EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS 2025, Vol. 18, Issue 2, Article Number 5871 ISSN 1307-5543 – ejpam.com Published by New York Business Global Computational Illustration of Fractional Inequalities via 2D Graphs with Application Ahsan Mehmood1,∗, Muhammad Younis1, Ahmad Aloqaily2, Dania Santina2, Muhammad Samraiz3, Gauhar Rahman4, Nabil Mlaiki2 1 School of Mathematical Sciences and Shanghai Key Laboratory of PMMP, East China Normal University, 500 Dongchuan Road, Shanghai 200241, Peoples Republic of China 2 Department of Mathematics and Sciences, Prince Sultan University, Riyadh 11586, Saudi Arabia 3 Department of Mathematics, University of Sargodha P.O. Box 40100, Sargodha, Pakistan 4 Department of Mathematics and Statistics, Hazara University, Mansehra 21300, Pakistan Abstract. In this article, we use generalized (k, s)-Riemann-Liouville fractional integral operator (GRLFIO) to explore the reverse forms of Minkowskis, Holder and Hermite-Hadamard-Fejer type inequalities within an interval-valued (ı.υ) (⋋s+1,℧) class of convexity. We comprise various ex- isting definitions and propose the novel concept of an ı.υ (⋋s+1,℧) convexity. Our findings show the remarkable adaptability by adjusting parameter bounds for (k, s)-GRLFIO within structure of an ı.υ (⋋s+1,℧) convexity presenting broader generalization and new perspective advancements to Hermite-Hadamard-Fejer and Pachpatte-type inequalities. In order to facilitate their applications, we examine the further consequences, constructed specific inequalities and illustrate them through graphical representations. Additionally, we validate the results using tables for various fractional orders. This study establishes the foundation for future research into the mathematical inequali- ties by emphasizing the importance of fractional integral operators and the expanded concept of convexity. 2020 Mathematics Subject Classifications: 26A33, 26A51, 26D15, 26D20 Key Words and Phrases: Hermite-Hadamard-inequality, Fractional calculus, Interval-valued, (k, s)-Riemann-Liouville fractional integral operator 1. Introduction Abel was the first scientist in the history of fractional calculus to use it to solve the Tautochrone problem [1]. To further improve the field, researchers have published their ∗Corresponding author. DOI: https://doi.org/10.29020/nybg.ejpam.v18i2.5871 Email addresses: mehmoodahsan154@gmail.com (A. Mehmood), younismuhammad303@gmail.com (M. Younis), maloqaily@psu.edu.sa (A. Aloqaily), dsantina@psu.edu.sa (D. Santina), muhammad.samraiz@uos.edu.pk (M. Samraiz), gauhar55uom@gmail.com (G. Rehman), nmlaiki@psu.edu.sa (N. Mlaiki) https://www.ejpam.com 1 Copyright: © 2025 The Author(s). (CC BY-NC 4.0) A. Mehmood et al. / Eur. J. Pure Appl. Math, 18 (2) (2025), 5871 2 of 26 work [2, 3]. These studies have made a significant impact on the applications and ac- complishments of fractional calculus in mathematical modeling and applied analysis [4, 5]. Without a non-singular kernel, Caputo et al. introduced the well-known Caputo deriva- tive in [6]. Still, there are a number of important research gaps in the idea. Several academics have created their own fractional operators with non-singular kernels to fill in these gaps [7–10]. Non-linear and non-singular extensions of fractional operators and their symmetric features were established by Wu et al. in [11]. A non-linear and non-singular fractional derivative was also introduced by Samraiz et al. in [12], who also examined its uses in applied analysis. In [13, 14], additional advancements in fractional operators using different kinds of kernel functions were introduced. The reader is referred to [15] for further information and applications concerning fractional operators. Convex analysis is based on the idea of convex functions (cf) and sets with convex epigraphs. Convexity is important in many areas of mathematics such as optimization, fixed point theory, topolog- ical spaces and advanced analysis. A positive second-order derivative indicates concavity, and derivatives can be used to assess the convexity of functions. Known for its real-world applications, the theory of inequalities encompasses several branches of mathematical anal- ysis. Error limitations for numerical quadrature methods, including the trapezoidal rule, midpoint rule, Ostrowski’s rule, and Simpson’s rules, are notably refined by integral in- equalities, especially when cf and their generalizations are applied. Additionally, these bounds show links between special functions, probability theory, information theory, and other fields. The notion of convexity can be used to generate a number of basic and Hermite-Hadamard-Fejer inequality. Let a function ∅ : [r1, r2] ⊂ R → R be continuous, so ∅ ( r1 + r2 2 ) ≤ 1 r2 − r1 ∫ r2 r1 ∅(x)dx ≤ ∅(r1) +∅(r2) 2 . One could consider this inequality to be an extra standard for cf . Additional information can be found in [16, 17]. Wu redpresented the unified form of convexity, explained as follows. Definition 1. [18] If a function ⋋ : ® → R be monotonic continuous-(mc) function, then ® ⊂ R is considered to be ⋋-convex set based on ⋋ if: ⋋−1 ((1− θ)⋋ (x) + θ ⋋ (y)) ∈ ®, for all x, y ∈ ® and θ ∈ [0, 1]. We, now resume the class ⋋-cf . Definition 2. A function ∅ : ® → R is stated to be ⋋-cf with respect to (w.r.t) strictly mc function ⋋ if: ∅(⋋−1((1− θ)⋋ (x) + θ ⋋ (y))) ≤ (1− θ)∅(x) + θ∅(y), for all x, y ∈ ® and θ ∈ [0, 1]. A. Mehmood et al. / Eur. J. Pure Appl. Math, 18 (2) (2025), 5871 3 of 26 In order to investigate the several relevant scientific domains, numerous authors have combined fractional calculus with ı.υ concepts. Working along these lines, Breckner de- veloped the idea of set-valued cf , as seen below. Definition 3. [19] A function ∅ : [r1, r2] → R+ i is stated to be ı.υ cf , if: ∅((1− θ)r1 + θr2) ⊇ (1− θ)∅(r1) + θ∅(r2), θ ∈ [0, 1]. The first person to apply inequalities to set-valued functions was Sadowska [20]. He investigated the Hermite-Hadamard-Fejer inequality in the context of set-valued cf . If function ∅ : [r1, r2] → R be an ı.υ cf , so ∅ ( r1 + r2 2 ) ⊇ 1 r2 − r1 ∫ r2 r1 ∅(θ)dθ ⊇ ∅(r1) +∅(r2) 2 . If B([r1, r2]) is a collection of all divisions of [r1, r2] and B(ρ1, [r1, r2]) be the family of all divisions P in a way that meshP < ρ1, then ∅ : [r1, r2] → R is referred to as ı.υ Riemann integrable on [r1, r2], if there exist →∈ R, and for each ϵ > 0 there exist ρ > 0 such that: d(S(∅, P, ρ),®) < ϵ, where S(∅, P, ρ) specifies the Riemann sum of Phi for any P ∈ B(ρ, [r1, r2]). The above expression represents that ® is the (IR)-integral of ∅ such that: ® = (IR) ∫ r2 r1 ∅(θ)dθ For the sake of brevity, we specify the space of Riemann integrable functions and ı.υ Riemann integration on[r1, r2] and by R[r1,r2] and IR[r1,r2] respectively. Theorem 1. [21] If a function ∅(θ) : [r1, r2] → R be an ı.υ continuous, then ∅(θ) = [∅∗(θ),∅∗(θ)],∅(θ) ∈ IR[r1,r2] ⇔ ∅∗(θ),∅∗(θ) ∈ IR[r1,r2], and (IR) ∫ r2 r1 ∅(θ)dθ = [ (R) ∫ r2 r1 ∅∗(θ)dθ, (R) ∫ r2 r1 ∅∗(θ)dθ ] . The Lebesgue integrable function (LI∅) space is defined by L[r1, r2]. Now, we retrieve the Riemann-Liouville fractional operator, which are provided below. Definition 4. [22] Let ∅(θ) ∈ [r1, r2], then Zβ r+1 ∅(r2) = 1 Γ(β) ∫ r2 r1 ∅(θ)(r2 − θ)β−1dθ, r1 < r2, β > 0. Similarly, the right side of the Riemann-Liouville fractional operator is given below Zβ r−2 ∅(r1) = 1 Γ(β) ∫ r2 r1 ∅(θ)(θ − r1) β−1dθ, r1 < r2, β > 0. A. Mehmood et al. / Eur. J. Pure Appl. Math, 18 (2) (2025), 5871 4 of 26 We, now replicated the ı.υ Riemann-Liouville fractional integral operator. Definition 5. [23] Let ∅(x) be ı.υ function such that ∅1(x),∅2(x) ∈ L[r1, r2], then Zβ x+∅(r2) = 1 Γ(β) ∫ r2 x ∅(θ)(r2 − θ)β−1dθ, x < r2, and Zβ y−∅(r1) = 1 Γ(β) ∫ y r1 ∅(θ)(θ − r1) β−1dθ, r1 < y, with β > 0, obviously Zβ x+∅(r2) = [ Zβ x+∅1(r2),Z β x+∅2(r2) ] , and Zβ y−∅(r1) = [ Zβ y−∅1(r1),Z β y−∅2(r1) ] . In recent years, ı.υ functions based on different partial and total ordered relations have been used to refine and generalize a number of integral inequalities. These works provided the groundwork for subsequent developments especially in mathematical inequal- ities pertaining to set-valued functions and represented the first attempts to improve the practical applications of inequalities. We will derive several fractional variations of re- verse Minkowski inequality, Holders inequality, Hermite-Hadamard inequality, its weighted form known as Fejer-Hermite-Hadamard inequality and some inequalities for the prod- uct of functions that are known to be Pachpatee’s type inclusions as applications of this class. Since the generic class of convexity and its implications in inequalities is a broader space of functions that contains redcf and non-redcf classes, it is the novel as- pect of the current proceeding. Our developed results allow for the characterization of large classes of functions. Our findings will also be useful tools for calculating different bounds for the ı.υ fractional operators. There are a few simulations for numerical ex- amples provided to verify the accuracy of the suggested findings. With this work, we intend to demonstrate further inequalities and related optimization issues to interested readers. Now, we introduce the idea of ı.υ-(⋋s+1,℧) redcf , it indicates how several new generalizations of convexity and the numerous classes of convexity that now exist can be produced as special instances. For our convenient the family of ı.υ-(⋋s+1,℧) redcf , ı.υ-(⋋s+1,℧) concave function, (⋋s+1,℧) redcf and (⋋s+1,℧) concave functions are rep- resented as SIGX([r1, r2], R + i ), SIGV ([r1, r2], R + i ), SGX([r1, r2], R) and SGV ([r1, r2], R) respectively. Definition 6. Let s ∈ R/{−1}, ⋋ be an increasing function and ∅ : [r1, r2] → R+ i be an ı.υ-(⋋s+1,℧) function fulfill the condition ∅(x) = [∅∗(x),∅∗(x)] and ℧ : [0, 1] → R be a positive function, so ∅ ( ⋋−1 ((1− θ)⋋s+1 (r1) + θ ⋋s+1 (r2)) 1 s+1 ) ⊇ ℧(1− θ)∅(r1) + ℧(θ)∅(r2), for all x ∈ [r1, r2] and θ ∈ [0, 1]. A. Mehmood et al. / Eur. J. Pure Appl. Math, 18 (2) (2025), 5871 5 of 26 Remark 1. (i) If we choose s = 0 and ⋋(θ) = θ in (6), then we obtain the following definition presented in [24]. ∅ ( ⋋−1 ((1− θ)⋋ (r1) + θ ⋋ (r2)) ) ⊇ ℧(1− θ)∅(r1) + ℧(θ)∅(r2), (ii) For the choice of s = 0, and ⋋(θ) = ℧(θ) = θ in (6) we acquire ⋋-ı.υ cf: ∅ ( ⋋−1 ((1− θ)⋋ (r1) + θ ⋋ (r2)) ) ⊇ (1− θ)∅(r1) + θ∅(r2). (iii) If we choose s = 0, ⋋(θ) = ℧(θ) = θ and ⋋(x) = 1 x in (6), then we acquire the ı.υ hc function define in [25]. ∅ ( r1r2 θr1 + (1− θ)r2 ) ⊇ (1− θ)∅(r1) + θ∅(r2). (iv) If we choose s = 0, ⋋(θ) = ℧(θ) = θ and ⋋(x) = xp in (6) then we obtain the ı.υ-p cf define in [26]. ∅ ( ((1− θ)rp1 + θrp2) 1 p ) ⊇ (1− θ)∅(r1) + θ∅(r2). (v) By selecting s = 0, ⋋(θ) = θ and ⋋(x) = x in (6), we retrieve the definition of ı.υ cf presented in [24]. Definition 7. If we fix s = 0, ⋋(θ) = θ and ℧(θ) = θs in (6), we retrieve the (⋋, s)-ı.υ cf ∅ ( ⋋−1 (θ ⋋ (r1) + (1− θ)⋋ (r2)) ) ⊇ (1− θ)s∅(r1) + θs∅(r2). Remark 2. The main definitional deductions will now be presented for Definition 7, presented in [24]. (i) Choosing ⋋(x) = 1 x , we retrieve the ı.υ harmonically s− cf , that is ∅ ( r1r2 θr1 + (1− θ)r2 ) ⊇ (1− θ)s∅(r1) + θs∅(r2). (ii) Choosing ⋋(x) = xp, p ≥ −1, we retrieve the ı.υ-p, s− cf , which are as follows: ∅ ( ((1− θ)rp1 + θrp2) 1 p ) ⊇ (1− θ)s∅(r1) + θs∅(r2). The authors in [27] used ı.υ cf and postquantum calculus to investigate novel repre- sentations of well-known results. In [28], different forms of trapezoid-type inequalities for non-cf linked sets on fuzzy domains were introduced. A generic convexity framework was used by Cortez et al. [29] to expand Jensen’s and Hermite-Hadamard inequalities. In- spired by these investigations, we provide ı.υ (⋋s+1,℧)-cf, defined by weighted arithmetic means, which have a strictly monotone function ⋋ and a non-negative function ℧. Using A. Mehmood et al. / Eur. J. Pure Appl. Math, 18 (2) (2025), 5871 6 of 26 suitable replacements for ⋋ and ℧, this framework creates new classes and unifies cur- rent convexity notions. In addition to inequalities for function products of the Pachpatte type, we derive several inequalities, such as Minkowski, Holder’s, Hermite-Hadamard, and Hermite-Hadamard-Fejer. The scope of convex and non-cf analysis is expanded by this generalized convexity framework, which offers methods for bounding ı.υ Riemann-Liouville fractional operators. Our results are supported by numerical examples and simulations, providing a basis for more inequality proofs and optimization problems. Theorem 2. [24] Let ∅ : [r1, r2] → R+ i be ı.υ function such that ∅(r1) = [∅∗,∅∗] with ∅∗ ≤ ∅∗ then, ∅ ∈ SIGX([r1, r2], R + i ), this implies ∅∗ ∈ SGX([r1, r2], R) and ∅∗ ∈ SGV ([r1, r2], R). Theorem 3. [24] Let ∅ ∈ SIGX([r1, r2], R + i ), then ∅ ( ⋋−1 ( 1 wn n∑ i=1 θi ⋋ (xi) )) ⊇ n∑ i=1 ℧ ( θi wn ) ∅(xi) for xi ∈ [r1, r2] and wn = ∑n i=1 θixi. Definition 8. [30] If a function ∅ is continuous on [a, b], s ∈ R/{−1} and k ≥ 0, then the (k, s)-Riemann-Liouville fractional integral operator for order β > 0 can be stated as s kZ β α+∅(m1) = (s+ 1)1− β k kΓk(β) ∫ m1 α+ (ms+1 1 − ns+1 1 ) β k −1ns 1∅(n1)dn1. (1) Now we are going to present generalized form of fractional operator (1). Definition 9. If a function ∅ is continuous on [a, b], s ∈ R/{−1}, k ≥ 0 and ⋋ be an increasing function then the (k, s)-GRLFIO for order β > 0 can be stated as s kZ β α∅(m1) = (s+ 1)1− β k kΓk(β) ∫ m1 α (⋋s+1(m1)−⋋s+1(n1)) β k −1 ⋋s (n1)⋋ ′ (n1)∅(n1)dn1. (2) Definition 10. [31] If a function ∅ is continuous on [a, b], s ∈ R/{−1}, k ≥ 0 and ⋋ be an increasing function then left and right sided (k, s)-GRLFIO for order β > 0 can be stated as s kZ β α+∅(m1) = (s+ 1)1− β k kΓk(β) ∫ m1 α+ (⋋s+1(m1)−⋋s+1(n1)) β k −1 ⋋s (n1)⋋ ′ (n1)∅(n1)dn1, m1 > α+ (3) and s kZ β ζ−∅(m1) = (s+ 1)1− β k kΓk(β) ∫ ζ m1 (⋋s+1(n1)−⋋s+1(m1)) β k −1 ⋋s (n1)⋋ ′ (n1)∅(n1)dn1, m1 < ζ. (4) A. Mehmood et al. / Eur. J. Pure Appl. Math, 18 (2) (2025), 5871 7 of 26 Now we discuss some applications of defined operators which we used later. Proposition 1. If a function ∅ is continuous on [a, b], s ∈ R/{−1}, k ≥ 0, ⋋ be an in- creasing function and C be a constant function then for left and right sided (k, s)-GRLFIO of order β > 0, we have the following results holds: s kZ β r+1 ⋋s+1 (r2) = (s+ 1)1− β k kΓk(β) ( ⋋s+1(r1)(⋋s+1(r2)−⋋s+1(r1)) β k β k (s+ 1) ) + (⋋s+1(r2)−⋋s+1(r1)) β k +1 β k (s+ 1)(βk + 1) s kZ β r−2 ⋋s+1 (r1) = (s+ 1)− β k βΓk(β) ( ⋋s+1 (r2)(⋋ s+1(r2)−⋋s+1(r1)) β k ) − (⋋s+1(r2)−⋋s+1(r1)) β k +1 (βk + 1) s kZ β r+1 C = C(s+ 1)− β k βΓk(β) ( ⋋s+1 (r2)−⋋s+1(r1) β k ) s kZ β r−2 C = C(s+ 1)− β k βΓk(β) ( ⋋s+1 (r2)−⋋s+1(r1) β k ) . 2. A Class of Some Results In this session, we will examine the reverse forms of Minkowski, Holder and Hermite- Hadamard-Fejer type inequalities via an ı.υ cf involving (k, s)-GRLFIO under the frame- work of convexity. In this result, we want to explore the Minkowski’s inequality. Theorem 4. Let s ∈ R/{−1}, k ≥ 0, also ∅, ϕ : [r1, r2] → R+ i be ı.υ functions such that ∅(χ) = [∅∗,∅∗] and ϕ(χ) = [ϕ∗, ϕ ∗], ( s kZ β r+1 ∅p(χ) ) < ∞ and ( s kZ β r+1 ϕp(χ) ) < ∞, then the expression (5) holds.[ 1 + ϑ(2 + µ) (1 + ϑ)(1 + µ) , 1 + µ(2 + ϑ) (1 + ϑ)(1 + µ) ][[ s kZ β r+1 (∅(χ) + ϕ(χ))p ] 1 p ] ⊇ [ s kZ β r+1 ∅p(χ) ] 1 p + [ s kZ β r+1 ϕp(χ) ] 1 p , (5) where, 0 < ϑ ≤ ∅∗(χ) ϕ∗(χ) ≤ µ and 0 < ϑ ≤ ∅∗(χ) ϕ∗(χ) ≤ µ for χ ∈ [r1, r2], p ≥ 1 with β > 0. Proof. Since ∅∗(χ) ϕ∗(χ) ≤ µ, this implies (µ+ 1)p∅∗p(χ) ≤ µp(∅∗(χ) + ϕ∗(χ))p. After multiplying by (s+1)1− β k kΓk(β) ( ⋋s+1 (r2) − ⋋s+1(χ) )β k −1 ⋋s (χ) ⋋ ′ (χ) and applying the integration w.r.t ”χ” over [r1, r2], we obtain the following expression. (µ+ 1)p (s+ 1)1− β k kΓk(β) ∫ r2 r1 ( ⋋s+1 (r2)−⋋s+1(χ) )β k −1 ⋋s (χ)⋋ ′ (χ)∅∗p(χ)dχ A. Mehmood et al. / Eur. J. Pure Appl. Math, 18 (2) (2025), 5871 8 of 26 ≤ µp (s+ 1)1− β k kΓk(β) ∫ r2 r1 ( ⋋s+1 (r2)−⋋s+1(χ) )β k −1 ⋋s (χ)⋋ ′ (χ)[∅∗(χ) + ϕ∗(χ)]pdχ. (6) By comparing the expressions (3) and (6), we have[s k Zβ r+1 ∅∗p(χ) ] 1 p ≤ µ µ+ 1 [s k Zβ r+1 [∅∗(χ) + ϕ∗(χ)]p ] 1 p . (7) For ϑ ≤ ∅∗(χ) ϕ∗(χ) , we can write ϑp[∅∗(χ) + ϕ∗(χ)] p ≤ (1 + ϑ)p(∅∗(χ)) p. Again multiplying by (s+1)1− β k kΓk(β) ( ⋋s+1 (r2) − ⋋s+1(χ) )β k −1 ⋋s (χ) ⋋ ′ (χ) and applying the integration over [r1, r2] w.r.t ”χ”, we have ϑp (s+ 1)1− β k kΓk(β) ∫ r2 r1 ( ⋋s+1 (r2)−⋋s+1(χ) )β k −1 ⋋s (χ)⋋ ′ (χ)[∅∗(χ) + ϕ∗(χ)] pdχ ≤ (ϑ+ 1)p (s+ 1)1− β k kΓk(β) ∫ r2 r1 ( ⋋s+1 (r2)−⋋s+1(χ) )β k −1 ⋋s (χ)⋋ ′ (χ)∅p ∗(χ)dχ. (8) Again by comparing the expressions (3) and (8), we have ϑ ϑ+ 1 [s k Zβ r+1 [∅∗(χ) + ϕ∗(χ)] p ] 1 p ≤ [s k Zβ r+1 ∅p ∗(χ) ] 1 p . (9) From the expressions (7) and (9)red, we have [s k Zβ r+1 ∅p(χ) ] 1 p = [[s k Zβ r+1 ∅p ∗(χ) ] 1 p , [s k Zβ r+1 ∅∗p(χ) ] 1 p ] ⊇ [ ϑ ϑ+ 1 [s k Zβ r+1 [∅∗(χ) + ϕ∗(χ)] p ] 1 p , µ µ+ 1 [s k Zβ r+1 [∅∗(χ) + ϕ∗(χ)]p ] 1 p ] = [ ϑ ϑ+ 1 , µ µ+ 1 ][s k Zβ r+1 [∅(χ) + ϕ(χ)]p ] 1 p . (10) Continuing the same procedure for ϑ ≤ ∅∗(χ) ϕ∗(χ) , then we have [s k Zβ r+1 ϕ∗p(χ) ] 1 p ≤ 1 ϑ+ 1 [s k Zβ r+1 [∅∗(χ) + ϕ∗(χ)]p ] 1 p . (11) Also for ∅∗(χ) ϕ∗(χ) ≤ µ, we have 1 µ+ 1 [s k Zβ r+1 [∅∗(χ) + ϕ∗(χ)] p ] 1 p ≤ [s k Zβ r+1 ϕp ∗(χ) ] 1 p . (12) A. Mehmood et al. / Eur. J. Pure Appl. Math, 18 (2) (2025), 5871 9 of 26 From the expressions (11) and (12), we have [s k Zβ r+1 ϕp(χ) ] 1 p = [[s k Zβ r+1 ϕp ∗(χ) ] 1 p , [s k Zβ r+1 ϕ∗p(χ) ] 1 p ] ⊇ [ 1 µ+ 1 [s k Zβ r+1 [∅∗(χ) + ϕ∗(χ)] p ] 1 p , 1 ϑ+ 1 [s k Zβ r+1 [∅∗(χ) + ϕ∗(χ)]p ] 1 p ] = [ 1 µ+ 1 , 1 ϑ+ 1 ][s k Zβ r+1 [∅(χ) + ϕ(χ)]p ] 1 p (13) Hence by adding (10) and (13) we get the required result (5). In this result, redwe are going to present the fractional reverse Holder’s inequality involving (k, s)-GRLFIO. Theorem 5. Let s ∈ R/{−1}, k ≥ 0, also ∅, ϕ : [r1, r2] → R+ i be ı.υ functions such that ∅(χ) = [∅∗,∅∗] and ϕ(χ) = [ϕ∗, ϕ ∗], ( s kZ β r+1 ∅p(χ) ) < ∞ and ( s kZ β r+1 ϕp(χ) ) < ∞, then the expression (14) holds.[( ϑ µ ) 1 pq , ( µ ϑ ) 1 pq ] [skZ β r+1 [∅ 1 p (χ)ϕ 1 q (χ)]] ⊇ [skZ β r+1 ∅(χ)] 1 p [skZ β r+1 ϕ(χ)] 1 q , (14) where, 0 < ϑ ≤ ∅∗(χ) ϕ∗(χ) ≤ µ and 0 < ϑ ≤ ∅∗(χ) ϕ∗(χ) ≤ µ for χ ∈ [r1, r2], p > 1 and 1 p + 1 q = 1 with β > 0. Proof. Since ∅∗(χ) ϕ∗(χ) ≤ µ, implies ∅∗(χ) ≤ µ 1 q∅∗ 1 p (χ)ϕ ∗ 1 q (χ). Multiplying by (s+1)1− β k kΓk(β) ( ⋋s+1(r2)−⋋s+1(χ) )β k −1 ⋋s(χ)⋋ ′ (χ) and applying the integration over [r1, r2] w.r.t ”χ”red, so we have [skZ β r+1 ∅∗(χ)] 1 p ≤ µ 1 pq [skZ β r+1 ] [ ∅∗ 1 p (χ)ϕ ∗ 1 q (χ) ] 1 p . (15) Analogously for ∅∗(χ) ϕ∗(χ) ≥ ϑ, we have [skZ β r+1 ∅∗(χ)] 1 p ≥ ϑ 1 pq [skZ β r+1 ] [ ∅ 1 p ∗ (χ)ϕ 1 q ∗ (χ) ] 1 p . (16) From the expressions (15) and (16), we can write [s k Zβ r+1 ∅(χ) ] 1 p = [[s k Zβ r+1 ∅∗(χ) ] 1 p , [s k Zβ r+1 ∅∗(χ) ] 1 p ] ⊇ [ ϑ 1 pq [skZ β r+1 ] [ ∅ 1 p ∗ (χ)ϕ 1 q ∗ (χ) ] 1 p , µ 1 pq [skZ β r+1 ] [ ∅∗ 1 p (χ)ϕ ∗ 1 q (χ) ] 1 p ] A. Mehmood et al. / Eur. J. Pure Appl. Math, 18 (2) (2025), 5871 10 of 26 = [ ϑ 1 pq , µ 1 pq ][s k Zβ r+1 [∅ 1 p (χ)ϕ 1 q (χ)] 1 p ] . (17) We can retrieve the following confinement for [s k Zβ r+1 ϕ(χ) ] 1 q . [s k Zβ r+1 ϕ(χ) ] 1 q = [[s k Zβ r+1 ϕ∗(χ) ] 1 q , [s k Zβ r+1 ϕ∗(χ) ] 1 q ] ⊇ [ 1 µ 1 pq [skZ β r+1 ] [ ∅ 1 p ∗ (χ)ϕ 1 q ∗ (χ) ] 1 q , 1 ϑ 1 pq [skZ β r+1 ] [ ∅∗ 1 p (χ)ϕ ∗ 1 q (χ) ] 1 q ] = [ 1 µ 1 pq , 1 ϑ 1 pq ][s k Zβ r+1 [∅ 1 p (χ)ϕ 1 q (χ)] 1 q ] . (18) Adding expressions (17) and (18) we get our required relation (14). Theorem 6. Let s ∈ R/{−1}, k ≥ 0 and ∅ ∈ SIGX([r1, r2], R + i ), then for β > 0 the following redinequality hold (s+ 1)− β k ℧(12)βΓk(β) ∅ ( ⋋−1 ( ⋋s+1(r1) +⋋s+1(r2) 2 ) 1 s+1 ) ⊇ 1( ⋋s+1 (r2)−⋋s+1(r1) )β k [ s kZ β r+1 ∅(r2) + s kZ β r−2 ∅ ( r1) ] ⊇ [ ∅(r1) +∅(r2) ](s+ 1)− β k kΓk(β) ∫ 1 0 θ β k −1[℧(θ) + ℧(1− θ)]dθ,∀x, y ∈ [r1, r2]. Proof. Since ∅ is ı.υ-(⋋s+1,℧) cf and for θ = 1 2 , we have ∅ ( ⋋−1 ( ⋋s+1(x) +⋋s+1(y) 2 ) 1 s+1 ) ⊇ ℧( 1 2 ) [ ∅(x) +∅(y) ] . (19) By substituting x = ⋋−1(θ⋋s+1 (r1)+(1−θ)⋋s+1 (r2)) 1 s+1 and y = ⋋−1((1−θ)⋋s+1 (r1)+ θ⋋s+1 (r2)) 1 s+1 in above inequality 19red. Also by multiplying both sides by (s+1)− β k kΓk(β) θ β k −1 also taking the integration over [0, 1] w.r.t ”θ”red, we have 1 ℧(12) (s+ 1)− β k kΓk(β) ∫ 1 0 θ β k −1∅ ( ⋋−1 ( ⋋s+1(r1) +⋋s+1(r2) 2 ) 1 s+1 ) dθ ⊇ (s+ 1)− β k kΓk(β) ∫ 1 0 θ β k −1∅ ( ⋋−1 (θ ⋋s+1 (r1) + (1− θ)⋋s+1 (r2)) 1 s+1 ) dθ + (s+ 1)− β k kΓk(β) ∫ 1 0 θ β k −1∅ ( ⋋−1 ((1− θ)⋋s+1 (r1) + θ ⋋s+1 (r2)) 1 s+1 ) dθ. (20) A. Mehmood et al. / Eur. J. Pure Appl. Math, 18 (2) (2025), 5871 11 of 26 Now we are proceeding with the left part of the inclusion (20). (s+ 1)− β k kΓk(β) ∫ 1 0 θ β k −1∅ ( ⋋−1 ( ⋋s+1(r1) +⋋s+1(r2) 2 ) 1 s+1 ) dθ = [ (s+ 1)1− β k kΓk(β) ∫ 1 0 θ β k −1∅∗ ( ⋋−1 ( ⋋s+1(r1) +⋋s+1(r2) 2 ) 1 s+1 ) dθ, (s+ 1)− β k kΓk(β) ∫ 1 0 θ β k −1∅∗ ( ⋋−1 ( ⋋s+1(r1) +⋋s+1(r2) 2 ) 1 s+1 ) dθ ] = [ (s+ 1)− β k βΓk(β) ∅∗ ( ⋋−1 ( ⋋s+1(r1) +⋋s+1(r2) 2 ) 1 s+1 ) + (s+ 1)− β k βΓk(β) ∅∗ ( ⋋−1 ( ⋋s+1(r1) +⋋s+1(r2) 2 ) 1 s+1 )] = (s+ 1)− β k βΓk(β) ∅ ( ⋋−1 ( ⋋s+1(r1) +⋋s+1(r2) 2 ) 1 s+1 ) . (21) For the right part of the inclusion (20). (s+ 1)− β k kΓk(β) ∫ 1 0 θ β k −1∅ ( ⋋−1 (θ ⋋s+1 (r1) + (1− θ)⋋s+1 (r2)) 1 s+1 ) dθ + (s+ 1)− β k kΓk(β) ∫ 1 0 θ β k −1∅ ( ⋋−1 ((1− θ)⋋s+1 (r1) + θ ⋋s+1 (r2)) 1 s+1 ) dθ = [ (s+ 1)1− β k kΓk(β) ∫ 1 0 θ β k −1∅∗ ( ⋋−1 (θ ⋋s+1 (r1) + (1− θ)⋋s+1 (r2)) 1 s+1 ) dθ + (s+ 1)− β k kΓk(β) ∫ 1 0 θ β k −1∅∗ ( ⋋−1 ((1− θ)⋋s+1 (r1) + θ ⋋s+1 (r2)) 1 s+1 ) dθ, (s+ 1)− β k kΓk(β) ∫ 1 0 θ β k −1∅∗ ( ⋋−1 (θ ⋋s+1 (r1) + (1− θ)⋋s+1 (r2)) 1 s+1 ) dθ + (s+ 1)− β k kΓk(β) ∫ 1 0 θ β k −1∅∗ ( ⋋−1 ((1− θ)⋋s+1 (r1) + θ ⋋s+1 (r2)) 1 s+1 ) dθ ] . By substituting ⋋s+1(χ) = θ ⋋s+1 (r1) + (1− θ)⋋s+1 (r2)red, we have = 1( ⋋s+1 (r2)−⋋s+1(r1) )β k × [ (s+ 1)1− β k kΓk(β) ∫ r2 r1 ( ⋋s+1 (r2)−⋋s+1(χ) )β k −1 ⋋s (χ)⋋ ′ (χ)∅(χ)dχ + (s+ 1)1− β k kΓk(β) ∫ r2 r1 (⋋s+1(χ)−⋋s+1(r1)) β k −1 ⋋s (χ)⋋ ′ (χ)∅(χ)dχ ] . A. Mehmood et al. / Eur. J. Pure Appl. Math, 18 (2) (2025), 5871 12 of 26 This implies = 1( ⋋s+1 (r2)−⋋s+1(r1) )β k [ s kZ β r+1 ∅(r2) + s kZ β r−2 ∅(r1) ] . (22) We get first half of our relation from the inclusions (21) and (22) for the second half employing ı.υ-(⋋s+1,℧) convexity of function ∅, we have ∅ ( ⋋−1 ( θ ⋋s+1 (r1) + (1− θ)⋋s+1 (r2) ) 1 s+1 ) ⊇ ℧(θ)∅(r1) + ℧(1− θ)∅(r2), (23) and ∅ ( ⋋−1 ( (1− θ)⋋s+1 (r1) + θ ⋋s+1 (r2) ) 1 s+1 ) ⊇ ℧(1− θ)∅(r1) + ℧(θ)∅(r2). (24) Adding (23) and (24) inclusions and multiplying both sides by (s+1)− β k kΓk(β) θ β k −1 also taking the integration over [0, 1] w.r.t ”θ”red, then we acquired our required relation. Corollary 1. If we fix ℧(θ) = θ in Theorem 6, we possess 2(s+ 1)− β k βΓk(β) ∅ ( ⋋−1 ( ⋋s+1(r1) +⋋s+1(r2) 2 ) 1 s+1 ) ⊇ 1( ⋋s+1 (r2)−⋋s+1(r1) )β k [ s kZ β r+1 ∅(r2) + s kZ β r−2 ∅(r1) ] ⊇ (s+ 1)− β k βΓk(β) [ ∅(r1) +∅(r2) ] . (25) Example 1. If we choose ∅(r) = [4 − ⋋s+1(r), 8 + ⋋s+1(r)] and ⋋(r) = r in (25) and using proposition 1, we have 2(s+ 1)− β k βΓk(β) ( 4− rs+1 1 + rs+1 2 2 , 8 + rs+1 1 + rs+1 2 2 ) ⊇ (s+ 1) −β k βΓk(β) ( 8− (rs+1 1 + rs+1 2 ), 16 + (rs+1 1 + rs+1 2 ) ) ⊇ (s+ 1)− β k βΓk(β) ( 8− (rs+1 1 + rs+1 2 ), 16 + (rs+1 1 + rs+1 2 ) ) . Example 2. For graphical representation if we choose ∅(r) = [4−⋋s+1(r), 8+⋋s+1(r)], ℧(r) = 1 8 and ⋋(r) = sin r in (25) and using the proposition 1, we have 8(s+ 1)− β k (β)Γk(β) ( 4− sins+1(r1) + sins+1(r2) 2 , 8 + sins+1(r1) + sins+1(r2) 2 ) A. Mehmood et al. / Eur. J. Pure Appl. Math, 18 (2) (2025), 5871 13 of 26 ⊇ (s+ 1) −β k βΓk(β) ( 8− (sins+1(r1) + sins+1(r2)), 16 + (sins+1(r1) + sins+1(r2)) ) ⊇ (s+ 1)− β k 4βΓk(β) ( 8− (sins+1(r1) + sins+1(r2)), 16 + (sins+1(r1) + sins+1(r2)) ) . Figure 1: Graphical representation of Theorem 6 corresponding to the choice of parameters r1 = 0.6, r1 < r2 ≤ π 2 , k = 1, s = 3 and β = 0.8. Also for tabular form we have r2 (a1, b1) (a2, b2) (a3, b3) 0.610000 (11.037156, 22.964070) (2.759289, 5.741017) (0.6898221.435254) 0.802159 (10.811418, 23.189807) (2.702855, 5.797452) (0.675714, 1.449363) 0.994319 (10.489794, 23.511432) (2.622448, 5.877858) (0.655612, 1.469465) 1.186478 (10.143322, 23.857903) (2.535831, 5.964476) (0.633958, 1.491119) 1.378637 (9.874478, 24.126747) (2.468620, 6.031687) (0.617155, 1.507922) 1.570796 (9.773019, 24.228207) (2.443255, 6.057052) (0.610814, 1.514263) Table 1: Interval bounds of Theorem 6 corresponding to the choice of parameters r1 = 0.6, r1 < r2 ≤ π 2 , k = 1, s = 3 and β = 0.8 Corollary 2. If we fix s = 0, k = 1, ⋋(r) = r and ℧(θ) = θ in Theorem 6, we possess Hermite-Hadamard-Fejer inequality for ı.υ-(⋋s+1,℧) cf that is provided in [23]. 1 βΓ(β) ∅ ( r1 + r2 2 ) ⊇ 1( r2 − r1 )β [Zβ r+1 ∅(r2) + Zβ r−2 ∅(r1) ] ⊇ 1 βΓ(β) [ ∅(r1) +∅(r2) ] . A. Mehmood et al. / Eur. J. Pure Appl. Math, 18 (2) (2025), 5871 14 of 26 Theorem 7. Let s ∈ R/{−1}, k ≥ 0, ∅ ∈ SIGX([r1, r2], R + i ), and ⅁ : [r1, r2] → R be symmetric w.r.t r1+r2 2 , then for β > 0 the following inclusions holds. 1 2℧(12) ∅ ( ⋋−1 ( ⋋s+1(r1) +⋋s+1(r2) 2 ) 1 s+1 )[ s kZ β r+1 ⅁(r2) + s kZ β r−2 ⅁(r1) ] ⊇ [ s kZ β r+1 ⅁∅(r2) + s kZ β r−2 ⅁∅(r1) ] ⊇ (s+ 1)− β k kΓk(β) (∫ 1 0 θ β k −1[℧(θ) + ℧(1− θ)] [ ⅁(⋋−1(θ ⋋s+1 (r1) + (1− θ)⋋s+1 (r2)) 1 s+1 ) ] × [∅(r1) +∅(r2)]dθ ) . Proof. Since ∅ is ı.υ-(⋋s+1,℧) cf and for θ = 1 2 , we have ∅ ( ⋋−1 ( ⋋s+1(x) +⋋s+1(y) 2 ) 1 s+1 ) ⊇ ℧( 1 2 ) [ ∅(x) +∅(y) ] . By substituting x = ⋋−1(θ⋋s+1 (r1)+(1−θ)⋋s+1 (r2)) 1 s+1 and y = ⋋−1((1−θ)⋋s+1 (r1)+ θ⋋s+1 (r2)) 1 s+1 in above expression. Multiplying both sides by (s+1)− β k kΓk(β) θ β k −1 [ ⅁(⋋−1(θ⋋s+1 (r1) + (1− θ)⋋s+1 (r2)) 1 s+1 ) ] and taking the integration over [0, 1] w.r.t ”θ”, we have 1 ℧(12) (s+ 1)− β k kΓk(β) ∫ 1 0 θ β k −1 [ ⅁(⋋−1(θ ⋋s+1 (r1) + (1− θ)⋋s+1 (r2)) 1 s+1 ) ] ×∅ ( ⋋−1 ( ⋋s+1(r1) +⋋s+1(r2) 2 ) 1 s+1 ) dθ ⊇ (s+ 1)− β k kΓk(β) ∫ 1 0 θ β k −1 [ ⅁(⋋−1(θ ⋋s+1 (r1) + (1− θ)⋋s+1 (r2)) 1 s+1 ) ] ×∅ ( ⋋−1 (θ ⋋s+1 (r1) + (1− θ)⋋s+1 (r2)) 1 s+1 ) dθ + (s+ 1)− β k kΓk(β) ∫ 1 0 θ β k −1 [ ⅁(⋋−1(θ ⋋s+1 (r1) + (1− θ)⋋s+1 (r2)) 1 s+1 ) ] ×∅ ( ⋋−1 ((1− θ)⋋s+1 (r1) + θ ⋋s+1 (r2)) 1 s+1 ) dθ. (26) Taking benefit of the fact ⅁(⋋−1(θ⋋s+1 (r1)+ (1− θ)⋋s+1 (r2)) 1 s+1 ) = ⅁(⋋−1((1− θ)⋋s+1 (r1) + θ ⋋s+1 (r2)) 1 s+1 ) for the left part of (26). (s+ 1)− β k kΓk(β) ∫ 1 0 θ β k −1 [ ⅁(⋋−1(θ ⋋s+1 (r1) + (1− θ)⋋s+1 (r2)) 1 s+1 ) ] A. Mehmood et al. / Eur. J. Pure Appl. Math, 18 (2) (2025), 5871 15 of 26 ×∅ ( ⋋−1 ( ⋋s+1(r1) +⋋s+1(r2) 2 ) 1 s+1 ) dθ = 1 2 [ (s+ 1)− β k kΓk(β) ∫ 1 0 θ β k −1 [ ⅁(⋋−1(θ ⋋s+1 (r1) + (1− θ)⋋s+1 (r2)) 1 s+1 ) ] ×∅ ( ⋋−1 ( ⋋s+1(r1) +⋋s+1(r2) 2 ) 1 s+1 ) dθ + (s+ 1)− β k kΓk(β) ∫ 1 0 θ β k −1 [ ⅁(⋋−1((1− θ)⋋s+1 (r1) + θ ⋋s+1 (r2)) 1 s+1 ) ] ×∅ ( ⋋−1 ( ⋋s+1(r1) +⋋s+1(r2) 2 ) 1 s+1 ) dθ ] . By substituting ⋋s+1(χ) = θ ⋋s+1 (r1) + (1− θ)⋋s+1 (r2)red, we have = 1 2(⋋s+1(r2)−⋋s+1(r1)) β k ∅ ( ⋋−1 ( ⋋s+1(r1) +⋋s+1(r2) 2 ) 1 s+1 ) × [ (s+ 1)1− β k kΓk(β) ∫ r2 r1 ( ⋋s+1 (r2)−⋋s+1(χ) )β k −1 ⋋s (χ)⋋ ′ (χ)⅁(χ)dχ + (s+ 1)1− β k kΓk(β) ∫ r2 r1 (⋋s+1(χ)−⋋s+1(r1)) β k −1 ⋋s (χ)⋋ ′ (χ)⅁(χ)dχ ] = 1 2(⋋s+1(r2)−⋋s+1(r1)) β k ∅ ( ⋋−1 ( ⋋s+1(r1) +⋋s+1(r2) 2 ) 1 s+1 ) × [ s kZ β r+1 ⅁(r2) + s kZ β r−2 ⅁(r1) ] . (27) For the right part of (26)red, we have (s+ 1)− β k kΓk(β) ∫ 1 0 θ β k −1 [ ⅁(⋋−1(θ ⋋s+1 (r1) + (1− θ)⋋s+1 (r2)) 1 s+1 ) ] ×∅ ( ⋋−1 (θ ⋋s+1 (r1) + (1− θ)⋋s+1 (r2)) 1 s+1 ) dθ + (s+ 1)− β k kΓk(β) ∫ 1 0 θ β k −1 [ ⅁(⋋−1(θ ⋋s+1 (r1) + (1− θ)⋋s+1 (r2)) 1 s+1 ) ] ×∅ ( ⋋−1 ((1− θ)⋋s+1 (r1) + θ ⋋s+1 (r2)) 1 s+1 ) dθ. Substituting ⋋s+1(χ) = θ ⋋s+1 (r1) + (1− θ)⋋s+1 (r2)red, we have = 1 (⋋s+1(r2)−⋋s+1(r1)) β k [ s kZ β r+1 ⅁∅(r2) + s kZ β r−2 ⅁∅(r1) ] . (28) A. Mehmood et al. / Eur. J. Pure Appl. Math, 18 (2) (2025), 5871 16 of 26 We achieve our first desired inclusion from (27) and (28). In this way for the other inclusion employing ı.υ-(⋋s+1,℧) convexity of function ∅, we have ∅ ( ⋋−1 ( θ ⋋s+1 (r1) + (1− θ)⋋s+1 (r2) ) ⊇ ℧(θ)∅(r1) + ℧(1− θ)∅(r2), (29) and ∅ ( ⋋−1 ( (1− θ)⋋s+1 (r1) + θ ⋋s+1 (r2) ) ⊇ ℧(1− θ)∅(r1) + ℧(θ)∅(r2). (30) Adding (29) and (30) inclusions and multiplying both sides by (s+1)− β k kΓk(β) θ β k −1 [ ⅁(⋋−1(θ⋋s+1 (r1) + (1 − θ) ⋋s+1 (r2)) 1 s+1 ) ] also taking the integration over [0, 1] w.r.t ”θ”red, then we acquired our required relation. Corollary 3. If we fix ℧(θ) = θ in Theorem 7, we possess ∅ ( ⋋−1 ( ⋋s+1(r1) +⋋s+1(r2) 2 ) 1 s+1 )[ s kZ β r+1 ⅁(r2) + s kZ β r−2 ⅁(r1) ] ⊇ [ s kZ β r+1 ⅁∅(r2) + s kZ β r−2 ⅁∅(r1) ] ⊇ [∅(r1) +∅(r2)] [ s kZ β r+1 ⅁(r2) + s kZ β r−2 ⅁(r1) ] . (31) Example 3. If we choose ∅(r) = [4−⋋s+1(r), 8 +⋋s+1(r)] and ⋋(r) = r and ⅁(r) = 1 in (31) and utilizing proposition 1, we have 8(s+ 1)− β k βΓk(β) ( 4− rs+1 1 + rs+1 2 2 , 8 + rs+1 1 + rs+1 2 2 )( rs+1 2 − rs+1 1 )β k ⊇ (s+ 1) −β k βΓk(β) ( rs+1 2 − rs+1 1 )β k ( 8− (rs+1 1 + rs+1 2 ), 16 + (rs+1 1 + rs+1 2 ) ) ⊇ (s+ 1)− β k 2δΓk(β) ( 8− (rs+1 1 + rs+1 2 ), 16 + (rs+1 1 + rs+1 2 ) )( rs+1 2 − rs+1 1 )β k . Example 4. For graphical representation if we choose ∅(r) = [4−⋋s+1(r), 8+⋋s+1(r)], ℧(r) = 1 8 and ⋋(r) = sin r and ⅁(r) = 1 in (31) and utilizing proposition 1, we have 4(s+ 1)− β k βΓk(β) ( 4− sins+1(r1) + sins+1(r2) 2 , 8 + sins+1(r1) + sins+1(r2) 2 )( sins+1(r2)− sins+1(r1) )β k ⊇ (s+ 1) −β k βΓk(β) ( sins+1(r2)− sins+1(r1) )β k ( 8− (sins+1(r1) + sins+1(r2)), 16 + (sins+1(r1) + sins+1(r2)) ) ⊇ (s+ 1)− β k 4δΓk(β) ( 8− (sins+1(r1) + sins+1(r2)), 16 + (sins+1(r1) + sins+1(r2)) )( sins+1(r2)− sins+1(r1) )β k . A. Mehmood et al. / Eur. J. Pure Appl. Math, 18 (2) (2025), 5871 17 of 26 Figure 2: Graphical representation of Theorem 7 corresponding to the choice of parameters r1 = 0.6, r1 < r2 ≤ π 2 , k = 1, s = 3 and β = 0.8. r2 (a1, b1) (a2, b2) (a3, b3) 0.6100 (0.0928, 0.1930) (0.0464, 0.0965) (0.0116, 0.0241) 0.8022 (1.2813, 2.7484) (0.6407, 1.3742) (0.1602, 0.3435) 0.9943 (2.4816, 5.5622) (1.2408, 2.7811) (0.3102, 0.6953) 1.1865 (3.5355, 8.3157) (1.7677, 4.1578) (0.4419, 1.0395) 1.3786 (4.2401, 10.3601) (2.1201, 5.1800) (0.5300, 1.2950) 1.5708 (4.4849, 11.1186) (2.2425, 5.5593) (0.5606, 1.3898) Table 2: Interval bounds of Theorem 7 corresponding to the choice of parameters r1 = 0.6, r1 < r2 ≤ π 2 , k = 1, s = 3 and β = 0.8. For tabular form we have Corollary 4. If we fix s = 0, k = 1, ⋋(r) = r and ℧(θ) = θ in Theorem 7, we possess ∅ ( r1 + r2 2 )[ Zβ r+1 ⅁(r2) + Zβ r−2 ⅁(r1) ] ⊇ [ Zβ r+1 ⅁∅(r2) + Zβ r−2 ⅁∅(r1) ] ⊇ [∅(r1) +∅(r2)] [ Zβ r+1 ⅁(r2) + Zβ r−2 ⅁(r1) ] . Theorem 8. Let s ∈ R/{−1}, ∅,⅁ ∈ SIGX([r1, r2], R + i ) and k ≥ 0, then for β > 0 the following redinequality fulfilled 1 (⋋s+1(r2)−⋋s+1(r1)) β k [ s kZ β r+1 ∅⅁(r2) + s kZ β r−2 ∅⅁(r1) ] ⊇ P (r1, r2) (s+ 1)− β k kΓk(β) ∫ 1 0 θ β k −1[℧1(θ)℧2(θ) + ℧1(1− θ)℧2(1− θ)]dθ +Q(r1, r2) (s+ 1)− β k kΓk(β) ∫ 1 0 θ β k −1[℧1(θ)℧2(1− θ) + ℧1(1− θ)℧2(θ)]dθ, where, P (r1, r2) = ∅(r1)⅁(r1) +∅(r2)⅁(r2), Q(r1, r2) = ∅(r1)⅁(r2) +∅(r2)⅁(r1). A. Mehmood et al. / Eur. J. Pure Appl. Math, 18 (2) (2025), 5871 18 of 26 Proof. Since ∅,⅁ ∈ SIGX([r1, r2], R +), then ∅ ( ⋋−1 (θ ⋋s+1 (r1) + (1− θ)⋋s+1 (r2)) 1 s+1 ) ⊇ ℧1(θ)∅(r1) + ℧1(1− θ)∅(r2) ⅁ ( ⋋−1 (θ ⋋s+1 (r1) + (1− θ)⋋s+1 (r2)) 1 s+1 ⊇ ℧2(θ)⅁(r1) + ℧2(1− θ)⅁(r2). By multiplying, we have ∅ ( ⋋−1 (θ ⋋s+1 (r1) + (1− θ)⋋s+1 (r2)) 1 s+1 ) ⅁ ( ⋋−1 (θ ⋋s+1 (r1) + (1− θ)⋋s+1 (r2)) 1 s+1 ) ⊇ ℧1(θ)℧2(θ)∅(r1)⅁(r1) + ℧1(θ)℧2(1− θ)∅(r1)⅁(r2) + ℧1(1− θ)℧2(θ)∅(r2)⅁(r1) + ℧1(1− θ)℧2(1− θ)∅(r2)⅁(r2). (32) Similarly, ∅ ( ⋋−1 ((1− θ)⋋s+1 (r1) + θ ⋋s+1 (r2)) 1 s+1 ) ⅁ ( ⋋−1 ((1− θ)⋋s+1 (r1) + θ ⋋s+1 (r2)) 1 s+1 ) ⊇ ℧1(1− θ)℧2(1− θ)∅(r1)⅁(r1) + ℧1(1− θ)℧2(θ)∅(r1)⅁(r2) + ℧1(θ)℧2(1− θ)∅(r2)⅁(r1) + ℧1(θ)℧2(θ)∅(r2)⅁(r2). (33) Adding (32) and (33) inclusions and multiplying both sides by (s+1)− β k kΓk(β) θ β k −1 also taking the integration over [0, 1] w.r.t ”θ”red, then we have (s+ 1)− β k kΓk(β) ∫ 1 0 θ β k −1∅ ( ⋋−1 (θ ⋋s+1 (r1) + (1− θ)⋋s+1 (r2)) 1 s+1 ) × ⅁ ( ⋋−1 (θ ⋋s+1 (r1) + (1− θ)⋋s+1 (r2)) 1 s+1 ) dθ + (s+ 1)− β k kΓk(β) ∫ 1 0 θ β k −1∅ ( ⋋−1 ((1− θ)⋋s+1 (r1) + θ ⋋s+1 (r2)) 1 s+1 ) × ⅁ ( ⋋−1 ((1− θ)⋋s+1 (r1) + θ ⋋s+1 (r2)) 1 s+1 ) dθ ⊇ (s+ 1)− β k kΓk(β) ∫ 1 0 θ β k −1[℧1(θ)℧2(θ) + ℧1(1− θ)℧2(1− θ)][∅(r1)⅁(r1) +∅(r2)⅁(r2)]dθ + (s+ 1)− β k kΓk(β) ∫ 1 0 θ β k −1[℧1(θ)℧2(1− θ) + ℧1(1− θ)℧2(θ)][∅(r1)⅁(r2) +∅(r2)⅁(r1)]dθ. Substituting ⋋s+1(χ) = θ ⋋s+1 (r1) + (1− θ)⋋s+1 (r2)red, then we have = 1 (⋋s+1(r2)−⋋s+1(r1)) β k [ s kZ β r+1 ∅⅁(r2) + s kZ β r−2 ∅⅁(r1) ] ⊇ P (r1, r2) (s+ 1)− β k kΓk(β) ∫ 1 0 θ β k −1[℧1(θ)℧2(θ) + ℧1(1− θ)℧2(1− θ)]dθ +Q(r1, r2) (s+ 1)− β k kΓk(β) ∫ 1 0 θ β k −1[℧1(θ)℧2(1− θ) + ℧1(1− θ)℧2(θ)]dθ. This complete our required relation. A. Mehmood et al. / Eur. J. Pure Appl. Math, 18 (2) (2025), 5871 19 of 26 Corollary 5. If we fix ℧(θ) = θ in Theorem 8, we possess 1 (⋋s+1(r2)−⋋s+1(r1)) β k [ s kZ β r+1 ∅⅁(r2) + s kZ β r−2 ∅⅁(r1) ] ⊇ (s+ 1)− β k kΓk(β) [ P (r1, r2) β2 k2 + β k + 1 (βk + 2)(βk + 1)(βk ) +Q(r1, r2) 2 (βk + 2)(βk + 1) ] , where, P (r1, r2) = ∅(r1)⅁(r1) +∅(r2)⅁(r2), Q(r1, r2) = ∅(r1)⅁(r2) +∅(r2)⅁(r1). Example 5. If we choose ∅(r) = [4 − ⋋s+1(r), 8 + ⋋s+1(r)] and ⋋(r) = r, ⅁(r) = 1, ℧(θ) = 1 4 in Theorem 8 and utilizing proposition 1, we have (s+ 1) −β k βΓk(β) ( 8− (rs+1 1 + rs+1 2 ), 16 + (rs+1 1 + rs+1 2 ) ) ⊇ (s+ 1)− β k 8βΓk(β) ( 8− (rs+1 1 + rs+1 2 ), 16 + (rs+1 1 + rs+1 2 ) ) . Example 6. For graphical representation if we choose ∅(r) = [4−⋋s+1(r), 8 +⋋s+1(r)] and ⋋(r) = sin r, ⅁(r) = 1, ℧(θ) = 1 8 in Theorem 8 and utilizing proposition 1, we have (s+ 1) −β k βΓk(β) ( 8− (sins+1(r1) + sins+1(r2)), 16 + (sins+1(r1) + sins+1(r2)) ) ⊇ (s+ 1)− β k 32βΓk(β) ( 8− (sins+1(r1) + sins+1(r2)), 16 + (sins+1(r1) + sins+1(r2)) ) . Figure 3: Graphical representation of Theorem 8 corresponding to the choice of parameters r1 = 0.6, r1 < r2 ≤ π 2 , k = 1, s = 3 and β = 0.8. For tabular form we have Corollary 6. If we fix s = 0, k = 1, ⋋(θ) = θ, and ℧(θ) = θ in Theorem 8, we possess 1 (r2 − r1)β [ Zβ r+1 ∅⅁(r2) + Zβ r−2 ∅⅁(r1) ] ⊇ 1 Γ(β) [ P (r1, r2) β2 + β + 1 (β + 2)(β + 1)(β) +Q(r1, r2) 2 (β + 2)(β + 1) ] . A. Mehmood et al. / Eur. J. Pure Appl. Math, 18 (2) (2025), 5871 20 of 26 r2 (a1, b1) (a2, b2) 0.6100 (2.7593, 5.7410) (0.0862, 0.1794) 0.8022 (2.7029, 5.7975) (0.0845, 0.1812) 0.9943 (2.6224, 5.8779) (0.0820, 0.1837) 1.1865 (2.5358, 5.9645) (0.0792, 0.1864) 1.3786 (2.4686, 6.0317) (0.0771, 0.1885) 1.5708 (2.4433, 6.0571) (0.0764, 0.1893) Table 3: Interval bounds of Theorem 8 corresponding to the choice of parameters r1 = 0.6, r1 < r2 ≤ π 2 , k = 1, s = 3 and β = 0.8. Corollary 7. If we fix ⋋(θ) = θ and s+ 1 ≥ 0 in Theorem 8, we possess 1 (rs+1 2 − rs+1 1 ) β k [ s kZ β (rs+1 1 )+ ∅⅁(rs+1 2 ) + s kZ β (rs+1 2 )− ∅⅁(rs+1 2 ) ] ⊇ P (r1, r2) (s+ 1)− β k kΓk(β) ∫ 1 0 θ β k −1[℧1(θ)℧2(θ) + ℧1(1− θ)℧2(1− θ)]dθ +Q(r1, r2) (s+ 1)− β k kΓk(β) ∫ 1 0 θ β k −1[℧1(θ)℧2(1− θ) + ℧1(1− θ)℧2(θ)]dθ Corollary 8. If we fix ⋋(θ) = θ, s+ 1 ≥ 0 and ℧(θ) = θ in Theorem 8, we possess 1 (rs+1 2 − rs+1 1 ) β k [ s kZ β (rs+1 1 )+ ∅⅁(rs+1 2 ) + s kZ β (rs+1 2 )− ∅⅁(rs+1 2 ) ] ⊇ (s+ 1)− β k kΓk(β) [ P (r1, r2) β2 k2 + β k + 1 (βk + 2)(βk + 1)(βk ) +Q(r1, r2) 2 (βk + 2)(βk + 1) ] . Theorem 9. Let s ∈ R/{−1}, ∅,⅁ ∈ SIGX([r1, r2], R + i ) and k ≥ 0, then for β > 0 the following redinequality fulfill (s+ 1)− β k βΓk(β) ∅ ( ⋋−1 ( ⋋s+1(r1) +⋋s+1(r2) 2 ) 1 s+1 ) ⅁ ( ⋋−1 ( ⋋s+1(r1) +⋋s+1(r2) 2 ) 1 s+1 ) ⊇ ℧1( 1 2)℧2( 1 2) (⋋s+1(r2)−⋋s+1(r1)) β k [ s kZ β r+1 ∅⅁(r2) + s kZ β r−2 ∅⅁(r1) ] + ℧1( 1 2 )℧2( 1 2 ) ( P (r1, r2) (s+ 1)− β k kΓk(β) ∫ 1 0 θ β k −1[℧1(θ)℧2(1− θ) + ℧1(1− θ)℧2(θ)]dθ +Q(r1, r2) (s+ 1)− β k kΓk(β) ∫ 1 0 θ β k −1[℧1(θ)℧2(θ) + ℧1(1− θ)℧2(1− θ)]dθ ) , where, P (r1, r2) = ∅(r1)⅁(r1) +∅(r2)⅁(r2), Q(r1, r2) = ∅(r1)⅁(r2) +∅(r2)⅁(r1). Proof. Since ∅,⅁ ∈ SIGX([r1, r2], R +), then ∅ ( ⋋−1 ( ⋋s+1(r1) +⋋s+1(r2) 2 ) 1 s+1 ) ⅁ ( ⋋−1 ( ⋋s+1(r1) +⋋s+1(r2) 2 ) 1 s+1 ) A. Mehmood et al. / Eur. J. Pure Appl. Math, 18 (2) (2025), 5871 21 of 26 ⊇ ℧1( 1 2 )℧2( 1 2 )[ ∅ ( ⋋−1 (θ ⋋s+1 (r1) + (1− θ)⋋s+1 (r2)) 1 s+1 ) ⅁ ( ⋋−1 (θ ⋋s+1 (r1) + ((1− θ)⋋s+1 (r2)) 1 s+1 ) +∅ ( ⋋−1 (θ ⋋s+1 (r1) + (1− θ)⋋s+1 (r2)) 1 s+1 ) ⅁ ( ⋋−1 ((1− θ)⋋s+1 (r1) + θ ⋋s+1 (r2)) 1 s+1 ) +∅ ( ⋋−1 ((1− θ)⋋s+1 (r1) + θ ⋋s+1 (r2)) 1 s+1 ) ⅁ ( ⋋−1 (θ(r1) + (1− θ)r2) 1 s+1 ) +∅ ( ⋋−1 ((1− θ)⋋s+1 (r1) + θ ⋋s+1 (r2)) 1 s+1 ) ⅁ ( ⋋−1 ((1− θ)⋋s+1 (r1) + θ ⋋s+1 (r2)) 1 s+1 )] . Multiplying preceding inclusion by (s+1)− β k kΓk(β) θ β k −1 and taking the integration on [0, 1] w.r.t ”θ”red, we have (s+ 1)− β k kΓk(β) ∫ 1 0 θ β k −1∅ ( ⋋−1 ( ⋋s+1(r1) +⋋s+1(r2) 2 ) 1 s+1 ) ⅁ ( ⋋−1 ( ⋋s+1(r1) +⋋s+1(r2) 2 ) 1 s+1 ) dθ ⊇ ℧1( 1 2 )℧2( 1 2 ) [ (s+ 1)− β k kΓk(β) ∫ 1 0 θ β k −1∅ ( ⋋−1 (θ ⋋s+1 (r1) + (1− θ)⋋s+1 (r2)) 1 s+1 ) × ⅁ ( ⋋−1 (θ ⋋s+1 (r1) + (1− θ)⋋s+1 (r2)) 1 s+1 ) dθ + (s+ 1)− β k kΓk(β) ∫ 1 0 θ β k −1∅ ( ⋋−1 (θ ⋋s+1 (r1) + (1− θ)⋋s+1 (r2)) 1 s+1 ) × ⅁ ( ⋋−1 ((1− θ)⋋s+1 (r1) + θ ⋋s+1 r2) 1 s+1 ) dθ + (s+ 1)− β k kΓk(β) ∫ 1 0 θ β k −1∅ ( ⋋−1 ((1− θ)⋋s+1 (r1) + θ ⋋s+1 (r2)) 1 s+1 ) × ⅁ ( ⋋−1 (θ ⋋s+1 (r1) + (1− θ)⋋s+1 (r2)) 1 s+1 ) dθ + (s+ 1)− β k kΓk(β) ∫ 1 0 θ β k −1∅ ( ⋋−1 ((1− θ)⋋s+1 (r1) + θ ⋋s+1 (r2)) 1 s+1 ) × ⅁ ( ⋋−1 ((1− θ)⋋s+1 (r1) + θ ⋋s+1 (r2)) 1 s+1 ) dθ ] . By definition of ı.υ-(⋋s+1,℧) and substituting ⋋s+1(χ) = θ⋋s+1 (r1)+(1−θ)⋋s+1 (r2)red, we have (s+ 1)− β k βΓk(β) ∅ ( ⋋−1 ( ⋋s+1(r1) +⋋s+1(r2) 2 ) 1 s+1 ) ⅁ ( ⋋−1 ( ⋋s+1(r1) +⋋s+1(r2) 2 ) 1 s+1 ) ⊇ ℧1( 1 2)℧2( 1 2) (⋋s+1(r2)−⋋s+1(r1)) β k [ s kZ β r+1 ∅⅁(r2) + s kZ β r−2 ∅⅁(r1) ] + ℧1( 1 2 )℧2( 1 2 ) ( P (r1, r2) (s+ 1)− β k kΓk(β) ∫ 1 0 θ β k −1[℧1(θ)℧2(1− θ) + ℧1(1− θ)℧2(θ)]dθ A. Mehmood et al. / Eur. J. Pure Appl. Math, 18 (2) (2025), 5871 22 of 26 +Q(r1, r2) (s+ 1)− β k kΓk(β) ∫ 1 0 θ β k −1[℧1(θ)℧2(θ) + ℧1(1− θ)℧2(1− θ)]dθ ) . Hence we have proved our result. Corollary 9. If we fix ℧(θ) = θ in Theorem 9, we possess (s+ 1)− β k βΓk(β) ∅ ( ⋋−1 ( ⋋s+1(r1) +⋋s+1(r2) 2 ) 1 s+1 ) ⅁ ( ⋋−1 ( ⋋s+1(r1) +⋋s+1(r2) 2 ) 1 s+1 ) ⊇ 1 (⋋s+1(r2)−⋋s+1(r1)) β k [ s kZ β r+1 ∅⅁(r2) + s kZ β r−2 ∅⅁(r1) ] + (s+ 1)− β k kΓk(β) [ P (r1, r2) 2 (β + 2)(β + 1) +Q(r1, r2) β2 + β + 1 (β + 2)(β + 1)(β) ] . Example 7. If we choose ∅(r) = [4 − ⋋s+1(r), 8 + ⋋s+1(r)] and ⋋(r) = r, ⅁(r) = 1, ℧1(r) = ℧2(r) = 1 4 in Theorem 9 and utilizing proposition 1, we have (s+ 1) −β k βΓk(β) ( 4− rs+1 1 + rs+1 2 2 , 8 + rs+1 1 + rs+1 2 2 ) ⊇ (s+ 1)− β k δΓk(β) ( 1 64 ( rs+1 2 − rs+1 1 )β k −1 ( 8− (rs+1 1 + rs+1 2 ), 16 + (rs+1 1 + rs+1 2 ) ) + 1 8 ( 8− (rs+1 1 + rs+1 2 ), 16 + (rs+1 1 + rs+1 2 ) )) . Example 8. For graphical representation if we choose ∅(r) = [4−⋋s+1(r), 8+⋋s+1(r)], ⋋(r) = sin r, ⅁(r) = 1 and ℧1(r) = ℧2(r) = 1 4 in Theorem 9 and utilizing proposition 1, we have (s+ 1) −β k βΓk(β) ( 4− sins+1(r1) + sins+1(r2) 2 , 8 + sins+1(r1) + sins+1(r2) 2 ) ⊇ (s+ 1)− β k 2048βΓk(β) (( 8− (sins+1(r1) + sins+1(r2)), 16 + (sins+1(r1) + sins+1(r2)) ) + ( 8− (sins+1(r1) + sins+1(r2)), 16 + (sins+1(r1) + sins+1(r2)) )) . For tabular value Corollary 10. If we fix s = 0, k = 1, ⋋(r) = r and ℧(θ) = θ in Theorem 9, we possess 1 βΓ(β) ∅ ( r1 + r2 2 ) ⅁ ( r1 + r2 2 ) A. Mehmood et al. / Eur. J. Pure Appl. Math, 18 (2) (2025), 5871 23 of 26 Figure 4: Graphical representation of Theorem 9 corresponding to the choice of parameters r1 = 0.6, r1 < r2 ≤ π 2 , k = 1, s = 3 and β = 0.8. r2 (a1, b1) (a2, b2) 0.6100 (1.3796e+ 00, 2.8705e+ 00) (2.6946e− 03, 5.6065e− 03) 0.8022 (1.3514e+ 00, 2.8987e+ 00) (2.6395e− 03, 5.6616e− 03) 0.9943 (1.3112e+ 00, 2.9389e+ 00) (2.5610e− 03, 5.7401e− 03) 1.1865 (1.2679e+ 00, 2.9822e+ 00) (2.4764e− 03, 5.8247e− 03) 1.3786 (1.2343e+ 00, 3.0158e+ 00) (2.4108e− 03, 5.8903e− 03) 1.5708 (1.2216e+ 00, 3.0285e+ 00) (2.3860e− 03, 5.9151e− 03) Table 4: Interval bounds of Theorem 9 corresponding to the choice of parameters r1 = 0.6, r1 < r2 ≤ π 2 , k = 1, s = 3 and β = 0.8. ⊇ 1 4(r2 − r1)β [ Zβ r+1 ∅⅁(r2) + Zβ r−2 ∅⅁(r1) ] + 1 kΓ(β) [ P (r1, r2) 2 (β + 2)(β + 1) +Q(r1, r2) β2 + β + 1 (β + 2)(β + 1)(β) ] . 3. Conclusion redFractional inequalities, which extend classical inequalities to fractional-order set- tings, play a crucial role in various industrial applications by providing precise analytical tools for optimization, stability analysis, and error estimation. In control systems and automation, fractional inequalities help establish stability criteria for fractional-order con- trollers, improving system robustness in robotics, aerospace and process control industries. In signal processing and telecommunications, they aid in error bounds estimation and per- formance analysis of fractional filters, enhancing data transmission and noise reduction. In materials science and mechanical engineering, fractional inequalities contribute to mod- eling stress-strain relationships in viscoelastic and complex materials, optimizing designs in structural engineering and manufacturing. The energy sector benefits from these in- equalities in analyzing fractional diffusion processes, optimizing heat conduction models, and improving energy storage systems like batteries and supercapacitors. Additionally, in biomedical engineering, they assist in developing fractional-order models for physiological systems, ensuring accurate predictions in drug delivery and neural activity analysis. By A. Mehmood et al. / Eur. J. Pure Appl. Math, 18 (2) (2025), 5871 24 of 26 refining analytical bounds and improving modeling accuracy, fractional inequalities signif- icantly contribute to industrial advancements across multiple domains. In this study, we introduce the novel Hermite-Hadamard-Fejer type inequalities within the structure of ı.υ (⋋s+1,℧) class of convexity. Our analysis provides comprehensive bounds for several well- known fractional problems. Specifically, we investigate the interplay between the classical Hermite-Hadamard-Fejer inequality and unified forms of Minkowskis and Holder inequal- ities within the class of convexity. We provide a versatile structure for mathematical inequalities related to generalized fractional operators by extending and generalizing the reverse forms of these inequalities with in the unified class of convexity. To enhance their practical applications, we investigate the additional implications, derive specific inequali- ties, and illustrate them through graphical representations. We also check the results by using tables for different fractional orders. The sharpness of our inequalities is confirmed by these graphical comparison. We urge readers to investigate more generalized fractional operators in order to create bigger classes of inequalities for future research. Also, by comparing recently hypothesized disparities with those that already exist, future research could evaluate the adaptability of their findings by graphical analysis. Acknowledgements The authors A. Aloqaily, D. Santina and N. Mlaiki would like to thank Prince Sultan University for paying the APC and for the support through the TAS research lab. Declarations: Availability of data and material The data used to support the findings of this study are available from the corresponding author upon request. Authors’ contributions All authors contributed equally and significantly in writing this article. All authors read and approved the final version. Competing interests The authors declare that they have no conflicts of interest. References [1] N. Abel. Solution de quelques problèmes à l’aide d’intégrales définies. Magasin for Naturvidenskaberne, 1:11–17, 1823. [2] A. A. Kilbas, H. M. Srivastava, and J. J. Trujillo. Theory and applications of fractional differential equations. 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