J. Adv. Math. Stud

J. Adv. Math. Stud. Vol. 9(2016), No. 1, 07-16 http://journal.fairpartners.ro

SOME NEWTON’S TYPE INEQUALITIES FOR HARMONIC CONVEX FUNCTIONS MUHAMMAD ASLAM NOOR, KHALIDA INAYAT NOOR AND SABAH IFTIKHAR Abstract. In this paper, some new Newton’s type inequalities for harmonic convex functions are obtained. Ideas and techniques of this may stimulate further research.

1. INTRODUCTION Convexity plays an essential role in the development of various fields of pure and applied sciences. In recent years, much attention has been given to derive Hermite-Hadamard type inequalities for various types of convex functions. For recent developments, see [6, 8, 9, 10, 12, 13]. Recently theory of convex sets and convex functions have been generalized and extended in several directions using novel and innovative techniques. A significant generalization of convex functions is that of harmonic convex functions which was introduced by Anderson et al. [2] and Iscan [8]. A function f : I = [a, b] ⊆ R \ {0} → R is harmonic convex function, if and only if, ¶ · µ ¶ µ ¶¸ µ Z b 2ab 1 4ab 4ab ab f (x) f ≤ f +f ≤ dx a+b 2 a + 3b 3a + b b − a a x2 · µ ¶ ¸ 2ab f (a) + f (b) 1 1 f + ≤ [f (a) + f (b)]. (1.1) ≤ 2 a+b 2 2 The inequality (1.1) is known as Hermite-Hadamard inequality for harmonic convex function. Both inequalities hold in the reversed direction if f is harmonic concave, see [6]. Gao et al. [7] and Noor et al. [14] have obtained Newton’s type inequalities for different types of convex functions. To the best of out knowledge, such type of inequalities have not been considered for harmonic convex functions. Inspired and motivated by this on going research, we first derive a new integral identity for differentiable harmonic convex function. This auxiliary result is used to obtain some new Newton’s type inequalities for harmonic convex functions, which is the main motivation of our work. Now we recall some basic concept. Received: April 22, 2015. Revised: September 12, 2015. 1991 Mathematics Subject Classification: 26D15, 26D10, 90C23 Key words and phrases: Harmonic convex functions, Newton’s inequality, Hermite-Hadamard inequality. c °2012 Fair Partners Team for the Promotion of Science & Fair Partners Publishers

1

2

Muhammad Aslam Noor, Khalida Inayat Noor and Sabah Iftikhar

Definition 1.1. [16]. A set I ⊆ R \ {0} is said to be a harmonic convex set, if xy ∈ I, ∀x, y ∈ I, t ∈ [0, 1]. tx + (1 − t)y Definition 1.2. [8]. A function f : I = [a, b] ⊆ R \ {0} → R is said to be a harmonic convex function, if and only if, µ ¶ xy f ≤ (1 − t)f (x) + tf (y), ∀x, y ∈ I, t ∈ [0, 1]. tx + (1 − t)y 2. MAIN RESULTS First of all we prove following Lemma which plays an important role in order to prove our main results. Lemma 2.1. Let f : I = [a, b] ⊆ R \ {0} − → R be a differentiable function on the interior I o 0 of I. If f ∈ L[a, b], then · µ ¶ µ ¶ ¸ Z b 3ab 3ab ab f (x) 1 f (a) + 3f + 3f + f (b) − dx 8 a + 2b 2a + b b − a a x2 ¶ µ Z 1 ab µ(t) 0 dt = ab(b − a) f 2 ta + (1 − t)b 0 [ta + (1 − t)b] where

  t − 18 , t ∈ [0, 31 ) t − 12 , t ∈ [ 31 , 23 ) µ(t) =  t − 78 , t ∈ [ 32 , 1].

Proof. Let

µ ¶ ab µ(t) 0 ab(b − a) f dt 2 ta + (1 − t)b 0 [ta + (1 − t)b] µ ¶ Z 31 t − 18 ab 0 f ab(b − a) dt 2 ta + (1 − t)b 0 [ta + (1 − t)b] µ ¶ Z 32 t − 12 ab 0 +ab(b − a) f dt 1 [ta + (1 − t)b]2 ta + (1 − t)b 3 µ ¶ Z 1 t − 78 ab 0 f +ab(b − a) dt 2 [ta + (1 − t)b]2 ta + (1 − t)b 3 Z

I

= =

= Thus

I1 + I2 + I3

µ ¶ t − 18 ab 0 = ab(b − a) f dt 2 ta + (1 − t)b 0 [ta + (1 − t)b] ¯ µ ¶¯ 13 Z 1 µ ¶ 3 ¯ ¯¡ 1¢ ab ab ¯ ¯ − f dt = ¯ t− f 8 ta + (1 − t)b ¯0 ta + (1 − t)b 0 µ ¶ Z 1 µ ¶ 3 ab 1 5 3ab f = f (a) + f − dt 8 24 a + 2b ta + (1 − t)b 0 Z

I1

1

1 3

Some Newton’s type inequalities for harmonic convex functions

3

Similarly, one can obtain µ ¶ t − 12 ab 0 f dt 1 [ta + (1 − t)b]2 ta + (1 − t)b 3 µ ¶ µ ¶ Z 2 µ ¶ 3 1 3ab 1 3ab ab f + f − f dt 1 6 2a + b 6 a + 2b ta + (1 − t)b 3 Z

I2

= ab(b − a) =

and

2 3

¶ µ t − 87 ab 0 = ab(b − a) f dt 2 [ta + (1 − t)b]2 ta + (1 − t)b 3 µ ¶ Z 1 µ ¶ ab 1 5 3ab f = f (b) + f − dt 2 8 24 2a + b ta + (1 − t)b 3 Z

I1

1

ab Now adding I1 , I2 and I3 , and change the variable x = ta+(1−t)b , we have · µ ¶ µ ¶ ¸ Z b 1 3ab 3ab ab f (x) I = I1 + I2 + I3 = f (a) + 3f + 3f + f (b) − dx, 8 a + 2b 2a + b b − a a x2

which is the required result.

¤

→ R be a differentiable function on the interior Theorem 2.1. Let f : I = [a, b] ⊆ R \ {0} − I o of I. If f 0 ∈ L[a, b] and |f 0 |q is harmonic convex function on I, q ≥ 1, then ¯ · ¯ µ ¶ µ ¶ ¸ Z b ¯1 3ab 3ab ab f (x) ¯¯ ¯ f (a) + 3f + 3f + f (b) − dx ¯8 ¯ a + 2b 2a + b b − a a x2 © 1 1 ≤ ab(b − a) (C1 (a, b))1− q [C4 (a, b)|f 0 (a)|q + C5 (a, b)|f 0 (b)|q ] q 1

1

+(C2 (a, b))1− q [C6 (a, b)|f 0 (a)|q + C7 (b, a)|f 0 (b)|q ] q

1 1ª +(C3 (b, a))1− q [C8 (b, a)|f 0 (a)|q + C9 (b, a)|f 0 (b)|q ] q ,

where Z C1 (a, b) = 0

1 3

¯ ¯ ¯t − 1 ¯ 8

£ ¤2 dt ta + (1 − t)b

a2 − 4ab + 3b2 1 = − ln 8b(a + 2b)(a − b)2 (a − b)2 ¯ ¯ Z 32 ¯t − 1 ¯ 2 C2 (a, b) = ¤2 dt £ 1 ta + (1 − t)b 3

µ

¶ 3(a + 7b)2 . 64b(a + 2b)

1 2(2a2 + 2b2 + 5ab) − ln 2(a − b)(a + 2b)(2a + b) (a − b)2 (2a + b)(a + 2b) ¯ ¯ Z 1 ¯t − 7 ¯ 8 C3 (b, a) = £ ¤2 dt 2 ta + (1 − t)b 3 µ ¶ 1 3(7a + b)2 3a2 − 4ab + b2 − ln = . 8a(2a + b)(a − b)2 (a − b)2 64a(2a + b) =

µ

¶ 3(a + b) . 2(a + 2b)(2a + b)

4


Z C4 (a, b)

= 0

1 3

¯ ¯ ¯t − 1 ¯(1 − t) 8 £ ¤2 dt ta + (1 − t)b

¶ µ 3a3 + 4b3 + 7ab2 − 14a2 b 42b2 + 75ab + 27a2 3(a + 7b)2 = . (2.1) − ln 24b(2a + b)(a − b)3 24(2a + b)(a − b)3 64b(a + 2b) ¯ ¯ Z 31 ¯t − 1 ¯t 8 C5 (a, b) = £ ¤2 dt 0 ta + (1 − t)b ¶ µ −a2 − 13b2 + 12ab 3a2 + 51ab + 90b2 3(a + 7b)2 = + ln . (2.2) 24(a + 2b)(a − b)3 24(a + 2b)(a − b)3 64b(a + 2b) ¯ Z 32 ¯¯ t − 12 ¯(1 − t) C6 (a, b) = £ ¤2 dt 1 ta + (1 − t)b 3 µ ¶ a3 − 2a2 b + ab2 6a3 + 2b3 + 11ab2 + 17a2 b 9(a + b)2 = − ln (2.3). 2(2a + b)(a + 2b)(a − b)3 2(2a + b)(a + 2b)(a − b)3 4b(a + 2b)(2a + b) ¯ ¯ Z 32 ¯t − 1 ¯t 2 C7 (b, a) = £ ¤2 dt 1 ta + (1 − t)b 3 µ ¶ −b3 + 2b2 a − ba2 6b3 + 2a3 + 11ba2 + 17b2 a 9(a + b)2 = + ln (2.4) . 2(2a + b)(a + 2b)(a − b)3 2(2a + b)(a + 2b)(a − b)3 4a(a + 2b)(2a + b) ¯ Z 1 ¯¯ t − 78 ¯(1 − t) C8 (b, a) = £ ¤2 dt 2 ta + (1 − t)b 3 µ ¶ 3(7a + b)2 13a2 + b2 − 12ab 3b2 + 51ab + 90a2 = − ln . (2.5) 24a(2a + b)(a − b)3 24(2a + b)(a − b)3 64a(2a + b) ¯ ¯ Z 1 ¯t − 7 ¯t 8 C9 (b, a) = £ ¤2 dt 2 ta + (1 − t)b 3 µ ¶ 4a3 + 3b3 + 7a2 b − 14ab2 42a2 + 75ab + 27b2 3(7a + b)2 = − + ln . (2.6) 24a(a + 2b)(a − b)3 24(a + 2b)(b − a)3 64a(2a + b) Proof. Using Lemma 2.1 and the power mean inequality, we have

≤

≤

¯ ¯ · µ ¶ µ ¶ ¸ Z b ¯1 3ab 3ab ab f (x) ¯¯ ¯ f (a) + 3f + 3f + f (b) − dx ¯ ¯8 a + 2b 2a + b b − a a x2 ¯ ¯ µ ·Z 1 µ ¶¯ ¶¯ 2 Z 3 3 ¯ 0 ¯ 0 ¯ ¯ |t − 18 | |t − 21 | ab ab ¯ ¯ ¯ ¯dt ab(b − a) f f dt + ¯ ¯ ¯ ¯ 2 2 1 ta + (1 − t)b [ta + (1 − t)b] ta + (1 − t)b 0 [ta + (1 − t)b] 3 ¯ µ ¸ ¶¯ Z 1 ¯ 0 ¯ |t − 78 | ab ¯f ¯dt + ¯ ¯ 2 [ta + (1 − t)b]2 ta + (1 − t)b 3 1 ¯ µ ·µ Z 1 ¶1− q µ Z 1 ¶¯q ¶ q1 3 3 ¯ 0 ¯ |t − 81 | |t − 18 | ab ¯ ¯ dt f ab(b − a) dt ¯ 2 2 ta + (1 − t)b ¯ 0 [ta + (1 − t)b] 0 [ta + (1 − t)b] ¯ µ ¶1− q1 µ Z 2 ¶¯q ¶ q1 µZ 2 3 3 ¯ 0 ¯ |t − 12 | |t − 12 | ab ¯ ¯ dt f dt + ¯ ¯ 2 2 1 1 [ta + (1 − t)b] [ta + (1 − t)b] ta + (1 − t)b 3 3


¯ µ ¶1− q1 µ Z 1 ¶¯q ¶ q1 ¸ ¯ 0 ¯ |t − 78 | |t − 87 | ab ¯ ¯ dt ¯f ta + (1 − t)b ¯ dt 2 [ta + (1 − t)b]2 2 [ta + (1 − t)b]2 3 3 £ ¤ ¶1 ·µ Z 1 ¶1− q1 µ Z 1 1 0 q 0 q q 3 3 |t − |t − 81 | 8 | (1 − t)|f (a)| + t|f (b)| ab(b − a) dt dt 2 2 [ta + (1 − t)b] 0 [ta + (1 − t)b] 0 ¤ ¶1 £ µZ 2 ¶1− q1 µ Z 2 1 1 0 q 0 q q 3 3 |t − |t − 2 | 2 | (1 − t)|f (a)| + t|f (b)| + dt dt 2 2 1 1 [ta + (1 − t)b] [ta + (1 − t)b] 3 3 £ ¤ ¶1 ¸ 1µZ µZ 1 ¶ 1− 1 q q |t − 78 | (1 − t)|f 0 (a)|q + t|f 0 (b)|q |t − 78 | + dt dt 2 2 2 [ta + (1 − t)b] 2 [ta + (1 − t)b] 3 3 1µZ 1 ·µ Z 1 ¶ ¶ q1 Z 31 1− q 3 3 |t − 81 | |t − 18 |(1 − t) 0 |t − 81 |t q 0 q ab(b − a) dt |f (a)| dt + |f (b)| dt 2 2 2 0 [ta + (1 − t)b] 0 [ta + (1 − t)b] 0 [ta + (1 − t)b] µZ 2 ¶1− q1 µ Z 2 ¶ q1 Z 32 3 3 |t − 12 | |t − 12 |(1 − t) 0 |t − 12 |t q 0 q dt |f (a)| dt + |f (b)| dt + 1 1 1 [ta + (1 − t)b]2 [ta + (1 − t)b]2 [ta + (1 − t)b]2 3 3 3 µZ 1 ¶1− q1 µ Z 1 ¶ q1 ¸ Z 1 |t − 78 | |t − 78 |(1 − t) 0 |t − 78 |t q 0 q + dt |f (a)| dt + |f (b)| dt 2 [ta + (1 − t)b]2 2 [ta + (1 − t)b]2 2 [ta + (1 − t)b]2 3 3 3 © 1 1 ab(b − a) (C1 (a, b))1− q [C4 (a, b)|f 0 (a)|q + C5 (a, b)|f 0 (b)|q ] q µZ +

≤

=

=

5

1

1

1

+(C2 (a, b))1− q [C6 (a, b)|f 0 (a)|q + C7 (b, a)|f 0 (b)|q ] q

1 1ª +(C3 (b, a))1− q [C8 (b, a)|f 0 (a)|q + C9 (b, a)|f 0 (b)|q ] q ,


¤

If q = 1, then Theorem 2.1 reduces to the following result. → R be a differentiable function on the interior Corollary 2.1. Let f : I = [a, b] ⊆ R \ {0} − I o of I. If f 0 ∈ L[a, b] and |f 0 | is harmonic convex function on I, then

≤

¯ · ¯ µ ¶ µ ¶ ¸ Z b ¯1 3ab 3ab ab f (x) ¯¯ ¯ f (a) + 3f + 3f + f (b) − dx¯ ¯8 a + 2b 2a + b b − a a x2 ab(b − a){[C4 (a, b) + C6 (a, b) + C8 (b, a)]|f 0 (a)| +[C5 (a, b) + C7 (b, a) + C9 (b, a)]|f 0 (b)|},

where C4 (a, b) − C9 (b, a) are given by (2.1) − (2.6) respectively. Theorem 2.2. Let f : I = [a, b] ⊆ R \ {0} − → R be a differentiable function on the interior I o of I. If f 0 ∈ L[a, b] and |f 0 |q is harmonic convex function on I, p, q > 1 and p1 + 1q = 1, then ¯ ¯ · µ ¶ µ ¶ ¸ Z b ¯1 f (x) ¯¯ 3ab 3ab ab ¯ f (a) + 3f dx¯ + 3f + f (b) − ¯8 a + 2b 2a + b b − a a x2 ¡ ¢ · µ 0 ¶1 3ab |f (a)|q + |f 0 2a+b |q q 1 ≤ ab(b − a) (C10 (p; a, b)) p 6

6


µ +

(C11 (p; a, b))

1 p

µ +

1

(C12 (p; b, a)) p

where

|f 0 |f

0

Z C10 (p; a, b)

¡ ¡

6 ¢q ¶1 ¸ | + |f 0 (b)|q q , 6

2 3

|t − 12 |p £ ¤2p dt ta + (1 − t)b

1

|t − 87 |p £ ¤2p dt. ta + (1 − t)b

0

C12 (p; b, a) =

3ab 2a+b

|t − 18 |p £ ¤2p dt ta + (1 − t)b

Z Z

¢q ¡ 3ab ¢ q ¶ 1 | + |f 0 2a+b | q

1 3

=

C11 (p; a, b) =

3ab a+2b

1 3

2 3

(2.7)

Proof. Using Lemma 2.1 and the Holder’s integral inequality, we have ¯ · ¯ µ ¶ µ ¶ ¸ Z b ¯1 3ab 3ab ab f (x) ¯¯ ¯ f (a) + 3f + 3f + f (b) − dx ¯8 ¯ a + 2b 2a + b b − a a x2 ¯ ¯ ·Z 1 ¯ µ ¶¯ 3 ¯ ¯ ¯¯ 0 t − 18 ab ¯ ¯ ¯¯ ≤ ab(b − a) ¯ [ta + (1 − t)b]2 ¯¯f ta + (1 − t)b ¯dt 0 ¯¯ µ ¶¯ Z 32 ¯ ¯ ¯ ¯¯ 0 t − 12 ab ¯dt ¯ ¯ ¯ + f ¯ ¯ ¯ 1 [ta + (1 − t)b]2 ta + (1 − t)b ¯ 3 ¯¯ µ ¶¯ ¸ Z 1¯ ¯ ¯¯ 0 ¯ t − 78 ab ¯ ¯¯f ¯dt + ¯ ¯ ¯ 2 2 [ta + (1 − t)b] ta + (1 − t)b ¯ 3 ¯p ¶ p1 µ Z 1 ¯ µ ·µ Z 1 ¯ ¶¯q ¶ q1 3 ¯ 3 ¯ ¯ ¯ t − 18 ab 0 ¯ ¯ ¯ ¯ ≤ ab(b − a) ¯ [ta + (1 − t)b]2 ¯ dt ¯f ta + (1 − t)b ¯ dt 0 0 ¯p ¶ p1 µ Z 2 ¯ µ µZ 2 ¯ ¶¯q ¶ q1 3 ¯ 3 ¯ ¯ ¯ t − 12 ab ¯ ¯ dt ¯f 0 ¯ dt + ¯ ¯ ¯ 2 1 1 [ta + (1 − t)b] ta + (1 − t)b ¯ 3 3 ¯p ¶ p1 µ Z 1 ¯ µ µZ 1 ¯ ¶¯q ¶ q1 ¸ ¯ ¯ ¯ 0 ¯ t − 78 ab ¯ ¯ ¯ ¯ + ¯ [ta + (1 − t)b]2 ¯ dt ¯f ta + (1 − t)b ¯ dt 2 2 3 3 ¶ p1 µ ¶ q1 ·µ Z 1 3ab Z a+2b 3 |t − 18 |p |f 0 (x)|q bp − ap ab dx = £ ¤2p dt p b−a a x2 0 ta + (1 − t)b µZ 2 ¶ p1 µ ¶ q1 3ab Z 2a+b 3 |t − 12 |p ab |f 0 (x)|q + dx £ ¤2p dt 1 3ab b − a a+2b x2 ta + (1 − t)b 3 ¶ p1 µ ¶ q1 ¸ µZ 1 Z b |t − 78 |p ab |f 0 (x)|q dx + (2.8) £ ¤2p dt 2 3ab b − a 2a+b x2 ta + (1 − t)b 3 Using the p-convexity of |f 0 |q , we obtain the following inequalities from inequality (1.1) ¡ 3ab ¢ q 3ab Z a+2b |f 0 (a)|q + |f 0 a+2b | ab |f 0 (x)|q dx ≤ , (2.9) 2 b−a a x 6


ab b−a and ab b−a

Z

Z

3ab 2a+b 3ab a+2b

b 3ab 2a+b

|f 0 |f 0 (x)|q dx ≤ x2

|f 0 |f 0 (x)|q dx ≤ x2

¡

¡

7

3ab a+2b

3ab 2a+b

¢q ¡ 3ab ¤ p1 ¢ q | | + |f 0 2a+b 6

¢q | + |f 0 (b)|q 6

,

.

(2.10)

(2.11)

A combination of (2.8)-(2.11) gives the required inequality (2.2).

¤

Theorem 2.3. Let f : I = [a, b] ⊆ R \ {0} − → R be a differentiable function on the interior I o of I. If f 0 ∈ L[a, b] and |f 0 |q is harmonic convex function on I, p, q > 1 and p1 + 1q = 1, then ¯ · ¯ µ ¶ µ ¶ ¸ Z b ¯1 3ab 3ab ab f (x) ¯¯ ¯ f (a) + 3f + 3f + f (b) − dx ¯8 ¯ a + 2b 2a + b b − a a x2 ¶1 ·µ p+1 3 + 5p+1 p ¡ ab(b − a) C13 (q; a, b)|f 0 (a)|q ≤ −9(2q − 1)(2q − 2)(a − b)2 24p (p + 1) µ ¶ p1 ¡ ¢1 1 0 q q C15 (q; a, b)|f 0 (a)|q +C14 (q; a, b)|f (b)| + p+1 6 (p + 1) µ p+1 ¶1 ¢1 3 + 5p+1 p ¡ 0 q q +C16 (q; b, a)|f (b)| + C17 (q; b, a)|f 0 (a)|q 24p (p + 1) ¸ ¢1 0 q q , +C18 (q; b, a)|f (b)| where C13 (q; a, b)

=

32q (a + 2b)−2q [(−5a2 + 4b2 − 8ab) + 4q(a2 − 2b2 + ab)] +9b−2q+1 [2a − b + 2q(b − a)].

C14 (q; a, b)

=

32q (a + 2b)−2q [(−a2 + 8b2 + 2ab) + 2q(a2 − 2b2 + ab)] + 9b−2q+2 .

C15 (q; a, b)

=

32q (2a + b)−2q [(−8a2 + b2 − 2ab) + 2q(2a2 − b2 − ab)] +32q (a + 2b)−2q [(5a2 − 4b2 + 8ab) + 2q(−2a2 + 4b2 − 2ab)].

C16 (q; b, a)

=

32q (a + 2b)−2q [(−8b2 + a2 − 2ab) + 2q(2b2 − a2 − ab)] +32q (2a + b)−2q [(5b2 − 4a2 + 8ab) + 2q(−2b2 + 4a2 − 2ab)].

C17 (q; b, a)

=

32q (2a + b)−2q [(−b2 + 8a2 + 2ab) + 2q(b2 − 2a2 + ab)] + 9a−2q+2 .

C18 (q; b, a)

=

32q (2a + b)−2q [(−5b2 + 4a2 − 8ab) + 4q(b2 − 2a2 + ab)] −9a−2q+1 [a − 2b − 2q(b − a)].

Proof. Using Lemma 2.1 and the Holder’s integral inequality, we have ¯ ¯ · µ ¶ µ ¶ ¸ Z b ¯1 f (x) ¯¯ 3ab 3ab ab ¯ f (a) + 3f dx¯ + 3f + f (b) − ¯8 a + 2b 2a + b b − a a x2 ¯ ·Z 1 µ ¶¯ 3 ¯ ¯¯ ¯ 1 ab 0 ¯ t − 1 ¯¯ ¯dt ≤ ab(b − a) f ¯ 2 8 [ta + (1 − t)b] ta + (1 − t)b ¯ 0 ¯ µ ¶¯ Z 32 ¯ ¯ 1 ab 1 ¯¯¯¯ 0 ¯dt ¯ + f t− ¯ ¯ 2 1 2 [ta + (1 − t)b] ta + (1 − t)b 3

8


¯ µ ¶¯ ¸ ¯ ¯¯ ¯ 1 ab 0 ¯t − 7 ¯¯ ¯dt f ¯ 2 2 8 [ta + (1 − t)b] ta + (1 − t)b ¯ 3 ·µ Z 1 ¶ p1 µ Z 1 ¯ ¶¯q ¶ q1 µ 3 ¯ 3 ¯ ¯ 1 ¯¯p 1 ab 0 ¯ ¯ ¯ ab(b − a) dt t− ¯ [ta + (1 − t)b]2 f ta + (1 − t)b ¯ dt 8 0 0 ¶ p1 µ Z 2 ¯ ¶¯q ¶ q1 µZ 2 µ 3 ¯ 3 ¯ ¯ ¯ 1 ab 0 ¯t − 1 ¯p dt ¯ ¯ dt + f ¯ 2 1 1 2 [ta + (1 − t)b] ta + (1 − t)b ¯ 3 3 µZ 1 ¶¯q ¶ q1 ¸ ¶ p1 µ Z 1 ¯ µ ¯ ¯ ¯ 7 ¯¯p 1 ab 0 ¯ ¯ ¯ + t− dt ¯ [ta + (1 − t)b]2 f ta + (1 − t)b ¯ dt 2 2 8 3 3 ¯ µ ·µ Z 1 ¶ p1 µ Z 1 ¶¯q ¶ q1 3 ¯ 3 ¯p ¯ ¯ 0 1 1 ab ¯t − ¯ dt ¯f ¯ dt ab(b − a) ¯ 2q 8 ta + (1 − t)b ¯ 0 0 [ta + (1 − t)b] ¯ µ µZ 2 ¶ p1 µ Z 2 ¶¯q ¶ q1 3 ¯ 3 ¯ 0 ¯ 1 ¯¯p 1 ab ¯ ¯ ¯ dt t− + dt f ¯ ¯ 2q 1 1 2 [ta + (1 − t)b] ta + (1 − t)b 3 3 ¯ µ µZ 1 ¶ p1 µ Z 1 ¶¯q ¶ q1 ¸ ¯ 0 ¯ ¯ ¯ ab 1 ¯f ¯ dt ¯t − 7 ¯p dt + ¯ 2q 2 2 [ta + (1 − t)b] 8 ta + (1 − t)b ¯ 3 3 ¶ p1 µ Z 1 ·µ Z 1 ¶ q1 3 ¯ 3 £ ¤ 1 ¯¯p 1 0 q 0 q ¯ t− dt ab(b − a) (1 − t)|f (a)| + t|f (b)| dt 2q 8 0 0 [ta + (1 − t)b] ¶ p1 µ Z 2 µZ 2 ¶ q1 3 ¯ 3 ¯ £ ¤ 1 0 q 0 q ¯t − 1 ¯p dt + (1 − t)|f (a)| + t|f (b)| dt 1 1 2 [ta + (1 − t)b]2q 3 3 ¶ p1 µ Z 1 µZ 1 ¶ q1 ¸ ¯ £ ¤ 7 ¯¯p 1 0 q 0 q ¯ dt + t− (1 − t)|f (a)| + t|f (b)| dt 2 2 [ta + (1 − t)b]2q 8 3 3 ·µ p+1 ¶1 µZ 1 ¶ q1 Z 31 3 (1 − t) t 3 + 5p+1 p 0 q 0 q ab(b − a) |f (a)| dt + |f (b)| dt 2q 2q 24p (p + 1) 0 [ta + (1 − t)b] 0 [ta + (1 − t)b] µ ¶ q1 ¶ p1 µ Z 2 Z 23 3 1 (1 − t) t 0 q 0 q + p+1 |f (a)| dt + |f (b)| dt 1 1 6 (p + 1) [ta + (1 − t)b]2q [ta + (1 − t)b]2q 3 3 µ p+1 ¶1 µZ 1 ¶ q1 ¸ Z 1 3 + 5p+1 p (1 − t) t 0 q 0 q + |f (a)| dt + |f (b)| dt 2 [ta + (1 − t)b]2q 2 [ta + (1 − t)b]2q 24p (p + 1) 3 3 µ ·µ p+1 ¶1 ¶ p1 p+1 p ¡ ¢1 1 3 +5 0 q 0 q q + ab(b − a) C13 (q; a, b)|f (a)| + C14 (q; a, b)|f (b)| 24p (p + 1) 6p+1 (p + 1) 1 µ p+1 ¶ ¡ ¢1 3 + 5p+1 p ¡ C15 (q; a, b)|f 0 (a)|q + C16 (q; b, a)|f 0 (b)|q q + C17 (q; b, a)|f 0 (a)|q p 24 (p + 1) ¸ ¢1 C18 (q; b, a)|f 0 (b)|q q , Z

1

+

≤

=

≤

=

=

+


¤


9

Acknowledgment. The authors would like to thank Dr. S. M. Junaid Zaidi, Rector, COMSATS Institute of Information Technology, Pakistan, for providing excellent research and academic environments. REFERENCES [1] M. Alomari, M. Darus and S. S. Dragomir: New inequalities of Simpson’s type for s-convex functions with applications, RGMIA Res. Rep. Coll., 12(2009), No. 4, Article 9, 18 pp., http://rgmia.org/papers/vi2n4/Simpson.pdf. [2] G. D. Anderson, M. K. Vamanamurthy and M. Vuorinen: Generalized convexity and inequalities, J. Math. Anal. Appl., 335(2007), 1294-1308. [3] G. Cristescu and L. Lupsa: Non-connected Convexities and Applications, Kluwer Academic Publishers, Dordrechet, Holland, (2002). [4] S.S. Dragomir, J. Pecaric and L. E. Persson: Some inequalities of Hadamard type, Soochow J. Math., 21(1995), 335-341. [5] S.S. Dragomir, R. P. Agarwal and P. Cerone: On Simpson Inequality and Applications, J. Inequal. Appl., 5(2000), 533-579. [6] S.S. Dragomir: Inequalities of Hermite-Hadamard type for HA-convex functions, RGMIA Res. Rep. Coll. 18(2015), Art. 38, 19 pp., http://rgmia.org/papers/v18/v18a38.pdf. [7] S. Gao and W. Shi: On new ineaulities of Newton’s type for functions whose secong derivatives absolute values are convex, Int. J. Pure Appl. Math., 74(2012), No. 1, 33-41. [8] Imdat Iscan. Hermite-Hadamard type inequalities for harmonically convex functions. Hacet. J. Math. Stats., 43(6)(2014), 935-942. [9] Imdat Iscan: Hermite-Hadamard and Simpson like inequalities for differentiable harmonically convex functions, J. Math., 2014(2014), Article ID 346305, 10 pp, http://dx.doi.org/10.1155/2014/346305. [10] M. A. Noor, K. I. Noor and M. U. Awan: Generalized Fractional Hermite-Hadamard Inequalities for twice differentiable s-Convex Functions, Filomat 29(2015), No. 4, 807-815. [11] M. V. Mihai, M. A. Noor, K. I. Noor and M. U. Awan: Some integral inequalities for harmonic h-convex functions involving hypergeometric functions, Appl. Math. Comput., 252(2015), 257-262. [12] M. A. Noor, K. I. Noor and M. U. Awan: Some integral inequalities for harmonivally h-convex functions, Politehn. Univ. Bucharest Sci. Bull, Ser. A Appl. Math. Phys., 77(2015), No. 1, 5-16. [13] M. A. Noor, K. I. Noor and M. U. Awan: Integral inequalities for coordinated Harmonically convex functions, Complex. Var. Elliptic. Equ., 60(2015), No. 6, 776-786. [14] M. A. Noor, K. I. Noor and M. U. Awan: Some Newton’s type inequalities for geometrically relative convex functions, Malays. J. Math. Sci., 2(2015). [15] C. P. Niculescu and L. E. Persson: Convex Functions and Their Applications, Springer-Verlag, New York, (2006). [16] H. N. Shi and Zhang: Some new judgement theorems of Schur geometric and Schur harmonic convexities for class of symmetric functions, J. Inequal. Appl., 527(2013), 9 pp., dio:10.1186/1029-242X-2013527. COMSATS Institute of Information Technology Department of Mathematics Park Road, Islamabad, Pakistan E-mail address: [email protected] COMSATS Institute of Information Technology Department of Mathematics Park Road, Islamabad, Pakistan E-mail address: [email protected] COMSATS Institute of Information Technology Department of Mathematics Park Road, Islamabad, Pakistan E-mail address: [email protected]