Journal of Inequalities in Pure and
Applied Mathematics
NOTE ON SOME HADAMARD-TYPE INEQUALITIES
M. KLARIČIĆ BAKULA AND J. PEČARIĆ
Department of Mathematics
Faculty of Natural Sciences, Mathematics and Education
University of Split , Teslina 12
21 000 Split, Croatia.
EMail: milica@pmfst.hr
Faculty of Textile Technology
University of Zagreb
Pierottijeva 6, 10000 Zagreb
Croatia.
EMail: pecaric@hazu.hr
URL: http://mahazu.hazu.hr/DepMPCS/indexJP.html
volume 5, issue 3, article 74,
2004.
Received 20 January, 2004;
accepted 05 April, 2004.
Communicated by: C. Giordano
Abstract
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c
2000
Victoria University
ISSN (electronic): 1443-5756
018-04
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Abstract
Some Hadamard-type inequalities involving the product of two convex functions
are obtained. Our results generalize the corresponding results of B.G.Pachpatte.
2000 Mathematics Subject Classification: 26D15, 26D20.
Key words: Integral inequalities, Hadamard’s inequality
Note on Some Hadamard-Type
Inequalities
Contents
1
Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
2
Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
References
3
6
M. Klaričić Bakula and J. Pečarić
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1.
Introduction
Let f be a convex function on [a, b] ⊂ R. The following double inequality:
Z b
a+b
1
f (a) + f (b)
(1.1)
f
≤
f (x)dx ≤
2
b−a a
2
is known in the literature as Hadamard’s inequality [1, p. 137], [2, p. 10] for
convex functions.
Recently B.G.Pachpatte [3] considered some new integral inequalities, analogous to that of Hadamard, involving the product of two convex functions. In
[3] the following theorem has been proved:
Note on Some Hadamard-Type
Inequalities
M. Klaričić Bakula and J. Pečarić
Theorem 1.1. Let f and g be nonnegative, convex functions on [a, b] ⊂ R. Then
Title Page
(i)
(1.2)
1
b−a
Z
a
b
1
1
f (x) g (x) dx ≤ M (a, b) + N (a, b) ,
3
6
(ii)
(1.3) 2f
a+b
2
a+b
g
2
Z b
1
1
1
≤
f (x) g (x) dx + M (a, b) + N (a, b) ,
b−a a
6
3
where M (a, b) = f (a) g (a) + f (b) g (b) and N (a, b) = f (a) g (b) +
f (b) g (a). Inequalities (1.2) and (1.3) are sharp in the sense that equalities hold for some f (x) and g (x) on [a, b].
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In the following Theorem 1.2 we give a variant of the corresponding Theorem 2 in [3].
Theorem 1.2. Let f and g be nonnegative, convex functions on [a, b] ⊂ R. Then
(i)
3
(1.4)
2 (b − a)2
Z bZ bZ
1
f (tx + (1 − t) y) g (tx + (1 − t) y) dtdxdy
a
a
0
1
≤
b−a
Z
b
f (x) g (x) dx +
a
1
[M (a, b) + N (a, b)] ;
8
Note on Some Hadamard-Type
Inequalities
M. Klaričić Bakula and J. Pečarić
(ii)
3
(1.5)
b−a
Z bZ
a
1
a+b
f tx + (1 − t)
2
0
a+b
× g tx + (1 − t)
dtdx
2
Z b
1
1
≤
f (x) g (x) dx + [M (a, b) + N (a, b)] ,
b−a a
2
where M (a, b) and N (a, b) are as in Theorem 1.1.
It should be noted that in [3, Theorem 2] inequalities (3) and (4) are established. Inequality (3) from [3, Theorem 2] is a variant of our inequality (1.4) in
which
1 M (a, b) + N (a, b)
8
(b − a)2
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stands in place of the term 18 [M (a, b) + N (a, b)]. Analogously, inequality (4)
from [3, Theorem 2] is a variant of our inequality (1.5) in which
1 1+b−a
[M (a, b) + N (a, b)]
4
b−a
stands in place of the term 21 [M (a, b) + N (a, b)] .
However, one can compare inequalities (3) and (4) with (1.4) and (1.5), respectively, to find out that estimates given by (1.4) and (1.5) are better (worse)
than those given by (3) and (4) in [3, Theorem 2] in case of b−a < 1 (b − a > 1).
But on careful inspection of the proof in [3, Theorem 2], the reader can
notice some errors in Pachpatte’s calculation, so inequalities (3) and (4) in [3,
Theorem 2] are in fact incorrect.
The aim of this paper is to prove some simple generalizations of Theorem
1.1 and Theorem 1.2, which additionally involve weight functions and also nonlinear transformations of the base interval [a, b]. Those generalizations are established in Theorem 2.1 and Theorem 2.2. The above cited Theorem 1.1 is a
special case of Theorem 2.1, while the above Theorem 1.2 is a special case of
our Theorem 2.2 (see Remark 2.2).
Note on Some Hadamard-Type
Inequalities
M. Klaričić Bakula and J. Pečarić
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2.
Results
Throughout the rest of the paper we shall use the following notation
[h; x, y] =
h (y) − h (x)
,
y−x
x 6= y
e
h (t) = th (α + β − t) ,
b
h (t) = th (t)
where h : [α, β] → R is a function, [α, β] ⊂ R, x, y, t ∈ [α, β]. Note that from
the above equalities we get
h
i βh (α) − αh (β)
e
,
h; α, β =
β−α
h
i βh (β) − αh (α)
b
h; α, β =
,
β−α
and, by simple calculation,
h
i h
i
b
(2.1)
h; α, β − e
h; α, β = (α + β) [h; α, β] .
Note on Some Hadamard-Type
Inequalities
M. Klaričić Bakula and J. Pečarić
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The following results are valid:
Theorem 2.1. Let f be a nonnegative convex function on [m1 , M1 ], g a nonnegative convex function on [m2 , M2 ], u : [a, b] → [m1 , M1 ] and v : [a, b] →
[m2 , M2 ] continuous functions, and p : [a, b] → R a positive integrable function. Then
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(i)
1
(2.2)
P
b
Z
p (x) f (u (x)) g (v (x)) dx
a
1
≤ [f ; m1 , M1 ] [g; m2 , M2 ]
P
Z
b
p (x) u (x) v (x) dx
Z
1 b
+ [f ; m1 , M1 ] [e
g ; m2 , M2 ]
p (x) u (x) dx
P a
Z b
h
i
1
+ fe; m1 , M1 [g; m2 , M2 ]
p (x) v (x) dx
P a
h
i
e
+ f ; m1 , M1 [e
g ; m2 , M2 ] .
a
Note on Some Hadamard-Type
Inequalities
M. Klaričić Bakula and J. Pečarić
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(ii)
Contents
m 2 + M2
m 1 + M1
(2.3) f
g
2
2
Z b
1
≤
p (x) f (u (x)) g (v (x)) dx
4P a
Z b
+
p (x) f (M1 + m1 − u (x)) g (M2 + m2 − v (x)) dx
a
Z b
1
+
−2 [f ; m1 , M1 ] [g; m2 , M2 ]
p (x) u (x) v (x) dx
4P
a
Z b
+ ([b
g ; m2 , M2 ] − [e
g ; m2 , M2 ]) [f ; m1 , M1 ]
p (x) u (x) dx
a
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Z b
i h
i
b
e
+ f ; m1 , M1 − f ; m1 , M1 [g; m2 , M2 ]
p (x) v (x) dx
a
i
h
i
1 h e
f ; m1 , M1 [b
g ; m2 , M2 ] + fb; m1 , M1 [e
g ; m2 , M2 ] ,
+
4
Rb
where P = a p (x) dx.
h
Proof. For any x ∈ [a, b] we can write
M1 − u (x)
u (x) − m1
u (x) =
m1 +
M1
M1 − m 1
M1 − m 1
(2.4)
Note on Some Hadamard-Type
Inequalities
M. Klaričić Bakula and J. Pečarić
and
v (x) =
(2.5)
M2 − v (x)
v (x) − m2
m2 +
M2 .
M2 − m 2
M2 − m 2
Since f and g are convex functions we have
M1 − u (x)
u (x) − m1
f (m1 ) +
f (M1 )
M1 − m 1
M1 − m 1
M1 f (m1 ) − m1 f (M1 )
u (x)
(f (M1 ) − f (m1 )) +
=
M1 − m 1
M1 − m 1
h
i
= [f ; m1 , M1 ] u (x) + fe; m1 , M1
f (u (x)) ≤
(2.6)
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and
M2 − v (x)
v (x) − m2
g (m2 ) +
g (M2 )
M2 − m 2
M2 − m 2
v (x)
M2 g (m2 ) − m2 g (M2 )
=
(g (M2 ) − g (m2 )) +
M2 − m 2
M2 − m 2
= [g; m2 , M2 ] v (x) + [e
g ; m2 , M2 ] .
g (v (x)) ≤
(2.7)
Functions f and g are nonnegative by assumption, so after multiplying (2.6)
and (2.7) we obtain
(2.8) f (u (x)) g (v (x))
≤ [f ; m1 , M1 ] [g; m2 , M2 ] u (x) v (x) + [f ; m1 , M1 ] [e
g ; m2 , M2 ] u (x)
h
i
h
i
+ [g; m2 , M2 ] fe; m1 , M1 v (x) + fe; m1 , M1 [e
g ; m2 , M2 ] .
Now, multiplying (2.8) by weight p (x), integrating over [a, b] and dividing by
P > 0 we get (i).
To obtain (ii) we can write
m 1 + M1
1 M1 − u (x)
u (x) − m1
=
m1 +
M1
2
2 M1 − m 1
M1 − m 1
u (x) − m1
M1 − u (x)
+
m1 +
M1 ,
M1 − m 1
M1 − m 1
m 2 + M2
1 M2 − v (x)
v (x) − m2
=
m2 +
M2
2
2 M2 − m 2
M2 − m 2
M2 − v (x)
v (x) − m2
+
m2 +
M2 .
M2 − m 2
M2 − m 2
Note on Some Hadamard-Type
Inequalities
M. Klaričić Bakula and J. Pečarić
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Using the Hadamard inequality (1.1) and the convexity of functions f and g,
we get
f
m 2 + M2
g
2
M1 − u (x)
u (x) − m1
1
≤
f
m1 +
M1
4
M1 − m 1
M1 − m 1
u (x) − m1
M1 − u (x)
+f
m1 +
M1
M1 − m 1
M1 − m 1
M2 − v (x)
v (x) − m2
× g
m2 +
M2
M2 − m 2
M2 − m 2
v (x) − m2
M2 − v (x)
+g
m2 +
M2 .
M2 − m 2
M2 − m 2
m 1 + M1
2
According to (2.4) and (2.5) , after some simple calculus we obtain
(2.9) f
m 2 + M2
m 1 + M1
g
2
2
1
≤ [f (u (x)) g (v (x)) + f (M1 + m1 − u (x)) g (M2 + m2 − v (x))]
4
1
M1 − u (x)
u (x) − m1
+
f
m1 +
M1
4
M1 − m 1
M1 − m 1
v (x) − m2
M2 − v (x)
×g
m2 +
M2
M2 − m 2
M2 − m 2
Note on Some Hadamard-Type
Inequalities
M. Klaričić Bakula and J. Pečarić
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+f
u (x) − m1
M1 − u (x)
m1 +
M1
M1 − m 1
M1 − m 1
M2 − v (x)
v (x) − m2
×g
m2 +
M2 .
M2 − m 2
M2 − m 2
Using the convexity of functions f and g, from inequality (2.9) we get
(2.10) f
≤
m 1 + M1
2
m 2 + M2
g
2
1
[f (u (x)) g (v (x)) + f (M1 + m1 − u (x)) g (M2 + m2 − v (x))]
4
M1 − u (x)
u (x) − m1
1
+
f (m1 ) +
f (M1 )
4
M1 − m 1
M1 − m 1
v (x) − m2
M2 − v (x)
×
g (m2 ) +
g (M2 )
M2 − m 2
M2 − m 2
u (x) − m1
M1 − u (x)
+
f (m1 ) +
f (M1 )
M1 − m 1
M1 − m 1
M2 − v (x)
v (x) − m2
×
g (m2 ) +
g (M2 ) .
M2 − m 2
M2 − m 2
With respect to the notation introduced at the beginning of this section, inequality (2.10) becomes
m 1 + M1
m 2 + M2
g
f
2
2
Note on Some Hadamard-Type
Inequalities
M. Klaričić Bakula and J. Pečarić
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1n
≤
f (u (x)) g (v (x))
4
o
+ f (M1 + m1 − u (x)) g (M2 + m2 − v (x))
(
h
i
1 +
[f ; m1 , M1 ] u (x) + fe; m1 , M1
4
× [b
g ; m2 , M2 ] − [g; m2 , M2 ] v (x)
h
i
b
+ f ; m1 , M1 − [f ; m1 , M1 ] u (x)
)
× [g; m2 , M2 ] v (x) + [e
g ; m2 , M2 ]
Note on Some Hadamard-Type
Inequalities
M. Klaričić Bakula and J. Pečarić
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1n
=
f (u (x)) g (v (x))
4
o
+ f (M1 + m1 − u (x)) g (M2 + m2 − v (x))
(
1
+
− 2 [f ; m1 , M1 ] [g; m2 , M2 ] u (x) v (x)
4
+ [b
g ; m2 , M2 ] − [e
g ; m2 , M2 ] [f ; m1 , M1 ] u (x)
h
i h
i
b
e
+ f ; m1 , M1 − f ; m1 , M1 [g; m2 , M2 ] v (x)
(2.11)
h
i
h
i
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+ fe; m1 , M1 [b
g ; m2 , M2 ] + fb; m1 , M1 [e
g ; m2 , M2 ]
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Now we multiply both sides of (2.11) by p (x), integrate over [a, b] and
divide by P . We thus obtain (ii) and the proof is completed.
Remark 2.1. Pachpatte’s results (1.2) and (1.3) can be obtained from (2.2)
and (2.3) respectively if we put p (x) = 1, u (x) = v (x) = x for all x ∈ [a, b]
(then we have m1 = m2 = a and M1 = M2 = b). In the case of g (x) ≡ 1
inequality (i) becomes the right side of Hadamard’s inequality (1.1) .
Theorem 2.2. Let f be a nonnegative convex function on [m1 , M1 ], g a nonnegative convex function on [m2 , M2 ], u : [a, b] → [m1 , M1 ] and v : [a, b] →
[m2 , M2 ] continuous functions, and p, q : [a, b] → R positive integrable functions. Then
(i)
Note on Some Hadamard-Type
Inequalities
M. Klaričić Bakula and J. Pečarić
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1
PQ
Z bZ bZ
1
p (x) q (y) f (tu (x) + (1 − t) u (y))
a
a
0
× g (tv (x) + (1 − t) v (y)) dtdxdy
Z b
1
Q
≤
f (u (x)) g (v (x)) p (x) dx
3P Q
a
Z b
+P
f (u (y)) g (v (y)) q (y) dy
a
Z b
Z b
1
+
p (x) f (u (x)) dx
q (y) g (v (y)) dy;
3P Q a
a
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(ii)
1
P
Z bZ
1
p (x) f (tu (x) + (1 − t) u) g (tv (x) + (1 − t) v) dtdx
Z b
1
1
≤
p (x) f (u (x)) g (v (x)) dx + f (u) g (v)
3P a
3
Z b
Z b
1
+
g (v)
p (x) f (u (x)) dx + f (u)
p (x) g (v (x)) dx ,
6P
a
a
Rb
Rb
where u = P1 a p (x) u (x) dx, v = Q1 a q (x) v (x) dx.
a
0
Proof. Since f and g are convex functions, for t ∈ [0, 1] we have
f (tu (x) + (1 − t) u (y)) ≤ tf (u (x)) + (1 − t) f (u (y))
g (tv (x) + (1 − t) v (y)) ≤ tg (v (x)) + (1 − t) g (v (y)) .
(2.12)
(2.13)
Functions f and g are nonnegative, so multiplying (2.12) and (2.13) we get
(2.14) f (tu (x) + (1 − t) u (y)) g (tv (x) + (1 − t) v (y))
≤ t2 f (u (x)) g (v (x)) + (1 − t)2 f (u (y)) g (v (y))
+ t (1 − t) [f (u (x)) g (v (y)) + f (u (y)) g (v (x))] .
Integrating (2.14) over [0, 1] we obtain
Z 1
(2.15)
f (tu (x) + (1 − t) u (y)) g (tv (x) + (1 − t) v (y)) dt
Note on Some Hadamard-Type
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0
≤
1
[f (u (x)) g (v (x)) + f (u (y)) g (v (y))]
3
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+
1
[f (u (x)) g (v (y)) + f (u (y)) g (v (x))] .
6
Now we multiply (2.15) by p (x) q (y), integrate over [a, b] × [a, b] and divide
by P Q, where
Z b
Z b
P =
p (x) dx, Q =
q (x) dx,
a
a
so we get
Z bZ bZ 1
1
p (x) q (y) f (tu (x) + (1 − t) u (y))
PQ a a 0
× g (tv (x) + (1 − t) v (y)) dtdxdy
Z b
Z b
1
≤
p (x) f (u (x)) g (v (x)) dx
q (y) dy
3P Q a
a
Z b
Z b
+
q (y) f (u (y)) g (v (y)) dy
p (x) dx
a
a
Z b
Z b
1
p (x) f (u (x)) dx
+
q (y) g (v (y)) dy
6P Q a
a
Z b
Z b
+
p (y) f (u (y)) dy
q (x) g (v (x)) dx
a
a
Z b
Z b
1
=
p (x) f (u (x)) g (v (x)) dx
q (y) dy
3P Q a
a
Note on Some Hadamard-Type
Inequalities
M. Klaričić Bakula and J. Pečarić
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Z
b
Z
b
+
q (y) f (u (y)) g (v (y)) dy
p (x) dx
a
Z b
Z b
1
+
p (x) f (u (x)) dx
q (y) g (v (y)) dy.
3P Q a
a
a
(2.16)
This is the desired inequality (i).
To prove inequality (ii), in (2.12) and (2.13) we substitute u (y) and v (y)
with u and v respectively.
Then we obtain
(2.17) f (tu (x) + (1 − t) u) g (tv (x) + (1 − t) v)
≤ t2 f (u (x)) g (v (x)) + (1 − t)2 f (u) g (v)
+ t (1 − t) [f (u (x)) g (v) + f (u) g (v (x))] .
Note on Some Hadamard-Type
Inequalities
M. Klaričić Bakula and J. Pečarić
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Integrating (2.17) in respect to t over [0, 1] we obtain
Z
1
f (tu (x) + (1 − t) u) g (tv (x) + (1 − t) v) dt
(2.18)
0
1
≤ [f (u (x)) g (v (x)) + f (u) g (v)]
3
1
+ [f (u (x)) g (v) + f (u) g (v (x))] .
6
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Similarly as before, from (2.18) we get
1
P
Z bZ
1
p (x) f (tu (x) + (1 − t) u) g (tv (x) + (1 − t) v) dtdx
Z b
1
1
≤
p (x) f (u (x)) g (v (x)) dx + f (u) g (v)
3P a
3
Z b
Z b
1
+
g (v)
p (x) f (u (x)) dx + f (u)
p (x) g (v (x)) dx .
6P
a
a
a
0
This completes the proof.
Note on Some Hadamard-Type
Inequalities
M. Klaričić Bakula and J. Pečarić
Remark 2.2. If in (i) we put u (x) = v (x) = x for all x ∈ [a, b], it becomes
1
(2.19)
PQ
Z bZ bZ
a
a
1
≤
Q
3P Q
0
Z b
a
Title Page
1
p (x) q (y) f (tx + (1 − t) y) g (tx + (1 − t) y) dtdxdy
Z b
p (x) f (x) g (x) dx + P
q (y) f (y) g (y) dy
a
Z b
Z b
1
+
p (x) f (x) dx
q (y) g (y) dy,
3P Q a
a
so by using a generalization of Hadamard’s inequality [1, p.138]
Z
a+b
1 b
f (a) + f (b)
≤
p (x) f (x) dx ≤
(2.20)
f
2
P a
2
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which holds for p (a + t) = p (b − t) , 0 ≤ t ≤ 12 (a + b) , we obtain from (2.19)
the following inequality
Z bZ bZ 1
1
p (x) q (y) f (tx + (1 − t) y) g (tx + (1 − t) y) dtdxdy
PQ a a 0
Z b
Z b
1
≤
Q
p (x) f (x) g (x) dx + P
q (y) f (y) g (y) dy
3P Q
a
a
1 f (a) + f (b) g (a) + g (b)
+
3
2
2
Z b
Z b
1
=
Q
p (x) f (x) g (x) dx + P
q (y) f (y) g (y) dy
3P Q
a
a
1
(2.21)
+
[M (a, b) + N (a, b)] .
12
Now it is easy to observe that if p (x) = q (x) = 1 for all x ∈ [a, b] inequality
(2.21) becomes the corrected Pachpatte’s result (1.4).
If we do the same in (ii) we get
1
P
Z bZ
1
a+b
g tx + (1 − t)
dtdx
p (x) f
2
a
0
Z b
1
1
a+b
a+b
≤
p (x) f (x) g (x) dx + f
g
3P a
3
2
2
Z b
Z b
1
a+b
a+b
+
g
p (x) f (x) dx + f
p (x) g (x) dx .
6P
2
2
a
a
a+b
tx + (1 − t)
2
Note on Some Hadamard-Type
Inequalities
M. Klaričić Bakula and J. Pečarić
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Using again (2.20) we obtain
Z Z
1 b 1
a+b
a+b
p (x) f tx + (1 − t)
g tx + (1 − t)
dtdx
P a 0
2
2
Z b
1
1 f (a) + f (b) g (a) + g (b)
≤
p (x) f (x) g (x) dx +
3P a
3
2
2
1 g (a) + g (b) f (a) + f (b) f (a) + f (b) g (a) + g (b)
+
+
6
2
2
2
2
Z b
1
1
=
p (x) f (x) g (x) dx + (M (a, b) + N (a, b))
3P a
6
Note on Some Hadamard-Type
Inequalities
M. Klaričić Bakula and J. Pečarić
Furthermore, in the case p (x) = 1 for all x ∈ [a, b] we get
Title Page
1
b−a
Z bZ
a
0
1
a+b
a+b
f tx + (1 − t)
g tx + (1 − t)
dtdx
2
2
Z b
1
1
≤
f (x) g (x) dx + (M (a, b) + N (a, b)) ,
3 (b − a) a
6
which is the corrected Pachpatte’s result (1.5).
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References
[1] J.E. PEČARIĆ, F. PROSCHAN AND Y.L. TONG, Convex Functions, Partial Orderings, and Statistical Applications, Academic Press, Inc. (1992).
[2] D.S. MITRINOVIĆ, J.E. PEČARIĆ AND A.M. FINK, Classical and New
Inequalities in Analysis, Kluwer Academic Publishers, 1993.
[3] B.G. PACHPATTE, On some inequalities for convex functions, RGMIA
Res. Rep. Coll., 6(E) (2003). [ONLINE http://rgmia.vu.edu.au/
v6(E).html].
Note on Some Hadamard-Type
Inequalities
M. Klaričić Bakula and J. Pečarić
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