International Journal of Analysis and Applications ISSN 2291-8639 Volume 13, Number 1 (2017), 1-14 http://www.etamaths.com SOME FEJÉR TYPE INEQUALITIES FOR HARMONICALLY-CONVEX FUNCTIONS WITH APPLICATIONS TO SPECIAL MEANS M. A. LATIF1,∗, S. S. DRAGOMIR1,2 AND E. MOMONIAT1 Abstract. In this paper, the notion of harmonic symmetricity of functions is introduced. A new identity involving harmonically symmetric functions is established and some new Fejér type integral inequalities are presented for the class of harmonically convex functions. The results presented in this paper are better than those established in recent literature concerning harmonically convex functions. Applications of our results to special means of positive real numbers are given as well. 1. Introduction The theory of convexity has been subject to extensive research during the past few years due it its utility in various branches of pure and applied mathematics. Many inequalities have been established by a number of researchers for convex functions but one of the most interesting inequalities is the Hermite-Hadamard inequality which provides a necessary and sufficient condition for a functions to be convex. Let f : I ⊆ R → R, a,b ∈ I with a < b f ( a + b 2 ) ≤ 1 b−a ∫ b a f(x)dx ≤ f(a) + f(b) 2 (1.1) holds if and only if f is convex. The inequalities (1.1) hold in reversed direction if f is concave. Many researchers have generalized the classical convexity in a number of ways and the inequality (1.1) has been generalized or extended for many classes of convex functions in numerous ways, see for instance [2–21] and the references therein. Let us recall some known concepts which will be used in the sequel of the paper. Definition 1.1. [9] Let I ⊂ R\{0} be a real interval. A function f : I → R is said to be harmonically convex, if f ( xy tx + (1 − t) y ) ≤ tf (y) + (1 − t) f (x) (1.2) for all x, y ∈ I and t ∈ [0, 1]. If the inequality in (1.2) is reversed, then f is said to be harmonically concave. Proposition 1.1. [9] Let I ⊂ R\{0} be a real interval and f : I → R is function, then: • if I ⊂ (0,∞) and f is convex and nondecreasing function then f is harmonically convex. • if I ⊂ (0,∞) and f is harmonically convex and nonincreasing function then f is convex. • if I ⊂ (−∞, 0) and f is harmonically convex and nondecreasing function then f is convex. • if I ⊂ (−∞, 0) and f is convex and nonincreasing function then f is harmonically convex. In [9], İşcan has also proved the following results for harmonically convex functions. Received 29th June, 2016; accepted 19th September, 2016; published 3rd January, 2017. 2010 Mathematics Subject Classification. Primary 26D15, Secondary 26A51, 26E60, 41A55. Key words and phrases. Hermite-Hadamard’s inequality; Fejér’s inequality; convex function; harmonically-convex function; Hölder’s inequality; power mean inequality. c©2017 Authors retain the copyrights of their papers, and all open access articles are distributed under the terms of the Creative Commons Attribution License. 1 2 LATIF, DRAGOMIR AND MOMONIAT Theorem 1.1. [9] Let I ⊂ R\{0} be a harmonically convex function and a, b ∈ I with a < b. If f ∈ L ([a,b]) then the following inequalities hold f ( 2ab a + b ) ≤ ab b−a ∫ b a f (x) x2 dx ≤ f (a) + f (b) 2 . The above inequalities are sharp. Theorem 1.2. [9] Let f : (0,∞) → R be a differentiable function on I◦, a, b ∈ I◦ with a < b, and f ′ ∈ L ([a,b]). If ∣∣∣f′∣∣∣q is harmonically convex [a,b] for q ≥ 1, then∣∣∣∣∣f (a) + f (b)2 − abb−a ∫ b a f (x) x2 dx ∣∣∣∣∣ ≤ ab (b−a)2 λ1−1q1 [ λ2 ∣∣∣f′ (a)∣∣∣q + λ3 ∣∣∣f′ (b)∣∣∣q]1q , where λ1 = 1 ab − 2 (b−a)2 ln ( (a + b) 2 4ab ) , λ2 = − 1 b (b−a) + 3a + b (b−a)3 ln ( (a + b) 2 4ab ) , λ2 = 1 a (b−a) − 3b + a (b−a)3 ln ( (a + b) 2 4ab ) = λ1 −λ2. Theorem 1.3. [9] Let f : (0,∞) → R be a differentiable function on I◦, a, b ∈ I◦ with a < b, and f ′ ∈ L ([a,b]). If ∣∣∣f′∣∣∣q is harmonically convex [a,b] for q > 1, 1p + 1q = 1, then∣∣∣∣∣f (a) + f (b)2 − abb−a ∫ b a f (x) x2 dx ∣∣∣∣∣ ≤ ab (b−a)2 ( 1 p + 1 )1 p [ µ1 ∣∣∣f′ (a)∣∣∣q + µ1 ∣∣∣f′ (b)∣∣∣q]1q , where µ1 = a2−2q + b1−2q [(b−a) (1 − 2q) −a] 2 (b−a)2 (1 −q) (1 − 2q) , µ2 = b2−2q −a1−2q [(b−a) (1 − 2q) + b] 2 (b−a)2 (1 −q) (1 − 2q) . Some applications of the above results can also be found in [9]. Chen and Wu [2] established the following Fejér type inequality for harmonically convex functions which provides a weighted generalization of the result given in Theorem 1.1. Theorem 1.4. [2] Let f : I ⊆ R\{0} → R be a harmonically convex function and a, b ∈ I with a < b. If f ∈ L ([a,b]), them one has be continuous f ( 2ab a + b )∫ b a g (x) x2 dx ≤ ∫ b a f (x) g (x) x2 dx ≤ f (a) + f (b) 2 ∫ b a g (x) x2 dx, (1.3) g : [a,b] → R is nonnegative, integrable and satisfies g ( ab x ) = g ( ab a + b−x ) . The main goal of this 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 and hence our results will provide a better weighted generalization of the results proved in Theorem 1.2 and Theorem 1.3. Some applications of our results to special means of positive real numbers will also be provided in Section 3. We believe that our findings are novel, new and better than those already exist and will open new ways for further research in this filed. FEJÉR TYPE INEQUALITIES FOR HARMONICALLY-CONVEX FUNCTIONS 3 2. Main Results Throughout this section we take U (t) = 2ab (1−t)a+(1+t)b and L (t) = 2ab (1+t)a+(1−t)b. The Beta function, the Gamma function and the integral from of the hypergeometric function are defined as follows to be used in the sequel of the paper B (α,β) = ∫ 1 0 tα−1 (1 − t)β−1 dt,α > 0,β > 0, Γ (α) = ∫ ∞ 0 tα−1e−tdt,α > 0 and 2F1 (α,β; γ; z) = 1 B (β,γ −β) ∫ 1 0 tβ−1 (1 − t)γ−β−1 (1 −zt)−α dt for |z| < 1,γ > β > 0. The notion of harmonically symmetric functions is given in following definition. Definition 2.1. A function g : [a,b] ⊆ R\{0}→ R is said to be 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 prove a weighted integral identity which will be used in establishing our main results. Lemma 2.1. 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 equality holds f (b) + f (a) 2 ∫ b a g (x) x2 dx− ∫ b a f (x) g (x) x2 dx = ( b−a 4ab )∫ 1 0 (∫ U(t) L(t) g (x) x2 dx )[ (U (t)) 2 f ′ (U (t)) − (L (t))2 f ′ (L (t)) ] dt. (2.1) Proof. Let I1 = ∫ 1 0 (∫ U(t) L(t) g (x) x2 dx ) (U (t)) 2 f ′ (U (t)) dt and I2 = ∫ 1 0 (∫ U(t) L(t) g (x) x2 dx ) (L (t)) 2 f ′ (L (t)) dt. Since g : [a,b] → [0,∞) is harmonically symmetric to 2ab a+b , then g (U (t)) = g (L (t)) for all t ∈ [0, 1]. Hence, we have I1 = ∫ 1 0 (∫ U(t) L(t) g (x) x2 dx ) (U (t)) 2 f ′ (U (t)) dt = −2ab b−a ∫ 1 0 (∫ U(t) L(t) g (x) x2 dx ) d [f (U (t))] = −2ab b−a (∫ U(t) L(t) g (x) x2 dx ) f (U (t)) ∣∣∣∣∣ 1 0 − ∫ 1 0 [g (U (t)) + g (L (t))] f (U (t)) dt = 2ab b−a f (a) ∫ b a g (x) x2 dx− 2 ∫ 1 0 g (U (t)) f (U (t)) dt = 2ab b−a f (a) ∫ b a g (x) x2 dx− 4ab b−a ∫ 2ab a+b a g (x) f (x) x2 dx. (2.2) 4 LATIF, DRAGOMIR AND MOMONIAT Analogously, we have − I2 = 2ab b−a f (b) ∫ b a g (x) x2 dx− 4ab b−a ∫ b 2ab a+b g (x) f (x) x2 dx. (2.3) Adding (2.2) and (2.3) and multiplying the result by b−a 4ab , we get the required identity. This completes the proof of the Lemma. � Lemma 2.2. For v > u > 0, we have∫ 1 0 t [ 2uv (1 − t) u + (1 + t) v ]2 dt = ( 2uv v + u )2 λ1 (u,v) , ∫ 1 0 t [ 2uv (1 + t) u + (1 − t) v ]2 dt = ( 2uv v + u )2 λ1 (v,u) , ∫ 1 0 t2 [ 2uv (1 − t) u + (1 + t) v ]2 dt = ( 2uv v + u )2 λ2 (u,v) , ∫ 1 0 t2 [ 2uv (1 + t) u + (1 − t) v ]2 dt = ( 2uv v + u )2 λ2 (v,u) , where λ1 (u,v) ∆ = ln ( 2v u + v ) + u−v 2v and λ2 (u,v) ∆ = ( v + u v −u )[ 2v u + v − u + v 2v − 2 ln ( 2v u + v )] . Proof. The proof follows from a straightforward computation. � Lemma 2.3. For v > u > 0 and p > 1, we have∫ 1 0 (1 + t) [ 2uv (1 − t) u + (1 + t) v ]2p dt = ( 2uv v + u )2p ζ1 (u,v; p) , ∫ 1 0 (1 − t) [ 2uv (1 − t) u + (1 + t) v ]2p dt = ( 2uv v + u )2p ζ2 (u,v; p) , ∫ 1 0 (1 + t) [ 2uv (1 + t) u + (1 − t) v ]2p dt = ( 2uv v + u )2p ζ1 (v,u; p) , ∫ 1 0 (1 − t) [ 2uv (1 + t) u + (1 − t) v ]2p dt = ( 2uv v + u )2p ζ2 (v,u; p) , where ζ1 (u,v; p) ∆ = 21−2pv ( v u+v )−2p [(1 − 2p) (v −u) −u] (v −u)2 (p− 1) (2p− 1) − (u + v) [(1 − 2p) (v −u) − 2u] 2 (v −u)2 (p− 1) (2p− 1) and ζ2 (u,v; p) ∆ = 41−pv2 ( v u+v )−2p + (u + v) [(2p− 1) (v −u) − 2v] 2 (v −u)2 (p− 1) (2p− 1) . Proof. The proof follows from a straightforward computation. � Now we present new Fejér type inequalities for harmonically-convex functions, which provide weighted generalization of some of the results established in recent literature. FEJÉR TYPE INEQUALITIES FOR HARMONICALLY-CONVEX FUNCTIONS 5 Theorem 2.1. Let f : I ⊆ (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 such that f ′ ∈ L ([a,b]). If ∣∣∣f′∣∣∣q is harmonically-convex on [a,b] for q ≥ 1, then the following inequality holds ∣∣∣∣∣f (b) + f (a)2 ∫ b a g (x) x2 dx− ∫ b a f (x) g (x) x2 dx ∣∣∣∣∣ ≤ ( b−a b + a )2 ( 1 2 )1/q ‖g‖∞ × { [λ1 (a,b)] 1−1/q [ ξ1 (a,b) ∣∣∣f′ (a)∣∣∣q + ξ2 (a,b) ∣∣∣f′ (b)∣∣∣q]1/q + [λ1 (b,a)] 1−1/q [ ξ2 (b,a) ∣∣∣f′ (a)∣∣∣q + ξ1 (b,a) ∣∣∣f′ (b)∣∣∣q]1/q} , (2.4) where ‖g‖∞ = supx∈[a,b] g (x) < ∞, ξ1 (a,b) ∆ = λ1 (a,b) + λ2 (a,b) , ξ2 (a,b) ∆ = λ1 (a,b) −λ2 (a,b) and λ1 (·, ·), λ2 (·, ·) are defined in Lemma 2.2. Proof. From Lemma 2.1, we get f (b) + f (a) 2 ∫ b a g (x) x2 dx− ∫ b a f (x) g (x) x2 dx ≤− ( b−a 2ab )2 ‖g‖∞ ∫ 1 0 [ t (U (t)) 2 f ′ (U (t)) − t (L (t))2 f ′ (L (t)) ] dt. (2.5) Now taking modulus on both sides of (2.5) and using Hölder’s inequality, we have ∣∣∣∣∣f (b) + f (a)2 ∫ b a g (x) x2 dx− ∫ b a f (x) g (x) x2 dx ∣∣∣∣∣ ≤ ( b−a 2ab )2 ‖g‖∞ {(∫ 1 0 t (U (t)) 2 dt )1−1/q (∫ 1 0 t (U (t)) 2 ∣∣∣f′ (U (t))∣∣∣q dt)1/q + (∫ 1 0 t (L (t)) 2 dt )1−1/q (∫ 1 0 t (L (t)) 2 ∣∣∣f′ (L (t))∣∣∣q dt)1/q } . (2.6) By the harmonic-convexity of ∣∣∣f′∣∣∣q on [a,b] for q ≥ 1 and by using Lemma 2.2, we have ∫ 1 0 t (U (t)) 2 ∣∣∣f′ (U (t))∣∣∣q dt = ∫ 1 0 t [ 2ab (1 − t) a + (1 + t) b ]2 × ∣∣∣∣f′ ( 2ab (1 − t) a + (1 + t) b )∣∣∣∣q dt ≤ 12 ∣∣∣f′ (a)∣∣∣q ∫ 1 0 t (1 + t) [ 2ab (1 − t) a + (1 + t) b ]2 dt + 1 2 ∣∣∣f′ (b)∣∣∣q ∫ 1 0 t (1 − t) [ 2ab (1 − t) a + (1 + t) b ]2 dt = 1 2 ( 2ab b + a )2 { [λ1 (a,b) + λ2 (a,b)] ∣∣∣f′ (a)∣∣∣q + [λ1 (a,b) −λ2 (a,b)] ∣∣∣f′ (b)∣∣∣q} (2.7) 6 LATIF, DRAGOMIR AND MOMONIAT and∫ 1 0 t (L (t)) 2 ∣∣∣f′ (L (t))∣∣∣q dt = ∫ 1 0 t [ 2ab (1 + t) a + (1 − t) b ]2 × ∣∣∣∣f′ ( 2ab (1 + t) a + (1 − t) b )∣∣∣∣q dt ≤ 12 ∣∣∣f′ (a)∣∣∣q ∫ 1 0 t (1 − t) [ 2ab (1 + t) a + (1 − t) b ]2 dt + 1 2 ∣∣∣f′ (b)∣∣∣q ∫ 1 0 t (1 + t) [ 2ab (1 + t) a + (1 − t) b ]2 dt = 1 2 ( 2ab b + a )2 × { [λ1 (b,a) −λ2 (b,a)] ∣∣∣f′ (a)∣∣∣q + [λ1 (b,a) + λ2 (b,a)] ∣∣∣f′ (b)∣∣∣q} . (2.8) A combination of (2.6), (2.7) and (2.8) gives the required result. This completes the proof of the theorem. � Corollary 2.1. Suppose the assumptions of Theorem 2.1 are satisfied. If q = 1, then the following inequality holds∣∣∣∣∣f (b) + f (a)2 ∫ b a g (x) x2 dx− ∫ b a f (x) g (x) x2 dx ∣∣∣∣∣ ≤ ( 1 2 )( b−a b + a )2 ‖g‖∞ { [ξ1 (a,b) + ξ2 (b,a)] ∣∣∣f′ (a)∣∣∣ + [ξ2 (a,b) + ξ1 (b,a)] ∣∣∣f′ (b)∣∣∣} , (2.9) where ‖g‖∞ = supx∈[a,b] g (x) < ∞ and ξ1 (·, ·), ξ2 (·, ·) are defined in Theorem 2.1. Corollary 2.2. If g (x) = ab b−a for all x ∈ [a,b] in Theorem 2.1, then∣∣∣∣∣f (b) + f (a)2 − abb−a ∫ b a f (x) x2 dx ∣∣∣∣∣ ≤ ( 1 2 )1/q ( b−a ab )( ab b + a )2 × { [λ1 (a,b)] 1−1/q [ ξ1 (a,b) ∣∣∣f′ (a)∣∣∣q + ξ2 (a,b) ∣∣∣f′ (b)∣∣∣q]1/q + [λ1 (b,a)] 1−1/q [ ξ2 (b,a) ∣∣∣f′ (a)∣∣∣q + ξ1 (b,a) ∣∣∣f′ (b)∣∣∣q]1/q} , (2.10) where ξ1 (·, ·), ξ2 (·, ·) are defined in Theorem 2.1 and λ1 (·, ·), λ2 (·, ·) are defined in Lemma 2.2. Corollary 2.3. If q = 1 in Corollary 2.2, then we get the following inequality∣∣∣∣∣f (b) + f (a)2 − abb−a ∫ b a f (x) x2 dx ∣∣∣∣∣ ≤ ( b−a 2ab )( ab b−a )2 { [ξ1 (a,b) + ξ2 (b,a)] ∣∣∣f′ (a)∣∣∣ + [ξ2 (a,b) + ξ1 (b,a)] ∣∣∣f′ (b)∣∣∣} , (2.11) where ξ1 (·, ·), ξ2 (·, ·) are defined in Theorem 2.1. Theorem 2.2. Let f : I ⊆ (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 such that f ′ ∈ L ([a,b]). If ∣∣∣f′∣∣∣q is harmonically-convex on [a,b] for q > 1, then the following inequality holds∣∣∣∣∣f (b) + f (a)2 ∫ b a g (x) x2 dx− ∫ b a f (x) g (x) x2 dx ∣∣∣∣∣ ≤‖g‖∞ ( b−a b + a )2 ( 1 2 )1/q ( q − 1 2q − 1 )1−1/q {[ ζ1 (a,b; q) ∣∣∣f′ (a)∣∣∣q + ζ2 (a,b; q) ∣∣∣f′ (b)∣∣∣q]1/q + [ ζ2 (b,a; q) ∣∣∣f′ (a)∣∣∣q + ζ1 (b,a; q) ∣∣∣f′ (b)∣∣∣q]1/q} , (2.12) FEJÉR TYPE INEQUALITIES FOR HARMONICALLY-CONVEX FUNCTIONS 7 where ζ1 (·, ·; ·) and ζ2 (·, ·; ·) are defined in Lemma 2.3. Proof. From (2.5) and Hölder’s inequality, we have∣∣∣∣∣f (b) + f (a)2 ∫ b a g (x) x2 dx− ∫ b a f (x) g (x) x2 dx ∣∣∣∣∣ ≤ ( b−a 2ab )2 ‖g‖∞ (∫ 1 0 tq/(q−1)dt )1−1/q × {(∫ 1 0 (U (t)) 2q ∣∣∣f′ (U (t))∣∣∣q dt)1/q + (∫ 1 0 (L (t)) 2q ∣∣∣f′ (L (t))∣∣∣q dt)1/q } . (2.13) Since ∣∣∣f′∣∣∣q is harmonically-convex on [a,b], we obtain ∫ 1 0 [U (t)] 2q ∣∣∣f′ (U (t))∣∣∣q dt = ∫ 1 0 [ 2ab (1 − t) a + (1 + t) b ]2q × ∣∣∣∣f′ ( 2ab (1 − t) a + (1 + t) b )∣∣∣∣q dt ≤ 12 ∣∣∣f′ (a)∣∣∣q ∫ 1 0 (1 + t) [ 2ab (1 − t) a + (1 + t) b ]2q dt + 1 2 ∣∣∣f′ (b)∣∣∣q ∫ 1 0 (1 − t) [ 2ab (1 − t) a + (1 + t) b ]2q dt (2.14) and∫ 1 0 [L (t)] 2q ∣∣∣f′ (L (t))∣∣∣q dt = ∫ 1 0 [ 2ab (1 + t) a + (1 − t) b ]2q × ∣∣∣∣f′ ( 2ab (1 + t) a + (1 − t) b )∣∣∣∣q dt ≤ 12 ∣∣∣f′ (a)∣∣∣q ∫ 1 0 (1 − t) [ 2ab (1 + t) a + (1 − t) b ]2q dt + 1 2 ∣∣∣f′ (b)∣∣∣q ∫ 1 0 (1 + t) [ 2ab (1 + t) a + (1 − t) b ]2q dt. (2.15) By applying Lemma 2.3 in inequalities (2.14) and (2.15) and then using the resulting inequalities in (2.13), we get the required inequality. � Corollary 2.4. If the assumptions of Theorem 2.2 are satisfied and if g (x) = ab b−a for all x ∈ [a,b], then the following inequality holds∣∣∣∣∣f (b) + f (a)2 − abb−a ∫ b a f (x) x2 dx ∣∣∣∣∣ ≤ ( ab b−a )( b−a b + a )2 ( 1 2 )1/q ( q − 1 2q − 1 )1−1/q {[ ζ1 (a,b; q) ∣∣∣f′ (a)∣∣∣q + ζ2 (a,b; q) ∣∣∣f′ (b)∣∣∣q]1/q + [ ζ2 (b,a; q) ∣∣∣f′ (a)∣∣∣q + ζ1 (b,a; q) ∣∣∣f′ (b)∣∣∣q]1/q} , (2.16) where ζ1 (·, ·; ·) and ζ2 (·, ·; ·) are defined in Lemma 2.3. Theorem 2.3. Let f : I ⊆ (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 such that f ′ ∈ L ([a,b]). If ∣∣∣f′∣∣∣q is harmonically-convex on [a,b] for q > 1, then the following inequality holds∣∣∣∣∣f (b) + f (a)2 ∫ b a g (x) x2 dx− ∫ b a f (x) g (x) x2 dx ∣∣∣∣∣ ≤ ( b−a b + a )2 ( 1 2 )2/q−1 ( q − 1 2q − 1 )1−1/q ‖g‖∞ × { [ζ1 (a,b; q) + ζ2 (b,a; q)] ∣∣∣f′ (a)∣∣∣q + [ζ2 (a,b; q) + ζ1 (b,a; q)] ∣∣∣f′ (b)∣∣∣q}1/q , (2.17) 8 LATIF, DRAGOMIR AND MOMONIAT where ζ1 (·, ·; ·) and ζ2 (·, ·; ·) are defined in Lemma 2.3. Proof. From the inequality 2.5 and Hölder’s inequality, we have∣∣∣∣∣f (b) + f (a)2 ∫ b a g (x) x2 dx− ∫ b a f (x) g (x) x2 dx ∣∣∣∣∣ ≤ ( b−a 2ab )2 ‖g‖∞ (∫ 1 0 tq/(q−1)dt )1−1/q × {(∫ 1 0 [U (t)] 2q ∣∣∣f′ (U (t))∣∣∣q dt)1/q + (∫ 1 0 [L (t)] 2q ∣∣∣f′ (L (t))∣∣∣q dt)1/q } . (2.18) 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))∣∣∣2q dt)1/q + (∫ 1 0 [L (t)] 2q ∣∣∣f′ (L (t))∣∣∣2q dt)1/q ≤ 21−1/q (∫ 1 0 [U (t)] 2q ∣∣∣f′ (U (t))∣∣∣q dt + ∫ 1 0 [L (t)] 2q ∣∣∣f′ (L (t))∣∣∣q dt)1/q . (2.19) Since ∣∣∣f′∣∣∣q is harmonically-convex on [a,b] for q > 1, we obtain ∫ 1 0 [U (t)] 2q ∣∣∣f′ (U (t))∣∣∣q dt + ∫ 1 0 [L (t)] 2q ∣∣∣f′ (L (t))∣∣∣q dt ≤ 1 2 ∣∣∣f′ (a)∣∣∣q ∫ 1 0 (1 + t) [ 2ab (1 − t) a + (1 + t) b ]2q dt + 1 2 ∣∣∣f′ (b)∣∣∣q ∫ 1 0 (1 − t) [ 2ab (1 − t) a + (1 + t) b ]2q dt + 1 2 ∣∣∣f′ (a)∣∣∣q ∫ 1 0 (1 − t) [ 2ab (1 + t) a + (1 − t) b ]2q dt + 1 2 ∣∣∣f′ (b)∣∣∣q ∫ 1 0 (1 + t) [ 2ab (1 + t) a + (1 − t) b ]2q dt = 1 2 ( 2ab b + a )2q { [ζ1 (a,b; q) + ζ2 (b,a; q)] ∣∣∣f′ (a)∣∣∣q + [ζ2 (a,b; q) + ζ1 (b,a; q)] ∣∣∣f′ (b)∣∣∣q} . (2.20) Using (2.19) in (2.20), we get (∫ 1 0 [U (t)] 2q ∣∣∣f′ (U (t))∣∣∣q dt)1/q + (∫ 1 0 [L (t)] 2q ∣∣∣f′ (L (t))∣∣∣q dt)1/q ≤ 21−2/q ( 2ab b + a )2 { [ζ1 (a,b; q) + ζ2 (b,a; q)] ∣∣∣f′ (a)∣∣∣q + [ζ2 (a,b; q) + ζ1 (b,a; q)] ∣∣∣f′ (b)∣∣∣q}1/q . (2.21) Applying (2.21) in (2.18), we obtain the required inequality (2.17). � Corollary 2.5. If the assumptions of Theorem 2.3 are satisfied and if g (x) = ab b−a for all x ∈ [a,b], then the following inequality holds∣∣∣∣∣f (b) + f (a)2 − abb−a ∫ b a f (x) x2 dx ∣∣∣∣∣ ≤ ab b−a ( b−a b + a )2 ( 1 2 )1−2/q ( q − 1 2q − 1 )1−1/q × { [ζ1 (a,b; q) + ζ2 (b,a; q)] ∣∣∣f′ (a)∣∣∣q + [ζ2 (a,b; q) + ζ1 (b,a; q)] ∣∣∣f′ (b)∣∣∣q}1/q , (2.22) where ζ1 (·, ·; ·) and ζ2 (·, ·; ·) are defined in Lemma 2.3. FEJÉR TYPE INEQUALITIES FOR HARMONICALLY-CONVEX FUNCTIONS 9 Theorem 2.4. Let f : I ⊆ (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 such that f ′ ∈ L ([a,b]). If ∣∣∣f′∣∣∣ is harmonically-convex on [a,b], then the following inequality holds for q > 1 ∣∣∣∣∣f (b) + f (a)2 ∫ b a g (x) x2 dx− ∫ b a f (x) g (x) x2 dx ∣∣∣∣∣ ≤ 1 2 ( b−a a + b )2 ‖g‖∞ ( [ς (a,b; q)] 1−1/q { [B (q + 1,q + 1)] 1/q ∣∣∣f′ (b)∣∣∣ + [ 2F1 (−q,q + 1; q + 2;−1) · 1 q + 1 ]1/q ∣∣∣f′ (a)∣∣∣ } + [ς (b,a; q)] 1−1/q × { [B (q + 1,q + 1)] 1/q ∣∣∣f′ (a)∣∣∣ + [2F1 (−q,q + 1; q + 2;−1) · 1 q + 1 ]1/q ∣∣∣f′ (b)∣∣∣ }) , (2.23) where B (·, ·) is the Beta function, 2F1 (·, ·; ·; ·) is the hypergeometric function and ς (a,b; q) ∆ = (q − 1) [ (a + b) −q+1 q−1 − (2b)− q+1 q−1 ] (q + 1) (b−a) (a + b)− 2q q−1 . Proof. We continue from (2.5) and by using the harmonic-convexity of ∣∣∣f′∣∣∣ on [a,b], we have ∣∣∣∣∣f (b) + f (a)2 ∫ b a g (x) x2 dx− ∫ b a f (x) g (x) x2 dx ∣∣∣∣∣ ≤ ( b−a 2ab )2 ‖g‖∞ ∫ 1 0 [ t (U (t)) 2 ∣∣∣f′ (U (t))∣∣∣ + t (L (t))2 ∣∣∣f′ (L (t))∣∣∣]dt ≤ ( b−a 2ab )2 ‖g‖∞ {∫ 1 0 (U (t)) 2 [ t ( 1 + t 2 )∣∣∣f′ (a)∣∣∣ + t(1 − t 2 )∣∣∣f′ (b)∣∣∣]dt + ∫ 1 0 (L (t)) 2 [ t ( 1 − t 2 )∣∣∣f′ (a)∣∣∣ + t(1 + t 2 )∣∣∣f′ (b)∣∣∣]dt} . (2.24) Using Hölder integral inequality, we have ∫ 1 0 [ 2ab (1 − t) a + (1 + t) b ]2 [ t ( 1 + t 2 )∣∣∣f′ (a)∣∣∣ + t(1 − t 2 )∣∣∣f′ (b)∣∣∣]dt ≤ (∫ 1 0 [ 2ab (1 − t) a + (1 + t) b ]2q/(q−1) dt )1−1/q × {[∫ 1 0 tq ( 1 + t 2 )q dt ]1/q ∣∣∣f′ (a)∣∣∣ + [∫ 1 0 tq ( 1 − t 2 )q dt ]1/q ∣∣∣f′ (b)∣∣∣ } = 1 2 ( 2ab a + b )2 [ς (a,b; q)] 1−1/q { [B (q + 1,q + 1)] 1/q ∣∣∣f′ (b)∣∣∣ + [ 2F1 (−q,q + 1; q + 2;−1) · 1 q + 1 ]1/q ∣∣∣f′ (a)∣∣∣ } . (2.25) 10 LATIF, DRAGOMIR AND MOMONIAT Similarly, one has ∫ 1 0 [ 2ab (1 + t) a + (1 − t) b ]2 [ t ( 1 − t 2 )∣∣∣f′ (a)∣∣∣ + t(1 + t 2 )∣∣∣f′ (b)∣∣∣]dt ≤ (∫ 1 0 [ 2ab (1 + t) a + (1 − t) b ]2q/(q−1) dt )1−1/q × {[∫ 1 0 tq ( 1 − t 2 )q dt ]1/q ∣∣∣f′ (a)∣∣∣ + [∫ 1 0 tq ( 1 + t 2 )q dt ]1/q ∣∣∣f′ (b)∣∣∣ } = 1 2 ( 2ab a + b )2 [ς (b,a; q)] 1−1/q { [B (q + 1,q + 1)] 1/q ∣∣∣f′ (a)∣∣∣ + [ 2F1 (−q,q + 1; q + 2;−1) · 1 q + 1 ]1/q ∣∣∣f′ (b)∣∣∣ } . (2.26) Using (2.25) and (2.26) in (2.24), we obtain the required inequality (2.23). � Corollary 2.6. Under the assumptions of Theorem 2.4, if g (x) = ab b−a for all x ∈ [a,b], then the following inequality holds ∣∣∣∣∣f (b) + f (a)2 − abb−a ∫ b a f (x) x2 dx ∣∣∣∣∣ ≤ 1 2 ( b−a a + b )2 ab b−a ( [ς (a,b; q)] 1−1/q { [B (q + 1,q + 1)] 1/q ∣∣∣f′ (b)∣∣∣ + [ 2F1 (−q,q + 1; q + 2;−1) · 1 q + 1 ]1/q ∣∣∣f′ (a)∣∣∣ } + [ς (b,a; q)] 1−1/q { [B (q + 1,q + 1)] 1/q ∣∣∣f′ (a)∣∣∣ + [ 2F1 (−q,q + 1; q + 2;−1) · 1 q + 1 ]1/q ∣∣∣f′ (b)∣∣∣ }) , (2.27) where B (·, ·) is the Beta function, 2F1 (·, ·; ·; ·) is the hypergeometric function and ς (·, ·; ·) is defined in Theorem 2.4. Theorem 2.5. Let f : I ⊆ (0,∞) → R be a differentiable function on I◦ and a, b ∈ I◦ with 0 < a < b < 1 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-convex on [a,b], then the following inequality holds for q > 1 ∣∣∣∣∣f (b) + f (a)2 ∫ b a g (x) x2 dx− ∫ b a f (x) g (x) x2 dx ∣∣∣∣∣ ≤ 12 ( b−a a + b )2 ‖g‖∞ × { [ν (a,b; q)] 1−1/q [( 1 q + 1 )1/q ∣∣∣f′ (b)∣∣∣ + (2q+1 − 1 q + 1 )1/q ∣∣∣f′ (a)∣∣∣ ] + [ν (b,a; q)] 1−1/q [( 1 q + 1 )1/q ∣∣∣f′ (a)∣∣∣ + (2q+1 − 1 q + 1 )1/q ∣∣∣f′ (b)∣∣∣ ]} , (2.28) FEJÉR TYPE INEQUALITIES FOR HARMONICALLY-CONVEX FUNCTIONS 11 where ν (a,b; q) = Γ ( 2q−1 q−1 ) Γ ( 3q−2 q−1 ) [b2q−1q−1 2F1 ( 2q q − 1 , 2q − 1 q − 1 ; 2q − 1 q − 1 ; b (a− b) a + b ) −a 2q−1 q−1 2F1 ( 2q q − 1 , 2q − 1 q − 1 ; 2q − 1 q − 1 ; a (a− b) a + b )] , Γ (·) is the Gamma function and 2F1 (·, ·; ·; ·) is the hypergeometric function. Proof. From (2.5) and by using the harmonic-convexity of ∣∣∣f′∣∣∣ on [a,b], we have ∣∣∣∣∣f (b) + f (a)2 ∫ b a g (x) x2 dx− ∫ b a f (x) g (x) x2 dx ∣∣∣∣∣ ≤ ( b−a 2ab )2 ‖g‖∞ ∫ 1 0 [ t (U (t)) 2 ∣∣∣f′ (U (t))∣∣∣ + t (L (t))2 ∣∣∣f′ (L (t))∣∣∣]dt ≤ ( b−a 2ab )2 ‖g‖∞ {∫ 1 0 (U (t)) 2 [ t ( 1 + t 2 )∣∣∣f′ (a)∣∣∣ + t(1 − t 2 )∣∣∣f′ (b)∣∣∣]dt + ∫ 1 0 (L (t)) 2 [ t ( 1 − t 2 )∣∣∣f′ (a)∣∣∣ + t(1 + t 2 )∣∣∣f′ (b)∣∣∣]dt} . (2.29) Application of Hölder integral inequality yields ∫ 1 0 [ 2ab (1 − t) a + (1 + t) b ]2 [ t ( 1 + t 2 )∣∣∣f′ (a)∣∣∣ + t(1 − t 2 )∣∣∣f′ (b)∣∣∣]dt ≤ (∫ 1 0 tq/(q−1) [ 2ab (1 − t) a + (1 + t) b ]2q/(q−1) dt )1−1/q × {[∫ 1 0 ( 1 + t 2 )q dt ]1/q ∣∣∣f′ (a)∣∣∣ + [∫ 1 0 ( 1 − t 2 )q dt ]1/q ∣∣∣f′ (b)∣∣∣ } = 1 2 ( 2ab a + b )2 [ν (a,b; q)] 1−1/q [( 1 q + 1 )1/q ∣∣∣f′ (b)∣∣∣ + (2q+1 − 1 q + 1 )1/q ∣∣∣f′ (a)∣∣∣ ] . (2.30) Similarly, one has ∫ 1 0 [ 2ab (1 + t) a + (1 − t) b ]2 [ t ( 1 − t 2 )∣∣∣f′ (a)∣∣∣ + t(1 + t 2 )∣∣∣f′ (b)∣∣∣]dt ≤ (∫ 1 0 tq/(q−1) [ 2ab (1 + t) a + (1 − t) b ]2q/(q−1) dt )1−1/q × {[∫ 1 0 ( 1 − t 2 )q dt ]1/q ∣∣∣f′ (a)∣∣∣ + [∫ 1 0 ( 1 + t 2 )q dt ]1/q ∣∣∣f′ (b)∣∣∣ } = 1 2 ( 2ab a + b )2 [ν (b,a; q)] 1−1/q [( 1 q + 1 )1/q ∣∣∣f′ (a)∣∣∣ + (2q+1 − 1 q + 1 )1/q ∣∣∣f′ (b)∣∣∣ ] . (2.31) Using (2.30) and (2.31) in (2.29), we obtain the required inequality (2.28). � 12 LATIF, DRAGOMIR AND MOMONIAT Corollary 2.7. Suppose the assumptions of Theorem 2.4 are satisfied and if g (x) = ab b−a for all x ∈ [a,b], then the following inequality holds∣∣∣∣∣f (b) + f (a)2 − abb−a ∫ b a f (x) x2 dx ∣∣∣∣∣ ≤ 12 ( b−a a + b )2 ( ab b−a ) × { [ν (a,b; q)] 1−1/q [( 1 q + 1 )1/q ∣∣∣f′ (b)∣∣∣ + (2q+1 − 1 q + 1 )1/q ∣∣∣f′ (a)∣∣∣ ] + [ν (b,a; q)] 1−1/q [( 1 q + 1 )1/q ∣∣∣f′ (a)∣∣∣ + (2q+1 − 1 q + 1 )1/q ∣∣∣f′ (b)∣∣∣ ]} , (2.32) where ν (·, ·; ·) is defined in Theorem 2.5. Remark 2.1. Some further results can be obtained from (2.24) but we omit the details for the interested readers. 3. Applications to Special Means In this section we apply some of the above established inequalities of Hermite-Hadamard type involving the product of a harmonically convex function and a harmonically symmetric function to construct inequalities for special means. For positive numbers a > 0 and b > 0 with a 6= b A (a,b) = a + b 2 , L (a,b) = b−a ln b− ln a , G (a,b) = √ ab, H (a,b) = 2ab a + b and Lp (a,b) =   [ bp+1−ap+1 (p+1)(b−a) ]1 p , p 6= −1, 0 L (a,b) , p = −1 1 e ( bb aa ) 1 b−a , p = 0 are the arithmetic mean, the logarithmic mean, geometric mean, harmonic mean and the generalized logarithmic mean of order p ∈ R respectively. For further information on means, we refer the readers to [1] and the references therein. Let g : [a,b] → R0 be defined as g (x) = ( a + b 2ab − 1 x )2 ,x ∈ [a,b] . It is obvious that g ( 1 1 a + 1 b − 1 x ) = g (x) for all x ∈ [a,b]. Hence g (x) = ( a+b 2ab − 1 x )2 , x ∈ [a,b] is harmonically symmetric with respect to x = 2ab a+b . Throughout in this section we will also assume that µ (a,b) = b−a 2ab . Now applications of our results are given in the following theorems to come. Theorem 3.1. Let 0 < a < b. Then the following inequality holds∣∣∣∣A2 (a,b) + 2G2 (a,b)3G2 (a,b) − A (a,b)L (a,b) ∣∣∣∣ ≤ (b−a)2 µ (a,b) H2 (a,b) [ ln ( G (a,b) A (a,b) ) + µ2 (a,b) G2 (a,b) ] . (3.1) FEJÉR TYPE INEQUALITIES FOR HARMONICALLY-CONVEX FUNCTIONS 13 Proof. Applying Theorem 2.1 to the functions f(x) = x for x > 0 and g (x) = ( a + b 2ab − 1 x )2 ,x ∈ [a,b] we get the desired result. � Theorem 3.2. Let 0 < a < b. Then for q ≥ 1, we have the following inequality holds ∣∣A(a2,b2)−G2 (a,b)∣∣ ≤ (1 2 )1/q µ (a,b) H2 (a,b) × { [λ1 (a,b)] 1−1/q [ 2λ1 (a,b) A (a q,bq) −q (b−a) λ2 (a,b) L q−1 q−1 (a,b) ]1/q + [λ1 (b,a)] 1−1/q [ 2λ1 (b,a) A (a q,bq) + q (b−a) λ2 (b,a) L q−1 q−1 (a,b) ]1/q} . (3.2) where λ1 (·, ·) and λ2 (·, ·) are defined in are defined in Lemma 2.2. Proof. The assertion follows from the inequality proved in Corollary 2.2 for f(x) = x2 for x > 0. � Corollary 3.1. If we take q = 1 in Corollary 3.1, then the following inequality holds valid ∣∣A(a2,b2)−G2 (a,b)∣∣ ≤ 2µ (a,b) H2 (a,b) A (a,b) [3 ln (G (a,b) A (a,b) ) + 2µ2 (a,b) G2 (a,b) ] . (3.3) Theorem 3.3. Let 0 < a < b and q > 1. Then∣∣∣∣A (a,b) − G2 (a,b)L (a,b) ∣∣∣∣ ≤ (2q − 2)1/q−1 µ (a,b) (2q − 1) (b−a)1/q × {[ A (a,b) H2q (a,b) −a2qb ]1/q + [ ab2q −A (a,b) H2q (a,b) ]1/q} . (3.4) Proof. Applying Corollary 2.4 to the function f(x) = x for x > 0, we get the desired result. � Theorem 3.4. Let 0 < a < b and r ∈ (−1,∞)\{0}. Then∣∣A(ar+2,br+2)−G2 (a,b) Lrr (a,b)∣∣ ≤ (r + 2) µ (a,b) H2 (a,b) × { A ( ar+2,br+2 )[ ln ( G (a,b) A (a,b) ) + G2 (a,b) µ2 (a,b) ] + (r + 1) A (a,b) Lrr (a,b) [ 2 ln ( G (a,b) A (a,b) ) + G2 (a,b) µ2 (a,b) ]} . (3.5) Proof. Applying Corollary 2.3 to the function f(x) = xr+2 for x > 0,r ∈ (−1,∞)\{0} , we get the required result. � Theorem 3.5. Let 0 < a < b and q > 1. Then ∣∣∣∣A2 (a,b) + 2G2 (a,b)3G2 (a,b) − A (a,b)L (a,b) ∣∣∣∣ ≤ ( q − 1 2q − 1 )1−1/q ( b−a b + a )3 G2/q (a,b) L2−2/q2q−q (a,b) H2 (a,b) . (3.6) 14 LATIF, DRAGOMIR AND MOMONIAT Proof. Applying Theorem 2.3 to the functions f(x) = x for x > 0 and g (x) = ( a + b 2ab − 1 x )2 ,x ∈ [a,b] we get the desired result. � References [1] P. S. Bullen, Handbook of Means and Their Inequalities, Mathematics and its Applications, Volume 560, Kluwer Academic Publishers,Dordrecht/Boston/London, 2003. [2] F. Chen and S. Wu, Hermite-Hadamard type inequalities for harmonically s-convex functions, Sci. World J. 2014 (2014), Article ID 279158. [3] S. S. Dragomir, R.P. Agarwal, Two inequalities for differentiable mappings and applications to special means of real numbers and to trapezoidal formula, Appl. Math. Lett. 11 (5) (1998), 91–95. [4] S. S. Dragomir, C.E.M. 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Qi, Integral inequalities of Hermite-Hadamard type for harmonically quasi-convex functions, Proceedings of the Jangjeon Mathematical Society, 16(3) (2013), 399-407. 1School of Computer Science and Applied Mathematics, University of the Witwatersrand, Private Bag 3, Wits 2050, Johannesburg, South Africa 2School of Engineering and Science, Victoria University, PO Box 14428 Melbourne City, MC 8001, Australia ∗Corresponding author: m amer latif@hotmail.com 1. Introduction 2. Main Results 3. Applications to Special Means References