Hindawi Publishing Corporation
Journal of Inequalities and Applications
Volume 2007, Article ID 72173, 9 pages
doi:10.1155/2007/72173
Research Article
Improvement of Aczél’s Inequality and Popoviciu’s Inequality
Shanhe Wu
Received 30 December 2006; Accepted 24 April 2007
Recommended by Laszlo I. Losonczi
We generalize and sharpen Acz
´
el’s inequality and Popoviciu’s inequality by means of
two classical inequalities, a unified improvement of Acz
´
el’s inequality and Popoviciu’s
inequality is given. As application, an integral inequality of Acz
´
el-Popoviciu type is es-
tablished.
Copyright © 2007 Shanhe Wu. This is an open access article distributed under the Cre-
ative Commons Attribution License, which permits unrestricted use, distribution, and
reproduction in any medium, provided the original work is properly cited.
1. Introduction
In 1956, Acz
´
el [1] proved the following result:
a
2
1
−
n
i=2
a
2
i
b
2
1
−
n
i=2
b
2
i
≤
a
1
b
1
−
n
i=2
a
i
b
i
2
, (1.1)
where a
i
, b
i
(i=1,2, ,n) are positive numbers such that a
2
1
−
n
i
=2
a
2
i
>0orb
2
1
−
n
i
=2
b
2
i
>
0. This inequality is called Acz
´
el’s inequality.
It is well known that Acz
´
el’s inequality has important applications in the theory of
functional equations in non-Euclidean geometry. In recent years, this inequality has at-
tracted the interest of many mathematicians and has motivated a large number of re-
search papers involving different proofs, various generalizations, improvements, and ap-
plications (see [2–11] and references therein). We state here a brief history on improve-
ment of Acz
´
el’s inequality.
Popoviciu [12] first presented an exponential extension of Acz
´
el’s inequality, as fol-
lows.
2 Journal of Inequalities and Applications
Theorem 1.1. Let p>0, q>0, 1/p+1/q
= 1,andleta
i
, b
i
(i = 1,2, ,n) be positive
numbers such that a
p
1
−
n
i
=2
a
p
i
> 0 and b
q
1
−
n
i
=2
b
q
i
> 0. Then
a
p
1
−
n
i=2
a
p
i
1/p
b
q
1
−
n
i=2
b
q
i
1/q
≤ a
1
b
1
−
n
i=2
a
i
b
i
. (1.2)
Wu and Debnath [13] generalized inequality (1.2) in the following form.
Theorem 1.2. Let p>0, q>0,andleta
i
, b
i
(i = 1,2, ,n) be positive numbers such that
a
p
1
−
n
i
=2
a
p
i
> 0 and b
q
1
−
n
i
=2
b
q
i
> 0. Then
a
p
1
−
n
i=2
a
p
i
1/p
b
q
1
−
n
i=2
b
q
i
1/q
≤ n
1−min{p
−1
+q
−1
,1}
a
1
b
1
−
n
i=2
a
i
b
i
. (1.3)
In a recent paper [14], Wu established a sharp and generalized version of Popoviciu’s
inequality as follows.
Theorem 1.3. Let p>0, q>0, 1/p+1/q
≥ 1,andleta
i
, b
i
(i = 1,2, ,n) be positive
numbers such that a
p
1
−
n
i
=2
a
p
i
> 0 and b
q
1
−
n
i
=2
b
q
i
> 0. Then
a
p
1
−
n
i=2
a
p
i
1/p
b
q
1
−
n
i=2
b
q
i
1/q
≤ a
1
b
1
−
n
i=2
a
i
b
i
−
a
1
b
1
max{p,q,1}
n
i=2
a
p
i
a
p
1
−
b
q
i
b
q
1
2
.
(1.4)
In this paper, we show a new sharp and generalized version of Popoviciu’s inequal-
ity, which is a unified improvement of Acz
´
el’s inequality and Popoviciu’s inequality. In
Section 4, the obtained result will be used to establish an integr al inequality of Acz
´
el-
Popoviciu type.
2. Lemmas
In order to prove the theorem in Section 3, we first introduce the following lemmas.
Lemma 2.1 (generalized H
¨
older inequality [15, page 20]). Let a
ij
> 0, λ
j
≥ 0(i = 1,2, ,
n, j
= 1,2, ,m),andletλ
1
+ λ
2
+ ···+ λ
m
= 1. Then
m
j=1
n
i=1
a
ij
λ
j
≥
n
i=1
m
j=1
a
λ
j
ij
(2.1)
with equality holding if and only if a
11
/a
1j
= a
21
/a
2j
= ··· = a
n1
/a
nj
( j = 2, 3, , m) for
λ
1
λ
2
···λ
n
= 0.
Lemma 2.2 (mean value inequality [16, page 17]). Let x
i
> 0, λ
i
> 0(i = 1,2, ,n) and let
λ
1
+ λ
2
+ ···+ λ
n
= 1. Then
n
i=1
λ
i
x
i
≥
n
i=1
x
λ
i
i
(2.2)
with equality holding if and only if x
1
= x
2
=···=x
n
.
Shanhe Wu 3
Lemma 2.3. Le t p
1
≥ p
2
≥ ··· ≥ p
m
> 0, 1/p
1
+1/p
2
+ ···+1/p
m
= 1, 0 <x
j
< 1(j =
1,2, ,m),andletx
m+1
= x
1
, p
m+1
= p
1
. Then
m
j=1
x
j
+
m
j=1
1 − x
p
j
j
1/p
j
≤ 1 −
1
2p
1
m
j=1
x
p
j
j
− x
p
j+1
j+1
2
(2.3)
with equality holding if and only if x
p
1
1
= x
p
2
2
=···=x
p
m
m
.
Proof. From hypotheses in Lemma 2.3,itiseasytoverifythat
1
p
m
≥
1
p
m−1
≥···≥
1
p
2
≥
1
p
1
> 0,
1
2p
2
−
1
2p
1
≥ 0,
1
2p
3
−
1
2p
2
≥ 0, ,
1
2p
m
−
1
2p
m−1
≥ 0,
1
2p
m
−
1
2p
1
≥ 0,
1
2p
1
+
1
2p
1
+
1
2p
2
−
1
2p
1
+
1
2p
2
+
1
2p
2
+
1
2p
3
−
1
2p
2
+ ···+
1
2p
m−2
+
1
2p
m−2
+
1
2p
m−1
−
1
2p
m−2
+
1
2p
m−1
+
1
2p
m−1
+
1
2p
m
−
1
2p
m−1
+
1
2p
1
+
1
2p
1
+
1
2p
m
−
1
2p
1
=
1
p
1
+
1
p
2
+ ···+
1
p
m
= 1.
(2.4)
Hence, by using Lemma 2.1 we obtain
x
p
1
1
+
1 − x
p
2
2
1/2p
1
x
p
2
2
+
1 − x
p
1
1
1/2p
1
x
p
2
2
+
1 − x
p
2
2
1/2p
2
−1/2p
1
×
x
p
2
2
+
1 − x
p
3
3
1/2p
2
x
p
3
3
+
1 − x
p
2
2
1/2p
2
x
p
3
3
+
1 − x
p
3
3
1/2p
3
−1/2p
2
.
.
.
×
x
p
m−2
m−2
+
1 − x
p
m−1
m−1
1/2p
m−2
×
x
p
m−1
m−1
+
1 − x
p
m−2
m−2
1/2p
m−2
x
p
m−1
m−1
+
1 − x
p
m−1
m−1
1/2p
m−1
−1/2p
m−2
×
x
p
m−1
m−1
+
1 − x
p
m
m
1/2p
m−1
x
p
m
m
+
1 − x
p
m−1
m−1
1/2p
m−1
x
p
m
m
+
1 − x
p
m
m
1/2p
m
−1/2p
m−1
×
x
p
m
m
+
1 − x
p
1
1
1/2p
1
x
p
1
1
+
1 − x
p
m
m
1/2p
1
x
p
m
m
+
1 − x
p
m
m
1/2p
m
−1/2p
1
≥ x
p
1
/2p
1
1
x
p
2
/2p
1
2
x
p
2
/2p
2
−p
2
/2p
1
2
x
p
2
/2p
2
2
···x
p
m−1
/2p
m−2
m−1
x
p
m−1
/2p
m−1
−p
m−1
/2p
m−2
m−1
x
p
m−1
/2p
m−1
m−1
× x
p
m
/2p
m−1
m
x
p
m
/2p
m
−p
m
/2p
m−1
m
x
p
m
/2p
1
m
x
p
m
/2p
m
−p
m
/2p
1
m
x
p
1
/2p
1
1
+
1 − x
p
1
1
1/2p
1
1 − x
p
2
2
1/2p
1
1 − x
p
2
2
1/2p
2
−1/2p
1
1 − x
p
2
2
1/2p
2
···
1 − x
p
m−1
m−1
1/2p
m−2
1 − x
p
m−1
m−1
1/2p
m−1
−1/2p
m−2
1 − x
p
m−1
m−1
1/2p
m−1
×
1 − x
p
m
m
1/2p
m−1
1 − x
p
m
m
1/2p
m
−1/2p
m−1
1 − x
p
m
m
1/2p
1
1 − x
p
m
m
1/2p
m
−1/2p
1
×
1 − x
p
1
1
1/2p
1
,
(2.5)
4 Journal of Inequalities and Applications
which is equivalent to
1 −
x
p
1
1
− x
p
2
2
2
1/2p
1
1 −
x
p
2
2
− x
p
3
3
2
1/2p
2
···
1 − (x
p
m−1
m−1
− x
p
m
m
2
1/2p
m−1
1 −
x
p
m
m
− x
p
1
1
2
1/2p
1
≥ x
1
x
2
···x
m
+
1 − x
p
1
1
1/p
1
1 − x
p
2
2
1/p
2
···
1 − x
p
m
m
1/p
m
.
(2.6)
On the other hand, it follows from Lemma 2.2 that
1
2p
1
1 −
x
p
1
1
− x
p
2
2
2
+
1
2p
2
1 −
x
p
2
2
− x
p
3
3
2
+ ···+
1
2p
m−1
1 −
x
p
m−1
m−1
− x
p
m
m
2
+
1
2p
1
1 −
x
p
m
m
− x
p
1
1
2
+
1
2p
2
+
1
2p
3
+ ···+
1
2p
m−1
+
1
p
m
·
1
≥
1 −
x
p
1
1
− x
p
2
2
2
1/2p
1
1 −
x
p
2
2
− x
p
3
3
2
1/2p
2
···
1 −
x
p
m−1
m−1
− x
p
m
m
2
1/2p
m−1
1 −
x
p
m
m
− x
p
1
1
2
1/2p
1
,
(2.7)
this y ields
1−
x
p
1
1
−x
p
2
2
2
1/2p
1
1−
x
p
2
2
−x
p
3
3
2
1/2p
2
···
1−
x
p
m−1
m−1
−x
p
m
m
2
1/2p
m−1
1−
x
p
m
m
−x
p
1
1
2
1/2p
1
≤
1
p
1
+
1
p
2
+ ···+
1
p
m
−
1
2p
1
x
p
1
1
− x
p
2
2
2
−
1
2p
2
x
p
2
2
− x
p
3
3
2
−···−
1
2p
m−1
x
p
m−1
m−1
− x
p
m
m
2
−
1
2p
1
x
p
m
m
− x
p
1
1
2
≤ 1 −
1
2p
1
x
p
1
1
− x
p
2
2
2
+
x
p
2
2
− x
p
3
3
2
+ ···+
x
p
m−1
m−1
+ x
p
m
m
2
+
x
p
m
m
− x
p
1
1
2
.
(2.8)
Combining inequalities (2.6)and(2.8) leads to inequality (2.3). In addition, from
Lemmas 2.1 and 2.2, we can easily deduce that the equality holds in both (2.6)and(2.8)
if and only if x
p
1
1
= x
p
2
2
=···=x
p
m
m
, and thus we obtain the condition of equality in (2.3).
The proof of Lemma 2.3 is complete.
3. Improvement of Acz
´
el’s inequality and Popov i ciu’s inequality
Theorem 3.1. Let p
1
≥ p
2
≥ ··· ≥ p
m
> 0, 1/p
1
+1/p
2
+ ···+1/p
m
= 1, a
ij
> 0, a
p
j
1j
−
n
i
=2
a
p
j
ij
> 0(i = 1, 2, , n, j = 1, 2, , m),andletp
m+1
= p
1
, a
im+1
= a
i1
(i = 1,2, ,n).
Then one has the following inequality:
m
j=1
a
p
j
1j
−
n
i=2
a
p
j
ij
1/p
j
≤
m
j=1
a
1j
−
n
i=2
m
j=1
a
ij
−
a
11
a
12
···a
1m
2p
1
m
j=1
n
i=2
a
p
j
ij
a
p
j
1j
−
a
p
j+1
ij+1
a
p
j+1
1j+1
2
.
(3.1)
Equality holds in (3.1)ifandonlyifa
p
1
11
/a
p
j
1j
= a
p
1
21
/a
p
j
2j
=···=a
p
1
n1
/a
p
j
nj
( j = 2,3, ,m).
Shanhe Wu 5
Proof. Since by hypotheses in Theorem 3.1 we have
0 <
a
p
j
1j
−
n
i
=2
a
p
j
ij
1/p
j
a
p
j
1j
1/p
j
< 1(j = 1,2, ,m), (3.2)
it follows from Lemma 2.3, with a substitution x
j
= (a
p
j
1j
−
n
i
=2
a
p
j
ij
)
1/p
j
/(a
p
j
1j
)
1/p
j
( j =
1,2, ,m)in(2.3), that
m
j=1
a
p
j
1j
−
n
i
=2
a
p
j
ij
a
p
j
1j
1/p
j
+
m
j=1
n
i
=2
a
p
j
ij
a
p
j
1j
1/p
j
≤ 1 −
1
2p
1
m
j=1
a
p
j
1j
−
n
i
=2
a
p
j
ij
a
p
j
1j
−
a
p
j+1
1j+1
−
n
i
=2
a
p
j+1
ij+1
a
p
j+1
1j+1
2
,
(3.3)
which is equivalent to
m
j=1
a
p
j
1j
−
n
i=2
a
p
j
ij
1/p
j
≤
m
j=1
a
1j
−
m
j=1
n
i=2
a
p
j
ij
1/p
j
−
a
11
a
12
···a
1m
2p
1
m
j=1
n
i=2
a
p
j
ij
a
p
j
1j
−
a
p
j+1
ij+1
a
p
j+1
1j+1
2
,
(3.4)
where equality holds if and only if (
n
i
=2
a
p
j
ij
)/a
p
j
1j
= (
n
i
=2
a
p
j+1
ij+1
)/a
p
j+1
1j+1
( j = 1,2, ,m), that
is, if and only if a
p
1
11
/a
p
j
1j
= (
n
i
=2
a
p
1
i1
)/(
n
i
=2
a
p
j
ij
)(j = 2, 3, ,m).
On the other hand, using Lemma 2.1 gives
m
j=1
n
i=2
a
p
j
ij
1/p
j
≥
n
i=2
m
j=1
a
ij
, (3.5)
where equality holds if and only if a
p
1
21
/a
p
j
2j
= a
p
1
31
/a
p
j
3j
=···=a
p
1
n1
/a
p
j
nj
( j = 2,3, ,m).
Combining inequalities (3.4)and(3.5) leads to the desired inequality (3.1). By means
of the conditions of equality in (3.4)and(3.5), it is easy to conclude that there is equality
in (3.1)ifandonlyifa
p
1
11
/a
p
j
1j
= a
p
1
21
/a
p
j
2j
= ··· = a
p
1
n1
/a
p
j
nj
( j = 2,3, ,m). This completes
the proof of Theorem 3.1.
As a consequence of Theorem 3.1,puttingm = 2, p
1
= p, p
2
= q, a
i1
= a
i
, a
i2
= b
i
(i =
1,2, ,n)in(3.1), we get the fol low ing.
Corollary 3.2. Let p
≥ q>0, 1/p+1/q = 1,andleta
i
, b
i
(i = 1,2, ,n) be positive num-
bers such that a
p
1
−
n
i
=2
a
p
i
> 0 and b
q
1
−
n
i
=2
b
q
i
> 0. Then
a
p
1
−
n
i=2
a
p
i
1/p
b
q
1
−
n
i=2
b
q
i
1/q
≤ a
1
b
1
−
n
i=2
a
i
b
i
−
a
1
b
1
p
n
i=2
a
p
i
a
p
1
−
b
q
i
b
q
1
2
(3.6)
with equality holding if and only if a
p
1
/b
q
1
= a
p
2
/b
q
2
=···=a
p
n
/b
q
n
.
6 Journal of Inequalities and Applications
A simple application of Corollary 3.2 yields the following sharp version of Popoviciu’s
inequality.
Corollary 3.3. Let p>0, q>0, 1/p+1/q
= 1,andleta
i
, b
i
(i = 1, 2, , n) be positive
numbers such that a
p
1
−
n
i
=2
a
p
i
> 0 and b
q
1
−
n
i
=2
b
q
i
> 0. Then
a
p
1
−
n
i=2
a
p
i
1/p
b
q
1
−
n
i=2
b
q
i
1/q
≤ a
1
b
1
−
n
i=2
a
i
b
i
−
a
1
b
1
max{p,q}
n
i=2
a
p
i
a
p
1
−
b
q
i
b
q
1
2
,
(3.7)
with equality holding if and only if a
p
1
/b
q
1
= a
p
2
/b
q
2
=···=a
p
n
/b
q
n
.
Obviously, inequalities (3.1), (3.6), and (3.7) are the improvement of Acz
´
el’s inequality
and Popoviciu’s inequality.
4. Integral version of Acz
´
el-Popoviciu-type inequality
As application of Theorem 3.1, we establish here an interesting integral inequality of
Acz
´
el-Popoviciu type.
Theorem 4.1. Let p
1
≥ p
2
≥ ··· ≥ p
m
> 0, 1/p
1
+1/p
2
+ ··· +1/p
m
= 1, B
j
> 0(j =
1,2, ,m),let f
j
be positive Riemann integrable functions on [a,b] such that B
p
j
j
−
b
a
f
p
j
j
(x) dx > 0 for all j = 1,2, ,m,andletB
m+1
= B
1
, p
m+1
= p
1
, f
m+1
= f
1
.Thenone
has the following inequality:
m
j=1
B
p
j
j
−
b
a
f
p
j
j
(x)dx
1/p
j
≤
m
j=1
B
j
−
b
a
m
j=1
f
j
(x)
dx −
B
1
B
2
···B
m
2p
1
m
j=1
b
a
f
p
j
j
(x)
B
p
j
j
−
f
p
j+1
j+1
(x)
B
p
j+1
j+1
dx
2
.
(4.1)
Proof. For any positive integer n, we choose an equidistant partition of [a,b]as
a<a+
b
− a
n
<
···<a+
b
− a
n
i<
···<a+
b
− a
n
(n
− 1) <b,
Δx
i
=
b − a
n
, i
= 1,2, ,n.
(4.2)
Since the hypothesis B
p
j
j
−
b
a
f
p
j
j
(x) dx > 0(j = 1,2, ,m) implies that
B
p
j
j
− lim
n→∞
n
i=1
f
p
j
j
a +
i(b
− a)
n
b − a
n
> 0(j
= 1,2, ,m), (4.3)
there exists a positive integer N such that
B
p
j
j
−
n
i=1
f
p
j
j
a +
i(b
− a)
n
b − a
n
> 0
∀n>N, j = 1,2, ,m. (4.4)
Shanhe Wu 7
Applying Theorem 3.1, one obtains for any n>N the following inequalit y:
m
j=1
B
p
j
j
−
n
i=1
f
p
j
j
a +
i(b
− a)
n
b − a
n
1/p
j
≤
m
j=1
B
j
−
n
i=1
m
j=1
f
j
a +
i(b
− a)
n
b − a
n
1/p
1
+1/p
2
+···+1/p
m
−
B
1
B
2
···B
m
2p
1
m
j=1
n
i=1
1
B
p
j
j
f
p
j
j
a +
i(b
− a)
n
b − a
n
−
1
B
p
j+1
j+1
f
p
j+1
j+1
a +
i(b
− a)
n
b − a
n
2
.
(4.5)
Note that 1/p
1
+1/p
2
+ ···+1/p
m
= 1, the above inequality can be transformed to
m
j=1
B
p
j
j
−
n
i=1
f
p
j
j
a +
i(b
− a)
n
b − a
n
1/p
j
≤
m
j=1
B
j
−
n
i=1
m
j=1
f
j
a +
i(b
− a)
n
b − a
n
−
B
1
B
2
···B
m
2p
1
m
j=1
n
i=1
1
B
p
j
j
f
p
j
j
a +
i(b
− a)
n
−
1
B
p
j+1
j+1
f
p
j+1
j+1
a +
i(b
− a)
n
b − a
n
2
,
(4.6)
where equality holds if and only if f
p
j
j
(a + i(b − a)/n)/B
p
j
j
= f
p
j+1
j+1
(a + i(b − a)/n)/B
p
j+1
j+1
for
all i
= 1,2, ,n ( j = 1,2, ,m).
In view of the hypotheses that f
j
are positive Riemann integ rable functions on [a,b]
and p
j
> 0(j = 1,2, ,m), we conclude that
m
j
=1
f
j
and f
p
j
j
( j = 1,2, ,m)arealso
integrable on [a,b]. Passing the limit as n
→∞in both sides of inequality (4.6), we obtain
the inequality (4.1). The proof of Theorem 4.1 is complete.
Remark 4.2. Motivated by the proof of Theorem 4.1, we propose here a conjecture.
Conjecture 4.3. Suppose that p
1
≥ p
2
≥··· ≥ p
m
> 0, 1/p
1
+1/p
2
+ ···+1/p
m
= 1, B
j
>
0(j
= 1,2, ,m), suppose also that f
j
∈ L
p
j
[a,b], B
p
j
j
−
b
a
| f
j
(x)|
p
j
dx > 0forallj =
1,2, ,m,letB
m+1
= B
1
, p
m+1
= p
1
, f
m+1
= f
1
. Then the following inequality holds true:
m
j=1
B
p
j
j
−
b
a
f
j
(x)
p
j
dx
1/p
j
≤
m
j=1
B
j
−
b
a
m
j=1
f
j
(x)
dx−
B
1
B
2
···B
m
2p
1
m
j=1
b
a
f
j
(x)
p
j
B
p
j
j
−
f
j+1
(x)
p
j+1
B
p
j+1
j+1
dx
2
(4.7)
8 Journal of Inequalities and Applications
with equality holding if and only if
| f
j
(x)|
p
j
/B
p
j
j
=|f
j+1
(x)|
p
j+1
/B
p
j+1
j+1
( j = 1,2, ,m)al-
most everywhere on [a,b].
As a consequence of Theorem 4.1,puttingm
= 2, p
1
= p, p
2
= q, B
1
= A, B
2
= B, f
1
=
f , f
2
= g in (4.1), we obtain the following.
Corollary 4.4. Let p
≥ q>0, 1/p+1/q = 1, A>0, B>0,andlet f , g be positive Riemann
integrable functions on [a,b] such that A
p
−
b
a
f
p
(x)dx > 0 and B
q
−
b
a
g
q
(x)dx > 0. Then
A
p
−
b
a
f
p
(x) dx
1/p
B
q
−
b
a
g
q
(x) dx
1/q
≤ AB −
b
a
f (x)g(x)dx −
AB
p
b
a
f
p
(x)
A
p
−
g
q
(x)
B
q
dx
2
.
(4.8)
Further , from Corollary 4.4 we have the following.
Corollary 4.5. Let p>0, q>0, 1/p +1/q
= 1, A>0, B>0,andlet f , g be positive
Riemann integrable functions on [a, b] such that A
p
−
b
a
f
p
(x)dx > 0 and B
q
−
b
a
g
q
(x) dx >
0. Then
A
p
−
b
a
f
p
(x) dx
1/p
B
q
−
b
a
g
q
(x) dx
1/q
≤ AB −
b
a
f (x)g(x)dx −
AB
max{p,q}
b
a
f
p
(x)
A
p
−
g
q
(x)
B
q
dx
2
.
(4.9)
Acknowledgment
The author would like to express hearty thanks to the anonymous referees for valuable
comments on this paper.
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Shanhe Wu: Department of Mathematics, Longyan College, Longyan, Fujian 364012, China
Email address: