1. Introduction A function f is convex on an interval [a,b] if for all x1, x2∈[a,b] where 0≤λ≤1,(1)fλx1+1λx2≤λfx1+1λfx2.The most classical result for convex functions can be seen in the following theorem.
Theorem 1. Let f:I→R be a convex function defined on the interval I of real numbers and a,b∈I with a<b. Then following double inequality (2)fa+b2≤1ba∫abfxdx≤fa+fb2holds.
It is known as the Hadamard inequality. History of this inequality begins with the papers of Hermite [1] and Hadamard [2] in the years 1883–1893 (see [3]). A rich literature of mathematical inequalities is due to convex functions equally determined by the Hadamard inequality, inspired from this inequality many closely related results have been established which have their applications in approximately all fields of mathematical analysis up to some extent (see, [3–8]).
In 1935 Grüss proved an inequality well known as the Grüss inequality stated in the following theorem [9].
Theorem 2. Let f,g:[a,b]→R be integrable functions such that ϕ≤f(x)≤Φ and γ≤g(x)≤Γ for all x∈[a,b], where ϕ, Φ, γ, Γ are constants. Then we have
(3)
1
b

a
∫
a
b
f
x
g
x
d
x

1
b

a
∫
a
b
f
x
d
x
·
1
b

a
∫
a
b
g
x
d
x
≤
1
4
Φ

ϕ
Γ

γ
,
where the constant 1/4 is sharp.
In 1938, Ostrowski established the following inequality known as the Ostrowski inequality stated.
Theorem 3. Let f:I→R, where I is an interval in R, be a mapping differentiable in Io the interior of I and a,b∈Io, a<b. If f′(t)≤M, for all t∈[a,b], then we have(4)fx1ba∫abftdt≤14+xa+b/22ba2baM,for all x∈[a,b].
By using Grüss inequality some results have been established which are well known as OstrowskiGrüss or OstrowskiGrüss type inequalities (see [10–12] and references therein). Several quadrature rules of numerical integration have been estimated using Ostrowski and OstrowskiGrüss type inequalities (see [4, 5, 8, 10, 12]).
In [4] Cerone and Dragomir have estimated differences of the Hadamard inequality as follows.
Theorem 4. Suppose that f:[a,b]→R be a twice differentiable function on (a,b) and suppose that γ≤f′′(t)≤Γ for all t∈(a,b). Then we have the double inequality:(5)γba224≤1ba∫abftdtfa+b2≤Γba224.
Theorem 5. Under the assumptions of Theorem 4 we have(6)γba212≤fa+fb21ba∫abftdt≤Γba212.
Ujević in [8] also estimated differences of the Hadamard inequality.
Theorem 6. Suppose that f:[a,b]→R be a twice differentiable function on (a,b) and suppose that γ≤f′′(t)≤Γ for all t∈(a,b). Then we have (7)3S2Γ24ba2≤1ba∫abftdtfa+b2≤3S2γ24ba2,where S=(f′(b)f′(a))/(ba).
Theorem 7. Under the assumptions of Theorem 6 we have(8)3SΓ24ba2≤fa+fb21ba∫abftdt≤3Sγ24ba2,where S=(f′bf′a)/(ba).
The aim of this paper is in fact to establish proof of well known Ostrowski inequality in a very straightforward way, and to establish bounds of a difference of the Hadamard inequality given in [4, 8] in very simple way, here there is no need to define a two variable kernel. In the last by involving a parameter a similar but general result have been found and some particular bounds of a difference of the Hadamard inequality are calculated, also an OstrowskiGrüss type inequality is obtained by elementary calculation.
3. Some Related Results In this section we give some more results in a very simple way. First by involving a parameter, we prove a general result that provides bounds of a nonnegative difference of the Hadamard inequality and gives particular bounds, and then an OstrowskiGrüss type inequality is proved.
Theorem 8. Suppose that f:[a,b]→R be a twice differentiable function on (a,b) and suppose that γ≤f′′(t)≤Γ for all t∈(a,b). Then we have (28)24+αγ24S24αba2≤1ba∫abftdtfa+b2≤24+αΓ24S24αba2,where S=(f′(b)f′a)/(ba).
Proof. It is clear that (29)∫abf′′tγdt≥0.From which one has (30)f′bf′ababaγba≥0.Further we can say for some α>0(31)Sγαba2≥0.Adding (21) and (31) one has(32)1ba∫abftdtfa+b2≥24+αγ24S24αba2.On the other hand ∫abΓf′′tdt≥0, which gives for α>0(33)SΓαba2≤0.Adding (24) and (33) one has(34)1ba∫abftdtfa+b2≤24+αΓ24S24αba2.From (32) and (34) we have the required inequality.
Corollary 9. If one selects, for example, α=24 and α=48 in Theorem 8, then (35)2γS24ba2≤1ba∫abftdtfa+b2≤2ΓS24ba2,3γS48ba2≤1ba∫abftdtfa+b2≤3ΓS48ba2.
In the following, adopting the pattern of proofs we give the following OstrowskiGrüss type inequality. It is remarkable to mention here that in [11] Cheng has proved an improved result adopting a comparatively different method.
Theorem 10. Let f:[a,b]→R, where I is an interval in R be a mapping differentiable in Io, the interior of I and a,b∈Io, a<b. If γ≤f′(t)≤Γ for all t∈(a,b), then we have
(36)
1
2
f
x

x

b
f
b

x

a
f
a
2
b

a

1
b

a
∫
a
b
f
t
d
t
≤
x

a
2
+
b

x
2
4
b

a
Γ

γ
,
for all x∈[a,b].
Proof. It is clear that (xt)(f′(t)γ)≥0 and (ta)(Γf′(t))≥0 for all t∈[a,x]. Therefore (37)∫axxtf′tγdt+∫axtaΓf′tdt≥0.After some computation we have(38)xafx+xafa2∫axftdt≤12xa2Γγ.Also it is easy to see (tx)(Γf′(t))≥0 and (bt)(f′(t)γ)≥0 for all t∈[x,b]. Therefore (39)∫xbtxΓf′tdt+∫xbbtf′tγdt≥0.After some computation it can be seen(40)bxfxxbfb2∫xbftdt≤12bx2Γγ.By adding (38) and (40) one has(41)12fxxbfbxafa2ba1ba∫abftdt≤xa2+bx24baΓγ.On the other hand we have (42)∫axtaf′tγdt+∫axxtΓf′tdt≥0;t∈a,x.This gives(43)xafx+xafa2∫axftdt≥12xa2Γγ.Also we have (44)∫xbbtΓf′tdt+∫xbtxf′tγdt≥0;t∈x,b,which gives(45)bxfxxbfb2∫xbftdt≥12bx2Γγ.Adding (43) and (45), then combining with (41) one can get (36).