 Next Article in Journal
Conformal and Geodesic Mappings onto Some Special Spaces
Next Article in Special Issue
A New Gronwall–Bellman Inequality in Frame of Generalized Proportional Fractional Derivative
Previous Article in Journal
Some New Inequalities for Dual Continuous g-Frames
Previous Article in Special Issue
Hermite-Hadamard Type Inequalities for Interval (h1, h2)-Convex Functions
Article

# Refinements of Majorization Inequality Involving Convex Functions via Taylor’s Theorem with Mean Value form of the Remainder

1
Department of Mathematics, Longyan University, Longyan 364012, China
2
Department of Mathematics, University of Peshawar, Peshawar 25000, Pakistan
*
Author to whom correspondence should be addressed.
Mathematics 2019, 7(8), 663; https://doi.org/10.3390/math7080663
Received: 16 June 2019 / Revised: 19 July 2019 / Accepted: 22 July 2019 / Published: 24 July 2019

## Abstract

The aim of this paper is to establish some refined versions of majorization inequality involving twice differentiable convex functions by using Taylor theorem with mean-value form of the remainder. Our results improve several results obtained in earlier literatures. As an application, the result is used for deriving a new fractional inequality.

## 1. Introduction

The notion of majorization was introduced in the celebrated monograph  by Hardy, Littlewood and Pólya, which was used as a measure of the diversity of the components of an n-dimensional vector.
Let $ν = ( ν 1 , ν 2 , … , ν n$) and $ϑ = ( ϑ 1 , ϑ 2 , … , ϑ n$) be two n-tuples. The n-tuple $ν$ is said to be majorized by $ϑ$ (in symbols $ν ≺ ϑ$) if $∑ i = 1 k ν [ i ] ≤ ∑ i = 1 k ϑ [ i ]$ for $k = 1 , 2 , … , n − 1$ and $∑ i = 1 n ν i = ∑ i = 1 n ϑ i$, where $ν [ 1 ] ≥ ν [ 2 ] ≥ ⋯ ≥ ν [ n ]$ and $ϑ [ 1 ] ≥ ϑ [ 2 ] ≥ ⋯ ≥ ϑ [ n ]$ are rearrangements of $ν$ and $ϑ$ in a descending order.
The majorization has been found many applications in different fields of mathematics. A survey of the applications of majorization and relevant results can be found in the monograph of Marshall and Olkin . Recently, the authors have given considerable attention to the generalizations and applications of the majorization and related inequalities, for details, we refer the reader to our papers [3,4,5,6,7,8,9,10,11,12,13].
In this paper we focus on a type of majorization inequality involving convex functions, which reveals the correlations among majorization, convex functions and inequalities. Now, let us recall briefly this type of majorization inequality.
The following classical majorization inequality can be found in the monographs of Marshall and Olkin  and Pečarić et al. .
Theorem 1.
Let $ν = ( ν 1 , ν 2 , … , ν n$), $ϑ = ( ϑ 1 , ϑ 2 , … , ϑ n$) be two n-tuples, $ν i , ϑ i ∈ I ( i = 1 , 2 , … , n$), I is an interval. Then
$∑ i = 1 n Ψ ( ν i ) ≤ ∑ i = 1 n Ψ ( ϑ i )$
holds for every continuous convex function $Ψ : I → R$ if and only if $ν ≺ ϑ$ holds.
Fuchs  gave a weighted generalization of the majorization theorem, as follows:
Theorem 2.
Let $ν = ( ν 1 , ν 2 , … , ν n$), $ϑ = ( ϑ 1 , ϑ 2 , … , ϑ n$) be two decreasing n-tuples, $ν i , ϑ i ∈ I ( i = 1 , 2 , … , n$), I is an interval. Suppose $ℓ 1 , ℓ 2 , … , ℓ n$ are real numbers such that $∑ i = 1 k ℓ i ν i ≤ ∑ i = 1 k ℓ i ϑ i$ for $k = 1 , 2 , … , n − 1$ and $∑ i = 1 n ℓ i ν i = ∑ i = 1 n ℓ i ϑ i$. Then
$∑ i = 1 n ℓ i Ψ ( ν i ) ≤ ∑ i = 1 n ℓ i Ψ ( ϑ i )$
holds for any continuous convex function $Ψ : I → R$.
Bullen, Vasić, and Stanković  presented a result similar to the above result, in which the condition of the tuples $ν , ϑ$ is relaxed and the condition of the function $Ψ$ is intensified.
Theorem 3.
Let $ν = ( ν 1 , ν 2 , … , ν n$), $ϑ = ( ϑ 1 , ϑ 2 , … , ϑ n$) be two decreasing n-tuples, $ν i , ϑ i ∈ I ( i = 1 , 2 , … , n$), I is an interval. Suppose $ℓ 1 , ℓ 2 , … , ℓ n$ are real numbers such that $∑ i = 1 k ℓ i ν i ≤ ∑ i = 1 k ℓ i ϑ i$ for $k = 1 , 2 , … , n$. If $Ψ : I → R$ is a continuous increasing convex function, then
$∑ i = 1 n ℓ i Ψ ( ν i ) ≤ ∑ i = 1 n ℓ i Ψ ( ϑ i ) .$
The aim of this paper is to establish the refinements of majorization inequalities of Theorems 1–3. To achieve this, we will first establish an equality by using Taylor theorem with mean-value form of the remainder, which enables us to deduce the refined versions of majorization inequalities mentioned above.

## 2. Lemma

Lemma 1.
Let $ν = ( ν 1 , ν 2 , … , ν n$), $ϑ = ( ϑ 1 , ϑ 2 , … , ϑ n$) be two n-tuples, $ν i , ϑ i ∈ ( a , b ) ( i = 1 , 2 , … , n$), and let $ℓ 1 , ℓ 2 , … , ℓ n$ be real numbers. If $Ψ : [ a , b ] → R$ is a function such that $Ψ ′ ∈ C [ a , b ]$ and $Ψ ″$ exists on ($a , b ) ,$ then there exists $τ i$ between $ν i$ and $ϑ i$ satisfying
$∑ i = 1 n ℓ i Ψ ( ϑ i ) − ∑ i = 1 n ℓ i Ψ ( ν i ) = ∑ i = 1 n Ψ ′ ( ν i ) ℓ i ( ϑ i − ν i ) + ∑ i = 1 n Ψ ″ ( τ i ) 2 ℓ i ( ϑ i − ν i ) 2 .$
Proof.
Using the Taylor’s formula with the Lagrange remainder (mean-value form of the remainder) gives
$Ψ ( ϑ i ) = Ψ ( ν i ) + Ψ ′ ( ν i ) 1 ! ( ϑ i − ν i ) + Ψ ″ ( τ i ) 2 ! ( ϑ i − ν i ) 2 ,$
where $ν i , ϑ i ∈ ( a , b ) ,$ $τ i$ is a real number between $ν i$ and $ϑ i$ ($i = 1 , 2 , … , n$).
Multiplying both sides of (5) by $ℓ i$ and taking summation over i ($i = 1 , 2 , … , n$), we get
$∑ i = 1 n ℓ i Ψ ( ϑ i ) = ∑ i = 1 n ℓ i Ψ ( ν i ) + ∑ i = 1 n Ψ ′ ( ν i ) ℓ i ( ϑ i − ν i ) + ∑ i = 1 n Ψ ″ ( τ i ) 2 ℓ i ( ϑ i − ν i ) 2 ,$
which is the desired equality (4). The proof of Lemma 1 is complete. □

## 3. Main Results

In this section, we establish some refinements of the majorization inequality.
Theorem 4.
Let $ν = ( ν 1 , ν 2 , … , ν n$), $ϑ = ( ϑ 1 , ϑ 2 , … , ϑ n$) be two n-tuples, $ν i , ϑ i ∈ ( a , b ) ( i = 1 , 2 , … , n$). If $ν ≺ ϑ$ and $Ψ : [ a , b ] → R$ is a twice differentiable convex function, then there exists a real number $τ i$ between $ν [ i ]$ and $ϑ [ i ]$ ($i = 1 , 2 , … , n$) such that
$∑ i = 1 n Ψ ( ϑ i ) − ∑ i = 1 n Ψ ( ν i ) ≥ ∑ i = 1 n Ψ ″ ( τ i ) 2 ( ϑ [ i ] − ν [ i ] ) 2 .$
where $ν [ 1 ] ≥ ν [ 2 ] ≥ ⋯ ≥ ν [ n ]$ and $ϑ [ 1 ] ≥ ϑ [ 2 ] ≥ ⋯ ≥ ϑ [ n ]$ are rearrangements of ν and ϑ in a descending order.
Proof.
Using Lemma 1 with $ℓ i = 1$, $ν i = ν [ i ]$, $ϑ i = ϑ [ i ]$ ($i = 1 , 2 , … , n$), one has
$∑ i = 1 n Ψ ( ϑ [ i ] ) − ∑ i = 1 n Ψ ( ν [ i ] ) = ∑ i = 1 n Ψ ′ ( ν [ i ] ) ( ϑ [ i ] − ν [ i ] ) + ∑ i = 1 n Ψ ″ ( τ i ) 2 ( ϑ [ i ] − ν [ i ] ) 2 ,$
that is
$∑ i = 1 n Ψ ( ϑ i ) − ∑ i = 1 n Ψ ( ν i ) = ∑ i = 1 n Ψ ′ ( ν [ i ] ) ( ϑ [ i ] − ν [ i ] ) + ∑ i = 1 n Ψ ″ ( τ i ) 2 ( ϑ [ i ] − ν [ i ] ) 2 ,$
where $ν i , ϑ i ∈ ( a , b )$, $τ i$ is a real number between $ν [ i ]$ and $ϑ [ i ]$ ($i = 1 , 2 , … , n$).
Let
$A k = ∑ i = 1 k ϑ [ i ] , B k = ∑ i = 1 k ν [ i ] ( k = 1 , 2 , … , n ) , A 0 = B 0 = 0 .$
Considering the first term in the right hand side of (7), we have
$∑ i = 1 n Ψ ′ ( ν [ i ] ) ( ϑ [ i ] − ν [ i ] ) = ∑ i = 1 n Ψ ′ ( ν [ i ] ) ( A i − A i − 1 − B i + B i − 1 ) = ∑ i = 1 n Ψ ′ ( ν [ i ] ) ( A i − B i ) − ∑ i = 1 n Ψ ′ ( ν [ i ] ) ( A i − 1 − B i − 1 ) = Ψ ′ ( ν [ n ] ) ( A n − B n ) + ∑ i = 1 n − 1 ( Ψ ′ ( ν [ i ] ) − Ψ ′ ( ν [ i + 1 ] ) ) ( A i − B i ) .$
It follows from $ν ≺ ϑ$ that $A n − B n = 0$ and $A i − B i ≥ 0$ for $i = 1 , 2 , … , n − 1$.
Additionally, since $Ψ$ is a continuous convex function on $[ a , b ]$, we deduce from $ν [ i ] ≥ ν [ i + 1 ]$ ($i = 1 , 2 , … , n − 1$) that
$Ψ ′ ( ν [ i ] ) − Ψ ′ ( ν [ i + 1 ] ) ≥ 0 for i = 1 , 2 , … , n − 1 .$
Hence
$∑ i = 1 n Ψ ′ ( ν [ i ] ) ( ϑ [ i ] − ν [ i ] ) ≥ 0 ,$
which, along with the equality (7), leads to the required inequality (6). This completes the proof of Theorem 4. □
Remark 1.
The inequality of Theorem 4 is a refinement of the inequality of Theorem 1, since the term $∑ i = 1 n Ψ ″ ( τ i ) 2 ( ϑ [ i ] − ν [ i ] ) 2$ in inequality (6) is nonnegative.
In the following, we provide two refinements of majorization inequality by keeping one of the tuples decreasing (increasing).
Theorem 5.
Let $ν = ( ν 1 , ν 2 , … , ν n$), $ϑ = ( ϑ 1 , ϑ 2 , … , ϑ n$) be two n-tuples, $ν i , ϑ i ∈ ( a , b ) ( i = 1 , 2 , … , n$), let $Ψ : [ a , b ] → R$ be a twice differentiable convex function, and let $ℓ 1 , ℓ 2 , … , ℓ n$ be real numbers such that $∑ i = 1 k ℓ i ν i ≤ ∑ i = 1 k ℓ i ϑ i$ for $k = 1 , 2 , … , n − 1$ and $∑ i = 1 n ℓ i ν i = ∑ i = 1 n ℓ i ϑ i$.
(i) If ν is a decreasing n-tuple, then there exists a real number $τ i$ between $ν [ i ]$ and $ϑ [ i ]$ ($i = 1 , 2 , … , n$) such that
$∑ i = 1 n ℓ i Ψ ( ϑ i ) − ∑ i = 1 n ℓ i Ψ ( ν i ) ≥ ∑ i = 1 n Ψ ″ ( τ i ) 2 ℓ i ( ϑ i − ν i ) 2 .$
(ii) If ϑ is a increasing n-tuple, then there exists another real number $σ i$ between $ν [ i ]$ and $ϑ [ i ]$ ($i = 1 , 2 , … , n$) such that
$∑ i = 1 n ℓ i Ψ ( ν i ) − ∑ i = 1 n ℓ i Ψ ( ϑ i ) ≥ ∑ i = 1 n Ψ ″ ( σ i ) 2 ℓ i ( ϑ i − ν i ) 2 .$
Proof.
(i) It follows from Lemma 1 that
$∑ i = 1 n ℓ i Ψ ( ϑ i ) − ∑ i = 1 n ℓ i Ψ ( ν i ) = ∑ i = 1 n Ψ ′ ( ν i ) ℓ i ( ϑ i − ν i ) + ∑ i = 1 n Ψ ″ ( τ i ) 2 ℓ i ( ϑ i − ν i ) 2 ,$
where $ν i , ϑ i ∈ ( a , b ) ,$ $τ i$ is a real number between $ν i$ and $ϑ i$ ($i = 1 , 2 , … , n$). Let
$A k = ∑ i = 1 k ℓ i ϑ i , B k = ∑ i = 1 k ℓ i ν i ( k = 1 , 2 , … , n ) , A 0 = B 0 = 0 .$
Then, we have $A i ≥ B i$ ($i = 1 , 2 , … , n − 1$), $A n = B n$, and
$∑ i = 1 n Ψ ′ ( ν i ) ℓ i ( ϑ i − ν i ) = ∑ i = 1 n Ψ ′ ( ν i ) ( A i − A i − 1 − B i + B i − 1 ) = Ψ ′ ( ν n ) ( A n − B n ) + ∑ i = 1 n − 1 ( Ψ ′ ( ν i ) − Ψ ′ ( ν i + 1 ) ) ( A i − B i ) . = ∑ i = 1 n − 1 ( Ψ ′ ( ν i ) − Ψ ′ ( ν i + 1 ) ) ( A i − B i ) .$
Noting that $Ψ$ is a continuous convex function on $[ a , b ]$, and $ν$ is a decreasing n-tuple, we obtain $Ψ ′ ( ν i ) − Ψ ′ ( ν i + 1 ) ≥ 0$ for $i = 1 , 2 , … , n − 1$.
Hence
$∑ i = 1 n Ψ ′ ( ν i ) ℓ i ( ϑ i − ν i ) ≥ 0 ,$
which, together with inequality (10), leads to the required inequality (8).
(ii) Similarly, we can prove the inequality (9) under the condition that $ϑ$ is an increasing n-tuple. The proof of Theorem 5 is complete. □
Remark 2.
The inequality (8) of Theorem 5 is a refinement of the inequality (2) of Theorem 2 in the case when $ℓ 1 , ℓ 2 , … , ℓ n$ are positive numbers.
Theorem 6.
Let $ν = ( ν 1 , ν 2 , … , ν n$), $ϑ = ( ϑ 1 , ϑ 2 , … , ϑ n$) be two n-tuples, $ν i , ϑ i ∈ ( a , b ) ( i = 1 , 2 , … , n$), let $Ψ : [ a , b ] → R$ be a twice differentiable and increasing convex function, and let $ℓ 1 , ℓ 2 , … , ℓ n$ be real numbers such that $∑ i = 1 k ℓ i ν i ≤ ∑ i = 1 k ℓ i ϑ i$ for $k = 1 , 2 , … , n$. If ν is a decreasing n-tuple, then there exists a real number $τ i$ between $ν [ i ]$ and $ϑ [ i ]$ ($i = 1 , 2 , … , n$) such that
$∑ i = 1 n ℓ i Ψ ( ϑ i ) − ∑ i = 1 n ℓ i Ψ ( ν i ) ≥ ∑ i = 1 n Ψ ″ ( τ i ) 2 ℓ i ( ϑ i − ν i ) 2 .$
Proof.
Let
$A k = ∑ i = 1 k ℓ i ϑ i , B k = ∑ i = 1 k ℓ i ν i ( k = 1 , 2 , … , n ) , A 0 = B 0 = 0 .$
By Lemma 1, for any $ν i , ϑ i ∈ ( a , b )$ ($i = 1 , 2 , … , n$), there exists a real number between $ν i$ and $ϑ i$ such that
$∑ i = 1 n ℓ i Ψ ( ϑ i ) − ∑ i = 1 n ℓ i Ψ ( ν i ) = ∑ i = 1 n Ψ ′ ( ν i ) ℓ i ( ϑ i − ν i ) + ∑ i = 1 n Ψ ″ ( τ i ) 2 ℓ i ( ϑ i − ν i ) 2 = ∑ i = 1 n Ψ ′ ( ν i ) ( A i − A i − 1 − B i + B i − 1 ) + ∑ i = 1 n Ψ ″ ( τ i ) 2 ℓ i ( ϑ i − ν i ) 2 = Ψ ′ ( ν n ) ( A n − B n ) + ∑ i = 1 n − 1 ( Ψ ′ ( ν i ) − Ψ ′ ( ν i + 1 ) ) ( A i − B i ) + ∑ i = 1 n Ψ ″ ( τ i ) 2 ℓ i ( ϑ i − ν i ) 2 .$
Since $Ψ$ is a continuous convex function on $[ a , b ]$, and $ν$ is a decreasing n-tuple, we obtain $Ψ ′ ( ν i ) − Ψ ′ ( ν i + 1 ) ≥ 0$ for $i = 1 , 2 , … , n − 1$. In addition, since $Ψ$ is an increasing function on $[ a , b ]$, we get $Ψ ′ ( ν n ) ≥ 0$. Now, by using the assumption conditions $A i ≥ B i$ ($k = 1 , 2 , … , n$), we conclude that
$Ψ ′ ( ν n ) ( A n − B n ) + ∑ i = 1 n − 1 ( Ψ ′ ( ν i ) − Ψ ′ ( ν i + 1 ) ) ( A i − B i ) ≥ 0 .$
Therefore, we have
$∑ i = 1 n ℓ i Ψ ( ϑ i ) − ∑ i = 1 n ℓ i Ψ ( ν i ) ≥ ∑ i = 1 n Ψ ″ ( τ i ) 2 ℓ i ( ϑ i − ν i ) 2 .$
The Theorem 6 is proved. □
Remark 3.
The inequality (11) of Theorem 6 is a refinement of the inequality (3) of Theorem 3 in the case when $ℓ 1 , ℓ 2 , … , ℓ n$ are positive numbers.
Theorem 7.
Let $ν = ( ν 1 , ν 2 , … , ν n$), $ϑ = ( ϑ 1 , ϑ 2 , … , ϑ n$) be two n-tuples, $ν i , ϑ i ∈ ( a , b ) ( i = 1 , 2 , … , n$), let $Ψ : [ a , b ] → R$ be a twice differentiable convex function, and let $ℓ 1 , ℓ 2 , … , ℓ n$ be positive numbers. If ν and $ϑ − ν$ are monotonic in the same sense, then there exists a real number $τ i$ between $ν [ i ]$ and $ϑ [ i ]$ ($i = 1 , 2 , … , n$) such that
$∑ i = 1 n ℓ i Ψ ( ϑ i ) − ∑ i = 1 n ℓ i Ψ ( ν i ) ≥ 1 ℓ 1 + ℓ 2 + ⋯ + ℓ n ∑ i = 1 n ℓ i Ψ ′ ( ν i ) ∑ i = 1 n ℓ i ( ϑ i − ν i ) + ∑ i = 1 n Ψ ″ ( τ i ) 2 ℓ i ( ϑ i − ν i ) 2 .$
Proof.
Since $Ψ$ is convex function, and tuple $ν$ and tuple $ϑ − ν$ are monotonic in the same sense, we conclude that $Ψ ′ ( ν$) and $ϑ − ν$ are monotonic in the same sense.
Using the Chebyshev’s inequality for weights $ℓ 1 , ℓ 2 , … , ℓ n$, we obtain
$( ∑ i = 1 n ℓ i ) ∑ i = 1 n ℓ i Ψ ′ ( ν i ) ( ϑ i − ν i ) ≥ ∑ i = 1 n ℓ i Ψ ′ ( ν i ) ∑ i = 1 n ℓ i ( ϑ i − ν i ) .$
On the other hand, by Lemma 1, for any $ν i , ϑ i ∈ ( a , b )$ ($i = 1 , 2 , … , n$), there exists a real number $τ i$ between $ν i$ and $ϑ i$ such that
$∑ i = 1 n ℓ i Ψ ( ϑ i ) − ∑ i = 1 n ℓ i Ψ ( ν i ) = ∑ i = 1 n Ψ ′ ( ν i ) ℓ i ( ϑ i − ν i ) + ∑ i = 1 n Ψ ″ ( τ i ) 2 ℓ i ( ϑ i − ν i ) 2 .$
Hence, we get
$∑ i = 1 n ℓ i Ψ ( ϑ i ) − ∑ i = 1 n ℓ i Ψ ( ν i ) ≥ 1 ℓ 1 + ℓ 2 + ⋯ + ℓ n ∑ i = 1 n ℓ i Ψ ′ ( ν i ) ∑ i = 1 n ℓ i ( ϑ i − ν i ) + ∑ i = 1 n Ψ ″ ( τ i ) 2 ℓ i ( ϑ i − ν i ) 2 .$
This proves the required inequality (12) in Theorem 7. □
Applying an additional condition $∑ i = 1 n ℓ i ν i ≤ ∑ i = 1 n ℓ i ϑ i$ to inequality (12), we obtain the following result.
Corollary 1.
Let $ν = ( ν 1 , ν 2 , … , ν n$), $ϑ = ( ϑ 1 , ϑ 2 , … , ϑ n$) be two n-tuples, $ν i , ϑ i ∈ ( a , b ) ( i = 1 , 2 , … , n$), let $Ψ : [ a , b ] → R$ be a twice differentiable and increasing convex function, and let $ℓ 1 , ℓ 2 , … , ℓ n$ be positive numbers. If ν and $ϑ − ν$ are monotonic in the same sense, and $∑ i = 1 n ℓ i ν i ≤ ∑ i = 1 n ℓ i ϑ i$, then there exists a real number $τ i$ between $ν [ i ]$ and $ϑ [ i ]$ ($i = 1 , 2 , … , n$) such that
$∑ i = 1 n ℓ i Ψ ( ϑ i ) − ∑ i = 1 n ℓ i Ψ ( ν i ) ≥ ∑ i = 1 n Ψ ″ ( τ i ) 2 ℓ i ( ϑ i − ν i ) 2 .$

## 4. An Application

In this section we establish a new fractional inequality to illustrate the application of our results.
Theorem 8.
Let $ξ 1 , ξ 2$, $ξ 3$ be positive numbers and $ξ 1 ≥ ξ 2 ≥ ξ 3$. Then we have the inequality
$1 2 ξ 1 + 1 2 ξ 2 + 1 2 ξ 3 − 1 ξ 1 + ξ 2 − 1 ξ 1 + ξ 3 − 1 ξ 2 + ξ 3 ≥ ( ξ 1 − ξ 2 ) 2 2 ξ 1 ( ξ 1 + ξ 2 ) 2 + ( 2 ξ 2 − ξ 1 − ξ 3 ) 2 2 ξ 2 ( ξ 1 + ξ 3 ) 2 + ( ξ 2 − ξ 3 ) 2 2 ξ 3 ( ξ 2 + ξ 3 ) 2 .$
Proof.
From the given condition $ξ 1 ≥ ξ 2 ≥ ξ 3$, it is easy to check that
$ξ 1 + ξ 2 ≥ ξ 1 + ξ 3 ≥ ξ 2 + ξ 3 , 2 ξ 1 ≥ 2 ξ 2 ≥ 2 ξ 3$
and
$( ξ 1 + ξ 2 , ξ 1 + ξ 3 , ξ 2 + ξ 3 ) ≺ ( 2 ξ 1 , 2 ξ 2 , 2 ξ 3 ) .$
Using Theorem 4 and taking $ν = ( ξ 1 + ξ 2 , ξ 1 + ξ 3 , ξ 2 + ξ 3 ) , ϑ = ( 2 ξ 1 , 2 ξ 2 , 2 ξ 3$), $Ψ ( x ) = 1 x , x ∈ ( 0 , + ∞$) in (6), we obtain that there exists a real number $τ i$ between $ν [ i ]$ and $ϑ [ i ]$ ($i = 1 , 2 , 3$) such that
$1 2 ξ 1 + 1 2 ξ 2 + 1 2 ξ 3 − 1 ξ 1 + ξ 2 − 1 ξ 1 + ξ 3 − 1 ξ 2 + ξ 3 ≥ 1 τ 1 3 ( ξ 1 − ξ 2 ) 2 + 1 τ 2 3 ( 2 ξ 2 − ξ 1 − ξ 3 ) 2 + 1 τ 3 3 ( ξ 2 − ξ 3 ) 2 .$
Further, by (5) we find that $τ 1 , τ 2 , τ 3$ satisfy
$1 2 ξ 1 − 1 ξ 1 + ξ 2 = − ξ 1 − ξ 2 ( ξ 1 + ξ 2 ) 2 + 1 τ 1 3 ( ξ 1 − ξ 2 ) 2 ,$
$1 2 ξ 2 − 1 ξ 1 + ξ 3 = − 2 ξ 2 − ξ 1 − ξ 3 ( ξ 1 + ξ 3 ) 2 + 1 τ 2 3 ( 2 ξ 2 − ξ 1 − ξ 3 ) 2 ,$
$1 2 ξ 3 − 1 ξ 2 + ξ 3 = − ξ 3 − ξ 2 ( ξ 2 + ξ 3 ) 2 + 1 τ 3 3 ( ξ 3 − ξ 2 ) 2 .$
From the above equations, we have
$τ 1 3 = 2 ξ 1 ( ξ 1 + ξ 2 ) 2 , τ 2 3 = 2 ξ 2 ( ξ 1 + ξ 3 ) 2 , τ 3 3 = 2 ξ 3 ( ξ 3 + ξ 2 ) 2 .$
Combining (15) and (16) leads to the desired inequality (14). The proof of Theorem 8 is complete. □

## Author Contributions

S.W. and M.A.K. finished the proofs of the main results and the writing work. H.U.H. gave lots of advice on the proofs of the main results and the writing work. All authors read and approved the final manuscript.

## Funding

This work was supported by the Teaching Reform Project of Longyan University (Grant No. 2017JZ02) and the Teaching Reform Project of Fujian Provincial Education Department (Grant No. FBJG20180120).

## Acknowledgments

The authors would like to express sincere appreciation to the anonymous reviewers for their helpful comments and suggestions.

## Conflicts of Interest

The authors declare no conflict of interest.

## References

1. Hardy, G.H.; Littlewood, J.E.; Pólya, G. Inequalities, 2nd ed.; Cambridge Univ. Press: Cambridge, UK, 1952. [Google Scholar]
2. Marshall, A.W.; Olkin, I. Inequalities: The Theory of Majorization and Its Applicaitons; Academic Press: New York, NY, USA, 1979. [Google Scholar]
3. Wu, S.H.; Adil Khan, M.; Basir, A.; Saadati, R. Some majorization integral inequalities for functions defined on rectangles. J. Inequal. Appl. 2018, 2018, 146. [Google Scholar] [CrossRef] [PubMed]
4. Adil Khan, M.; Wu, S.H.; Ullah, H.; Chu, Y.M. Discrete majorization type inequaliites for convex functions on rectangles. J. Inequal. Appl. 2019, 2019, 16. [Google Scholar] [CrossRef]
5. Wu, S.H.; Shi, H.N. A relation of weak majorization and its applications to certain inequalities for means. Math. Slov. 2009, 61, 561–570. [Google Scholar] [CrossRef]
6. Shi, H.N.; Wu, S.H. Majorized proof and refinement of the discrete Steffensen’s inequality. Taiwan. J. Math. 2007, 11, 1203–1208. [Google Scholar] [CrossRef]
7. Zaheer Ullah, S.; Adil Khan, M.; Abdulhameed Khan, Z.; Chu, Y.M. Integral majorization type inequalities for the functions in the sense of strongly convexity. J. Funct. Spaces 2019, 2019, 9487823. [Google Scholar]
8. Adil Khan, M.; Khalid, S.; Pečarić, J.E. Refinements of some majorization type inequalities. J. Math. Inequal. 2013, 7, 73–92. [Google Scholar] [CrossRef]
9. Adil Khan, M.; Khan, J.; Pečarić, J.E. Generalization of Jensen’s and Jensen-Steffensen’s inequalities by generalized majorization theorem. J. Math. Inequal. 2017, 11, 1049–1074. [Google Scholar] [CrossRef]
10. Adil Khan, M.; Latif, N.; Pečarić, J.E. Generalization of majorization theorem. J. Math. Inequal. 2015, 9, 847–872. [Google Scholar] [CrossRef]
11. Adil Khan, M.; Latif, N.; Pečarić, J.E.; Perić, I. On Majorization for Matrices. Math. Balk. 2013, 27, 3–19. [Google Scholar]
12. Khan, J.; Adil Khan, M.; Pečarić, J. On Jensen’s Type Inequalities via Generalized Majorization Inequalities. Filomat 2018, 32, 5719–5733. [Google Scholar] [CrossRef]
13. Zaheer Ullah, S.; Adil Khan, M.; Chu, Y.M. Majorization theorems for strongly convex functions. J. Inequal. Appl. 2019, 2019, 58. [Google Scholar] [CrossRef]
14. Pečarić, J.E.; Proschan, F.; Tong, Y.L. Convex Functions, Partial Orderings and Statistical Applications; Academic Press: New York, NY, USA, 1992. [Google Scholar]
15. Fuchs, L. A new proof of an inequality of Hardy-Littlewood-Pólya. Mat. Tidsskr. B 1947, 53–54, stable/24530530. [Google Scholar]
16. Bullen, P.S.; Vasić, P.M.; Stanković, L.J. A Problem of A. Oppenheim; University of Belgrade: Belgrade, Serbia, 1973. [Google Scholar]