On Improvements of Kantorovich Type Inequalities

Abstract

In the paper, we give some new improvements of the Kantorovich type inequalities by using Popoviciu’s, Hölder’s, Bellman’s and Minkowski’s inequalities. These results in special case yield Hao’s, reverse Cauchy’s and Minkowski’s inequalities.

1. Introduction

The Pólya–Szegö’s inequality can be stated as follows ([1] or ([2], p. 62)).

If $u_k$ and $v_k$ are non-negative real sequences, and $0<m_1\le u_k\le M_1,$ and $0<m_2\le v_k\le M_2$ for $k=1,2,\dots ,n$, then

(1)$\sum _{k=1}^n{u}_k^2\sum _{k=1}^n{v}_k^2\le \frac{{(M_1{M}_2+m_1{m}_2)}^2}{4m_1{m}_2{M}_1{M}_2}{\left(\sum _{k=1}^n{u}_k{v}_k\right)}^2.$

The Pólya–Szegö’s inequality was studied extensively and numerous variants, generalizations, and extensions appeared in the literature (see [3,4,5,6] and the references cited therein). The integral forms of Pólya–Szegö’s inequality were recently established in [7,8,9,10]. The weighted version of inequality (1) was proved in papers of Watson [11] and Greub and Rheinboldt [12]:

(2)$\sum _{k=1}^n{\omega }_k{u}_k^2·\sum _{k=1}^n{\omega }_k{v}_k^2\le \frac{{(M_1{M}_2+m_1{m}_2)}^2}{4m_1{m}_2{M}_1{M}_2}{\left(\sum _{k=1}^n{\omega }_k{u}_k{v}_k\right)}^2,$

where ${\omega }_k$ is a nonnegative n-tuple.

An interesting generalization of Kantorovich type inequality was given by Hao ([13], p. 122), so we shall give his result:

(3)${\left(\sum _{k=1}^n{\omega }_k{u}_k^2\right)}^{1/p}{\left(\sum _{k=1}^n{\omega }_k{v}_k^2\right)}^{1/q}\le \ell \left(\sum _{k=1}^n{\omega }_k{u}_k{v}_k\right),$

where $0<\frac{1}{q}\le \frac{1}{p}<1$ and $\frac{1}{p}+\frac{1}{q}=1,$ and

(4)$\ell =\frac{q{M}_1{M}_2+p{m}_1{m}_2}{pq{\left(m_1{M}_1\right)}^{1/q}{\left(m_2{M}_2\right)}^{1/p}}.$

We recall that, with the name “Kantorovich”, we also usually refer to some integral-type extension of classical inequalities, classical pointwise operators, and other mathematical tools—see, e.g., [14,15,16,17].

The first aim of this paper is to give a new improvement of the Kantorovich type inequality (3). We combine organically Popoviciu’s, Hölder’s, and Hao’s inequalities to derive a new inequality, which is a generalization of Label (3).

Corresponding to (3), we can obtain a reverse Minkowski’s inequality as follows:

(5)$\ell {\left(\sum _{k=1}^n{\omega }_k{(u_k+v_k)}^2\right)}^{1/p}\ge {\left(\sum _{k=1}^n{\omega }_k{u}_k^2\right)}^{1/p}+{\left(\sum _{k=1}^n{\omega }_k{v}_k^2\right)}^{1/p},$

where p, q, ${\omega }_k$, $u_k$, $v_k$ are as in (3), and ℓ is definied in (4).

Another aim of this paper is to give a new reverse Minkowski’s inequality. We combine organically Bellman’s and Minkowski’s inequalities to derive a new inequality, which is generalization of the reverse Minkowski’s inequality (5).

2. Results

We need the following Lemmas to prove our main results.

Lemma 1.

(Popoviciu’s inequality) ([18], p. 58) Let $p>0,q>0$, $\frac{1}{p}+\frac{1}{q}=1$, and $a=\{a_1,\dots ,a_n\}$ and $b=\{b_1,\dots ,b_n\}$ be two series of positive real numbers and such that $a_1^p-{\sum }_{i=2}^n{a}_i^p>0$ and $b_1^q-{\sum }_{i=2}^n{b}_i^q>0$. Then,

(6) ${\left(a_1^p-\sum _{i=2}^n{a}_i^p\right)}^{1/p}{\left(b_1^q-\sum _{i=2}^n{b}_i^q\right)}^{1/q}\le a_1{b}_1-\sum _{i=2}^n{a}_i{b}_i,$

with equality if and only if $a=\mu b$, where μ is a constant.

Lemma 2.

(Bellman’s inequality) ([19], p. 38) Let $a=\{a_1,\dots ,a_n\}$ and $b=\{b_1,\dots ,b_n\}$ be two series of positive real numbers and $p>1$ such that $a_1^p-{\sum }_{i=2}^n{a}_i^p>0$ and $b_1^p-{\sum }_{i=2}^n{b}_i^p>0,$ then

(7) ${\left(a_1^p-\sum _{i=2}^n{a}_i^p\right)}^{1/p}+{\left(b_1^p-\sum _{i=2}^n{b}_i^p\right)}^{1/p}\le {\left({(a_1+b_1)}^p-\sum _{i=2}^n{(a_i+b_i)}^p\right)}^{1/p},$

with equality if and only if $a=\upsilon b$, where υ is a constant.

Lemma 3.

(Hölder’s weighted inequality) ([13], p. 100) Let $p>0,q>0$, $\frac{1}{p}+\frac{1}{q}=1$, and $a_k,b_k$ and ${\omega }_k$ be non-negative real numbers, then

(8) $\sum _{k=1}^n{\omega }_k{a}_k{b}_k\le {\left(\sum _{k=1}^n{\omega }_k{a}_k^p\right)}^{1/p}{\left(\sum _{k=1}^n{\omega }_k{b}_k^q\right)}^{1/q}.$

Lemma 4.

Let $0<\frac{1}{q}\le \frac{1}{p}<1$ and $\frac{1}{p}+\frac{1}{q}=1.$ If $u_k$, $v\left(k\right)$ and ${\omega }_k$ are non-negative real sequences, and $0<m_1\le u_k\le M_1,$ and $0<m_2\le v_k\le M_2$ for $k=1,2,\dots ,n$, then

(9) $\ell {\left(\sum _{k=1}^n{\omega }_k{(u_k+v_k)}^2\right)}^{1/p}\ge {\left(\sum _{k=1}^n{\omega }_k{u}_k^2\right)}^{1/p}+{\left(\sum _{k=1}^n{\omega }_k{v}_k^2\right)}^{1/p},$

where ℓ is as in Label (4).

Proof. 

From (3), we have

$\begin{array}{ccc}\hfill \ell \sum _{k=1}^n{\omega }_k{(u_k+v_k)}^2& =& \ell \sum _{k=1}^n{\omega }_k{u}_k(u_k+v_k)+\ell \sum _{k=1}^n{\omega }_k{v}_k(u_k+v_k)\hfill \\ & \ge & {\left(\sum _{k=1}^n{\omega }_k{u}_k^2\right)}^{1/p}{\left(\sum _{k=1}^n{\omega }_k{(u_k+v_k)}^2\right)}^{1/q}+{\left(\sum _{k=1}^n{\omega }_k{v}_k^2\right)}^{1/p}{\left(\sum _{k=1}^n{\omega }_k{(u_k+v_k)}^2\right)}^{1/q}.\hfill \end{array}$

Hence,

$\ell {\left(\sum _{k=1}^n{\omega }_k{(u_k+v_k)}^2\right)}^{1/p}\ge {\left(\sum _{k=1}^n{\omega }_k{u}_k^2\right)}^{1/p}+{\left(\sum _{k=1}^n{\omega }_k{v}_k^2\right)}^{1/p}.$

This proof is complete. ☐

Our main results are given in the following theorems.

Theorem 1.

Let $m,n\in {\mathbb{N}}^{+}$, $0<\frac{1}{q}\le \frac{1}{p}<1$ and $\frac{1}{p}+\frac{1}{q}=1.$ Let $u_k,v_k$, $a_k$, $b_k$, ${\omega }_k$ and ${\mu }_k$ be non-negative real sequences such as ${\omega }_k{u}_k^2>m{\mu }_k{a}_k^p$ and ${\omega }_k{v}_k^2>m{\mu }_k{b}_k^q$, where $k=1,2,\dots ,n$. If $0<m_1\le u_k\le M_1$ and $0<m_2\le v_k\le M_2$, then

(10) $\sum _{k=1}^n\left(\ell {\omega }_k{u}_k{v}_k-m{\mu }_k{a}_k{b}_k\right)\ge {\left(\sum _{k=1}^n\left({\omega }_k{u}_k^2-m{\mu }_k{a}_k^p\right)\right)}^{1/p}{\left(\sum _{k=1}^n\left({\omega }_k{v}_k^2-m{\mu }_k{b}_k^q\right)\right)}^{1/q},$

where ℓ is as in (4).

Proof. 

Let’s prove this theorem by mathematical induction for m. First, we prove that (10) holds for $m=1$. From (3) and (8), we obtain

(11)$\ell \left(\sum _{k=1}^n{\omega }_k{u}_k{v}_k\right)\ge {\left(\sum _{k=1}^n{\omega }_k{u}_k^2\right)}^{1/p}{\left(\sum _{k=1}^n{\omega }_k{v}_k^2\right)}^{1/q},$

and

(12)$\left(\sum _{k=1}^n{\mu }_k{a}_k{b}_k\right)\le {\left(\sum _{k=1}^n{\mu }_k{a}_k^p\right)}^{1/p}{\left(\sum _{k=1}^n{\mu }_k{b}_k^q\right)}^{1/q}.$

From (11), (12) and, in view of the Popoviciu’s inequality, we have

$\begin{array}{ccc}\hfill \sum _{k=1}^n\left(\ell {\omega }_k{u}_k{v}_k-{\mu }_k{a}_k{b}_k\right)& \ge & {\left(\sum _{k=1}^n{\omega }_k{u}_k^2\right)}^{1/p}{\left(\sum _{k=1}^n{\omega }_k{v}_k^2\right)}^{1/q}-{\left(\sum _{k=1}^n{\mu }_k{a}_k^p\right)}^{1/p}{\left(\sum _{k=1}^n{\mu }_k{b}_k^q\right)}^{1/q}\hfill \\ & \ge & {\left(\sum _{k=1}^n({\omega }_k{u}_k^2-{\mu }_k{a}_k^p)\right)}^{1/p}{\left(\sum _{k=1}^n({\omega }_k{v}_k^2-{\mu }_k{b}_k^q)\right)}^{1/q}.\hfill \end{array}$

This shows (10) right for $m=1.$

Suppose that (10) holds when $m=r-1$; we have

(13)$\sum _{k=1}^n\left(\ell {\omega }_k{u}_k{v}_k-(r-1){\mu }_k{a}_k{b}_k\right)\ge {\left(\sum _{k=1}^n\left({\omega }_k{u}_k^2-(r-1){\mu }_k{a}_k^p\right)\right)}^{1/p}{\left(\sum _{k=1}^n\left({\omega }_k{v}_k^2-(r-1){\mu }_k{b}_k^q\right)\right)}^{1/q}.$

From (6), (12) and (13), we obtain

$\begin{array}{ccc}\hfill \sum _{k=1}^n\left(\ell {\omega }_k{u}_k{v}_k-r{\mu }_k{a}_k{b}_k\right)& \ge & {\left(\sum _{k=1}^n\left({\omega }_k{u}_k^2-(r-1){\mu }_k{a}_k^p\right)\right)}^{1/p}{\left(\sum _{k=1}^n\left({\omega }_k{v}_k^2-(r-1){\mu }_k{b}_k^q\right)\right)}^{1/q}\hfill \\ & -& {\left(\sum _{k=1}^n{\mu }_k{a}_k^p\right)}^{1/p}{\left(\sum _{k=1}^n{\mu }_k{b}_k^q\right)}^{1/q}\hfill \\ & \ge & {\left(\sum _{k=1}^n\left({\omega }_k{u}_k^2-r{\mu }_k{a}_k^p\right)\right)}^{1/p}{\left(\sum _{k=1}^n\left({\omega }_k{v}_k^2-r{\mu }_k{b}_k^q\right)\right)}^{1/q}.\hfill \end{array}$

This shows that (10) is correct if $m=r-1$, then $m=r$ is also correct. Hence, (10) is right for any $m\in {\mathbb{N}}^{+}$.

This proof is complete. ☐

Taking $m=1$ and ${\omega }_k={\mu }_k$ in Theorem 1, we have the following result.

Corollary 1.

Let $p,q,u_k,v_k,a_k$, $b_k$ and ${\omega }_k$ are as in Theorem 1, then

$\sum _{k=1}^n\left({\omega }_k(\ell u_k{v}_k-a_k{b}_k)\right)\ge {\left(\sum _{k=1}^n{\omega }_k(u_k^2-a_k^p)\right)}^{1/p}{\left(\sum _{k=1}^n{\omega }_k(v_k^2-b_k^q)\right)}^{1/q},$

where ℓ is as in (4).

Taking $m=1$, $p=q=2$ and ${\omega }_k={\mu }_k=1$ in Theorem 1, we have the following result.

Corollary 2.

Let $u_k,v_k,a_k$ and $b_k$ are as in Theorem 1, then

(14) $\sum _{k=1}^n\left(\frac{M_1{M}_2+m_1{m}_2}{2\sqrt{m_1{m}_2{M}_1{M}_2}}{u}_k{v}_k-a_k{b}_k\right)\ge {\left(\sum _{k=1}^n(u_k^2-a_k^2)\right)}^{1/2}{\left(\sum _{k=1}^n(v_k^2-b_k^2)\right)}^{1/2}.$

Taking for $a_k=0$ and $b_k=0$ in (14), we get the following interesting reverse Cauchy’s inequality.

$\frac{M_1{M}_2+m_1{m}_2}{2\sqrt{m_1{m}_2{M}_1{M}_2}}·\sum _{k=1}^n{u}_k{v}_k\ge {\left(\sum _{k=1}^n{u}_k^2\right)}^{1/2}{\left(\sum _{k=1}^n{v}_k^2\right)}^{1/2}.$

Theorem 2.

Let $m,n\in {\mathbb{N}}^{+}$, $0<\frac{1}{q}\le \frac{1}{p}<1$ and $\frac{1}{p}+\frac{1}{q}=1.$ Let $u_k,v_k$, $a_k$, $b_k$, ${\omega }_k$ and ${\mu }_k$ be non-negative real sequences such as ${\omega }_k{u}_k^2>m{a}_k^p$ and ${\omega }_k{v}_k>m{b}_k^q$, where $k=1,2,\dots ,n$. If $0<m_1\le u_k\le M_1$ and $0<m_2\le v_k\le M_2$, then

(15) ${\left(\sum _{k=1}^n\left({\ell }^p{\omega }_k{(u_k+v_k)}^2-m{(a_k+b_k)}^p\right)\right)}^{1/p}\ge {\left(\sum _{k=1}^n({\omega }_k{u}_k^2-m{a}_k^p)\right)}^{1/p}+{\left(\sum _{k=1}^n({\omega }_k{v}_k^2-m{b}_k^p)\right)}^{1/p},$

where ℓ is as in (4).

Proof. 

First, we prove that (15) holds for $m=1$. From (9) and in view of Minkowski’s inequality, it is easy to obtain

(16)$\ell {\left(\sum _{k=1}^n{\omega }_k{(u_k+v_k)}^2\right)}^{1/p}\ge {\left(\sum _{k=1}^n{\omega }_k{u}_k^2\right)}^{1/p}+{\left(\sum _{k=1}^n{\omega }_k{v}_k^2\right)}^{1/p},$

and

(17)${\left(\sum _{k=1}^n{(a_k+b_k)}^{p}dx\right)}^{1/p}\le {\left(\sum _{k=1}^n{a}_k^p\right)}^{1/p}+{\left(\sum _{k=1}^n{b}_k^p\right)}^{1/p}.$

From (16), (17) and the Bellman’s inequality, we have

$\begin{array}{l}{\left({\displaystyle \sum _{k=1}^n}\left({\ell }^p{\omega }_k{(u_k+v_k)}^2-{(a_k+b_k)}^p\right)\right)}^{1/p}\\ \ge {\left\{{\left[{\left({\displaystyle \sum _{k=1}^n}{\omega }_k{u}_k^2\right)}^{1/p}+{\left({\displaystyle \sum _{k=1}^n}{\omega }_k{v}_k^2\right)}^{1/p}\right]}^p-{\left[{\left({\displaystyle \sum _{k=1}^n}{a}_k^p\right)}^{1/p}+{\left({\displaystyle \sum _{k=1}^n}{b}_k^p\right)}^{1/p}\right]}^p\right\}}^{1/p}\\ \ge {\left({\displaystyle \sum _{k=1}^n}({\omega }_k{u}_k^2-a_k^p)\right)}^{1/p}+{\left({\displaystyle \sum _{k=1}^n}({\omega }_k{v}_k^2-b_k^p)\right)}^{1/p}.\end{array}$

This shows that (15) holds for $m=1$

Supposing that (15) holds when $m=r-1$, we have

(18)$\begin{array}{c}\hfill {\left(\sum _{k=1}^n\left({\ell }^p{\omega }_k{(u_k+v_k)}^2-(r-1){(a_k+b_k)}^2\right)\right)}^{1/p}\ge {\left(\sum _{k=1}^n({\omega }_k{u}_k^2-(r-1)a_k^p)\right)}^{1/p}\\ \hfill +{\left(\sum _{k=1}^n({\omega }_k{v}_k^2-(r-1)b_k^p)\right)}^{1/p}.\end{array}$

From (17), (18) and by using the Bellman’s inequality again, we obtain

$\begin{array}{ccc}\hfill {\left(\sum _{k=1}^n\left({\ell }^p{\omega }_k{(u_k+v_k)}^2-r{(a_k+b_k)}^2\right)\right)}^{1/p}& \ge & \left\{\right[{\left(\sum _{k=1}^n({\omega }_k{u}_k^2-(r-1)a_k^p)\right)}^{1/p}\hfill \\ & +& {\left(\sum _{k=1}^n({\omega }_k{v}_k^2-(r-1)b_k^p)\right)}^{1/p}{]}^p\hfill \\ & -& {\left[{\left(\sum _{k=1}^n{a}_k^p\right)}^{1/p}+{\left(\sum _{k=1}^n{b}_k^p\right)}^{1/p}\right]}^p{\}}^{1/p}\hfill \\ & \ge & {\left(\sum _{k=1}^n({\omega }_k{u}_k^2-r{a}_k^p)\right)}^{1/p}+{\left(\sum _{k=1}^n({\omega }_k{v}_k^2-r{b}_k^p)\right)}^{1/p}.\hfill \end{array}$

This shows that (15) is correct if $m=r-1$, then $m=r$ is also correct. Hence, (15) is right for any $m\in {\mathbb{N}}^{+}$.

This proof is complete. ☐

Taking for $m=1$, $p=2$ and ${\omega }_k=1$, we have the following result.

Corollary 3.

Let $u_k$, $v_k$, $a_k$, $b_k$, $m_1$, $m_2$, $M_1$, and $M_2$ be as in Theorem 2, then

${\left[\left(\sum _{k=1}^n\left(\hslash {(u_k+v_k)}^2-{(a_k+b_k)}^2\right)\right)\right]}^{1/2}\ge {\left(\sum _{k=1}^n(u_k^2-a_k^2)\right)}^{1/2}+{\left(\sum _{k=1}^n(v_k^2-b_k^2)\right)}^{1/2},$

where

$\hslash =\frac{{(M_1{M}_2+m_1{m}_2)}^2}{4m_1{m}_2{M}_1{M}_2}.$

Author Contributions

C.-J.Z. and W.-S.C. jointly contributed to the main results. All authors read and approved the nal manuscript.

Funding

Research is supported by National Natural Science Foundation of China (11371334, 10971205). Wing-Sum Cheung: Research is partially supported by the Research Grants Council of the Hong Kong SAR.

Acknowledgments

The authors express their thanks to the three referees for their excellent suggestions. The authors express their thanks to W. Li for his valuable help.

Conflicts of Interest

The authors declare no conflicts of interest.