New some Hadamard's type inequalities for co-ordinated convex ...

arXiv:1005.0700v1 [math.CA] 5 May 2010

NEW SOME HADAMARD’S TYPE INEQUALITIES FOR CO-ORDINATED CONVEX FUNCTIONS MEHMET ZEKI SARIKAYA⋆♣ , ERHAN. SET , M. EMIN OZDEMIR , AND SEVER S. DRAGOMIRH Abstract. In this paper, we establish new some Hermite-Hadamard’s type inequalities of convex functions of 2−variables on the co-ordinates.

1. Introduction Let f : I ⊆ R → R be a convex mapping defined on the interval I of real numbers and a, b ∈ I, with a < b. the following double inequality is well known in the literature as the Hermite-Hadamard inequality: Z b 1 a+b f (a) + f (b) ≤ f . f (x) dx ≤ 2 b−a a 2

Let us now consider a bidemensional interval ∆ =: [a, b] × [c, d] in R2 with a < b and c < d. A mapping f : ∆ → R is said to be convex on ∆ if the following inequality: f (tx + (1 − t) z, ty + (1 − t) w) ≤ tf (x, y) + (1 − t) f (z, w)

holds, for all (x, y) , (z, w) ∈ ∆ and t ∈ [0, 1] .A function f : ∆ → R is said to be on the co-ordinates on ∆ if the partial mappings fy : [a, b] → R, fy (u) = f (u, y) and fx : [c, d] → R, fx (v) = f (x, v) are convex where defined for all x ∈ [a, b] and y ∈ [c, d] (see [3]). A formal definition for co-ordinated convex function may be stated as follows: Definition 1. A function f : ∆ → R will be called co-ordinated canvex on ∆, for all t, s ∈ [0, 1] and (x, y), (u, v) ∈ ∆,if the following inequality holds: f (tx + (1 − t) y, su + (1 − s) v) ≤

tsf (x, u) + s(1 − t)f (y, u) + t(1 − s)f (x, v) + (1 − t)(1 − s)f (y, v).

Clearly, every convex function is co-ordinated convex. Furthermore, there exist co-ordinated convex function which is not convex, (see, [3]). For several recent results concerning Hermite-Hadamard’s inequality for some convex function on the co-ordinates on a rectangle from the plane R2 , we refer the reader to ([1]-[6]). Also, in [3], Dragomir establish the following similar inequality of Hadamard’s type for co-ordinated convex mapping on a rectangle from the plane R2 . 2000 Mathematics Subject Classification. 26A51, 26D15. Key words and phrases. convex function, co-ordinated convex mapping, Hermite-Hadamard inequality. ⋆ corresponding author. 1

MEHMET ZEKI SARIKAYA⋆♣ , ERHAN. SET , M. EMIN OZDEMIR , AND SEVER S. DRAGOMIRH 2

Theorem 1. Suppose that f : ∆ → R is co-ordinated convex on ∆. Then one has the inequalities:

(1.1)

a+b c+d , 2 2   b Z Zd 1 1 1 c+d a+b dx + , y dy  f x, f 2 b−a 2 d−c 2

f

≤

a

≤

≤

c

Zb Zd

1 f (x, y) dydx (b − a) (d − c) a c  Zb Zb 1 1 1 f (x, c) dx + f (x, d) dx 4 b−a b−a a a  Zd Zd 1 1 + f (a, y) dy + f (b, y) dy  d−c d−c c

≤

c

f (a, c) + f (a, d) + f (b, c) + f (b, d) . 4

The above inequalities are sharp. The main purpose of this paper is to establish new Hadamard-type inequalities of convex functions of 2-variables on the co-ordinates.

2. Inequalities for co-ordinated convex functions Lemma 1. Let f : ∆ ⊂ R2 → R2 be a partial differentiable mapping on ∆ := ∂2f ∈ L(∆), then the following [a, b] × [c, d] in R2 with a < b and c < d. If ∂t∂s equality holds:

(2.1)

Zb Zd f (a, c) + f (a, d) + f (b, c) + f (b, d) 1 + f (x, y) dydx 4 (b − a) (d − c) a c   Zb Zd 1 1 1 − [f (x, c) + f (x, d)] dx + [f (a, y) + f (b, y)] dy  2 b−a d−c a

=

(b − a) (d − c) 4

c

Z1 Z1 0

0

2

(1 − 2t)(1 − 2s)

∂ f (ta + (1 − t)b, sc + (1 − s)d) dtds. ∂t∂s

CO-ORDINATED CONVEX FUNCTIONS

3

Proof. By integration by parts, we get Z1 Z1

(2.2)

0

=

Z1

=

Z1

(1 − 2s)(1 − 2t)

∂2f (ta + (1 − t)b, sc + (1 − s)d) dtds ∂t∂s

0

(

1 1 ∂f (1 − 2s) (1 − 2t) (ta + (1 − t)b, sc + (1 − s)d) a − b ∂s 0 0  Z1  2 ∂f + (ta + (1 − t)b, sc + (1 − s)d) dt ds  a−b ∂s 0

1 ∂f 1 ∂f (1 − 2s) − (a, sc + (1 − s)d) − (b, sc + (1 − s)d) a − b ∂s a − b ∂s 0  Z1  2 ∂f + (ta + (1 − t)b, sc + (1 − s)d) dt ds  a−b ∂s 0

=

1 Z ∂f ∂f 1  (1 − 2s) (a, sc + (1 − s)d) + (b, sc + (1 − s)d) ds b−a ∂s ∂s 0  Z1 Z1  ∂f (1 − 2s) −2 (ta + (1 − t)b, sc + (1 − s)d) dtds .  ∂s 0

0

Thus, again by integration by parts in the right hand side of (2.2), it follows that Z1

(2.3)

(1 − 2s)

0

∂f ∂f (a, sc + (1 − s)d) + (b, sc + (1 − s)d) ds ∂s ∂s

∂f (ta + (1 − t)b, sc + (1 − s)d) dtds ∂s 0 0 1 (f (a, sc + (1 − s)d) + f (b, sc + (1 − s)d)) (1 − 2s) c−d 0 Z1 2 + (f (a, sc + (1 − s)d) + f (b, sc + (1 − s)d)) ds c−d −2

=

Z1 Z1

(1 − 2s)

0

1 Z1 ( f (ta + (1 − t)b, sc + (1 − s)d) (1 − 2s) −2 c−d 0 0  Z1  2 + f (ta + (1 − t)b, sc + (1 − s)d) ds dt  c−d 0


=

f (a, c) + f (b, c) f (a, d) + f (b, d) − c−d c−d Z1 2 (f (a, sc + (1 − s)d) + f (b, sc + (1 − s)d)) ds + c−d −

0

Z1

f (ta + (1 − t)b, c) f (ta + (1 − t)b, d) − c−d c−d 0  Z1  2 f (ta + (1 − t)b, sc + (1 − s)d) ds dt +  c−d

−2

−

0

=

f (a, c) + f (a, d) + f (b, c) + f (b, d) (d − c) Z1 Z1 4 + f (ta + (1 − t)b, sc + (1 − s)d) dsdt (d − c) 0 0 1 Z 2 (f (a, sc + (1 − s)d) + f (b, sc + (1 − s)d)) ds − (d − c)  0  Z1  + (f (ta + (1 − t)b, c) + f (ta + (1 − t)b, d)) dt .  0

Writing (2.3) in (2.2), using the change of the variable x = ta + (1 − t)b and , we y = sc + (1 − s)d for t, s ∈ [0, 1]2 , and multiplying the both sides by (b−a)(d−c) 4 obtain (2.1), which completes the proof. Theorem 2. Let f : ∆ ⊂ R2 → R2 be a partial mapping on ∆ := 2differentiable ∂ f is a convex function on the [a, b] × [c, d] in R2 with a < b and c < d. If ∂t∂s co-ordinates on ∆, then one has the inequalities: f (a, c) + f (a, d) + f (b, c) + f (b, d) (2.4) 4 Zb Zd 1 + f (x, y) dydx − A (b − a) (d − c) a c 2 2 2   2 ∂ f ∂ f ∂ f ∂ f (a, c) (a, d) (b, c) (b, d) + + + ∂s∂t ∂s∂t ∂s∂t (b − a) (d − c)  ∂s∂t  ≤ 16 4 where

  Zb Zd 1 1 1 A= [f (x, c) + f (x, d)] dx + [f (a, y) + f (b, y)] dy  . 2 b−a d−c a

c


5

Proof. From Lemma 1, we have f (a, c) + f (a, d) + f (b, c) + f (b, d) 4 b Z Zd 1 + f (x, y) dydx − A (b − a) (d − c) a

≤

c

(b − a) (d − c) 4 2 Z1 Z1 ∂ f × (ta + (1 − t)b, sc + (1 − s)d) dtds. |(1 − 2t)(1 − 2s)| ∂t∂s 0

0

Since f : ∆ → R is co-ordinated convex on ∆, then one has: f (a, c) + f (a, d) + f (b, c) + f (b, d) 4 Zb Zd 1 f (x, y) dydx − A + (b − a) (d − c) a

≤

c

(b − a) (d − c) 4  2 Z1 Z1 ∂ f ×  |(1 − 2t)(1 − 2s)| t (a, sc + (1 − s)d) ∂t∂s 0 0 2 ∂ f (b, sc + (1 − s)d) dt ds. +(1 − t) ∂t∂s

Firstly, by calculating the integral in above inequality, we have Z1 0 1

=

Z2

0

+

2 2 ∂ f ∂ f |1 − 2t| t (a, sc + (1 − s)d) + (1 − t) (b, sc + (1 − s)d) dt ∂t∂s ∂t∂s

2 2 ∂ f ∂ f (a, sc + (1 − s)d) + (1 − t) (b, sc + (1 − s)d) dt (1 − 2t) t ∂t∂s ∂t∂s

Z1 1 2

2 2 ∂ f ∂ f (2t − 1) t (a, sc + (1 − s)d) + (1 − t) (b, sc + (1 − s)d) dt ∂t∂s ∂t∂s 1 = 4

2 2 ∂ f ∂ f ∂t∂s (a, sc + (1 − s)d) + ∂t∂s (b, sc + (1 − s)d) .


Thus, we obtain

(2.5)

f (a, c) + f (a, d) + f (b, c) + f (b, d) 4 Zb Zd 1 + f (x, y) dydx − A (b − a) (d − c) a

≤

c

(b − a) (d − c) 16 2 2 Z1 ∂ f ∂ f × |1 − 2s| (a, sc + (1 − s)d) + (b, sc + (1 − s)d) ds. ∂t∂s ∂t∂s 0

A similar way for other integral, since f : ∆ → R is co-ordinated convex on ∆, we get Z1

(2.6)

2 2 ∂ f ∂ f (a, sc + (1 − s)d) + (b, sc + (1 − s)d) ds |1 − 2s| ∂t∂s ∂t∂s

0 1

=

Z2

0

2 2 ∂ f ∂ f (a, c) + (1 − s) (a, d) ds (1 − 2s) s ∂t∂s ∂t∂s 1

+

Z2

0

=

Z1 1 2

2 2 ∂ f ∂ f (b, c) + (1 − s) (b, d) ds (1 − 2s) s ∂t∂s ∂t∂s

2 2 ∂ f ∂ f (2s − 1) s (a, c) + (1 − s) (a, d) ds ∂t∂s ∂t∂s

+

Z1 1 2

=

2 2 ∂ f ∂ f (2s − 1) s (b, c) + (1 − s) (b, d) ds ∂t∂s ∂t∂s

2 2 2 2 ∂ f ∂ f ∂ f ∂ f ∂t∂s (a, c) + ∂t∂s (a, d) + ∂t∂s (b, c) + ∂t∂s (b, d) 4

By the (2.5) and (2.6), we get the inequality (2.4).

.

Theorem 3. Let f : ∆ ⊂ R2 → R2 be a partial 2 qdifferentiable mapping on ∆ := ∂ f , q > 1,is a convex function on [a, b] × [c, d] in R2 with a < b and c < d. If ∂t∂s


7

the co-ordinates on ∆, then one has the inequalities: f (a, c) + f (a, d) + f (b, c) + f (b, d) (2.7) 4 Zb Zd 1 + f (x, y) dydx − A (b − a) (d − c) a

≤

where

c

(b − a) (d − c) 2

4 (p + 1) p q 2 q 2 q 2 q  q1  2 ∂ f ∂ f ∂ f ∂ f (a, c) + ∂s∂t (a, d) + ∂s∂t (b, c) + ∂s∂t (b, d) ∂s∂t   × 4

  Zb Zd 1 1 1 [f (x, c) + f (x, d)] dx + [f (a, y) + f (b, y)] dy  . A= 2 b−a d−c a

and

1 p

+

1 q

c

= 1.

Proof. From Lemma 1, we have f (a, c) + f (a, d) + f (b, c) + f (b, d) 4 b Z Zd 1 + f (x, y) dydx − A (b − a) (d − c) a

≤

c

(b − a) (d − c) 4 2 Z1 Z1 ∂ f |(1 − 2t)(1 − 2s)| (ta + (1 − t)b, sc + (1 − s)d) dtds. × ∂t∂s 0

0

By using the well known H¨ older inequality for double integrals, f : ∆ → R is co-ordinated convex on ∆, then one has: f (a, c) + f (a, d) + f (b, c) + f (b, d) 4 b Z Zd 1 + f (x, y) dydx − A (b − a) (d − c) a c 1 1  p1 Z Z (b − a) (d − c)  p ≤ |(1 − 2t)(1 − 2s)| dtds 4 0

0

 q1 1 1 q Z Z 2 ∂ f  . × ∂t∂s (ta + (1 − t)b, sc + (1 − s)d) dtds 0

0


2 q ∂ f is convex function on the co-ordinates on ∆, we know that for t ∈ [0, 1] Since ∂t∂s

≤ and

q 2 ∂ f ∂t∂s (ta + (1 − t)b, sc + (1 − s)d) q q 2 2 ∂ f ∂ f (a, sc + (1 − s)d) + (1 − t) (b, sc + (1 − s)d) t ∂t∂s ∂t∂s 2 q ∂ f ∂t∂s (ta + (1 − t)b, sc + (1 − s)d) q q 2 2 ∂ f ∂ f (a, c) + t(1 − s) (a, d) ts ∂t∂s ∂t∂s q 2 q 2 ∂ f ∂ f (b, c) + (1 − t)(1 − s) (b, c) +(1 − t)s ∂t∂s ∂t∂s

≤

hence, it follows that

f (a, c) + f (a, d) + f (b, c) + f (b, d) 4 b Z Zd 1 + f (x, y) dydx − A (b − a) (d − c) a

≤

2

4 (p + 1) p 1 1 q q 2 Z Z 2 ∂ f ∂ f  × ts (a, c) + t(1 − s) (a, d) ∂t∂s ∂t∂s 0

=

c

(b − a) (d − c)

0

q q 2 2 q1 ∂ f ∂ f (b, c) + (1 − t)(1 − s) (b, d) dtds +(1 − t)s ∂t∂s ∂t∂s

(b − a) (d − c) 2

4 (p + 1) p q 2 q 2 q 2 q  q1  2 ∂ f ∂ f ∂ f ∂ f (a, c) (a, d) (b, c) (b, d) + + + ∂s∂t ∂s∂t ∂s∂t ∂s∂t  . × 4

Theorem 4. Let f : ∆ ⊂ R2 → R2 be a partial 2 qdifferentiable mapping on ∆ := ∂ f , q ≥ 1,is a convex function on [a, b] × [c, d] in R2 with a < b and c < d. If ∂t∂s


9

the co-ordinates on ∆, then one has the inequalities: f (a, c) + f (a, d) + f (b, c) + f (b, d) (2.8) 4 b Z Zd 1 + f (x, y) dydx − A (b − a) (d − c) a

≤

where

c

(b − a) (d − c) 16 q 2 q 2 q 2 q  1q  2 ∂ f ∂ f ∂ f ∂ f  ∂t∂s (a, c) + ∂t∂s (a, d) + ∂t∂s (b, c) + ∂t∂s (b, d)   ×   4

  Zb Zd 1 1 1 A= [f (x, c) + f (x, d)] dx + [f (a, y) + f (b, y)] dy  . 2 b−a d−c a

c

Proof. From Lemma 1, we have f (a, c) + f (a, d) + f (b, c) + f (b, d) 4 b Z Zd 1 f (x, y) dydx − A + (b − a) (d − c) a

≤

c

(b − a) (d − c) 4 2 Z1 Z1 ∂ f |(1 − 2t)(1 − 2s)| (ta + (1 − t)b, sc + (1 − s)d) dtds. × ∂t∂s 0

0

By using the well known power mean inequality for double integrals, f : ∆ → R is co-ordinated convex on ∆, then one has: f (a, c) + f (a, d) + f (b, c) + f (b, d) 4 b Z Zd 1 + f (x, y) dydx − A (b − a) (d − c) a c 1 1 1− q1 Z Z (b − a) (d − c)  ≤ |(1 − 2t)(1 − 2s)| dtds 4 0

0

 1q 1 1 q 2 Z Z ∂ f (ta + (1 − t)b, sc + (1 − s)d) dtds . × |(1 − 2t)(1 − 2s)| ∂t∂s 0

0

10 MEHMET ZEKI SARIKAYA⋆♣ , ERHAN. SET , M. EMIN OZDEMIR , AND SEVER S. DRAGOMIRH

2 q ∂ f is convex function on the co-ordinates on ∆, we know that for t ∈ [0, 1] Since ∂t∂s

≤

q 2 ∂ f (ta + (1 − t)b, sc + (1 − s)d) ∂t∂s q q 2 2 ∂ f ∂ f (a, sc + (1 − s)d) + (1 − t) (b, sc + (1 − s)d) t ∂t∂s ∂t∂s

and

≤

q 2 ∂ f ∂t∂s (ta + (1 − t)b, sc + (1 − s)d) 2 q q 2 ∂ f ∂ f ts (a, c) + t(1 − s) (a, d) ∂t∂s ∂t∂s q q 2 2 ∂ f ∂ f (b, c) + (1 − t)(1 − s) (b, c) +(1 − t)s ∂t∂s ∂t∂s

hence, it follows that

f (a, c) + f (a, d) + f (b, c) + f (b, d) 4 Zb Zd 1 + f (x, y) dydx − A (b − a) (d − c) a

≤

c

1− q1 (b − a) (d − c) 1 4 4 1 1 q q 2 2 Z Z ∂ f ∂ f  (a, c) + t(1 − s) (a, d) |(1 − 2t)(1 − 2s)| ts × ∂t∂s ∂t∂s 0

0

q q 2 2 q1 ∂ f ∂ f (b, c) + (1 − t)(1 − s) (b, d) dtds +(1 − t)s . ∂t∂s ∂t∂s


Firstly, by calculating the integral in above inequality, we have

Z1

q 2 q 2 ∂ f ∂ f |1 − 2t| ts (a, c) + t(1 − s) (a, d) ∂t∂s ∂t∂s 0 2 q q 2 ∂ f ∂ f +(1 − t)s (b, c) + (1 − t)(1 − s) (b, d) dt ∂t∂s ∂t∂s 1

=

Z2

q q 2 2 ∂ f ∂ f (a, c) + t(1 − s) (a, d) (1 − 2t) ts ∂t∂s ∂t∂s 0 2 q q 2 ∂ f ∂ f +(1 − t)s (b, c) + (1 − t)(1 − s) (b, d) dt ∂t∂s ∂t∂s q q 2 2 Z1 ∂ f ∂ f (a, c) + t(1 − s) (a, d) + (2t − 1) ts ∂t∂s ∂t∂s 1 2

q q 2 2 ∂ f ∂ f (b, c) + (1 − t)(1 − s) (b, d) dt +(1 − t)s ∂t∂s ∂t∂s 2 q q 2 ∂ f ∂ f s (a, c) (a, d) (1 − s) ∂t∂s ∂t∂s = + 24 q q 242 ∂ 2f ∂ f (b, c) (b, d) 5(1 − s) 5s ∂t∂s ∂t∂s + + 24 24 q q 2 2 ∂ f ∂ f (a, c) (a, d) 5(1 − s) 5s ∂t∂s ∂t∂s + + 24 224 2 q q ∂ f ∂ f (1 − s) s (b, c) (b, d) ∂t∂s ∂t∂s + + 24 24 2 q q 2 ∂ f ∂ f s (a, c) + (1 − s) (a, d) ∂t∂s ∂t∂s = 4 2 q 2 q ∂ f ∂ f s (b, c) + (1 − s) (b, d) ∂t∂s ∂t∂s + . 4

11

12 MEHMET ZEKI SARIKAYA⋆♣ , ERHAN. SET , M. EMIN OZDEMIR , AND SEVER S. DRAGOMIRH

Thus, we obtain f (a, c) + f (a, d) + f (b, c) + f (b, d) (2.9) 4 Zb Zd 1 + f (x, y) dydx − A (b − a) (d − c) a c 1 q 2 Z ∂ f (b − a) (d − c)  |1 − 2s| s (a, c) ≤ 16 ∂t∂s 0

2 q 2 q 2 q q1 ∂ f ∂ f ∂ f +(1 − s) (a, d) + s (b, c) + (1 − s) (b, d) ds . ∂t∂s ∂t∂s ∂t∂s

A similar way for other integral, since f : ∆ → R is co-ordinated convex on ∆, we get Z1

(2.10)

q q 2 2 ∂ f ∂ f |1 − 2s| s (a, c) + (1 − s) (a, d) ∂t∂s ∂t∂s 0 q q 2 2 ∂ f ∂ f (b, c) + (1 − s) (b, d) ds +s ∂t∂s ∂t∂s 1

=

Z2

q q 2 2 ∂ f ∂ f (1 − 2s) s (a, c) + (1 − s) (a, d) ∂t∂s ∂t∂s 0 q 2 q 2 ∂ f ∂ f (b, c) + (1 − s) (b, d) ds +s ∂t∂s ∂t∂s 1 q q 2 Z ∂ 2f ∂ f + (2s − 1) s (a, c) + (1 − s) (a, d) ∂t∂s ∂t∂s 1 2

=

=

q q 2 2 ∂ f ∂ f (b, c) + (1 − s) (b, d) ds +s ∂t∂s ∂t∂s q q q q 2 2 2 2 ∂ f ∂ f ∂ f ∂ f (a, d) (b, c) (b, d) 5 5 ∂t∂s (a, c) ∂t∂s ∂t∂s ∂t∂s + + + 24 24 24 q q q q 2 2 24 2 ∂ f ∂ f ∂ 2f ∂ f (a, c) (a, d) (b, c) (b, d) 5 5 ∂t∂s ∂t∂s ∂t∂s ∂t∂s + + + + 24 24 24 q 2 q 2 q 2 q24 2 ∂ f ∂ f ∂ f ∂ f ∂t∂s (a, c) + ∂t∂s (a, d) + ∂t∂s (b, c) + ∂t∂s (b, d) . 4

By the (2.9) and (2.10), we get the inequality (2.8). Remark 1. Since

1 4

1, the estimation given in Theorem 4 is

better than the one given in Theorem 3.


13

References [1] M. Alomari and M. Darus, Co-ordinated s-convex function in the first sense with some Hadamard-type inequalities, Int. J. Contemp. Math. Sciences, 3 (32) (2008), 1557-1567. [2] M. Alomari and M. Darus, On the Hadamard’s inequality for log-convex functions on the coordinates, J. of Inequal. and Appl, Article ID 283147, (2009), 13 pages. [3] S.S. Dragomir, On Hadamard’s inequality for convex functions on the co-ordinates in a rectangle from the plane, Taiwanese Journal of Mathematics, 4 (2001), 775-788. [4] M. A. Latif and M. Alomari, Hadamard-type inequalities for product two convex functions on the co-ordinetes, Int. Math. Forum, 4(47), 2009, 2327-2338. [5] M. A. Latif and M. Alomari, On the Hadamard-type inequalities for h-convex functions on the co-ordinetes, Int. J. of Math. Analysis, 3(33), 2009, 1645-1656. ¨ [6] M.E. Ozdemir, E. Set and M.Z. Sarikaya, New some Hadamard’s type inequalities for coordinated m-convex and (α, m)-convex functions, RGMIA, Res. Rep. Coll., 13 (2010), Supplement, Article 4. ♣ Department of Mathematics,Faculty of Science and Arts, Du ¨ zce University, Du ¨ zce, Turkey E-mail address: [email protected] Atatu ¨ rk University, K.K. Education Faculty, Department of Mathematics, 25240, Campus, Erzurum, Turkey E-mail address: [email protected] Graduate School of Natural and Applied Sciences, Ag ˙ ˘ rı Ibrahim C ¸ ec ¸ en University, ˘ Agrı, Turkey E-mail address: [email protected] H Research Group in Mathematical Inequalities & Applications, School of Engineering & Science, Victoria University, PO Box 14428, Melbourne City, MC 8001, Australia. E-mail address: [email protected] URL: http://rgmia.vu.edu.au/dragomir