# Some Integral Inequalities forn-Polynomial ζ-Preinvex Functions

Authors: Shanhe Wu; Muhammad Uzair Awan; Muhammad Ubaid Ullah; Sadia Talib; Artion Kashuri
Journal: Journal of Function Spaces
 (2021)

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## Abstract

In this paper, we study the properties ofn-polynomial ζ-preinvex functions and establish some integral inequalities of Hermite-Hadamard type via this class of convex functions. Moreover, we discuss some special cases which provide a significant complement to the integral estimations of preinvex functions. Applications of the obtained results to the inequalities for special means are also considered.


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## Body

## 1. Introduction and Preliminaries

The geometric inequalities involving volume, surface area, mean width, etc. in the Orlicz space have attracted considerable attention of researchers, and the convexity properties of functions have been a powerful tool for dealing with various problems of convex geometry (see [1, 2]). This suggests that it is a significant work to develop new inequalities for generalized convex functions. For this purpose, let us start with recalling some concepts and notations on the convexity of functions.A setC⊂ℝ is said to be convex if
(1)1−tx+ty∈C,for any x,y∈C and t∈0,1.A functionF:C↦ℝ is said to be convex if the inequality
(2)F1−tx+ty≤1−tFx+tFyholds for any x,y∈C and t∈0,1.In recent years, the classical concept of convexity has been extended and generalized in different directions. Mititelu [3] introduced the notion of invex set, as follows.Definition 1 [3].
LetX⊂ℝ be a nonempty set and η:ℝ×ℝ↦ℝ be a real-valued function. A set X is said to be invex with respect to η if
(3)x+tηy,x∈X,for all x,y∈X and t∈0,1.The invexity would reduce to the classical convexity ifηy,x=y−x. Weir and Mond [4] defined the class of preinvex functions as follows.Definition 2 [4].
LetX⊂ℝ be a nonempty invex set with respect to η:ℝ×ℝ↦ℝ. A function F:X↦ℝ is said to be preinvex with respect to η if the inequality
(4)Fx+tηy,x≤1−tFx+tFyholds for all x,y∈X and t∈0,1.
As a generalization of convex functions, Gordji et al. [5] introduced the notion of ζ-convex function.Definition 3 [5].
A functionF:I⊂ℝ↦ℝ is said to be ζ-convex function with respect to ζ:ℝ×ℝ↦ℝ if the inequality
(5)Ftx+1−ty≤Fy+tζFx,Fy
holds for allx,y∈I and t∈0,1.The properties of convexity have numerous applications in different fields of pure and applied mathematics; especially, the concept of convexity has close relation with the theory of inequalities. Many inequalities are direct consequences of the applications of classical convexity. As is known to us, the Hermite-Hadamard inequality is one of the most significant result associated with convex functions, it reads as follows.LetF:a,b⊂ℝ→ℝ be a convex function, then
(6)Fa+b2≤1b−a∫abFxdx≤Fa+Fb2.Noor [6] obtained a generalization of classical Hermite-Hadamard’s inequality using the class of preinvex functions, as follows.LetF:a,a+ηb,a→ℝ be a preinvex function, then
(7)F2a+ηb,a2≤1ηb,a∫aa+ηb,aFxdx≤Fa+Fb2.The result of Noor has inspired a lot of investigators to deal with new generalizations and refinements of Hermite-Hadamard’s inequality via preinvexity. For example, Barani et al. [7] obtained the generalizations of Hermite-Hadamard’s inequality for functions whose derivative absolute values are preinvex. Du et al. [8] and Noor et al. [9] obtained several generalizations of Hermite-Hadamard’s inequality via s,m-preinvex functions and h-preinvex functions, respectively. Park [10, 11] derived several variations of Hermite-Hadamard’s inequality from differentiable preinvex functions. Sarikaya et al. [12] and Wu et al. [13] established the Hermite-Hadamard-like type inequalities via log-preinvex functions and harmonically p,h,m-preinvex functions, respectively. Wang and Liu [14] and Li [15] obtained different refinements of Hermite-Hadamard’s inequality using s-preinvex functions. Deng et al. [16, 17] and Wu et al. [18] deduced some quantum Hermite-Hadamard-type inequalities by using generalized s,m-preinvex functions and strongly preinvex functions, respectively.Recently, Toplu et al. [19] proposed the concept of n-polynomial convex functions and investigated their properties.In this paper, we shall introduce a new class ofn-polynomial convex functions based on a different form of inequality in the definition compared with [19], which is convenient to the generalizations and applications of n-polynomial convexity. More specifically, we will define a class of convex functions called as n-polynomial ζ-preinvex functions. We then show that this class of convex functions contains a number of other classes of convex functions. Furthermore, we establish some new integral inequalities of Hermite-Hadamard type for n-polynomial ζ-preinvex functions. Finally, we apply the obtained inequalities to establish two inequalities for special means.Firstly, we introduce the notion ofn-polynomial ζ-preinvex functions.Definition 4.
Letn∈ℕ. A nonnegative function F:X↦ℝ is said to be n-polynomial ζ-preinvex with respect to bifunctions η,ζ:ℝ×ℝ↦ℝ if the inequality
(8)Fa+tηb,a≤Fa+1n∑s=1n1−1−tsζFb,Faholds for all a,b∈X and t∈0,1.
Note that if we taken=1, then we have 1-polynomial ζ-preinvexity, which is just the ζ-preinvex functions defined by the inequality
(9)Fa+tηb,a≤Fa+tζFb,Fa,∀a,b∈X,t∈0,1.
If we takeζFb,Fa=Fb−Fa, then we obtain the class of n-polynomial preinvex functions, which is defined by the inequality
(10)Fa+tηb,a≤1n∑s=1n1−tsFa+1n∑s=1n1−1−tsFb,∀a,b∈X,t∈0,1.
If we takeηb,a=b−a, then we get the class of n-polynomial ζ-convex functions, which is defined by the inequality
(11)Fa+tb−a≤Fa+1n∑s=1n1−1−tsζFb,Fa,∀a,b∈X,t∈0,1.
If we taken=1 in inequality (11), then we have the class of ζ-convex functions. Furthermore, we obtain the classical convex functions by setting ζFb,Fa=Fb−Fa.
If we taken=2 in Definition 4, then we have the class of 2-polynomial ζ-preinvex functions, which is defined by the following inequality:
(12)Fa+tηb,a≤Fa+3t−t22ζFb,Fa,∀a,b∈X,t∈0,1.Note that0≤t≤3t−t2/2, this shows that, for every nonnegative bifunction ζ, the ζ-preinvex function is also the 2-polynomial ζ-preinvex functions. More generally, we have the following result.Proposition 5.
For every nonnegative bifunctionζ and n≥2, if F:X↦ℝ is a n−1-polynomial ζ-preinvex function, then F is a n-polynomial ζ-preinvex function.
To verify the validity of Proposition5, it is enough to show that
(13)1n−1∑s=1n−11−1−ts≤1n∑s=1n1−1−ts,
for anyn≥2 and t∈0,1.
Direct computation gives(14)1n−1∑s=1n−11−1−ts−1n∑s=1n1−1−ts=1−tt1−1−tnn−1−1−tn−1n−1=1−tt1−tn+nt1−tn−1−1nn−1=1−tt1−tn+nt1−tn−1−1−t+tnnn−1=−1−ttCn2t21−tn−2+Cn3t31−tn−3+⋯+Cnntnnn−1≤0,which implies the required inequality (13).As a consequence, we obtain the following.Proposition 6.
For every nonnegative bifunctionζ, if F:X↦ℝ is a ζ-preinvex function, then F is a n-polynomial ζ-preinvex function.ChoosingζFb,Fa=Fb−Fa in Proposition 6 gives the following.Proposition 7.
IfF:a,a+ηb,a→ℝ is a preinvex function with Fb−Fa≥0, then F is a n-polynomial preinvex function.


## 2. Main Results

In this section, we establish some new Hermite-Hadamard-type inequalities using the class ofn-polynomial ζ-preinvex functions. We first need to introduce the notation called Condition C, which was presented by Mohan and Neogy in [20].Condition C.
LetX⊂ℝ be an invex set with respect to bifunction η.,., we say that the bifunction η.,. satisfies the Condition C, if for any x,y∈X and t∈0,1, we have
(15)ηx,x+tηy,x=−tηy,x,ηy,x+tηy,x=1−tηy,x.Note that for anyx,y∈X, t1,t2∈0,1 and from Condition C, we can deduce
(16)ηx+t2ηy,x,x+t1ηy,x=t2−t1ηy,x.Throughout the paper we assume that Condition C is satisfied for the domain with respect to bifunctionη.,. as a precondition.Theorem 8.
LetF:a,a+ηb,a→ℝ be a n-polynomial ζ-preinvex function. If ηb,a>0 and F∈La,a+ηb,a, then we have
(17)F2a+ηb,a2−n+2−n−1nMζ≤1ηb,a∫aa+ηb,aFxdx≤Fa+1n∑s=1nss+1ζFb,Fa,
whereMζ is the upper bound of bifunction ζ.Proof.
Using the definition ofn-polynomial ζ-preinvex function and Condition C, we have
(18)F2a+ηb,a2=Fa+1−tηb,a+12t−1−tηb,a=Fa+1−tηb,a+12ηa+tηb,a,a+1−tηb,a≤Fa+1−tηb,a+1n∑s=1n1−12sζFa+tηb,a,Fa+1−tηb,a≤Fa+1−tηb,a+n+2−n−1nMζ.
Hence, we obtain(19)Fa+1−tηb,a≥F2a+ηb,a2−n+2−n−1nMζ.
Integrating both sides of the above inequality with respect tot on 0,1, it follows that
(20)∫01Fa+1−tηb,adt≥F2a+ηb,a2−n+2−n−1nMζ,
that is,(21)1ηb,a∫aa+ηb,aFxdx≥F2a+ηb,a2−n+2−n−1nMζ.
The left-hand side inequality of (17) is proved.
On the other hand, from the definition ofn-polynomial ζ-preinvex function, one has
(22)Fa+tηb,a≤Fa+1n∑s=1n1−1−tsζFb,Fa.
Integrating both sides of the above inequality with respect tot on 0,1, we obtain
(23)1ηb,a∫aa+ηb,aFxdx=∫01Fa+tηb,adt≤∫01Fa+1n∑s=1n1−1−tsζFb,Fadt=Fa+1n∑s=1nss+1ζFb,Fa.
This proves the right-hand side inequality of (17). The proof of Theorem 8 is complete.Before we put forward another kind of integral inequality of Hermite-Hadamard type, we need to prove an auxiliary result, which will play a key role in deducing subsequent results. For the sake of simplicity, we letI=a,a+ηb,a and let I∘ be the interior of I.Lemma 9.
LetF:I↦ℝ be a differentiable function on I∘ with ηb,a>0, minλ,μ≥t>0. If F′∈LI, then
(24)μFa+λFa+ηb,aλ+μ−1ηb,a∫aa+ηb,aFxdx=ηb,aλ+μ2∫0μ−tF′a+μ−tλ+μηb,adt+∫0λtF′a+μ+tλ+μηb,adt.Proof.
Let(25)I=ηb,aλ+μ2∫0μ−tF′a+μ−tλ+μηb,adt+∫0λtF′a+μ+tλ+μηb,adt=I1+I2.
Integrating by parts yields(26)I1=ηb,aλ+μ2∫0μ−tF′a+μ−tλ+μηb,adt=1λ+μμFa−∫0μFa+μ−tλ+μηb,adt=μλ+μFa−1ηb,a∫aa+μλ+μηb,aFxdx.
Similarly,(27)I2=λλ+μFa+ηb,a−1ηb,a∫a+μ/λ+μηb,aa+ηb,aFxdx.
Substituting the formulations ofI1 and I2 in (25) leads to the desired identity (24). The proof of Lemma 9 is complete.We shall now give some estimations of bounds for Hermite-Hadamard-type inequalities.Theorem 10.
LetF:I⟶ℝ be a differentiable function on I∘ with ηb,a>0, λ>0, μ>0, and let F′∈LI. If ∣F′∣ is n-polynomial ζ-preinvex function, then
(28)μFa+λFa+ηb,aλ+μ−1ηb,a∫aa+ηb,aFxdx≤ηb,aλ+μ2λ2+μ22∣F′a∣+1n∑s=1nK1ζ∣F′b∣,∣F′a∣+1n∑s=1nK2ζ∣F′b∣,∣F′a∣,where
(29)K1=μ22−μs+2λ+μs+1−λ+μs+2+λs+2s+1s+2λ+μs,K2=λ22−λs+2s+1s+2λ+μs.Proof.
Using Lemma9 and the assumption that ∣F′∣ is n-polynomial ζ-preinvex function, we have
(30)μFa+λFa+ηb,aλ+μ−1ηb,a∫aa+ηb,aFxdx≤ηb,aλ+μ2∫0μ∣−t∣F′a+μ−tμ+ληb,adt+∫0λ∣t∣F′a+μ+tλ+μηb,adt≤ηb,aλ+μ2∫0μt∣F′a∣+1n∑s=1n1−1−μ−tμ+λsζ∣F′b∣,∣F′a∣dt+∫0λt∣F′a∣+1n∑s=1n1−1−μ+tμ+λsζ∣F′b∣,∣F′a∣dt=ηb,aλ+μ2∫0μt∣F′a∣dt+1n∑s=1n∫0μt1−λ+tλ+μsζ∣F′b∣,∣F′a∣dt+∫0λt∣F′a∣dt+1n∑s=1n∫0λt1−λ−tλ+μsζ∣F′b∣,∣F′a∣dt=ηb,aλ+μ2λ2+μ22∣F′a∣+1n∑s=1n∫0μt1−λ+tλ+μsζ∣F′b∣,∣F′a∣dt+1n∑s=1n∫0λt1−λ−tλ+μsζ∣F′b∣,∣F′a∣dt,
which implies the desired inequality (28) since
(31)∫0μt1−λ+tλ+μsdt=μ22−μs+2λ+μs+1−λ+μs+2+λs+2s+1s+2λ+μs=K1,∫0λt1−λ−tλ+μsdt=λ22−λs+2s+1s+2λ+μs=K2.
This completes the proof of Theorem10.Next, we discuss some special cases of Theorem10.
(I)
If we considerλ=μ=1 in Theorem 10, then we have(32)Fa+Fa+ηb,a2−1ηb,a∫aa+ηb,aFxdx≤ηb,a4∣F′a∣+1n∑s=1n2ss+1s+2−21+2s+1s2s+1s+1s+2ζ∣F′b∣,∣F′a∣+1n∑s=1n2ss+1s+2−2s+1s+22s+1ζ∣F′b∣,∣F′a∣.(II)
If we takeηb,a=b−a in Theorem 10, then we get(33)μFa+λFbλ+μ−1b−a∫abFxdx≤b−aλ+μ2λ2+μ22∣F′a∣+1n∑s=1nK1ζ∣F′b∣,∣F′a∣+1n∑s=1nK2ζ∣F′b∣,∣F′a∣.(III)
If we chooseζ∣F′b∣,∣F′a∣=∣F′b∣−∣F′a∣ in Theorem 10, then we obtain(34)μFa+λFa+ηb,aλ+μ−1ηb,a∫aa+ηb,aFxdx≤ηb,aλ+μ2λ2+μ22∣F′a∣+1n∑s=1nK1+K2∣F′b∣−∣F′a∣.Theorem 11.
LetF:I→ℝ be a differentiable function on I∘ with ηb,a>0, λ>0, μ>0, and let F′∈LI, 1/p+1/q=1, p>1,q>1. If F′q is n-polynomial ζ-preinvex function, then
(35)μFa+λFa+ηb,aλ+μ−1ηb,a∫aa+ηb,aFxdx≤ηb,aλ+μ2μp+1p+11/pμF′aq+1n∑s=1nK3ζF′bq,F′aq1/q+λp+1p+11/pλF′aq+1n∑s=1nK4ζF′bq,F′aq1/q,where
(36)K3=μ−λ+μs+1−λs+1s+1λ+μs,K4=λ−λs+1s+1λ+μs.Proof.
Using Lemma9, Hölder’s inequality, and the fact that F′q is n-polynomial ζ-preinvex function, it follows that
(37)μFa+λFa+ηb,aλ+μ−1ηb,a∫aa+ηb,aFxdx≤ηb,aλ+μ2∫0μ∣−t∣F′a+μ−tμ+ληb,adt+∫0λ∣t∣F′a+μ+tλ+μηb,adt≤ηb,aλ+μ2∫0μ−tpdt1/p∫0μF′a+μ−tλ+μηb,aqdt1/q+∫0λtpdt1/p∫0λF′a+μ+tλ+μηb,aqdt1/q≤ηb,aλ+μ2μp+1p+11/p∫0μF′aq+1n∑s=1n1−λ+tλ+μsζF′bq,F′aqdt1/q+λp+1p+11/p∫0λF′aq+1n∑s=1n1−λ−tλ+μsζF′bq,F′aqdt1/q=ηb,aλ+μ2μp+1p+11/pμF′aq+1n∑s=1nK3ζF′bq,F′aq1/q+λp+1p+11/pλF′aq+1n∑s=1nK4ζF′bq,F′aq1/q,where
(38)K3=∫0μ1−λ+tλ+μsdt=μ−λ+μs+1−λs+1s+1λ+μs,K4=∫0λ1−λ−tλ+μsdt=λ−λs+1s+1λ+μs.
The proof of Theorem11 is complete.We now discuss some special cases of Theorem11.
(I)
If we chooseλ=μ=1 in Theorem 11, then(39)Fa+Fa+ηb,a2−1ηb,a∫aa+ηb,aFxdx≤ηb,a4p+11/pF′aq+1n∑s=1n2ss+1−2s+1−12ss+1ζF′bq,F′aq1/q+F′aq+1n∑s=1n2ss+1−12ss+1ζF′bq,F′aq1/q.(II)
If we takeηb,a=b−a in Theorem 11, then(40)μFa+λFbλ+μ−1b−a∫abFxdx≤b−aλ+μ2μp+1p+11/pμF′aq+1n∑s=1nK3ζF′bq,F′aq1/q+λp+1p+11/pλF′aq+1n∑s=1nK4ζF′bq,F′aq1/q.(III)
If we putζF′bq,F′aq=F′bq−F′aq in Theorem 11, then(41)μFa+λFa+ηb,aλ+μ−1ηb,a∫aa+ηb,aFxdx≤ηb,aλ+μ2μp+1p+11/pμ−1n∑s=1nK3F′aq+1n∑s=1nK3F′bq1/q+λp+1p+11/pλ−1n∑s=1nK4F′aq+1n∑s=1nK4F′bq1/q.Theorem 12.
LetF:I→ℝ be a differentiable function on I∘ with ηb,a>0, λ>0, μ>0, and let F′∈LI, q≥1. If F′q is n-polynomial ζ-preinvex function, then
(42)μFa+λFa+ηb,aλ+μ−1ηb,a∫aa+ηb,aFxdx≤ηb,aλ+μ2μ221−1/qμ22F′aq+1n∑s=1nK1ζF′bq,F′aq1/q+λ221−1/qλ22F′aq+1n∑s=1nK2ζF′bq,F′aq1/q,
whereK1 and K2 are the expressions as described in Theorem 10.Proof.
Note thatF′q is n-polynomial ζ-preinvex function, by using the power mean inequality, we have
(43)μFa+λFa+ηb,aλ+μ−1ηb,a∫aa+ηb,aFxdx≤ηb,aλ+μ2∫0μ∣−t∣F′a+μ−tμ+ληb,adt+∫0λ∣t∣F′a+μ+tλ+μηb,adt≤ηb,aλ+μ2∫0μ∣−t∣dt1−1/q∫0μtF′a+μ−tμ+ληb,aqdt1/q+∫0λ∣t∣dt1−1/q∫0λtF′a+μ+tλ+μηb,aqdt1/q≤ηb,aλ+μ2μ221−1/q∫0μtF′aqdt+1n∑s=1n∫0μt1−λ+tλ+μsζF′bq,F′aqdt1/q+λ221−1/q∫0λtF′aqdt+1n∑s=1n∫0λt1−λ−tλ+μsζF′bq,F′aqdt1/q=ηb,aλ+μ2μ221−1/qμ22F′aq+1n∑s=1nK1ζF′bq,F′aq1/q+λ221−1/qλ22F′aq+1n∑s=1nK2ζF′bq,F′aq1/q.
Here,K1 and K2 are formulated as that of Theorem 10. This completes the proof of Theorem 12.We now discuss some special cases of Theorem12.
(I)
Choosingλ=μ=1 in Theorem 12, we get(44)Fa+Fa+ηb,a2−1ηb,a∫aa+ηb,aFxdx≤ηb,a23−1/qF′aq2+1n∑s=1ns+1s+22s−2s+2s−22s+1s+1s+2ζF′bq,F′aq1/q+F′aq2+1n∑s=1ns+1s+22s−2s+1s+22s+1ζF′bq,F′aq1/q.(II)
Choosingηb,a=b−a in Theorem 12, we have(45)μFa+λFbλ+μ−1b−a∫abFxdx≤b−aλ+μ2μ221−1/qμ22F′aq+1n∑s=1nK1ζF′bq,F′aq1/q+λ221−1/qλ22F′aq+1n∑s=1nK2ζF′bq,F′aq1/q.(III)
TakingζF′bq,F′aq=F′bq−F′aq in Theorem 12, we obtain(46)μFa+λFa+ηb,aλ+μ−1ηb,a∫aa+ηb,aFxdx≤ηb,aλ+μ2μ221−1/qμ22−1n∑s=1nK1F′aq+1n∑s=1nK1F′bq1/q+λ221−1/qλ22−1n∑s=1nK2F′aq+1n∑s=1nK2F′bq1/q.Theorem 13.
LetF:I→ℝ be a differentiable function on I∘ with ηb.aλ>0, λ>0, μ>0, and let F′∈LI, 1/p+1/q=1, p>1,q>1. If F′q is n-polynomial ζ-preinvex function, then
(47)μFa+λFa+ηb,aλ+μ−1ηb,a∫aa+ηb,aFxdx≤ηb,aλ+μ21μμp+2p+1p+21/pμ22F′aq+1n∑s=1nK5ζF′bq,F′aq1/q+μp+2p+21/pμ22F′aq+1n∑s=1nK6ζF′bq,F′aq1/q+1λλp+2p+1p+21/p×λ22F′aq+1n∑s=1nK7ζF′bq,F′aq1/q+λp+2p+21/pλ22F′aq+1n∑s=1nK8ζF′bq,F′aq1/q,
where(48)K5=μ22−μs+2s+2λ+μs,K6=μ22−μs+2s+1s+2λ+μs,K7=λ22−μ+λs+2−μs+2−λs+2μs+1s+1s+2λ+μs,K8=λ22−μ+λs+2−μs+2−λs+2λ+μs+1s+1s+2λ+μs.Proof.
Note thatF′q is n-polynomial ζ-preinvex function, by using the refined Hölder inequality (see [19]), we obtain
(49)μFa+λFa+ηb,aλ+μ-1ηb,a∫aa+ηb,aFxdx≤ηb,aλ+μ2∫0μ-tF′a+μ-tμ+ληb,adt+∫0λtF′a+μ+tλ+μηb,adt≤ηb,aλ+μ21μ∫0μμ-t-tpdt1/p∫0μμ-tF′a+μ-tλ+μηb,aqdt1/q+∫0μt-tpdt1/p∫0μtF′a+μ-tλ+μηb,aqdt1/q+1λ∫0λλ-ttpdt1/p∫0λλ-tF′a+μ-tλ+μηb,aqdt1/q+∫0λtF′a+μ-tλ+μηb,aqdt1/q≤ηb,aλ+μ21μμp+2p+1p+21/p∫0μμ-t×F′aq+1n∑s=1n1-μ-tλ+μsζF′bq,F′aqdt1/q+μp+2p+21/q∫0μtF′aq+1n∑s=1n1-μ-tλ+μsζF′bq,F′aqdt1/q+1λλp+2p+1p+21/p∫0λλ-t×F′aq+1n∑s=1n1-μ-tλ+μsζF′bq,F′aqdt1/q+λp+2p+21/q∫0λtF′aq+1n∑s=1n1-μ-tλ+μsζF′bq,F′aqdt1/q=ηb,aλ+μ21μμp+2p+1p+21/pμ22F′aq+1n∑s=1nK5ζF′bq,F′aq1/q+μp+2p+21/pμ22F′aq+1n∑s=1nK6ζF′bq,F′aq1/q+1λλp+2p+1p+21/pλ22F′aq+1n∑s=1nK7ζF′bq,F′aq1/q+μp+2p+21/pλ22F′aq+1n∑s=1nK8ζF′bq,F′aq1/q
A direct computation gives(50)K5=∫0μμ−t1−μ−tλ+μsdt=μ22−μs+2s+2λ+μs,K6=∫0μt1−μ−tλ+μsdt=μ22−μs+2s+1s+2λ+μs,K7=∫0λλ−t1−μ+tλ+μsdt=λ22−μ+λs+2−μs+2−λs+2μs+1s+1s+2λ+μs,K8=∫0λt1−μ+tλ+μsdt=λ22−μ+λs+2−μs+2−λs+2λ+μs+1s+1s+2λ+μs.
This completes the proof of Theorem13.Let us now discuss some special cases of Theorem13.
(I)
If we takeλ=μ=1 in Theorem 13, we get(51)Fa+Fa+ηb,a2−1ηb,a∫aa+ηb,aFxdx≤ηb,a41p+1p+21/pF′aq2+1n∑s=1n12−1s+22sζF′bq,F′aq1/q+F′aq2+1n∑s=1n12−1s+1s+22sζF′bq,F′aq1/q+1p+21/pF′aq2+1n∑s=1n12−2s+2−s−3s+1s+22sζF′bq,F′aq1/q+F′aq2+1n∑s=1n12+1+s2s+1s+1s+22sζF′bq,F′aq1/q.(II)
If we putηb,a=b−a in Theorem 13, then(52)μFa+λFbλ+μ−1b−a∫abFxdx≤b−aλ+μ21μμp+2p+1p+21/pμ22F′aq+1n∑s=1nK5ζF′bq,F′aq1/q+μp+2p+21/pμ22F′aq+1n∑s=1nK6ζF′bq,F′aq1/q+1λλp+2p+1p+21/pλ22F′aq+1n∑s=1nK7ζF′bq,F′aq1/q+λp+2p+21/pλ22F′aq+1n∑s=1nK8ζF′bq,F′aq1/q.(III)
If we chooseζF′bq,F′aq=F′bq−F′aq in Theorem 13, then(53)μFa+λFa+ηb,aλ+μ−1ηb,a∫aa+ηb,aFxdx≤ηb,aλ+μ21μμp+2p+1p+21/pμ22−1n∑s=1sK5F′aq+1n∑s=1nK5F′bq1/q+μp+2p+21/pμ22−1n∑s=1nK6F′aq+1n∑s=1nK6F′bq1/q+1λλp+2p+1p+21/pλ22−1n∑s=1nK7F′aq+1n∑s=1nK7F′bq1/q+λp+2p+21/pλ22−1n∑s=1nK8F′aq+1n∑s=1nK8F′bq1/q.


## 3. Application to Special Means

Let us recall the definitions of the arithmetic mean, weighted arithmetic mean, and the mean for functions, as follows:(1)
The arithmetic mean(54)Aa1,a2,⋯,an=a1+a2+⋯+ann.(2)
The weighted arithmetic mean(55)Aa1,a2,⋯,an;p1,p2,⋯,pn=p1a1+p2a2+⋯+pnanp1+p2+⋯+pn.(3) The mean of the functionΦ on a,b(56)AΦa,b=1b−a∫abΦxdx.We establish the following inequalities for special means.Proposition 14.
LetΦ:a,a+ηb,a→ℝ be a preinvex function with Φb−Φa≥0. If ηb,a>0 and Φ∈La,a+ηb,a, then we have the following inequality
(57)AΦa,a+ηb,a≤Φa+Φb−ΦaA12,23,⋯,nn+1.Proof.
TakingFx=Φx,x∈a,a+ηb,a and ζFb,Fa=Fb−Fa. Since Φx is a preinvex function with Φb−Φa≥0, we deduce from Proposition 7 that Φx is a n-polynomial preinvex function. Using Theorem 8, we obtain the desired inequality (57).Proposition 15.
LetΦ be a differentiable function, and let Φ′:a,a+ηb,a→ℝ be a preinvex function with ∣Φ′b∣−∣Φ′a∣≥0. If ηb,a>0, λ>0, μ>0, and ∣Φ′∣∈La,a+ηb,a, then we have the following inequality
(58)AΦa,Φa+ηb,a;μ,λ−AΦa,a+ηb,a≤ηb,aλ+μ2Φ′aAλ2,μ2+2nΦ′b−Φ′a∑s=1nAK1,K2,
whereK1 and K2 are the expressions as that described in Theorem 10.Proof.
ChoosingFx=Φx,x∈a,a+ηb,a and ζFb,Fa=Fb−Fa. Since ∣Φ′x∣ is a preinvex function with ∣Φ′b∣−∣Φ′a∣≥0, it follows from Proposition 7 that ∣Φ′x∣ is a n-polynomial preinvex function. Using Theorem 10 leads to the desired inequality (58).



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