggr) ^{2} \biggr\} ^{\frac{1}{p _{j _{2}}}} \prod_{j=1} ^{m} \Biggl( \sum_{i=1} ^{n} a _{ji} ^{p _{j}} \Biggr) ^{\frac{2}{p _{j}}} . \end{aligned}$$
Therefore (3.2) follows. The proof is complete.

4 Proof of Theorem 1.2

We may suppose \(p _{j _{2}} \leq p _{j _{1}}\) and let \(\sum_{j=1} ^{m} \frac{1}{p _{j}}=\Lambda\). We show
$$ \sum_{i=1} ^{n} \prod _{j=1} ^{m} a _{j i} \geq \biggl\{ 1- \biggl( \frac{\sum_{i=1} ^{n} a _{j _{1} i} ^{p _{j _{1}}} e _{i}}{\sum_{i=1} ^{n} a _{j _{1} i} ^{p _{j _{1}}}} - \frac{\sum_{k=1} ^{n} a _{j _{2} i} ^{p _{j _{2}}} e _{i}}{\sum_{k=1} ^{n} a _{j _{2} i} ^{p _{j _{1}}}} \biggr) ^{2} \biggr\} ^{\frac{1}{2 p _{j _{2}} }} \prod_{j=1} ^{m} \Biggl( \sum_{i=1} ^{n} a _{ji} ^{p _{j}} \Biggr) ^{\frac {1}{p _{j}}}. $$
(4.1)
It is simple to see that
$$\begin{aligned} \Biggl( \sum_{i=1} ^{n} \prod _{j=1} ^{m} a _{ji} \Biggr) ^{2} &= \Biggl\{ \sum_{i=1} ^{n} \sum_{k=1} ^{n} \prod _{l=1} ^{m} a _{lk} \prod _{j=1} ^{m} a _{ji} \Biggr\} ^{1- \Lambda} \Biggl\{ \sum_{i=1} ^{n} \sum_{k=1} ^{n} ( 1-e _{i} +e _{k} ) \prod_{l=1} ^{m} a _{lk} \prod_{j=1} ^{m} a _{ji} \Biggr\} ^{\Lambda} \\ &\geq\sum_{i=1} ^{n} \sum _{k=1} ^{n} \Biggl[ \Biggl\{ \prod _{l=1} ^{m} a _{lk} \prod _{j=1} ^{m} a _{ji} \Biggr\} ^{1- \Lambda} \Biggl\{ ( 1-e _{i} +e _{k} ) \prod _{l=1} ^{m} a _{lk} \prod _{j=1} ^{m} a _{ji} \Biggr\} ^{\Lambda} \Biggr] \\ &=\sum_{i=1} ^{n} \sum _{k=1} ^{n} \Biggl\{ ( 1-e _{i} +e _{k} )^{\Lambda} \prod_{l=1} ^{m} a _{lk} \prod_{j=1} ^{m} a _{ji} \Biggr\} \\ &=\sum_{i=1} ^{n} \sum _{k=1} ^{n} \prod_{l=1} ^{m} \bigl\{ a _{lk} ( 1-e _{i} +e _{k} )^{\frac{1}{p _{l}}} \bigr\} \prod_{j=1} ^{m} a _{j i} \\ &\geq\sum_{i=1} ^{n} \prod _{l=1} ^{m} \Biggl[ \sum _{k=1} ^{n} \bigl\{ a _{l k} ^{p _{l}} ( 1-e _{i} +e _{k} ) \bigr\} \Biggr] ^{\frac{1}{p _{l}}}\prod_{j=1} ^{m} a _{j i} \end{aligned}$$
by applying Lemma 2.1 and Lemma 2.2. On the other hand, it is straightforward to see that
$$\begin{aligned}& \Biggl[ \sum_{k=1} ^{n} \bigl\{ a _{j _{1} k} ^{p _{j _{1}}} ( 1-e _{i} +e _{k} ) \bigr\} \Biggr] ^{\frac{1}{p _{j _{1}}}} \Biggl[ \sum_{k=1} ^{n} \bigl\{ a _{j _{2} k} ^{p _{j _{2}}} ( 1-e _{i} +e _{k} ) \bigr\} \Biggr] ^{\frac{1}{p _{j _{2}}}}a _{j _{1} i} a _{j _{2} i} \\& \quad = \Biggl( \sum_{k=1} ^{n} A _{j _{1} j _{1}} \Biggr) ^{\frac{1}{p _{j _{1}}} - \frac{1}{p _{j _{2}}} } \Biggl( \sum _{k=1} ^{n} A _{j _{2} j _{1}} \Biggr) ^{\frac{1}{p _{j _{2}}}} \Biggl( \sum_{k=1} ^{n} A _{j _{1} j _{2}} \Biggr) ^{\frac{1}{p _{j _{2}}}} \end{aligned}$$
and
$$ \Biggl[ \sum_{k=1} ^{n} \bigl\{ a _{j k} ^{p _{j}} ( 1-e _{i} +e _{k} ) \bigr\} \Biggr] ^{\frac{1}{p _{j}}}a _{ji}= \Biggl( \sum _{k=1} ^{n} A _{jj} \Biggr) ^{\frac{1}{p _{j}}} , $$
where \(A _{j l} =A _{j l} ( i, k ) =a _{j k} ^{p _{j}} a _{l i} ^{p _{l}} ( 1-e _{i} +e _{k} )\) (\(j,l=1,2,3,\ldots,m\)). Since only one of the numbers \(\frac{1}{p _{j _{1}}} - \frac{1}{p _{j _{2}}}\), \(\frac{1}{p _{j _{2}}}\), \(\frac{1}{p _{j}}\) (\(j \ne j _{1} ,j _{2}\)) is positive, by applying Lemma 2.2 again,
$$\begin{aligned}& \sum_{i=1} ^{n} \prod _{l=1} ^{m} \Biggl[ \sum _{k=1} ^{n} \bigl\{ a _{l k} ^{p _{l}} ( 1-e _{i} +e _{k} ) \bigr\} \Biggr] ^{\frac{1}{p _{l}}}\prod_{j=1} ^{m} a _{j i} \\& \quad = \sum_{i=1} ^{n} \Biggl\{ \Biggl( \sum_{k=1} ^{n} A _{j _{1} j _{1}} \Biggr) ^{ \frac{1}{p _{j _{1}}} - \frac{1}{p _{j _{2}}} } \Biggl( \sum_{k=1} ^{n} A _{j _{2} j _{1}} \Biggr) ^{\frac{1}{p _{j _{2}}}} \Biggl( \sum _{k=1} ^{n} A _{j _{1} j _{2}} \Biggr) ^{\frac{1}{p _{j _{2}}}} \prod_{j \ne j _{1} ,j _{2}} \Biggl(\sum _{k=1} ^{n} A _{jj} \Biggr) ^{\frac{1}{p _{j}}} \Biggr\} \\& \quad \geq \Biggl( \sum_{i=1} ^{n} \sum _{k=1} ^{n} A _{j _{1} j _{1}} \Biggr) ^{ \frac{1}{p _{j _{1}}} - \frac{1}{p _{j _{2}}} } \Biggl( \sum_{i=1} ^{n} \sum_{k=1} ^{n} A _{j _{2} j _{1}} \Biggr) ^{\frac{1}{p _{j _{2}}}} \\& \qquad {}\times \Biggl( \sum_{i=1} ^{n} \sum_{k=1} ^{n} A _{j _{1} j _{2}} \Biggr) ^{\frac{1}{p _{j _{2}}}}\prod_{j \ne j _{1} ,j _{2}} \Biggl( \sum _{i=1} ^{n} \sum_{k=1} ^{n} A _{j j} \Biggr) ^{\frac{1}{p _{j}}}. \end{aligned}$$
On the other hand, it is also straightforward to see that
$$\begin{aligned}& \sum_{i=1} ^{n} \sum _{k=1} ^{n} A _{j _{2} j _{1}} \sum _{i=1} ^{n} \sum_{k=1} ^{n} A _{j _{1} j _{2}} \\& \quad = \Biggl( \sum_{i=1} ^{n} a _{j _{1} i} ^{p _{j _{1}}} \Biggr) ^{2} \Biggl( \sum_{k=1} ^{n} a _{j _{2} k} ^{p _{j _{2}}} \Biggr) ^{2} \biggl\{ 1- \biggl( \frac{\sum_{i=1} ^{n} a _{j _{1} i} ^{p _{j _{1}}} e _{i}}{\sum_{i=1} ^{n} a _{j _{1} i} ^{p _{j _{1}}}} - \frac{\sum_{k=1} ^{n} a _{j _{2} } ^{p _{j _{2}}} e _{k}}{\sum_{k=1} ^{n} a _{j _{2} k} ^{p _{j _{2}}}} \biggr) ^{2} \biggr\} \end{aligned}$$
and, for \(j=1, 2, 3, \ldots, m\), that
$$ \sum_{i=1} ^{n} \sum _{k=1} ^{n} A _{j j} = \Biggl( \sum _{i=1} ^{n} a _{j i} ^{p _{j}} \Biggr) ^{2}. $$
Hence,
$$\begin{aligned}& \Biggl( \sum_{i=1} ^{n} \sum _{k=1} ^{n} A _{j _{1} j _{1}} \Biggr) ^{\frac{1}{p _{j _{1}}} - \frac{1}{p _{j _{2}}} } \Biggl( \sum_{i=1} ^{n} \sum_{k=1} ^{n} A _{j _{2} j _{1}} \Biggr) ^{\frac{1}{p _{j _{2}}}} \\& \qquad {}\times\Biggl( \sum_{i=1} ^{n} \sum_{k=1} ^{n} A _{j _{1} j _{2}} \Biggr) ^{\frac{1}{p _{j _{2}}}} \prod_{j \ne j _{1} ,j _{2}} \Biggl( \sum _{i=1} ^{n} \sum_{k=1} ^{n} A _{jj} \Biggr) ^{\frac{1}{p _{j}}} \\& \quad = \Biggl( \sum_{i=1} ^{n} a _{j _{1} i} ^{p _{j _{1}}} \Biggr) ^{\frac {2}{p _{j _{1}}}} \Biggl( \sum _{i=1} ^{n} a _{j _{2} i} ^{p _{j _{2}}} \Biggr) ^{\frac{2}{p _{j _{2}}}} \\& \qquad {}\times \biggl\{ 1- \biggl( \frac{\sum_{i=1} ^{n} a _{j _{1} i} ^{p _{j _{1}}} e _{i}}{\sum_{i=1} ^{n} a _{j _{1} i} ^{p _{j _{1}}}} - \frac{\sum_{k=1} ^{n} a _{j _{2} i} ^{p _{j _{2}}} e _{i}}{\sum_{k=1} ^{n} a _{j _{2} i} ^{p _{j _{2}}}} \biggr) ^{2} \biggr\} ^{\frac{1}{p _{j _{2}}}} \prod _{j\ne j _{1} ,j _{2}} \Biggl( \sum_{i=1} ^{n} a _{j i} ^{p _{j}} \Biggr) ^{\frac{2}{p _{j}}}. \end{aligned}$$
Therefore (4.1) follows. The proof is complete.

5 Further observations

As a generalization of the reversed Hölder inequality, JF Tian obtained the following.

Theorem B

([6], Theorem 2.2)

Let \(a_{i}>0\), \(b_{i}>0\) (\(i=1,2, \ldots, n\)), \(1-e_{i}+e_{j}\geq0\) (\(i, j=1,2, \ldots, n\)), \(q<0\), and \(\frac {1}{p}+\frac{1}{q}\geq0\). Then
$$\begin{aligned} \sum_{i=1} ^{n} a _{i} b _{i} \geq& n ^{1-\max\{ \frac{1}{p} + \frac {1}{q} ,1 \} } \Biggl( \sum _{i=1} ^{n} a _{i} ^{p} \Biggr) ^{\frac{1}{p}} \Biggl( \sum_{i=1} ^{n} b _{i} ^{q} \Biggr) ^{\frac{1}{q}} \\ &{}\times\biggl\{ 1- \biggl( \frac{\sum_{i=1} ^{n} b _{i} ^{q} e _{i}}{ \sum_{i=1} ^{n} b _{i} ^{q} } - \frac{\sum_{i=1} ^{n} a_{i} b _{i} e _{i}}{\sum_{i=1} ^{n} a_{i} b _{i} } \biggr) ^{2} \biggr\} ^{\frac{\max \{ -1, \frac{1}{q} \}}{2}}. \end{aligned}$$

We extended Theorem B to multiple sequences as follows. The idea of the proof is essentially the same as that of Theorem B and is different from that of Theorem 1.2. Note that there is no implication between Theorem 1.2 and Theorem 5.1.

Theorem 5.1

Let \(a_{j i}>0\) (\(i=1, 2, \ldots, n\); \(j=1, 2, \ldots, m\)), \(1-e_{i}+e_{j}\geq0\) (\(i, j=1, 2, \ldots, n\)), \(p_{j}<0\) (\(j=1, 2, \ldots, m-1\)), \(\Lambda:=\sum_{j=1}^{m}\frac{1}{p_{j}} >0\), and \(\{ p_{j_{1}}\} \subset\{p _{1}, p _{2} , \ldots, p _{m-1}\}\). Then
$$\begin{aligned} \sum_{i=1} ^{n} \prod _{j=1}^{m} a _{j i} \geq& n ^{1-\max \{ \Lambda,1 \} } \biggl\{ 1- \biggl( \frac{\sum_{i=1} ^{n} a _{j_{1} i} ^{p_{j_{1}}} e _{i}}{\sum_{i=1} ^{n} a _{j_{1} i} ^{p_{j_{1}}} } - \frac{\sum_{i=1} ^{n} \prod_{j=1}^{m}a_{j i} e _{i}}{\sum_{i=1} ^{n} \prod_{j=1}^{m}a_{j i} } \biggr) ^{2} \biggr\} ^{\frac{1}{2 \min \{ -1, p_{j_{1}} \} }} \\ &{}\times\prod_{j=1}^{m} \Biggl( \sum_{i=1} ^{n} a_{j i}^{p_{j}} \Biggr) ^{\frac{1}{p_{j}}}. \end{aligned}$$

Proof

Case (1): \(-1 \leq p_{j_{1}} <0\). Since \(\sum_{j=1}^{m}\frac{1}{p_{j}} >0\), we have \(p_{m}>0\). Let \(q_{j}=-\frac{p_{j}}{p_{m}}>0\) (\(j=1, 2, \ldots, m-1\)), and \(q _{m} = \frac{{1}}{ {p _{m}}}>0\), then \(q _{m} \geq q _{j_{1}} >0\).

When \(0< \sum_{j=1} ^{m} \frac{1}{q _{j}} <1\), we obtain
$$\begin{aligned}& \sum_{i=1} ^{n} \Biggl( a _{1 i} ^{-p _{m}} \times \cdots \times a _{m-1 i} ^{-p _{m}} \times \prod_{j=1} ^{m} a _{ji} ^{p _{m}} \Biggr) \\& \quad \leq n ^{1-\min \{ \sum_{j=1} ^{m} \frac{1}{q _{j}} , 1 \}} \biggl\{ 1- \biggl( \frac{\sum_{i=1} ^{n} a _{j _{1} i} ^{p _{j _{1}}} e _{i}}{\sum_{i=1} ^{n} a _{j _{1} i} ^{p _{j _{1}}}} - \frac{\sum_{i=1} ^{n}\prod_{j=1} ^{m} a _{ji} e _{i}}{\sum_{i=1} ^{n} \prod_{j=1} ^{m} a _{j i}} \biggr) ^{2} \biggr\} ^{\frac{1}{2\max \{ q _{m} ,q _{j _{1}} \}}} \\& \qquad {}\times\Biggl\{ \prod_{j=1} ^{m-1} \Biggl( \sum _{i=1} ^{n} a _{j i} ^{p _{j}} \Biggr) ^{\frac{1}{q _{j}}} \Biggr\} \Biggl( \sum_{i=1} ^{n} \prod_{j=1} ^{m} a _{j i} \Biggr) ^{\frac{1}{q _{m}}} \end{aligned}$$
by using Theorem 1.1. That is,
$$\begin{aligned} \sum_{i=1} ^{n} a _{m i} ^{p _{m}} \leq& n ^{1- \sum_{j=1} ^{m} \frac{1}{q _{j}}} \biggl\{ 1- \biggl( \frac{\sum_{i=1} ^{n} a _{j _{1} i} ^{p _{j _{1}}} e _{i}}{\sum_{i=1} ^{n} a _{j _{1} i} ^{p _{j _{1}}}} - \frac{\sum_{i=1} ^{n} \prod_{j=1} ^{m} a _{j i} e _{i}}{\sum_{i=1} ^{n} \prod_{j=1} ^{m} a _{j i}} \biggr) ^{2} \biggr\} ^{\frac{p _{m}}{2}} \\ &{}\times \Biggl\{ \prod_{j=1} ^{m-1} \Biggl( \sum_{i=1} ^{n} a _{j i} ^{p _{j}} \Biggr) ^{- \frac{p _{m}}{p _{j}}} \Biggr\} \Biggl( \sum _{i=1} ^{n} \prod_{j=1} ^{m} a _{j i} \Biggr) ^{p _{m}}. \end{aligned}$$
Therefore
$$\begin{aligned} \Biggl( \sum_{i=1} ^{n} a _{m i} ^{p _{m}} \Biggr) ^{\frac{1}{p _{m}}} \leq& n ^{\sum_{j=1} ^{m} \frac{1}{p _{j}}-1} \biggl\{ 1- \biggl( \frac{\sum_{i=1} ^{n} a _{j _{1} i} ^{p _{j _{1}}} e _{i}}{\sum_{i=1} ^{n} a _{j _{1} i} ^{p _{j _{1}}}} - \frac{\sum_{i=1} ^{n} \prod_{j=1} ^{m} a _{j i} e _{i}}{\sum_{i=1} ^{n} \prod_{j=1} ^{m} a _{j i}} \biggr) ^{2} \biggr\} ^{\frac{1}{2}} \\ &{}\times \Biggl\{ \prod_{j=1} ^{m-1} \Biggl( \sum_{i=1} ^{n} a _{j i} ^{p _{j}} \Biggr) ^{- \frac{1}{p _{j}}} \Biggr\} \Biggl( \sum _{i=1} ^{n} \prod_{j=1} ^{m} a _{j i} \Biggr). \end{aligned}$$
Hence
$$\begin{aligned} \sum_{i=1} ^{n} \prod _{j=1} ^{m} a _{j i} \geq& n ^{1- \sum_{j=1} ^{m} \frac{1}{p _{j}}} \biggl\{ 1- \biggl( \frac{\sum_{i=1} ^{n} a _{j _{1} i} ^{p _{j _{1}}} e _{i}}{\sum_{i=1} ^{n} a _{j _{1} i} ^{p _{j _{1}}}} - \frac{\sum_{i=1} ^{n} \prod_{j=1} ^{m} a _{j i} e _{i}}{ \sum_{i=1} ^{n}\prod_{j=1} ^{m} a _{j i}} \biggr) ^{2} \biggr\} ^{- \frac{1}{2}} \\ &{}\times\prod _{j=1} ^{m} \Biggl( \sum _{i=1} ^{n} a _{ji} ^{p _{j}} \Biggr) ^{\frac{1}{p _{j}}} . \end{aligned}$$
From \(0< \sum_{j=1} ^{m} \frac{1}{q _{j}} <1\), we have \(\sum_{j=1} ^{m} \frac{1}{p _{j}} >1\). Consequently
$$\begin{aligned} \sum_{i=1} ^{n} \prod _{j=1}^{m} a _{j i} \geq& n ^{1-\max \{ \sum_{j=1} ^{m} \frac{1}{p_{j}} ,1 \} } \biggl\{ 1- \biggl( \frac{\sum_{i=1} ^{n} a _{j_{1} i} ^{p_{j_{1}}} e _{i}}{\sum_{i=1} ^{n} a _{j_{1} i} ^{p_{j_{1}}} } - \frac{\sum_{i=1} ^{n} \prod_{j=1}^{m}a_{j i} e _{i}}{ \sum_{i=1} ^{n} \prod_{j=1}^{m}a_{j i} } \biggr) ^{2} \biggr\} ^{\frac{1}{2 \min \{ -1, p_{j_{1}} \} }} \\ &{}\times\prod_{j=1}^{m} \Biggl( \sum_{i=1} ^{n} a_{j i}^{p_{j}} \Biggr) ^{\frac{1}{p_{j}}}. \end{aligned}$$
When \(\sum_{j=1} ^{m} \frac{1}{q _{j}} \geq1\), then \(\sum_{j=1} ^{m} \frac{1}{p _{j}} \leq1\), so that similarly we obtain
$$\begin{aligned} \sum_{i=1} ^{n} \prod _{j=1}^{m} a _{j i} \geq& \biggl\{ 1- \biggl( \frac{\sum_{i=1} ^{n} a _{j_{1} i} ^{p_{j_{1}}} e _{i}}{\sum_{i=1} ^{n} a _{j_{1} i} ^{p_{j_{1}}} } - \frac{\sum_{i=1} ^{n} \prod_{j=1}^{m}a_{j i} e _{i}}{\sum_{i=1} ^{n} \prod_{j=1}^{m}a_{j i} } \biggr) ^{2} \biggr\} ^{\frac{1}{2 \min \{ -1, p_{j_{1}} \} }}\prod_{j=1}^{m} \Biggl( \sum _{i=1} ^{n} a_{j i}^{p_{j}} \Biggr) ^{\frac{1}{p_{j}}} \\ =& n ^{1-\max \{ \sum_{j=1} ^{m} \frac{1}{p_{j}} ,1 \} } \biggl\{ 1- \biggl( \frac{\sum_{i=1} ^{n} a _{j_{1} i} ^{p_{j_{1}}} e _{i}}{\sum_{i=1} ^{n} a _{j_{1} i} ^{p_{j_{1}}} } - \frac{\sum_{i=1} ^{n} \prod_{j=1}^{m}a_{j i} e _{i}}{ \sum_{i=1} ^{n} \prod_{j=1}^{m}a_{j i} } \biggr) ^{2} \biggr\} ^{\frac{1}{2 \min \{ -1, p_{j_{1}} \} }} \\ &{}\times\prod _{j=1}^{m} \Biggl( \sum _{i=1} ^{n} a_{j i}^{p_{j}} \Biggr) ^{\frac{1}{p_{j}}}. \end{aligned}$$

Case (2): \(p _{j_{1}} < -1\). Since \(\sum_{j=1} ^{m} \frac{1}{p _{j}} >0\), we have \(p _{m}>0\). Let \(q _{j} =- \frac{p _{j}}{p _{m}} >0\) (\(j=1, 2, \ldots, m-1\)), and \(q _{m} = \frac{1}{p _{m}} >0\), then \(q _{j_{1}} > q _{m} >0\).

When \(0< \sum_{j=1} ^{m} \frac{1}{q _{j}} <1\), we obtain
$$\begin{aligned}& \sum_{i=1} ^{n} \Biggl( a _{1 i} ^{-p _{m}} \times a _{2 i} ^{-p _{m}} \times \cdots \times a _{m-1 i} ^{-p _{m}} \times \prod_{j=1} ^{m} a _{j i} ^{p _{m}} \Biggr) \\& \quad \leq n ^{1-\min \{ \sum_{j=1} ^{m} \frac{1}{q _{j}} ,1 \}} \biggl\{ 1- \biggl( \frac{\sum_{i=1} ^{n} a _{j _{1} i} ^{p _{j _{1}}} e _{i}}{\sum_{i=1} ^{n} a _{j _{1} i} ^{p _{j _{1}}}} - \frac {\sum_{i=1} ^{n} \prod_{j=1} ^{m} a _{j i} e _{i}}{\sum_{i=1} ^{n}\prod_{j=1} ^{m} a _{j i}} \biggr) ^{2} \biggr\} ^{\frac{1}{2\max \{ q _{m}, q _{j _{1}} \}}} \\& \qquad {}\times\Biggl\{ \prod_{j=1} ^{m-1} \Biggl( \sum _{i=1} ^{n} a _{j i} ^{p _{j}} \Biggr) ^{\frac{1}{q _{j}}} \Biggr\} \Biggl( \sum_{i=1} ^{n} \prod_{j=1} ^{m} a _{j i} \Biggr) ^{\frac{1}{q _{m}}} \end{aligned}$$
by using Theorem 1.1 again. That is,
$$\begin{aligned} \sum_{i=1} ^{n} a _{m i} ^{p _{m}} \leq& n ^{1- \sum_{j=1} ^{m} \frac{1}{q _{j}}} \biggl\{ 1- \biggl( \frac{\sum_{i=1} ^{n} a _{j _{1} i} ^{p _{j _{1}}} e _{i}}{\sum_{i=1} ^{n} a _{j _{1} i} ^{p _{j _{1}}}} - \frac{\sum_{i=1} ^{n} \prod_{j=1} ^{m} a _{j i} e _{i}}{ \sum_{i=1} ^{n} \prod_{j=1} ^{m} a _{j i}} \biggr) ^{2} \biggr\} ^{- \frac{p _{m}}{2p _{j _{1}}}} \\ &{}\times \Biggl\{ \prod_{j=1} ^{m-1} \Biggl( \sum_{i=1} ^{n} a _{j i} ^{p _{j}} \Biggr) ^{- \frac{p _{m}}{p _{j}}} \Biggr\} \Biggl( \sum _{i=1} ^{n} \prod_{j=1} ^{m} a _{j i} \Biggr) ^{p _{m}} . \end{aligned}$$
Hence
$$\sum_{i=1} ^{n} \prod _{j=1} ^{m} a _{j i} \geq n ^{1- \sum_{j=1} ^{m} \frac{1}{p _{j}}} \biggl\{ 1- \biggl( \frac{\sum_{i=1} ^{n} a _{j _{1} i} ^{p _{j _{1}}} e _{i}}{\sum_{i=1} ^{n} a _{j _{1} i} ^{p _{j _{1}}}} - \frac{\sum_{i=1} ^{n}\prod_{j=1} ^{m} a _{j i} e _{i}}{ \sum_{i=1} ^{n} \prod_{j=1} ^{m} a _{j i}} \biggr) ^{2} \biggr\} ^{\frac{1}{2p _{j _{1}}}} \prod _{j=1} ^{m} \Biggl( \sum _{i=1} ^{n} a _{j i} ^{p _{j}} \Biggr) ^{\frac{1}{p _{j}}} . $$
From \(0< \sum_{j=1} ^{m} \frac{1}{q _{j}} <1\), we have \(\sum_{j=1} ^{m} \frac{1}{p _{j}} >1\). Consequently
$$\begin{aligned} \sum_{i=1} ^{n} \prod _{j=1} ^{m} a _{j i} \geq& n ^{1-\max \{ \sum_{j=1} ^{m} \frac{1}{p _{j}}, 1 \}} \biggl\{ 1- \biggl( \frac {\sum_{i=1} ^{n} a _{j _{1} i} ^{p _{j _{1}}} e _{i}}{\sum_{i=1} ^{n} a _{j _{1} i} ^{p _{j _{1}}}} - \frac{\sum_{i=1} ^{n} \prod_{j=1} ^{m} a _{j i} e _{i}}{\sum_{i=1} ^{n} \prod_{j=1} ^{m} a _{j i}} \biggr) ^{2} \biggr\} ^{\frac{1}{2\min \{ -1, p _{j _{1}} \}}} \\ &{}\times\prod _{j=1} ^{m} \Biggl( \sum _{i=1} ^{n} a _{j i} ^{p _{j}} \Biggr) ^{\frac{1}{p _{j}}} . \end{aligned}$$
When \(\sum_{j=1} ^{m} \frac{1}{q _{j}} \geq1\), then \(\sum_{j=1} ^{m} \frac{1}{p _{j}} \leq1\), so that similarly we obtain
$$\begin{aligned} \sum_{i=1} ^{n} \prod _{j=1} ^{m} a _{j i} \geq& \biggl\{ 1- \biggl( \frac{\sum_{i=1} ^{n} a _{j _{1} i} ^{p _{j _{1}}} e _{i}}{\sum_{i=1} ^{n} a _{j _{1} i} ^{p _{j _{1}}}} - \frac {\sum_{i=1} ^{n} \prod_{j=1} ^{m} a _{j i} e _{i}}{\sum_{i=1} ^{n} \prod_{j=1} ^{m} a _{j i}} \biggr) ^{2} \biggr\} ^{\frac{1}{2\min \{ -1, p _{j _{1}} \}}} \prod_{j=1} ^{m} \Biggl( \sum_{i=1} ^{n} a _{j i} ^{p _{j}} \Biggr) ^{\frac{1}{p _{j}}} \\ = & n ^{1-\max \{ \sum_{j=1} ^{m} \frac{1}{p _{j}}, 1 \}} \biggl\{ 1- \biggl(\frac{\sum_{i=1} ^{n} a _{j _{1} i} ^{p _{j _{1}}} e _{i}}{\sum_{i=1} ^{n} a _{j _{1} i} ^{p _{j _{1}}}} - \frac {\sum_{i=1} ^{n} \prod_{j=1} ^{m} a _{j i} e _{i}}{\sum_{i=1} ^{n} \prod_{j=1} ^{m} a _{j i}} \biggr) ^{2} \biggr\} ^{\frac{1}{2\min \{ -1, p _{j _{1}} \}}} \\ &{}\times\prod _{j=1} ^{m} \Biggl( \sum _{i=1} ^{n} a _{j i} ^{p _{j}} \Biggr) ^{\frac{1}{p _{j}}}. \end{aligned}$$
The proof is complete. □

Declarations

Acknowledgements

This work was supported by NRF-2010-0021986.

Open Access This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/4.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly credited.

Authors’ Affiliations

(1)
Department of Mathematics Education, Andong National University, Andong, 760-749, Korea
(2)
Department of Mathematics, Graduate School, Andong National University, Andong, 760-749, Korea

References

  1. Hu, K: On an inequality and its applications. Sci. Sin. 24(8), 1047-1055 (1981) MATHGoogle Scholar
  2. Wu, S: A new sharpened and generalized version of Hölder’s inequality and its applications. Appl. Math. Comput. 197, 708-714 (2008) View ArticleMATHMathSciNetGoogle Scholar
  3. Wu, S: Generalization of a sharp Hölder’s inequality and its applications. J. Math. Anal. Appl. 332, 741-750 (2007) View ArticleMATHMathSciNetGoogle Scholar
  4. Beckenbach, EF, Bellman, R: Inequalities. Springer, Berlin (1983) Google Scholar
  5. Hardy, G, Littlewood, JE, Polya, G: Inequalities. Cambridge University Press, New York (1991) Google Scholar
  6. Tian, J-F: Extension of Hu Ke’s inequality and its applications. J. Inequal. Appl. 2011, 77 (2011) View ArticleGoogle Scholar

Copyright

Advertisement