International Journal of Analysis and Applications ISSN 2291-8639 Volume 12, Number 2 (2016), 188-197 http://www.etamaths.com FEJÉR TYPE INEQUALITIES FOR HARMONICALLY (s,m)-CONVEX FUNCTIONS IMRAN ABBAS BALOCH1,∗, İMDAT İŞCAN2 AND SILVESTRU SEVER DRAGOMIR3 Abstract. In this paper, a new weighted identity involving harmonically symmetric functions and differentiable functions is established. By using the notion of harmonic symmetricity, harmonic (s,m)-convexity, analysis and some auxiliary results, some new Fejér type integral inequalities are presented for the class of harmonically (s,m)-convex functions. 1. Introduction A function f : I ⊆ R → R is called convex function if f(λx + (1 − λ)y) ≤ λf(x) + (1 − λ)f(y) for all x,y ∈ I and λ ∈ [0, 1]. There are many results associated with convex functions in the area of inequalities, but one of them is the classical Hermite-Hadamard (see [21]) inequalities: (1.1) f (a + b 2 ) ≤ 1 b−a ∫ b a f(x)dx ≤ f(a) + f(b) 2 , for all a,b ∈ I, with a < b. The inequalities in (1.1) hold in reversed direction if f is a concave function. A vast literature have been produced by a number of mathematicians for convex functions but (1.1) is considered to be the most famous inequality for convex mappings due to its usefulness and many applications in various branches of pure and applied mathematics. The definition of classical or usual convex functions has been generalized in a variety of ways and as a consequence many researchers have established a number of Hermite-Hadamard type inequalities by using different generalizations of the classical convexity, see for instance [2]-[23] and the references mentioned in these papers. One of the generalizations of classical convexity is the harmonic (s,m)-convexity in second sense, which unifies the notion of Harmonically convex [12] and Harmonically s-convex functions in second sense [13] introduced by Imdat Iscan, as stated in the definition below. Definition 1. [1] The function f : I ⊂ (0,∞) → R is said to be harmonically (s,m)-convex in second sense, where s ∈ (0, 1] and m ∈ (0, 1] if f ( mxy mty + (1 − t)x ) = f ( ( t x + 1 − t my )−1 ) ≤ tsf(x) + m(1 − t)sf(y) ∀x,y ∈ I and t ∈ [0, 1]. Remark 1. Note that for s = 1,harmonic (s,m)-convexity reduces to harmonic m-convexity and for m = 1, harmonic (s,m)-convexity reduces to harmonic s-convexity in second sense (see [13]) and for s,m = 1, harmonic (s,m)-convexity reduces to ordinary harmonic convexity (see [12]). Proposition 1. Let f : (0,∞) → R be a function a) if f is (s,m)-convex function in second sense and non-decreasing, thenf is harmonically (s,m)- convex function in second sense. b) if f is harmonically (s,m)-convex function in second sense and non-increasing, then f is (s,m)- convex function in second sense. 2010 Mathematics Subject Classification. Primary 26D15; Secondary 26A51, 26E60, 41A55. Key words and phrases. Hermite-Hadamard’s inequality; Fejér’s inequality; convex function; harmonically (s,m)- convex function; Hölder’s inequality; power mean inequality. c©2016 Authors retain the copyrights of their papers, and all open access articles are distributed under the terms of the Creative Commons Attribution License. 188 HARMONICALLY (s,m)-CONVEX FUNCTIONS 189 Remark 2. According to proposition 1, every non-decreasing (s,m)-convex function in second sense is also harmonically (s,m)-convex function in second sense. Example 1. (see[3]) Let 0 < s < 1 and a,b,c ∈ R, then function f : (0,∞) → R defined by f(x) = { a, x = 0 bxs + c, x > 0 is non-decreasing s-convex function in second sense for b ≥ 0 and 0 ≤ c ≤ a. Hence, by proposition 1, f is harmonically (s, 1)-convex function. Proposition 2. Let s ∈ [0, 1], m ∈ (0, 1], f : [a,mb] ⊂ (0,∞) → R, be an increasing function and g : [a,mb] → [a,mb], g(x) = mab a+mb−x, a < mb. Then f is harmonically (s,m)-convex in second sense on [a,mb] if and only if fog is (s,m)-convex in second sense on [a,mb]. The following result of the Hermite-Hadamard type holds. Theorem 1. Let f : I ⊂ (0,∞) → R be a harmonically (s,m)-convex function in second sense with s ∈ [0, 1] and m ∈ (0, 1]. If 0 < a < b < ∞ and f ∈ L[a,b], then one has following inequality ab b−a ∫ b a f(x) x2 dx ≤ min [f(a) + mf( b m ) s + 1 , f(b) + mf( a m ) s + 1 ] Corollary 1. If we take m = 1 in Theorem 1, then we get ab b−a ∫ b a f(x) x2 dx ≤ f(a) + f(b) s + 1 Corollary 2. If we take s = 1 in Theorem 1, then we get ab b−a ∫ b a f(x) x2 dx ≤ min [f(a) + mf( b m ) 2 , f(b) + mf( a m ) 2 ] Chen and Wu [4], established the following weighted Fejér type inequality for the harmonically convex function as follow Theorem 2. [4] Let f : I ⊂ R\{0}→ R be a harmonically convex function and a,b ∈ I with a < b. If f ∈ L([a,b]), then one has (1.2) f ( 2ab a + b )∫ b a g(x) x2 dx ≤ ∫ b a g(x)f(x) x2 dx ≤ f(a) + f(b) 2 ∫ b a g(x) x2 dx, where g : [a,b] → R is non-negative, integrable and satisfies g (ab x ) = g ( ab a + b−x ) The main purpose of the present paper is to introduce a new notion of harmonically symmetric functions and to establish an identity involving a harmonically symmetric function and a differentiable function. We will prove some Fejér type inequalities by using this identity related with the second part of the inequality given above by (1.2).We believe that our findings are novel, new and better than those already exist and will open new ways for further research in this field. 2. Main Results Throughout this section, we take L(t) = 2ab (1−t)a+(1+t)b and U(t) = 2ab (1+t)a+(1−t)b. The Beta function, the Gamma function and the integral form of the hypergeometric function are defined as follows to be used in the sequel of paper B(α,β) = Γ(α)Γ(β) Γ(α + β) = ∫ 1 0 tα−1(1 − t)β−1dt, α,β > 0 Γ(α) = ∫ ∞ 0 tα−1e−tdt, α > 0 190 BALOCH, İŞCAN AND DRAGOMIR and 2F1(α,β; γ,z) = 1 B(β,γ −β) ∫ 1 0 tβ−1(1 − t)γ−β−1(1 −zt)−αdt, γ > β > 0, |z| < 1 The notion of harmonically symmetric functions is defined as follows: Definition 2. We say that a function g : [a,b] ⊆ R\{0}→ R is harmonically symmetric with respect to 2ab a+b if g(x) = g ( 1 1 a + 1 b − 1 x ) holds for all x ∈ [a,b]. Now, we give the weighted integral equality by using which we establish our results in this article. Lemma 1. Let f : I ⊆ R\{0}→ R be a differentiable function on I◦ and a,b ∈ I◦ with a < b and let g : [a,b] → [0,∞) be continuous positive mapping and harmonically symmetric to 2ab a+b . If f′ ∈ L([a,b]), then the following identity holds f(a) + f(b) 2 ∫ b a g(x) x2 dx− ∫ b a g(x)f(x) x2 dx = b−a 4ab ∫ 1 0 (∫ U(t) L(t) g(x) x2 dx )[ (U(t))2f′(U(t)) − (L(t))2f′(L(t)) ] dt Proof. Since, g : [a,b] → [0,∞) is harmonically symmetric to 2ab a+b , then g(U(t)) = g(L(t)). Consider I = b−a 4ab ∫ 1 0 (∫ U(t) L(t) g(x) x2 dx )[ (U(t))2f′(U(t)) − (L(t))2f′(L(t)) ] dt = 1 2 [∫ 1 0 (∫ U(t) L(t) g(x) x2 dx ) d[f(U(t)) + f(L(t))] ] = 1 2 [(∫ U(t) L(t) g(x) x2 dx ) (f(U(t)) + f(L(t))) ∣∣1 0 − b−a 2ab ∫ 1 0 (g(U(t)) + g(L(t)))(f(U(t)) + f(L(t)))dt ] = 1 2 [ (f(a) + f(b)) (∫ b a g(x) x2 dx ) − b−a ab ∫ 1 0 g(U(t)f(U(t)dt − b−a ab ∫ 1 0 g(L(t)f(L(t)dt ] = 1 2 [ (f(a) + f(b)) ∫ b a g(x) x2 dx− 2 ∫ 2ab a+b a g(x)f(x) x2 dx + 2 ∫ a 2ab a+b g(x)f(x) x2 dx ] = f(a) + f(b) 2 ∫ b a g(x) x2 dx− ∫ b a g(x)f(x) x2 dx � Now, we present new Fejér type inequalities for harmonically (s,m)-convex functions, which give the weighted generalization of some of the results established in resent literature. Theorem 3. Let f : I ⊆ (0,∞) → R be a differentiable function on I◦ and a, b m ∈ I◦, m ∈ (0, 1] with a < b and let g : [a,b] → [0,∞) be continuous positive mapping and harmonically symmetric to 2ab a+b such that f′ ∈ L([a,b]). If |f′|q is harmonically (s,m)-convex on [a, b m ] for q ≥ 1, then the following inequality holds ∣∣∣∣f(a) + f(b)2 ∫ b a g(x) x2 dx− ∫ b a g(x)f(x) x2 dx ∣∣∣∣ HARMONICALLY (s,m)-CONVEX FUNCTIONS 191 ≤ b−a 8ab a 2 q ‖g‖∞ { λ 1−1 q 1 (a,b) ({ 22B(s + 1, 2).2F1(2,s + 1,s + 3; b−a b ) − 21−sB(s + 1, 1).2F1(2,s + 1,s + 2; b−a 2b ) + 1 2s B(s + 2, 1).2F1(2,s + 2,s + 3; b−a 2b ) } |f′(b)|q + m22−sb2 (b + a)2 B(1,s + 2).2F1(2, 1,s + 3; b−a b + a )|f′( a m )|q )1 q + λ 1−1 q 2 (a,b) × ({ 22B(2,s + 1).2F1(2, 2,s + 3; b−a b ) − 22−sb2 (b + a)2 B(1,s + 1).2F1(2, 1,s + 2; b−a b + a ) − 22−sb2 (b + a)2 B(2,s + 1).2F1(2, 2,s + 3; b−a b + a ) } |f′(a)|q (2.1) + m 2s B(s + 2, 1).2F1(2,s + 2,s + 3; b−a 2b )|f′( b m )|q )1 q } Proof. From Lemma 1 and hölder’s inequality, we get∣∣∣∣f(a) + f(b)2 ∫ b a g(x) x2 dx− ∫ b a g(x)f(x) x2 dx ∣∣∣∣ ≤ b−a8ab ‖g‖∞ × {(∫ 1 0 (1 − t)(U(t))2dt )1−1 q (∫ 1 0 (1 − t)(U(t))2|f′(U(t))|qdt )1 q (2.2) + (∫ 1 0 (1 − t)(L(t))2dt )1−1 q (∫ 1 0 (1 − t)(L(t))2|f′(L(t))|qdt )1 q } By the harmonic (s,m)-convexity of |f′|q on [a,b] for q ≥ 1, we have∫ 1 0 (1 − t)(U(t))2|f′(U(t))|qdt = ∫ 1 0 (1 − t) ( 2ab (1 + t)a + (1 − t)b )2 × ∣∣∣∣f′( 2ab(1 + t)a + (1 − t)b) ∣∣∣∣qdt ≤ 12s |f′(b)|q ∫ 1 0 (1 − t)(1 + t)s ( 2ab (1 + t)a + (1 − t)b )2 dt + m 1 2s |f′( a m )|q ∫ 1 0 (1 − t)s+1 ( 2ab (1 + t)a + (1 − t)b )2 dt (2.3) = { 22a2B(s + 1, 2).2F1(2,s + 1,s + 3; b−a b ) − a2 2s−1 B(s + 1, 1).2F1(2,s + 1,s + 2; b−a 2b ) + a2 2s B(s+2, 1).2F1(2,s+2,s+3; b−a 2b ) } |f′(b)|q+ ma2b2 2s−2(b + a)2 B(1,s+2).2F1(2, 1,s+3; b−a b + a )|f′( a m )|q and ∫ 1 0 (1 − t)(L(t))2|f′(L(t))|qdt = ∫ 1 0 (1 − t) ( 2ab (1 − t)a + (1 + t)b )2 × ∣∣∣∣f′( 2ab(1 − t)a + (1 + t)b) ∣∣∣∣qdt ≤ 12s |f′(a)|q ∫ 1 0 (1 − t)(1 + t)s ( 2ab (1 − t)a + (1 + t)b )2 dt + m 1 2s |f′( b m )|q ∫ 1 0 (1 − t)s+1 ( 2ab (1 − t)a + (1 + t)b )2 dt (2.4) = { 22a2B(2,s + 1).2F1(2, 2,s + 3; b−a b ) − 22−sa2b2 (b + a)2 B(1,s + 1).2F1(2, 1,s + 2; b−a b + a ) − 22−sa2b2 (b + a)2 B(2,s+1).2F1(2, 2,s+3; b−a b + a ) } |f′(a)|q+ ma2 2s B(s+2, 1).2F1(2,s+2,s+3; b−a 2b )|f′( b m )|q 192 BALOCH, İŞCAN AND DRAGOMIR Moreover, (2.5) ∫ 1 0 (1 − t)(U(t))2dt = ∫ 1 0 (1 − t) ( 2ab (1 + t)a + (1 − t)b )2 dt = ( 2ab b−a )2 ln( a + b 2a ) − (2ab)2 b2 −a2 := λ1(a,b) and (2.6) ∫ 1 0 (1 − t)(L(t))2dt = ∫ 1 0 (1 − t) ( 2ab (1 − t)a + (1 + t)b )2 dt = (2ab)2 b2 −a2 + ( 2ab b−a )2 ln( a + b 2b ) := λ2(a,b) A combination of (2.2), (2.3), (2.4), (2.5) and (2.6) gives required result. This completes the proof. � Corollary 3. Suppose the assumptions of the Theorem 3 are satisfied. If g(x) = ab b−a for all x ∈ [a,b], then one has the following inequality∣∣∣∣f(a) + f(b)2 ∫ b a g(x) x2 dx− ∫ b a g(x)f(x) x2 dx ∣∣∣∣ ≤ a 2 q 8 { λ 1−1 q 1 (a,b) ({ 22B(s + 1, 2).2F1(2,s + 1,s + 3; b−a b ) − 21−sB(s + 1, 1).2F1(2,s + 1,s + 2; b−a 2b ) + 1 2s B(s + 2, 1).2F1(2,s + 2,s + 3; b−a 2b ) } |f′(b)|q + m22−sb2 (b + a)2 B(1,s + 2).2F1(2, 1,s + 3; b−a b + a )|f′( a m )|q )1 q + λ 1−1 q 2 (a,b) × ({ 22B(2,s + 1).2F1(2, 2,s + 3; b−a b ) − 22−sb2 (b + a)2 B(1,s + 1).2F1(2, 1,s + 2; b−a b + a ) − 22−sb2 (b + a)2 B(2,s + 1).2F1(2, 2,s + 3; b−a b + a ) } |f′(a)|q (2.7) + m 2s B(s + 2, 1).2F1(2,s + 2,s + 3; b−a 2b )|f′( b m )|q )1 q } Theorem 4. Let f : I ⊆ (0,∞) → R be a differentiable function on I◦ and a, b m ∈ I◦, m ∈ (0, 1] with a < b and let g : [a,b] → [0,∞) be continuous positive mapping and harmonically symmetric to 2ab a+b such that f′ ∈ L([a,b]). If |f′|q is harmonically (s,m)-convex on [a, b m ] for q > 1, then the following inequality holds ∣∣∣∣f(a) + f(b)2 ∫ b a g(x) x2 dx− ∫ b a g(x)f(x) x2 dx ∣∣∣∣ ≤ a(b−a)8b .‖g‖∞ × {({ 2B(s + 1, 1).2F1(2q,s + 1,s + 2; b−a b ) − 21−sB(s + 1, 1).2F1(2q,s + 1,s + 2; b−a 2b ) } |f′(b)|q +m22q−s( b b + a )2qB(1,s + 1).2F1(2q, 1,s + 2; b−a b + a )|f′( a m )|q )1 q + ({ 2B(1,s + 1).2F1(2q, 1,s + 2; b−a b ) − 22q−s( b b + a )2qB(1,s + 1).2F1(2q, 1,s + 2; b−a b + a ) } |f′(a)|q (2.8) + m 2s B(s + 1, 1).2F1(2q,s + 1,s + 2; b−a 2b )|f′( b m )|q )1 q HARMONICALLY (s,m)-CONVEX FUNCTIONS 193 Proof. From Lemma 1 and hölder’s inequality, we get∣∣∣∣f(a) + f(b)2 ∫ b a g(x) x2 dx− ∫ b a g(x)f(x) x2 dx ∣∣∣∣ ≤ b−a8ab ‖g‖∞ (∫ 1 0 (1 − t) q q−1 dt )1−1 q (2.9) × {(∫ 1 0 (U(t))2q|f′(U(t))|qdt )1 q + (∫ 1 0 (L(t))2q|f′(L(t))|qdt )1 q } By the harmonic (s,m)-convexity of |f′|q on [a,b] for q > 1, we have∫ 1 0 (U(t))2q|f′(U(t))|qdt = ∫ 1 0 ( 2ab (1 + t)a + (1 − t)b )2q × ∣∣∣∣f′( 2ab(1 + t)a + (1 − t)b) ∣∣∣∣qdt ≤ 12s |f′(b)|q ∫ 1 0 (1 + t)s ( 2ab (1 + t)a + (1 − t)b )2q dt + m 1 2s |f′( a m )|q ∫ 1 0 (1 − t)s ( 2ab (1 + t)a + (1 − t)b )2q dt (2.10) = a2q { 2B(s + 1, 1).2F1(2q,s + 1,s + 2; b−a b ) − 21−sB(s + 1, 1).2F1(2q,s + 1,s + 2; b−a 2b ) } |f′(b)|q + m22q−s( ab b + a )2qB(1,s + 1).2F1(2q, 1,s + 2; b−a b + a )|f′( a m )|q and ∫ 1 0 (L(t))2q|f′(L(t))|qdt = ∫ 1 0 ( 2ab (1 − t)a + (1 + t)b )2q × ∣∣∣∣f′( 2ab(1 − t)a + (1 + t)b) ∣∣∣∣qdt ≤ 12s |f′(a)|q ∫ 1 0 (1 + t)s ( 2ab (1 − t)a + (1 + t)b )2q dt + m 1 2s |f′( b m )|q ∫ 1 0 (1 − t)s ( 2ab (1 − t)a + (1 + t)b )2q dt (2.11) = a2q { 2B(1,s+ 1).2F1(2q, 1,s+ 2; b−a b )−22q−s( b b + a )2qB(1,s+ 1).2F1(2q, 1,s+ 2; b−a b + a ) } |f′(a)|q + ma2q 2s B(s + 1, 1).2F1(2q,s + 1,s + 2; b−a 2b )|f′( b m )|q By putting (2.10) and (2.11) in (2.9), we get desired result. � Corollary 4. Suppose the assumptions of the Theorem 3 are satisfied. If g(x) = ab b−a for all x ∈ [a,b], then one has the following inequality∣∣∣∣f(a) + f(b)2 ∫ b a g(x) x2 dx− ∫ b a g(x)f(x) x2 dx ∣∣∣∣ ≤ a28 × {({ 2B(s + 1, 1).2F1(2q,s + 1,s + 2; b−a b ) − 21−sB(s + 1, 1).2F1(2q,s + 1,s + 2; b−a 2b ) } |f′(b)|q +m22q−s( b b + a )2qB(1,s + 1).2F1(2q, 1,s + 2; b−a b + a )|f′( a m )|q )1 q + ({ 2B(1,s + 1).2F1(2q, 1,s + 2; b−a b ) − 22q−s( b b + a )2qB(1,s + 1).2F1(2q, 1,s + 2; b−a b + a ) } |f′(a)|q (2.12) + m 2s B(s + 1, 1).2F1(2q,s + 1,s + 2; b−a 2b )|f′( b m )|q )1 q 194 BALOCH, İŞCAN AND DRAGOMIR Theorem 5. Let f : I ⊆ (0,∞) → R be a differentiable function on I◦ and a, b m ∈ I◦, m ∈ (0, 1] with a < b and let g : [a,b] → [0,∞) be continuous positive mapping and harmonically symmetric to 2ab a+b such that f′ ∈ L([a,b]). If |f′|q is harmonically (s,m)-convex on [a, b m ] for q > 1, then the following inequality holds ∣∣∣∣f(a) + f(b)2 ∫ b a g(x) x2 dx− ∫ b a g(x)f(x) x2 dx ∣∣∣∣ ≤ 21−1q a(b−a)8b ‖g‖∞( 2B(s + 1, 1).2F1(2q,s + 1,s + 2; b−a b )|f′(b)|q + 2B(1,s + 1).2F1(2q, 1,s + 2; b−a b )|f′(a)|q + m|f′( b m )|q −|f′(b)|q 2s .B(s + 1, 1).2F1(2q,s + 1,s + 2; b−a 2b ) (2.13) +22q−s( b b + a )2q(m|f′( a m )|q −|f′(a)|q)B(1,s + 1).2F1(2q, 1,s + 2; b−a b + a ) )1 q Proof. From Lemma 1 and hölder’s inequality, we get∣∣∣∣f(a) + f(b)2 ∫ b a g(x) x2 dx− ∫ b a g(x)f(x) x2 dx ∣∣∣∣ ≤ b−a8ab ‖g‖∞ (∫ 1 0 (1 − t) q q−1 dt )1−1 q (2.14) × {(∫ 1 0 (U(t))2q|f′(U(t))|qdt )1 q + (∫ 1 0 (L(t))2q|f′(L(t))|qdt )1 q } By the power-mean inequality (ar + br ≤ 21−r(a + b)r for a > 0, b > 0 and r < 1), we have(∫ 1 0 (U(t))2q|f′(U(t))|qdt )1 q + (∫ 1 0 (L(t))2q|f′(L(t))|qdt )1 q (2.15) ≤ 21− 1 q (∫ 1 0 (U(t))2q|f′(U(t))|qdt + ∫ 1 0 (L(t))2q|f′(L(t))|qdt )1 q Since, |f′|q is harmonically (s,m)-convex on [a,b] for q > 1, we obtain∫ 1 0 (U(t))2q|f′(U(t))|qdt + ∫ 1 0 (L(t))2q|f′(L(t))|qdt ≤ 1 2s |f′(b)|q ∫ 1 0 (1 + t)s ( 2ab (1 + t)a + (1 − t)b )2q dt +m 1 2s |f′( a m )|q ∫ 1 0 (1 − t)s ( 2ab (1 + t)a + (1 − t)b )2q dt + 1 2s |f′(a)|q ∫ 1 0 (1 + t)s ( 2ab (1 − t)a + (1 + t)b )2q dt +m 1 2s |f′( b m )|q ∫ 1 0 (1 − t)s ( 2ab (1 − t)a + (1 + t)b )2q dt = a2q { 2B(s + 1, 1).2F1(2q,s + 1,s + 2; b−a b ) − 21−sB(s + 1, 1).2F1(2q,s + 1,s + 2; b−a 2b ) } |f′(b)|q +m22q−s( ab b + a )2qB(1,s + 1).2F1(2q, 1,s + 2; b−a b + a )|f′( a m )|q +a2q { 2B(1,s + 1).2F1(2q, 1,s + 2; b−a b ) − 22q−s( b b + a )2qB(1,s + 1).2F1(2q, 1,s + 2; b−a b + a ) } |f′(a)|q + ma2q 2s B(s + 1, 1).2F1(2q,s + 1,s + 2; b−a 2b )|f′( b m )|q HARMONICALLY (s,m)-CONVEX FUNCTIONS 195 using (2.15) in (2.14), we get(∫ 1 0 (U(t))2q|f′(U(t))|qdt )1 q + (∫ 1 0 (L(t))2q|f′(L(t))|qdt )1 q ≤ 21− 1 q a2 ( 2B(s + 1, 1).2F1(2q,s + 1,s + 2; b−a b )|f′(b)|q + 2B(1,s + 1).2F1(2q, 1,s + 2; b−a b )|f′(a)|q + m|f′( b m )|q −|f′(b)|q 2s .B(s + 1, 1).2F1(2q,s + 1,s + 2; b−a 2b ) (2.16) +22q−s( b b + a )2q(m|f′( a m )|q −|f′(a)|q)B(1,s + 1).2F1(2q, 1,s + 2; b−a b + a ) )1 q Applying (2.17) in (2.14), we obtain the required inequality. � Corollary 5. Suppose the assumptions of the Theorem 3 are satisfied. If g(x) = ab b−a for all x ∈ [a,b], then one has the following inequality∣∣∣∣f(a) + f(b)2 ∫ b a g(x) x2 dx− ∫ b a g(x)f(x) x2 dx ∣∣∣∣ ≤ 21−1q a28( 2B(s + 1, 1).2F1(2q,s + 1,s + 2; b−a b )|f′(b)|q + 2B(1,s + 1).2F1(2q, 1,s + 2; b−a b )|f′(a)|q + m|f′( b m )|q −|f′(b)|q 2s .B(s + 1, 1).2F1(2q,s + 1,s + 2; b−a 2b ) (2.17) +22q−s( b b + a )2q(m|f′( a m )|q −|f′(a)|q)B(1,s + 1).2F1(2q, 1,s + 2; b−a b + a ) )1 q Theorem 6. Let f : I ⊆ (0,∞) → R be a differentiable function on I◦ and a, b m ∈ I◦, m ∈ (0, 1] with a < b and let g : [a,b] → [0,∞) be continuous positive mapping and harmonically symmetric to 2ab a+b such that f′ ∈ L([a,b]). If |f′| is harmonically (s,m)-convex on [a, b m ], then the following inequality holds for q > 1 ∣∣∣∣f(a) + f(b)2 ∫ b a g(x) x2 dx− ∫ b a g(x)f(x) x2 dx ∣∣∣∣ ≤ (b−a)8ab ‖g‖∞( 1sq + 1)1q × { 22−s( ab b + a )2 ( (2sq+1 − 1)|f′(b)| + m|f′( a m )| )( B(1, 2q − 1 q − 1 ).2F1( 2q q − 1 , 1, 3q − 2 q − 1 ; b−a b + a ) )q−1 q (2.18) + a2 2s ( (2sq+1 − 1)|f′(a)| + m|f′( b m )| )( B( 2q − 1 q − 1 , 1).2F1( 2q q − 1 , 2q − 1 q − 1 , 3q − 2 q − 1 ; b−a b ) )q−1 q } Proof. From Lemma 1 and by using the harmonic (s,m)-convexity of |f′| on [a,b] , we get∣∣∣∣f(a) + f(b)2 ∫ b a g(x) x2 dx− ∫ b a g(x)f(x) x2 dx ∣∣∣∣ ≤ b−a8ab ‖g‖∞ × [∫ 1 0 (1 − t)(U(t))2|f′(U(t))|dt + ∫ 1 0 (1 − t)(L(t))2|f′(L(t))|dt ] ≤ b−a 8ab ‖g‖∞ {∫ 1 0 (1 − t)(U(t))2 [ ( 1 + t 2 )s|f′(b)| + m( 1 − t 2 )s|f′( a m )| ] (2.19) + ∫ 1 0 (1 − t)(L(t))2 [ ( 1 + t 2 )s|f′(a)| + m( 1 − t 2 )s|f′( b m )| ]} Now, by using hölder’s inequality, we get∫ 1 0 (1 − t)(U(t))2 [ ( 1 + t 2 )s|f′(b)| + m( 1 − t 2 )s|f′( a m )| ] dt 196 BALOCH, İŞCAN AND DRAGOMIR ≤ (∫ 1 0 (1 − t) q q−1 (U(t)) 2q q−1 dt )q−1 q × {(∫ 1 0 ( 1 + t 2 )sqdt )1 q |f′(b)| + m (∫ 1 0 ( 1 − t 2 )sqdt )1 q |f′( a m )| } (2.20) = 22−s( ab b + a )2( 1 sq + 1 ) 1 q ( (2sq+1−1)|f′(b)|+m|f′( a m )| )( B(1, 2q − 1 q − 1 ).2F1( 2q q − 1 , 1, 3q − 2 q − 1 ; b−a b + a ) )q−1 q . Similarly, one has ∫ 1 0 (1 − t)(L(t))2 [ ( 1 + t 2 )s|f′(a)| + m( 1 − t 2 )s|f′( b m )| ] (2.21) = a2 2s ( 1 sq + 1 ) 1 q ( (2sq+1 − 1)|f′(a)| + m|f′( b m )| )( B( 2q − 1 q − 1 , 1).2F1( 2q q − 1 , 2q − 1 q − 1 , 3q − 2 q − 1 ; b−a b ) )q−1 q . � Corollary 6. Suppose the assumptions of the Theorem 3 are satisfied. If g(x) = ab b−a for all x ∈ [a,b], then one has the following inequality∣∣∣∣f(a) + f(b)2 ∫ b a g(x) x2 dx− ∫ b a g(x)f(x) x2 dx ∣∣∣∣ ≤ 18( 1sq + 1)1q × { 22−s( ab b + a )2 ( (2sq+1 − 1)|f′(b)| + m|f′( a m )| )( B(1, 2q − 1 q − 1 ).2F1( 2q q − 1 , 1, 3q − 2 q − 1 ; b−a b + a ) )q−1 q (2.22) + a2 2s ( (2sq+1 − 1)|f′(a)| + m|f′( b m )| )( B( 2q − 1 q − 1 , 1).2F1( 2q q − 1 , 2q − 1 q − 1 , 3q − 2 q − 1 ; b−a b ) )q−1 q } 3. Competing interests The authors have no Competing interests regarding this article. 4. Funding This research article is partially supported by Higher Education Commission of Pakistan. 5. 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