# Unifications of Continuous and Discrete Fractional Inequalities of the Hermite–Hadamard–Jensen–Mercer Type via Majorization

Authors: Shah Faisal; Muhammad Adil Khan; Tahir Ullah Khan; Tareq Saeed; Zaid Mohammmad Mohammad Mahdi Sayed
Journal: Journal of Function Spaces
 (2022)

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## Abstract

The main objective of the paper is to develop an innovative idea of bringing continuous and discrete inequalities into a unified form. The desired objective is thus obtained by embedding majorization theory with the existing notion of continuous inequalities. These notions are applied to the latest generalized form of the inequalities, popularly known as the Hermite–Hadamard–Jensen–Mercer inequalities. Moreover, the frequently-used Caputo fractional operators are employed, which are rightly considered critical, especially for applied problems. Both weighted and unweighted forms of the developed results are discussed. In addition to this, some bounds are also provided for the absolute difference between the left- and right-sides of the main results.


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## Body

## 1. Introduction

The field of mathematical inequalities contributes to a wide area of research in mathematics. With the passage of time, this field has emerged as a separate discipline, despite the fact that it was being used as a tool earlier. The addition of the notion of convexity enriched its literature and stimulated a new trend among researchers. As a result, many new inequalities came to the surface. These inequalities are (but not limited to) Ostrowski inequalities [1], Jensen’s inequalities [2], the Jensen–Mercer inequalities [3], Fejér inequalities [4], Hermite–Hadamard inequalities [5], and their various variants. The Hermite–Hadamard inequality is believed to be the most widespread inequality in the literature and has received much attention in the last few years. This inequality is defined as follows:Ifϕ:I⟶ℝ is a convex function with ϑ,θ∈I such that ϑ<θ then(1)ϕϑ+θ2≤1θ−ϑ∫ϑθϕudu≤ϕϑ+ϕθ2.The direction of the inequality given in (1) reverses whenever the function ϕ is concave. This inequality has been established for different generalized convex functions, for example, s−convex [6], η−convex [7], strongly convex [8], and coordinate convex function [9]. Research works in this field have also been extended to the theory of fractional calculus. As there are multiple numbers of fractional operators but due to our interest, we limit ourselves to the well-known Caputo fractional operators. Their definition is given as follows:Definition 1 (Caputo fractional derivative operators).
Consider a functionϕ∈Cnϑ,θ (the space of functions whose nth-derivative exist and continuous on ϑ,θ), α>0, such that α∉1,2,3,… and n=α+1. Then the α ordered, Caputo fractional derivative operators are defined as [10](2)Dcϑ+αϕz=1Γn−α∫ϑzϕnuz−uα−n+1du,z>ϑ,Dcθ−αϕz=−1nΓn−α∫zθϕnuu−zα−n+1du,z<θ.WhereDcϑ+αϕz and Dcθ−αϕz stand for the left- and right-sided Caputo fractional derivative operators, respectively.It may be noted that if the usual derivative of the functionϕ of order n exists, then it coincides with Dcϑ+αϕz for α=n∈1,2,3,…. Also, for n=1 and α=0, we have(3)Dcϑ+αϕz=Dcθ−αϕz=ϕz.Some research work related to the Caputo fractional operators can be found in [11–13] and the references therein.The associated Hermite–Jensen–Mercer inequality in terms of Caputo fractional operators is defined as follows [14]:Theorem 1 (Hermite–Jensen–Mercer inequality).
Consider a functionϕ defined on the interval ϑ,θ, such that ϕ∈Cnϑ,θ and ϕn is convex on ϑ,θ with x1,y1⊂ϑ,θ, α>0, then we have(4)ϕnϑ+θ−x1+y12≤2n−α−1Γn−α+1y1−x1n−αDcϑ+θ−x1+y1/2+αϕϑ+θ−x1+−1nDcϑ+θ−x1+y1/2−αϕϑ+θ−y1≤ϕnϑ+ϕnθ−ϕnx1+ϕny12.The inequality (1) can be obtained from (4) when n=1, α=0, x1=ϑ and y1=θ. Some more work related to Hermite–Jensen–Mercer inequalities via fractional operators can be traced in [15–20]. Now, we state the definition of majorization in terms of which we want to present our results [21].Definition 2 (Majorization).
Let us considera=a1,…,al and b=b1,…,bl are two l−tuples of real numbers arranged in order al≤al−1≤⋯≤a1, and bl≤bl−1≤⋯≤b1, then a is said to majorize b (or b is said to be majorized by a), if for k=1,2,…,l−1, we have(5)∑s=1kbs≤∑s=1kas,and∑s=1las=∑s=1lbs.
Ifa majorizes b, then symbolically it is written as b≺a.Niezgoda [22] has used the concept of majorization and extended the Jensen–Mercer inequality given as follows:Theorem 2 (Majorized discrete Jensen–Mercer inequality).
Let us consider a convex functionϕ defined on the interval I, r×l real matrix xis , and l− tuple δ=δ1,…,δl such that δs,xis∈I for all i=1,2,…,r , s∈1,…,l with σi≥0, ∑i=1rσi=1. If δ majorizes every row of xis, then we have(6)ϕ∑s=1lδs−∑s=1l−1∑i=1rσixis≤∑s=1lϕδs−∑s=1l−1∑i=1rσiϕxis.The following lemmas will help us to prove our main results [23].Lemma 1.
Let us consider a convex functionϕ defined on the interval I, r×l real matrix xis, and two l− tuples δ=δ1,…,δl, p=p1,…,pl, such that δs, xis∈I, σi≥0, ∑i=1rσi=1, ps≥0, with pl≠0, η=1/pl, for all i=1,2,…,r,s∈1,…,l. If for each i=1,2,…,r, xi1,…,xim is a decreasing l−tuple and satisfying(7)∑s=1kpsxis≤∑s=1kpsδs,fork=1,2,…,l−1,∑s=1lpsδs=∑s=1lpsxis,
Then(8)ϕ∑s=1lηpsδs−∑s=1l−1∑i=1rησipsxis≤∑s=1lηpsϕδs−∑s=1l−1∑i=1rησipsϕxis.Lemma 2.
Let us consider a convex functionϕ defined on the interval I, r×l real matrix xis, and two l− tuples δ=δ1,…,δl, p=p1,…,pl, such that δs, xis∈I, σi≥0, ∑i=1rσi=1, ps≥0, with pl≠0, η=1/pl, for all i=1,2,…,r,s∈1,…,l. If for eachi=1,…,r, δs−xis and xis are monotonically in the same sense and(9)∑s=1lpsδs=∑s=1lpsxis,
Then(10)ϕ∑s=1lηpsδs−∑s=1l−1∑i=1rησipsxis≤∑s=1lηpsϕδs−∑s=1l−1∑i=1rησipsϕxis.The theory of majorization has been successful in drawing the attention of researchers working in various fields. It has been used as a key element in solving complicated optimization problems [24, 25]. Some more recent applications of majorization theory in signal processing and communication can be seen in [26, 27]. For further successive work carried out via the concept of majorization, one is referred to [28–34] and the references therein. In the present era, despite the existence of various diverse research fields, the shrinking of more than one research field into one is direly needed. The reason is that new ideas grow fast when they attract the attention of a maximum number of researchers. In our case, since inequalities have two main aspects, one is that of continuous inequalities and the other is of discrete inequalities. Both subfields have been absorbing the attention of many researchers at the same time. The fault is that the majority of the results are based only on simple conversions from discrete to continuous or vice versa. The concept of adding new ideas or strengthening an existing one is rarely utilized. In such a situation, there is a need for the provision of such a platform which can play the role of bringing researchers from the abovementioned subfields together and utilize their energies and efforts in one direction. The present attempt may be considered one of the endeavors in this regard.The present paper is summarized as follows: first of all, Theorem3 is devoted to the establishment of a new unified form of Hermite–Hadamard–Jensen–Mercer inequality. This objective is achieved by utilizing the majorized l−tuples in the context of Caputo fractional operators. A slightly different variant of Theorem 3 is presented in the form of Theorem 4. In order to verify and provide proof of the fact that the newly-obtained results are the unifications and generalizations of those already existing results, Remark 1 and Remark 2 are presented. In addition to this, weighted versions of the obtained results are also provided, taking the weighted generalized Mercer’s inequality into account. These weighted results can be traced to Theorem 5 and Theorem 6. Moreover, two new identities, connected with the right- and left-sides of Theorem 3 and Theorem 4, respectively, are discovered. Employing these lemmas, various bounds associated with the absolute difference of the two right- and left-most terms in the main results are obtained. These results are discussed in Theorem 7, Theorem 8, Theorem 9, Theorem 10, and Theorem 11. Remark 5, Remark 6, and Remark 7 show that the newly-derived identities also generalize those previously-defined identities, while Remark 8 discusses the previous version of Theorem 10. Corollary 1 gives details about a previous bound while Corollary 2, and Corollary 3 provide information about the classical integral versions of Theorem 9 and Theorem 11. At the end, conclusion of the overall attempt is presented.


## 2. Main Results

The following theorem presents the Hermite–Hadamard–Jensen–Mercer fractional inequality for the Caputo fractional operators.Theorem 3.
Let us consider a functionϕ∈CnI and δ=δ1,…,δl, x=x1,…,xl, y=y1,…,yl are three l-tuples, such that δs,xs,ys∈I, for all s∈1,…,l, xl>yl, α>0. If ϕn is a convex function on I, x≺δ, and y≺δ, then(11)ϕn∑s=1lδs−∑s=1l−1xs+ys2≤Γn−α+12∑s=1l−1ys−xsn−αDc∑s=1lδs−∑s=1l−1ys+αϕ∑s=1lδs−∑s=1l−1xs+−1nDc∑s=1lδs−∑s=1l−1xs−αϕ∑s=1lδs−∑s=1l−1ys≤ϕn∑s=1lδs−∑s=1l−1ys+ϕn∑s=1lδs−∑s=1l−1xs2≤∑s=1lϕnδs−∑s=1l−1ϕnxs+∑s=1l−1ϕnys2.Proof.
It can be written as(12)ϕn∑s=1lδs−∑s=1l−1xs+ys2=ϕn12∑s=1lδs−∑s=1l−1xs+∑s=1lδs−∑s=1l−1ys,=ϕn12t∑s=1lδs−∑s=1l−1xs+1−t∑s=1lδs−∑s=1l−1ys+t∑s=1lδs−∑s=1l−1ys+1−t∑s=1lδs−∑s=1l−1xs.
Sinceϕn is a convex function, therefore (12) gives the following inequality:(13)ϕn∑s=1lδs−∑s=1l−1xs+ys2≤12ϕnt∑s=1lδs−∑s=1l−1xs+1−t∑s=1lδs−∑s=1l−1ys+ϕnt∑s=1lδs−∑s=1l−1ys+1−t∑s=1lδs−∑s=1l−1xs.
By multiplying both sides of (13) by tn−α−1 and then integrating over t∈0,1, we get(14)1n−αϕn∑s=1lδs−∑s=1l−1xs+ys2≤12∫01tn−α−1ϕnt∑s=1lδs−∑s=1l−1xs+1−t∑s=1lδs−∑s=1l−1ysdt+∫01tn−α−1ϕnt∑s=1lδs−∑s=1l−1ys+1−t∑s=1lδs−∑s=1l−1xsdt=12∑s=1l−1ys−xsn−α∫∑s=1lδs−∑s=1l−1ys∑s=1lδs−∑s=1l−1xsϕnuu−∑s=1lδs−∑s=1l−1ysα−n+1du+∫∑s=1lδs−∑s=1l−1ys∑s=1lδs−∑s=1l−1xsϕnu∑s=1lδs−∑s=1l−1xs−uα−n+1du.
In order to apply the definition of the Caputo fractional operators in (14), first, we show that(15)∑s=1lδs−∑s=1l−1ys<∑s=1lδs−∑s=1l−1xs.
By the hypotheses, we havex≺δ and y≺δ, therefore(16)∑s=1l−1ys−∑s=1l−1xs=xl−yl.
Also,(17)xl>yl⇒xl−yl>0.
By substituting (17) in (16), and adding ∑s=1lδs to both sides, we get(18)∑s=1lδs−∑s=1l−1ys<∑s=1lδs−∑s=1l−1xs.
Now (14) implies(19)1n−αϕn∑s=1lδs−∑s=1l−1xs+ys2≤Γn−α2∑s=1l−1ys−xsn−αDc∑s=1lδs−∑δsl−1ys+αϕ∑s=1lδs−∑s=1l−1xs+−1nDc∑s=1lδs−∑s=1l−1xs−αϕ∑s=1lδs−∑s=1l−1ys,and so(20)ϕn∑s=1lδs−∑s=1l−1xs+ys2≤Γn−α+12∑s=1l−1ys−xsn−αDc∑s=1lδs−∑s=1l−1ys+αϕ∑s=1lδs−∑s=1l−1xs+−1nDc∑s=1lδs−∑s=1l−1xs−αϕ∑s=1lδs−∑s=1l−1ys.
Thus, the first inequality of (11) is completed. Now, using the convexity of ϕn, we obtain the second inequality in the following manner:(21)ϕnt∑s=1lδs−∑s=1l−1xs+1−t∑s=1lδs−∑s=1l−1ys≤tϕn∑s=1lδs−∑s=1l−1xs+1−tϕn∑s=1lδs−∑s=1l−1ys,(22)ϕnt∑s=1lδs−∑s=1l−1ys+1−t∑s=1lδs−∑s=1l−1xs≤tϕn∑s=1lδs−∑s=1l−1ys+1−tϕn∑s=1lδs−∑s=1l−1xs.
Adding (21) and (22), and then applying Theorem 2 for r=1 and σ1=1, we obtain(23)ϕnt∑s=1lδs−∑s=1l−1xs+1−t∑s=1lδs−∑s=1l−1ys+ϕnt∑s=1lδs−∑s=1l−1ys+1−t∑s=1lδs−∑s=1l−1xs≤ϕn∑s=1lδs−∑s=1l−1xs+ϕn∑s=1lδs−∑s=1l−1ys≤2∑s=1lϕnδs−∑s=1l−1ϕnxs+∑s=1l−1ϕnys.
By multiplying both sides of (23) by tn−α−1 and then integrating over t∈0,1, we get the second and third inequality in (11).Remark 1.
For the hypothesis of Theorem3, if l=2, then we get the following inequality:(24)ϕnδ1+δ2−x1+y12≤Γn−α+12y1−x1n−αDcδ1+δ2−y1+αϕδ1+δ2−x1+−1nDcδ1+δ2−x1−αϕδ1+δ2−y1,≤ϕnδ1+δ2−x1+ϕnδ1+δ2−y12≤ϕnδ1+ϕnδ2−ϕnx1+ϕny12.
Moreover, forn=1 and α=0, we obtain the result of Kian and Moslehian [35].Remark 2.
If we takex1=δ1, y1=δ2, then inequality (24) reduces to inequality (2.2) in [15].Another result for the Hermite–Hadamard–Jensen–Mercer fractional inequality is given as follows:Theorem 4.
Let all the conditions in the hypothesis of Theorem3 hold. Then,(25)ϕn∑s=1lδs−∑s=1l−1xs+ys2≤2n−α−1Γn−α+1∑s=1l−1ys−xsn−α−1nDc∑s=1lδs−∑s=1l−1xs+ys/2−αϕ∑s=1lδs−∑s=1l−1ys+Dc∑s=1lδs−∑s=1l−1xs+ys/2+αϕ∑s=1lδs−∑s=1l−1xs≤∑s=1lϕnδs−∑s=1l−1ϕnxs+∑s=1l−1ϕnys2.Proof.
Let us considert∈0,1. To prove the required result, we proceed as follows:(26)ϕn∑s=1lδs−∑s=1l−1xs+ys2=ϕn12∑s=1lδs−∑s=1l−1xs+∑s=1lδs−∑s=1l−1ys,=ϕn12∑s=1lδs−t2∑s=1l−1xs+2−t2∑s=1l−1ys+∑s=1lδs−t2∑s=1l−1ys+2−t2∑s=1l−1xs.
Sinceϕn is a convex function, therefore (26) gives the following inequality:(27)ϕn∑s=1lδs−∑s=1l−1xs+ys2≤12ϕn∑s=1lδs−t2∑s=1l−1xs+2−t2∑s=1l−1ys+ϕn∑s=1lδs−t2∑s=1l−1ys+2−t2∑s=1l−1xs.
By multiplying both sides of (27) by tn−α−1 and then integrating over t∈0,1, we obtain(28)1n−αϕn∑s=1lδs−∑s=1l−1xs+ys2≤12∫01tn−α−1ϕn∑s=1lδs−t2∑s=1l−1xs+2−t2∑s=1l−1ysdt+∫01tn−α−1ϕn∑s=1lδs−t2∑s=1l−1ys+2−t2∑s=1l−1xsdt=12∑s=1l−1ys−xs/2n−α∫∑s=1lδs−∑s=1l−1ys∑s=1lδs−∑s=1l−1xs+ys/2ϕnuu−∑s=1lδs−∑s=1l−1ysα−n+1du+∫∑s=1lδs−∑s=1l−1xs+ys/2∑s=1lδ∑s=1lδss−∑s=1l−1xsϕnu∑s=1lδs−∑s=1l−1xs−uα−n+1du.
Following the same procedure, as given in the proof of Theorem3, we can show that(29)∑s=1lδs−∑s=1l−1xs+ys2<∑s=1lδs−∑s=1l−1xs,and∑s=1lδs−∑s=1l−1xs+ys2>∑s=1lδs−∑s=1l−1ys.
Now, from (28), we deduce(30)1n−αϕn∑s=1lδs−∑s=1l−1xs+ys2=2n−α−1Γn−α∑s=1l−1ys−xsn−α×−1nDc∑s=1lδs−∑s=1l−1xs+ys/2−αϕ∑s=1lδs−∑s=1l−1ys+Dc∑s=1lδs−∑s=1l−1xs+ys/2+αϕ∑s=1lδs−∑s=1l−1xs.
So, we have(31)ϕn∑s=1lδs−∑s=1l−1xs+ys2≤2n−α−1Γn−α+1∑s=1l−1ys−xsn−α×−1nDc∑s=1lδs−∑s=1l−1xs+ys/2−αϕ∑s=1lδs−∑s=1l−1ys+Dc∑s=1lδs−∑s=1l−1xs+ys/2+αϕ∑s=1lδs−∑s=1l−1xs.
This proves the first inequality in (25).
In order to prove the second inequality of (25), we use Theorem 2 for r=2, σ1=t/2, and σ2=2−t/2 as follows:(32)ϕn∑s=1lδs−t2∑s=1l−1xs+2−t2∑s=1l−1ys≤∑s=1lϕnδs−t2∑s=1l−1ϕnxs+2−t2∑s=1l−1ϕnys,(33)ϕn∑s=1lδs−t2∑s=1l−1ys+2−t2∑s=1l−1xs≤∑s=1lϕnδs−t2∑s=1l−1ϕnys+2−t2∑s=1l−1ϕnxs.
Adding (32) and (33), we get(34)ϕn∑s=1lδs−t2∑s=1l−1xs+2−t2∑s=1l−1ys+ϕn∑s=1lδs−t2∑s=1l−1ys+2−t2∑s=1l−1xs≤2∑s=1lϕnδs−∑s=1l−1ϕnxs+∑s=1l−1ϕnys.
By multiplying both sides of (34) by tn−α−1 and then integrating over t∈0,1, we obtain the second inequality of (25).We establish the following result for the Caputo fractional operators on the basis of Lemma1.Theorem 5.
Let us consider a functionϕ∈CnI, such that ϕn is a convex function on I and δ=δ1,…,δl, x=x1,…,xl, y=y1,…,yl, and p=p1,…,pl are four l-tuples, such that δs,xs,ys∈I, ps≥0 with pl≠0 for all s∈1,…,l, η=1/pl, xl>yl, α>0. If x and y are decreasing l-tuples and(35)∑s=1kpsxs≤∑s=1kpsδs,∑s=1kpsys≤∑s=1kpsδs,fork=1,…,l−1,∑s=1lpsδs=∑s=1lpsxs,∑s=1lpsδs=∑s=1lpsys,then(36)ϕn∑s=1lηpsδs−η∑s=1l−1psxs+psys2≤Γn−α+12∑s=1l−1ηpsys−ηpsxsn−αDc∑s=1lηpsδs−∑s=1l−1ηpsys+αϕ∑s=1lηpsδs−∑s=1l−1ηpsxs+−1nDc∑s=1lηpsδs−∑s=1l−1ηpsxs−αϕ∑s=1lηpsδs−∑s=1l−1ηpsys≤ϕn∑s=1lηpsδs−∑s=1l−1ηpsys+ϕn∑s=1lηpsδs−∑s=1l−1ηpsxs2≤∑s=1lηpsϕnδs−∑s=1l−1ηpsϕnxs+∑s=1l−1ηpsϕnys2.Proof.
It can be written as(37)ϕn∑s=1lηpsδs−η∑s=1l−1psxs+ys2=ϕn12∑s=1lηpsδs−∑s=1l−1ηpsxs+∑s=1lηpsδs−∑s=1l−1ηpsys=ϕn12t∑s=1lηpsδs−∑s=1l−1ηpsxs+1−t∑s=1lηpsδs−∑s=1l−1ηpsys+t∑s=1lηpsδs−∑s=1l−1ηpsys+1−t∑s=1lηpsδs−∑s=1l−1ηpsxs.
Sinceϕn is a convex function, therefore (37) gives the following inequality:(38)ϕn∑s=1lηpsδs−η∑s=1l−1psxs+ys2≤12ϕnt∑s=1lηpsδs−∑s=1l−1ηpsxs+1−t∑s=1lηpsδs−∑s=1l−1ηpsys+ϕnt∑s=1lηpsδs−∑s=1l−1ηpsys+1−t∑s=1lηpsδs−∑s=1l−1ηpsxs.
By multiplying both sides of (38) by tn−α−1 and then integrating over t∈0,1, we get(39)1n−αϕn∑s=1lηpsδs−η∑s=1l−1psxs+ys2≤12∫01tn−α−1ϕnt∑s=1lηpsδs−∑s=1l−1ηpsxs+1−t∑s=1lηpsδs−∑s=1l−1ηpsysdt+∫01tn−α−1ϕnt∑s=1lηpsδs−∑s=1l−1ηpsys+1−t∑s=1lηpsδs−∑s=1l−1ηpsxsdt,=12∑s=1l−1ηpsys−ηpsxsn−α∫∑s=1lηpsδs−∑s=1l−1ηpsys∑s=1lηpsδs−∑s=1l−1ηpsxsϕnuu−∑s=1lηpsδs−∑s=1l−1ηpsysα−n+1du+∫∑s=1lηpsδs−∑s=1l−1ηpsys∑s=1lηpsδs−∑s=1l−1ηpsxsϕnu∑s=1lηpsδs−∑s=1l−1ηpsxs−uα−n+1du.
In order to apply the definition of the Caputo fractional operators in (39), first, we show that(40)∑s=1lηpsδs−∑s=1l−1ηpsys<∑s=1lηpsδs−∑s=1l−1ηpsxs.
Given that∑s=1lpsδs=∑s=1lpsxs and ∑s=1lpsδs=∑s=1lpsys, we have(41)∑s=1l−1psys−∑s=1l−1psxs=plxl−plyl.
Also,(42)xl>yl⇒plxl>plyl⇒plxl−plyl>0.
By substituting (42) in (41), and adding ∑s=1lηpsδs to both sides, we get(43)∑s=1lηpsδs−∑s=1l−1ηpsys<∑s=1lηpsδs−∑s=1l−1ηpsxs.
Now (39) implies(44)1n−αϕn∑s=1lηpsδs−η∑s=1l−1psxs+ys2≤Γn−α2∑s=1l−1ηpsys−ηpsxsn−α×Dc∑s=1lηpsδs−∑s=1l−1ηpsys+αϕ∑s=1lηpsδs−∑s=1l−1ηpsxs+−1nDc∑s=1lηpsδs−∑s=1l−1ηpsys−αϕ∑s=1lηpsδs−∑s=1l−1ηpsys,and so(45)ϕn∑s=1lηpsδs−η∑s=1l−1psxs+ys2≤Γn−α+12∑s=1l−1ηpsys−ηpsxsn−α×Dc∑s=1lηpsδs−∑s=1l−1ηpsys+αϕ∑s=1lηpsδs−∑s=1l−1ηpsxs+−1nDc∑s=1lηpsδs−∑s=1l−1ηpsxs−αϕ∑s=1lηpsδs−∑s=1l−1ηpsys.
Thus, we achieved the first inequality of (36).
To prove the second inequality, from the convexity ofϕn we may write that(46)ϕnt∑s=1lηpsδs−∑s=1l−1ηpsxs+1−t∑s=1lηpsδs−∑s=1l−1ηpsys≤tϕn∑s=1lηpsδs−∑s=1l−1ηpsxs+1−tϕn∑s=1lηpsδs−∑s=1l−1ηpsys,(47)ϕnt∑s=1lηpsδs−∑s=1l−1ηpsys+1−t∑s=1lηpsδs−∑s=1l−1ηpsxs≤tϕn∑s=1lηpsδs−∑s=1l−1ηpsys+1−tϕn∑s=1lηpsδs−∑s=1l−1ηpsxs.
Adding (46) and (47) and then using Lemma 1 for r=2, σ1=t, and σ2=1−t, we obtain(48)ϕnt∑s=1lηpsδs−∑s=1l−1ηpsxs+1−t∑s=1lηpsδs−∑s=1l−1ηpsys+ϕnt∑s=1lηpsδs−∑s=1l−1ηpsys+1−t∑s=1lηpsδs−∑s=1l−1ηpsxs≤ϕn∑s=1lηpsδs−∑s=1l−1ηpsxs+ϕn∑s=1lηpsδs−∑s=1l−1ηpsys≤2∑s=1lηpsϕnδs−∑s=1l−1ηpsϕnxs+∑s=1l−1ηpsϕnys.
By multiplying both sides of (48) by tn−α−1 and then integrating over t∈0,1, we get the second and third inequality in (36).We establish the following result for the Caputo fractional operators on the basis of Lemma2.Theorem 6.
Let us consider a functionϕ∈CnI, such that ϕn is a convex function on I and δ=δ1,…,δl, x=x1,…,xl, y=y1,…,yl, and p=p1,…,pl are four l-tuples, such that δs,xs,ys∈I, ps≥0 with pl≠0 for all s∈1,…,l, η=1/pl, xl>yl, α>0. If δ−x, x, δ−y, and y are monotonically in the same sense and(49)∑s=1lpsδs=∑s=1lpsxs,∑s=1lpsδs=∑s=1lpsys,then(50)ϕn∑s=1lηpsδs−η∑s=1l−1psxs+psys2≤Γn−α+12∑s=1l−1ηpsys−ηpsxsn−αDc∑s=1lηpsδs−∑s=1l−1ηpsys+αϕ∑s=1lηpsδs−∑s=1l−1ηpsxs+−1nDc∑s=1lηpsδs−∑s=1l−1ηpsxs−αϕ∑s=1lηpsδs−∑s=1l−1ηpsys≤ϕn∑s=1lηpsδs−∑s=1l−1ηpsys+ϕn∑s=1lηpsδs−∑s=1l−1ηpsxs2≤∑s=1lηpsϕnδs−∑s=1l−1ηpsϕnxs+∑s=1l−1ηpsϕnys2.Proof.
By using Lemma2 and following the procedure given in the proof of Theorem 5, we can obtain (50).Remark 3.
Theorem5 and Theorem 6 provide weighted forms of Theorem 3.Remark 4.
The weighted versions of Theorem4 can be obtained in a similar fashion.


## 3. Bounds Associated with the Main Results

In this section first, we discover two new identities associated with the right- and left-sides of the main results. Then utilizing these identities, we establish bounds for the absolute difference of the two right- and left-most terms of the main results.Lemma 3.
Let us consider a differentiable functionϕ defined on I, such that ϕ∈Cn+1I and δ=δ1,…,δl, x=x1,…,xl, and y=y1,…,yl are three l− tuples, such that δs,xs,ys∈I, for all s∈1,…,l, α>0, t∈0,1. If ϕn+1∈LI, then(51)ϕn∑s=1lδs−∑s=1l−1ys+ϕn∑s=1lδs−∑s=1l−1xs2−Γn−α+12∑s=1l−1ys−xsn−α×Dc∑s=1lδs−∑s=1l−1ys+αϕ∑s=1lδs−∑s=1l−1xs+−1nDc∑s=1lδs−∑s=1l−1xs−αϕ∑s=1lδs−∑s=1l−1ys=∑s=1l−1ys−xs2∫01tn−α−1−tn−αϕn+1∑s=1lδs−∑s=1l−1txs+1−tysdt.Proof.
To prove our required result, we consider that(52)I=∫01tn−α−1−tn−αϕn+1∑s=1lδs−∑s=1l−1txs+1−tysdt,=∫01tn−αϕn+1∑s=1lδs−∑s=1l−1txs+1−tysdt−∫011−tn−αϕn+1∑s=1lδs−∑s=1l−1txs+1−tysdt=I1−I2.
Assuming that∑s=1lδs−∑s=1l−1ys<∑s=1lδs−∑s=1l−1xs and using integration by parts formula, we obtain(53)I1=∫01tn−αϕn+1∑s=1lδs−∑s=1l−1txs+1−tysdt=tn−αϕn∑s=1lδs−∑s=1l−1txs+1−tys∑s=1l−1ys−xs01−n−α∑s=1l−1ys−xs×∫01tn−α−1ϕn∑s=1lδs−∑s=1l−1txs+1−tysdt,=ϕn∑s=1lδs−∑s=1l−1xs∑s=1l−1ys−xs−Γn−α+1∑s=1l−1ys−xsn−α+1−1nDc∑s=1lδs−∑s=1l−1xs−αϕ∑s=1lδs−∑s=1l−1ys.
Similarly,(54)I2=∫011−tn−αϕn+1∑s=1lδs−∑s=1l−1txs+1−tysdt=1−tn−αϕn∑s=1lδs−∑s=1l−1txs+1−tys∑s=1l−1ys−xs01+n−α∑s=1l−1ys−xs×∫011−tn−α−1ϕn∑s=1lδs−∑s=1l−1txs+1−tysdt=−ϕn∑s=1lδs−∑s=1l−1ys∑s=1l−1ys−xs+Γn−α+1∑s=1l−1ys−xsn−α+1Dc∑s=1lδs−∑s=1l−1ys+αϕ∑s=1lδs−∑s=1l−1xs.
Now, we have(55)I=ϕn∑s=1lδs−∑s=1l−1xs+ϕn∑s=1lδs−∑s=1l−1ys∑s=1l−1ys−xs−Γn−α+1∑s=1l−1ys−xsn−α+1×Dc∑s=1lδs−∑s=1l−1ys+αϕ∑s=1lδs−∑s=1l−1xs+−1nDc∑s=1lδs−∑s=1l−1xs−αϕ∑s=1lδs−∑s=1l−1ys.
Multiplying both sides by∑s=1l−1ys−xs/2, we get (51). □Remark 5.
Lemma3 gives the following equality for x1=δ1, y1=δ2, and l=2 proved by Farid et al. [15].(56)ϕnδ1+ϕnδ22−Γn−α+12δ2−δ1n−αDcδ1+αϕδ2+−1nDcδ2−αϕδ1=δ2−δ12∫01tn−α−1−tn−αϕn+1tδ2+1−tδ1dt.Remark 6.
If we takeα=0 and n=1 in Remark 5, then the equality (56) gives(57)ϕδ1+ϕδ22−1δ2−δ1∫δ1δ2ϕudu=δ2−δ12∫012t−1ϕ′tδ2+1−tδ1dt.
The equality (57) has been proved by Dragomir and Agarwal [5].The following results have been established on the basis of Lemma3:Theorem 7.
Let us consider a differentiable functionϕ defined on I, such that ϕ∈Cn+1I and δ=δ1,…,δl, x=x1,…,xl, y=y1,…,yl are three l− tuples, such that δs,xs,ys∈I, for all s∈1,…,l, xl>yl, α>0. If δ majorizes x, y, and ϕn+1 is convex on I, then(58)ϕn∑s=1lδs−∑s=1l−1ys+ϕn∑s=1lδs−∑s=1l−1xs2−Γn−α+12∑s=1l−1ys−xsn−α×Dc∑s=1lδs−∑s=1l−1ys+αϕ∑s=1lδs−∑s=1l−1xs+−1nDc∑s=1lδs−∑s=1l−1xs−αϕ∑s=1lδs−∑s=1l−1ys≤∑s=1l−1ys−xsn−α+11−12n−α∑s=1lϕn+1δs−∑s=1l−1ϕn+1xs+∑s=1l−1ϕn+1ys2.Proof.
From Lemma3, it follows that(59)ϕn∑s=1lδs−∑s=1l−1ys+ϕn∑s=1lδs−∑s=1l−1xs2−Γn−α+12∑s=1l−1ys−xsn−α×Dc∑s=1lδs−∑s=1l−1ys+αϕ∑s=1lδs−∑s=1l−1xs+−1nDc∑s=1lδs−∑s=1l−1xs−αϕ∑s=1lδs−∑s=1l−1ys=∑s=1l−1ys−xs2∫01tn−α−1−tn−αϕn+1∑s=1lδs−∑s=1l−1txs+1−tysdt≤∑s=1l−1ys−xs2∫01tn−α−1−tn−αϕn+1∑s=1lδs−∑s=1l−1txs+1−tysdt.
Using Theorem2 for r=2, σ1=t, and σ2=1−t in (59) as a consequence of the convexity of ϕn+1, we obtain(60)≤∑s=1l−1ys−xs2∫01tn−α−1−tn−α×∑s=1lϕn+1δs−t∑s=1l−1ϕn+1xs+1−t∑s=1l−1ϕn+1ysdt.=∑s=1l−1ys−xs2∫01/21−tn−α−tn−α∑s=1lϕn+1δs−t∑s=1l−1ϕn+1xs+1−t∑s=1l−1ϕn+1ysdt+∫1/21tn−α−1−tn−α×∑s=1lϕn+1δs−t∑s=1l−1ϕn+1xs+1−t∑s=1l−1ϕn+1ysdt.=∑s=1l−1ys−xs2C1+C2.
Now findingC1 and C2, we have(61)C1=∫01/21−tn−α−tn−α∑s=1lϕn+1δs−t∑s=1l−1ϕn+1xs+1−t∑s=1l−1ϕn+1ysdt=∑s=1lϕn+1δs∫0121−tn−α−tn−αdt−∑s=1l−1ϕn+1xs×∫012t1−tn−α−tn−αdt+∑s=1l−1ϕn+1ys∫01/21−tn−α−tn−α1−tdt=∑s=1lϕn+1δs1−2α−nn−α+1−∑s=1l−1ϕn+1xs∫01/2t1−tn−αdt−∫01/2tn−α+1dt+∑s=1l−1ϕn+1ys∫01/21−tn−α+1dt−∫01/21−ttn−αdt=∑s=1lϕn+1δs1−2α−nn−α+1−∑s=1l−1ϕn+1xs×1n−α+1n−α+2−2α−n−1n−α+2+∑s=1l−1ϕn+1ys1n−α+2−2α−n−1n−α+1,C2=∫1/21tn−α−1−tn−α∑s=1lϕn+1δs−t∑s=1l−1ϕn+1xs+1−t∑s=1l−1ϕn+1ysdt=∑s=1lϕn+1δs∫1/21tn−α−1−tn−αdt−∑s=1l−1ϕn+1xs×∫1/21ttn−α−1−tn−αdt+∑s=1l−1ϕn+1ys∫1/21tn−α−1−tn−α1−tdt=∑s=1lϕn+1δs1−2α−nn−α+1−∑s=1l−1ϕn+1xs∫1/21tn−α+1dt−∫1/21t1−tn−αdt+∑s=1l−1ϕn+1ys∫1211−ttn−αdt−∫1/211−tn−α+1dt=∑s=1lϕn+1δs1−2α−nn−α+1−∑s=1l−1ϕn+1xs1n−α+2−2α−n−1n−α+1+∑s=1l−1ϕn+1ys1n−α+1n−α+2−2α−n−1n−α+2.
AddingC1 and C2, we get(62)C1+C2=21−2α−nn−α+1∑s=1lϕn+1δs−∑s=1l−1ϕn+1xs+∑s=1l−1ϕn+1ys2.
Inserting (62) in (60), we achieve (58).Corollary 1.
If we takel=2, x1=δ1, and y1=δ2 in Theorem 7, then inequality (58) reduces to(63)ϕnδ1+ϕnδ22−Γn−α+12δ2−δ1n−αDcδ1+αϕδ2+−1nDcδ2−αϕδ1≤δ2−δ1n−α+11−12n−αϕn+1δ1+ϕn+1δ22,which is proved in [15].Theorem 8.
Let us consider a differentiable functionϕ defined on I, such that ϕ∈Cn+1I and δ=δ1,…,δl, x=x1,…,xl, y=y1,…,yl are three l− tuples, such that δs,xs,ys∈I, for all s∈1,…,l, xl>yl, α>0. If q>1, δ majorizes x, y, and ϕn+1q is convex on I, then(64)ϕn∑s=1lδs−∑s=1l−1ys+ϕn∑s=1lδs−∑s=1l−1xs2−Γn−α+12∑s=1l−1ys−xsn−α×Dc∑s=1lδs−∑s=1l−1ys+αϕ∑s=1lδs−∑s=1l−1xs+−1nDc∑s=1lδs−∑s=1l−1xs−αϕ∑s=1lδs−∑s=1l−1ys≤∑s=1l−1ys−xsn−α+11−12n−α∑s=1lϕn+1δsq−∑s=1l−1ϕn+1xsq+∑s=1l−1ϕn+1ysq21/q.Proof.
From Lemma3, it follows that(65)ϕn∑s=1lδs−∑s=1l−1ys+ϕn∑s=1lδs−∑s=1l−1xs2−Γn−α+12∑s=1l−1ys−xsn−α×Dc∑s=1lδs−∑s=1l−1ys+αϕ∑s=1lδs−∑s=1l−1xs+−1nDc∑s=1lδs−∑s=1l−1xs−αϕ∑s=1lδs−∑s=1l−1ys=∑s=1l−1ys−xs2∫01tn−α−1−tn−αϕn+1∑s=1lδs−∑s=1l−1txs+1−tysdt≤∑s=1l−1ys−xs2∫01tn−α−1−tn−αϕn+1∑s=1lδs−∑s=1l−1txs+1−tysdt.
By applying power mean inequality to the above integral, we obtain(66)≤∑s=1l−1ys−xs2∫01tn−α−1−tn−αdt1−1/q×∫01tn−α−1−tn−α×ϕn+1∑s=1lδs−∑s=1l−1txs+1−tysqdt1/q=∑s=1l−1ys−xs2∫01/21−tn−α−tn−αdt+∫1/21tn−α−1−tn−αdt1−1/q×∫01tn−α−1−tn−αϕn+1∑s=1lδs−∑s=1l−1txs+1−tysqdt1/q.
Sinceϕn+1q is convex, therefore using Theorem 2 for r=2, σ1=t, and σ2=1−t in (66), we obtain(67)=∑s=1l−1ys−xs2∫01/21−tn−α−tn−αdt+∫1/21tn−α−1−tn−αdt1−1/q×∫01tn−α−1−tn−α∑s=1lϕn+1δsq−t∑s=1l−1ϕn+1xsq+1−t∑s=1l−1ϕn+1ysqdt1/q=∑s=1l−1ys−xs2∫01/21−tn−α−tn−αdt+∫1/21tn−α−1−tn−αdt1−1/q×∫01/21−tn−α−tn−α∑s=1lϕn+1δsq−t∑s=1l−1ϕn+1xsq+1−t∑s=1l−1ϕn+1ysqdt+∫1/21tn−α−1−tn−α∑s=1lϕn+1δsq−t∑s=1l−1ϕn+1xsq+1−t∑s=1l−1ϕn+1ysqdt1/q.
By calculating these simple integrals, we get (64).Another lemma is established as follows:Lemma 4.
Let all the conditions in the hypothesis of Lemma3 hold. Then,(68)2n−α−1Γn−α+1∑s=1l−1ys−xsn−αDc∑s=1lδs−∑s=1l−1xs+ys/2+αϕ∑s=1lδs−∑s=1l−1xs+−1nDc∑s=1lδs−∑s=1l−1xs+ys/2−αϕ∑s=1lδs−∑s=1l−1ys−ϕn∑s=1lδs−∑s=1l−1xs+ys2=∑s=1l−1ys−xs4∫01tn−αϕn+1∑s=1lδs−2−t2∑s=1l−1xs+t2∑s=1l−1ysdt−∫01tn−αϕn+1∑s=1lδs−2−t2∑s=1l−1ys+t2∑s=1l−1xsdt.Proof.
It can be easily proved by following the procedure given in the proof of Lemma3.Remark 7.
When we takel=2, x1=δ1, and y1=δ2 in Lemma 4, then it reduces to the equality (3.1) in [16].The following results have been established on the basis of Lemma4.Theorem 9.
Let all the conditions in the hypothesis of Theorem7 hold. Then,(69)2n−α−1Γn−α+1∑s=1l−1ys−xsn−αDc∑s=1lδs−∑s=1l−1xs+ys/2+αϕ∑s=1lδs−∑s=1l−1xs+−1nDc∑s=1lδs−∑s=1l−1xs+ys/2−αϕ∑s=1lδs−∑s=1l−1ys−ϕn∑s=1lδs−∑s=1l−1xs+ys2≤∑s=1l−1ys−xs2n−α+1∑s=1lϕn+1δs−∑s=1l−1ϕn+1xs+∑s=1l−1ϕn+1ys2.Proof.
From Lemma4, it follows that(70)2n−α−1Γn−α+1∑s=1l−1ys−xsn−αDc∑s=1lδs−∑s=1l−1xs+ys/2+αϕ∑s=1lδs−∑s=1l−1xs+−1nDc∑s=1lδs−∑s=1l−1xs+ys/2−αϕ∑s=1lδs−∑s=1l−1ys−ϕn∑s=1lδs−∑s=1l−1xs+ys2=∑s=1l−1ys−xs4∫01tn−αϕn+1∑s=1lδs−2−t2∑s=1l−1xs+t2∑s=1l−1ysdt−∫01tn−αϕn+1∑s=1lδs−2−t2∑s=1l−1ys+t2∑s=1l−1xsdt,≤∑s=1l−1ys−xs4∫01tn−αϕn+1∑s=1lδs−2−t2∑s=1l−1xs+t2∑s=1l−1ysdt+∫01tn−αϕn+1∑s=1lδs−2−t2∑s=1l−1ys+t2∑s=1l−1xsdt.
By utilizing Theorem2 for r=2, σ1=2−t/2, and σ2=t/2 in (70), we obtain(71)≤∑s=1l−1ys−xs4∫01tn−α∑s=1lϕn+1δs−2−t2∑s=1l−1ϕn+1xs+t2∑s=1l−1ϕn+1ysdt+∫01tn−α∑s=1lϕn+1δs−2−t2∑s=1l−1ϕn+1ys+t2∑s=1l−1ϕn+1xs|dt=∑s=1l−1ys−xs4∑s=1lϕn+1δsn−α+1−∑s=1l−1ϕn+1xsn−α+1+∑s=1l−1ϕn+1xs2n−α+2−∑s=1l−1ϕn+1ys2n−α+2+∑s=1lϕn+1δsn−α+1−∑s=1l−1ϕn+1ysn−α+1+∑s=1l−1ϕn+1ys2n−α+2−∑s=1l−1ϕn+1xs2n−α+2=∑s=1l−1ys−xs2n−α+1∑s=1lϕn+1δs−∑s=1l−1ϕn+1xs+∑s=1l−1ϕn+1ys2.
This finishes the proof.Corollary 2.
If we takel=2, n=1, and α=0 in Theorem 9, then we get the following inequality:(72)1y1−x1∫δ1+δ2−y1δ1+δ2−x1ϕudu−ϕδ1+δ2−x1+y12≤y1−x14ϕ′δ1+ϕ′δ2−ϕ′x1+ϕ′y12.Theorem 10.
Let us consider a differentiable functionϕ defined on I, such that ϕ∈Cn+1I and δ=δ1,…,δl, x=x1,…,xl, y=y1,…,yl are three l− tuples, such that δs,xs,ys∈I, for all s∈1,…,l, xl>yl, α>0. If q>1 such that 1/p+1/q=1, δ majorizes x, y, and ϕn+1q is convex on I, then(73)2n−α−1Γn−α+1∑s=1l−1ys−xsn−αDc∑s=1lδs−∑s=1l−1xs+ys/2+αϕ∑s=1lδs−∑s=1l−1xs+−1nDc∑s=1lδs−∑s=1l−1xs+ys/2−αϕ∑s=1lδs−∑s=1l−1ys−ϕn∑s=1lδs−∑s=1l−1xs+ys2≤∑s=1l−1ys−xs164np−αp+11/p41/q·2∑s=1lϕn+1δs−31/q+1∑s=1l−1ϕn+1xs+∑s=1l−1ϕn+1ys.Proof.
From Lemma4, it follows that(74)2n−α−1Γn−α+1∑s=1l−1ys−xsn−αDc∑s=1lδs−∑s=1l−1xs+ys/2+αϕ∑s=1lδs−∑s=1l−1xs+−1nDc∑s=1lδs−∑s=1l−1xs+ys/2−αϕ∑s=1lδs−∑s=1l−1ys−ϕn∑s=1lδs−∑s=1l−1xs+ys2=∑s=1l−1ys−xs4∫01tn−αϕn+1∑s=1lδs−2−t2∑s=1l−1xs+t2∑s=1l−1ysdt−∫01tn−αϕn+1∑s=1lδs−2−t2∑s=1l−1ys+t2∑s=1l−1xsdt,≤∑s=1l−1ys−xs4∫01tn−αϕn+1∑s=1lδs−2−t2∑s=1l−1xs+t2∑s=1l−1ysdt+∫01tn−αϕn+1∑s=1lδs−2−t2∑s=1l−1ys+t2∑s=1l−1xsdt.
By applying Hölder’s inequality to the above integral, we have(75)≤∑s=1l−1ys−xs4∫01tn−αpdt1/p∫01ϕn+1∑s=1lδs−2−t2∑s=1l−1xs+t2∑s=1l−1ysqdt1/q+∫01tn−αpdt1/p∫01ϕn+1∑s=1lδs−2−t2∑s=1l−1ys+t2∑s=1l−1xsqdt1/q=∑s=1l−1ys−xs4∫01tn−αpdt1/p∫01ϕn+1∑s=1lδs−2−t2∑s=1l−1xs+t2∑s=1l−1ysqdt1/q+∫01ϕn+1∑s=1lδs−2−t2∑s=1l−1ys+t2∑s=1l−1xsqdt1/q.
Sinceϕn+1q is convex, therefore using Theorem 2 for r=2, σ1=2−t/2, and σ2=t/2 in (75), we obtain(76)=∑s=1l−1ys−xs41np−αp+11/p∫01∑s=1lϕn+1δsq−2−t2∑s=1l−1ϕn+1xsq+t2∑s=1l−1ϕn+1ysqdt1/q+∫01∑s=1lϕn+1δsq−2−t2∑s=1l−1ϕn+1ysq+t2∑s=1l−1ϕn+1xsqdt1/q=∑s=1l−1ys−xs41np−αp+11/p∑s=1lϕn+1δsq−143∑s=1l−1ϕn+1xsq+∑s=1l−1ϕn+1ysq1/q+∑s=1lϕn+1δsq−143∑s=1l−1ϕn+1ysq+∑s=1l−1ϕn+1xsq1/q.
By using Minkowski’s inequality, we get(77)=∑s=1l−1ys−xs164np−αp+11/p41/q·2∑s=1lϕn+1δs−31/q+1∑s=1l−1ϕn+1xs+∑s=1l−1ϕn+1ys.
This completes the proof.Remark 8.
If we choosel=2 in Theorem 10, then we get inequality (36) in [14].Theorem 11.
Let all the conditions in the hypothesis of Theorem8 hold. Then,(78)2n−α−1Γn−α+1∑s=1l−1ys−xsn−αDc∑s=1lδs−∑s=1l−1xs+ys/2+αϕ∑s=1lδs−∑s=1l−1xs+−1nDc∑s=1lδs−∑s=1l−1xs+ys/2−αϕ∑s=1lδs−∑s=1l−1ys−ϕn∑s=1lδs−∑s=1l−1xs+ys2≤∑s=1l−1ys−xs41n−α+11−1/q1n−α+1∑s=1lϕn+1δsq−n−α+32n−α+1n−α+2∑s=1l−1ϕn+1xsq+12n−α+2∑s=1l−1ϕn+1ysq1/q+1n−α+1∑s=1lϕn+1δsq−n−α+32n−α+1n−α+2∑s=1l−1ϕn+1ysq+12n−α+2∑s=1l−1ϕn+1xsq1/q.Proof.
From Lemma4, it follows that(79)2n−α−1Γn−α+1∑s=1l−1ys−xsn−αDc∑s=1lδs−∑s=1l−1xs+ys/2+αϕ∑s=1lδs−∑s=1l−1xs+−1nDc∑s=1lδs−∑s=1l−1xs+ys/2−αϕ∑s=1lδs−∑s=1l−1ys−ϕn∑s=1lδs−∑s=1l−1xs+ys2=∑s=1l−1ys−xs4∫01tn−αϕn+1∑s=1lδs−2−t2∑s=1l−1xs+t2∑s=1l−1ysdt−∫01tn−αϕn+1∑s=1lδs−2−t2∑s=1l−1ys+t2∑s=1l−1xsdt.≤∑s=1l−1ys−xs4∫01tn−αϕn+1∑s=1lδs−2−t2∑s=1l−1xs+t2∑s=1l−1ysdt+∫01tn−αϕn+1∑s=1lδs−2−t2∑s=1l−1ys+t2∑s=1l−1xsdt.
By applying power mean inequality to the above integral, we get(80)≤∑s=1l−1ys−xs4∫01tn−αdt1−1/q∫01tn−αϕn+1∑s=1lδs−2−t2∑s=1l−1xs+t2∑s=1l−1ysqdt1/q+∫01tn−αdt1−1/q∫01tn−αϕn+1∑s=1lδs−2−t2∑s=1l−1ys+t2∑s=1l−1xsqdt1/q=∑s=1l−1ys−xs41n−α+11−1/q∫01tn−αϕn+1∑s=1lδs−2−t2∑s=1l−1xs+t2∑s=1l−1ysqdt1/q+∫01tn−αϕn+1∑s=1lδs−2−t2∑s=1l−1ys+t2∑s=1l−1xsqdt1/q.
Due to the convexity ofϕn+1q, using Theorem 2 for r=2, σ1=2−t/2, and σ2=t/2 in (80), we have(81)=∑s=1l−1ys−xs41n−α+11−1/q∫01tn−α∑s=1lϕn+1δsq−2−t2∑s=1l−1ϕn+1xsq+t2∑s=1l−1ϕn+1ysqdt1/q+∫01tn−α∑s=1lϕn+1δsq−2−t2∑s=1l−1ϕn+1ysq+t2∑s=1l−1ϕn+1xsqdt1/q=∑s=1l−1ys−xs41n−α+11−1/q1n−α+1∑s=1lϕn+1δsq−n−α+32n−α+1n−α+2∑s=1l−1ϕn+1xsq+12n−α+2∑s=1l−1ϕn+1ysq1/q+1n−α+1∑s=1lϕn+1δsq−n−α+32n−α+1n−α+2∑s=1l−1ϕn+1ysq+12n−α+2∑s=1l−1ϕn+1xsq1/q.Hence, the proof is completed.Corollary 3.
Forl=2, n=1, and α=0, Theorem 11 gives the following inequality:(82)1y1−x1∫δ1+δ2−y1δ1+δ2−x1ϕudu−ϕδ1+δ2−x1+y12≤y1−x123q−1/qϕ′δ1+ϕ′δ22−2ϕ′x1q+ϕ′y1q21/q+ϕ′δ1+ϕ′δ22−2ϕ′y1q+ϕ′x1q21/q.Remark 9.
We can also obtain weighted versions for all the results derived in this section.


## 4. Conclusion

A new idea in the form of unified inequalities has been put forward. Tools that helped during the development of the main results are the notions of some existing inequalities, majorization theory, and various forms of convex functions. The results have been put up in the context of Hermite–Hadamard–Jensen–Mercer inequalities. The selection of the present areas of inequalities has been made on the basis of their consistent attraction for researchers and their vast applicability in enormous fields. Both the weighted and unweighted versions of the obtained results have been presented. Moreover, some new identities for differentiable functions have been derived. Using these identities and considering the convexity ofϕn+1 and ϕn+1qq>1, bounds for the absolute difference of the right- and left-sides of the main results have been provided.



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