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 Aczél’s inequality and Popoviciu’s inequality by means of
two classical inequalities, a unified improvement of Aczél’s inequality and Popoviciu’s
inequality is given. As application, an integral inequality of Aczél-Popoviciu type is established.
Copyright © 2007 Shanhe Wu. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and
reproduction in any medium, provided the original work is properly cited.
1. Introduction
In 1956, Aczél [1] proved the following result:
a21 −
n
i=2
a2i
b12 −
n
i=2
bi2
≤ a1 b1 −
n
2
ai bi
,
(1.1)
i =2
where ai , bi (i = 1,2,...,n) are positive numbers such that a21 − ni=2 a2i >0 or b12 − ni=2 bi2 >
0. This inequality is called Aczél’s inequality.
It is well known that Aczél’s inequality has important applications in the theory of
functional equations in non-Euclidean geometry. In recent years, this inequality has attracted the interest of many mathematicians and has motivated a large number of research papers involving different proofs, various generalizations, improvements, and applications (see [2–11] and references therein). We state here a brief history on improvement of Aczél’s inequality.
Popoviciu [12] first presented an exponential extension of Aczél’s inequality, as follows.
2
Journal of Inequalities and Applications
Theorem 1.1. Let p > 0, q > 0, 1/ p + 1/q = 1, and let ai , bi (i = 1,2,...,n) be positive
p
p
q
q
numbers such that a1 − ni=2 ai > 0 and b1 − ni=2 bi > 0. Then
p
a1
−
n
1/ p p
ai
q
b1
−
i=2
n
1/q
q
bi
≤ a1 b1 −
i=2
n
ai bi .
(1.2)
i =2
Wu and Debnath [13] generalized inequality (1.2) in the following form.
Theorem 1.2. Let p > 0, q > 0, and let ai , bi (i = 1,2,...,n) be positive numbers such that
p
p
q
q
a1 − ni=2 ai > 0 and b1 − ni=2 bi > 0. Then
p
a1 −
n
1/ p p
q
b1 −
ai
i=2
n
1/q
q
≤ n1−min{ p
bi
−1
+q−1 ,1}
i =2
a1 b1 −
n
ai bi .
(1.3)
i =2
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, and let ai , bi (i = 1,2,...,n) be positive
p
p
q
q
numbers such that a1 − ni=2 ai > 0 and b1 − ni=2 bi > 0. Then
p
a1
−
n
1/ p p
ai
i=2
q
b1
−
n
1/q
q
bi
≤ a1 b1 −
i=2
n
ai bi
i =2
p
q
n
ai
bi
a1 b1
−
p − q
max{ p, q,1} i=2 a1 b1
2
.
(1.4)
In this paper, we show a new sharp and generalized version of Popoviciu’s inequality, which is a unified improvement of Aczél’s inequality and Popoviciu’s inequality. In
Section 4, the obtained result will be used to establish an integral inequality of AczélPopoviciu type.
2. Lemmas
In order to prove the theorem in Section 3, we first introduce the following lemmas.
Lemma 2.1 (generalized Hölder inequality [15, page 20]). Let ai j > 0, λ j ≥ 0 (i = 1,2,...,
n, j = 1,2,...,m), and let λ1 + λ2 + · · · + λm = 1. Then
m n
j =1
λ j
ai j
i=1
≥
n m
λj
ai j
(2.1)
i=1 j =1
with equality holding if and only if a11 /a1 j = a21 /a2 j = · · · = an1 /an j ( j = 2,3,...,m) for
λ1 λ2 · · · λn = 0.
Lemma 2.2 (mean value inequality [16, page 17]). Let xi > 0, λi > 0 (i = 1,2,...,n) and let
λ1 + λ2 + · · · + λn = 1. Then
n
i=1
λ i xi ≥
n
xiλi
i =1
with equality holding if and only if x1 = x2 = · · · = xn .
(2.2)
Shanhe Wu 3
Lemma 2.3. Let p1 ≥ p2 ≥ · · · ≥ pm > 0, 1/ p1 + 1/ p2 + · · · + 1/ pm = 1, 0 < x j < 1 ( j =
1,2,...,m), and let xm+1 = x1 , pm+1 = p1 . Then
m
xj +
j =1
m
p j 1/ p j
1 − xj
≤ 1−
j =1
p
m
1 pj
p j+1 2
x j − x j+1
2p1 j =1
p
(2.3)
p
with equality holding if and only if x1 1 = x2 2 = · · · = xmm .
Proof. From hypotheses in Lemma 2.3, it is easy to verify that
1
1
1
1
≥
≥ ··· ≥
≥
> 0,
pm pm−1
p2 p1
1
1
1
1
1
1
1
1
−
≥ 0,
−
≥ 0,...,
−
≥ 0,
−
≥ 0,
2p2 2p1
2p3 2p2
2pm 2pm−1
2pm 2p1
1
1
1
1
1
1
1
1
1
1
+
+
−
+
+
+
−
+ ··· +
+
2p1 2p1
2p2 2p1
2p2 2p2
2p3 2p2
2pm−2 2pm−2
+
=
1
1
1
1
1
1
1
1
1
1
−
+
+
+
−
+
+
+
−
2pm−1 2pm−2 2pm−1 2pm−1 2pm 2pm−1 2p1 2p1
2pm 2p1
1
1
1
+ + ··· +
= 1.
p1 p2
pm
(2.4)
Hence, by using Lemma 2.1 we obtain
p
p
x1 1 + 1 − x2 2
1/2p1
p
p
x2 2 + 1 − x1 1
p2 p3 × x2 + 1 − x3
1/2p2
p
1/2p1
p
p
p
1/2p2 −1/2p1
p
x2 2 + 1 − x2 2
x3 3 + 1 − x2 2
1/2p2
p
x3 3 + 1 − x3 3
1/2p3 −1/2p2
..
.
pm−2 pm−1 1/2pm−2
× xm−2 + 1 − xm−1
1/2pm−1 −1/2pm−2
pm−1 pm−2 1/2pm−2
p
p
× xm−1 + 1 − xm−2
xmm−−11 + 1 − xmm−−11
1/2pm−1 pm pm−1 pm 1/2pm−1
p
p
p 1/2pm −1/2pm−1
× xm−1 + 1 − xm
xmm + 1 − xmm−−11
xm + 1 − xmm
pm p1 1/2p1
p
p 1/2p1
p
p 1/2pm −1/2p1
× xm + 1 − x1
x1 1 + 1 − xmm
xmm + 1 − xmm
p1 /2p1 p2 /2p1 p2 /2p2 − p2 /2p1 p2 /2p2
pm−1 /2pm−2 pm−1 /2pm−1 − pm−1 /2pm−2 pm−1 /2pm−1
x2
x2
x2
· · · xm−1
xm−1
xm−1
≥ x1
pm /2pm−1 pm /2pm − pm /2pm−1 pm /2p1 pm /2pm − pm /2p1 p1 /2p1
× xm
xm
xm
xm
x1
p1 1/2p1 p2 1/2p1 p2 1/2p2 −1/2p1 p 1/2p2
+ 1 − x1
1 − x2
1 − x2
1 − x2 2
1/2pm−1 −1/2pm−2 1/2pm−1
pm−1 1/2pm−2 p
p
· · · 1 − xm−1
1 − xmm−−11
1 − xmm−−11
pm 1/2pm−1 p 1/2pm −1/2pm−1 p 1/2p1 p 1/2pm −1/2p1
× 1 − xm
1 − xmm
1 − xmm
1 − xmm
p1 1/2p1
× 1 − x1
,
(2.5)
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Journal of Inequalities and Applications
which is equivalent to
p
p
1 − x1 1 − x2 2
2
1/2p1
p
p
1 − x2 2 − x3 3
2
1/2p2
p
pm−1
pm 2 1/2pm−1
p 2 1/2p1
· · · 1 − (xm−1 − xm
1 − xmm − x1 1
p1 1/ p1 p 1/ p2
pm 1/ pm
≥ x1 x2 · · · xm + 1 − x1
1 − x2 2
· · · 1 − xm
.
(2.6)
On the other hand, it follows from Lemma 2.2 that
p
p
p
1
1
1
p 2
p 2
p 2
1 − x1 1 − x2 2 +
1 − x2 2 − x3 3 + · · · +
1 − xmm−−11 − xmm
2p1
2p2
2pm−1
p
1
1
1
1
1
p 2
·1
+
1 − xmm − x1 1 +
+
+ ··· +
+
2p1
2p2 2p3
2pm−1 pm
≥ 1−
p
x1 1
··· 1 −
p2 2 1/2p1
− x2
p
xmm−−11
1−
p
x2 2
pm 2 1/2pm−1
− xm
p3 2 1/2p2
− x3
p
p
1 − xmm − x1 1
2
1/2p1
(2.7)
,
this yields
p
p
1− x1 1 −x2 2
2
1/2p1
p
p
1− x2 2 −x3 3
2
pm−1
p m 2
· · · 1− xm−1 − xm
1/2p2
1/2pm−1
p
p
1 − xmm − x1 1
2
1/2p1
1
1
1
1 p1
1 p2
p 2
p 2
+ + ··· +
x − x2 2 −
x − x3 3
−
p1 p2
pm
2p1 1
2p2 2
1 pm−1
1 pm
p 2
p 2
− ··· −
xm−1 − xmm −
xm − x1 1
2pm−1
2p1
p
1 p1
p2 2 p2
p3 2
p 2 p
p 2
≤ 1−
x − x2 + x2 − x3 + · · · + xmm−−11 + xmm + xmm − x1 1 .
2p1 1
≤
(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)
p
p
p
if and only if x1 1 = x2 2 = · · · = xmm , and thus we obtain the condition of equality in (2.3).
The proof of Lemma 2.3 is complete.
3. Improvement of Aczél’s inequality and Popoviciu’s inequality
pj
Theorem 3.1. Let p1 ≥ p2 ≥ · · · ≥ pm > 0, 1/ p1 + 1/ p2 + · · · + 1/ pm = 1, ai j > 0, a1 j −
n
pj
i=2 ai j > 0 (i = 1,2,...,n, j = 1,2,...,m), and let pm+1 = p1 , aim+1 = ai1 (i = 1,2,...,n).
Then one has the following inequality:
m
j =1
pj
a1 j
−
n
i=2
1/ p j
pj
ai j
m
n m
p
p
j
j+1
m
n
ai j+1
a a · · · a1m ai j
≤
a1 j −
ai j − 11 12
−
pj
p j+1
2p1
a1 j+1
j =1
i=2 j =1
j =1 i=2 a1 j
2
.
(3.1)
p
pj
p
pj
p
pj
Equality holds in (3.1) if and only if a111 /a1 j = a211 /a2 j = · · · = an11 /an j ( j = 2,3,...,m).
Shanhe Wu 5
Proof. Since by hypotheses in Theorem 3.1 we have
pj
a1 j
pj
i =2 a i j
−
0<
1/ p j
n
< 1 ( j = 1,2,...,m),
p j 1/ p j
a1 j
pj
it follows from Lemma 2.3, with a substitution x j = (a1 j −
1,2,...,m) in (2.3), that
m a p j − n a p j 1/ p j
1j
i=2 i j
(3.2)
n
p j 1/ p
pj
j /(a )1/ p j
1j
i =2 a i j )
n
p j 1/ p j
i=2 ai j
pj
pj
a1 j
a1 j
j =1
n p j
p j+1
p j+1 2
m pj
a1 j+1 − ni=2 ai j+1
1 a 1 j − i =2 a i j
≤ 1−
−
,
pj
p j+1
2p1 j =1
a1 j
a1 j+1
j =1
+
m
(j =
(3.3)
which is equivalent to
m
pj
n
a1 j −
j =1
1/ p j
pj
ai j
≤
i=2
m
m n
a1 j −
j =1
j =1
1/ p j
pj
−
ai j
i=2
p
p
j
j+1
m
n
ai j+1
a11 a12 · · · a1m ai j
−
pj
p j+1
2p1
a1 j+1
j =1 i=2 a1 j
2
,
(3.4)
pj
pj
p j+1
p j+1
where equality holds if and only if ( ni=2 ai j )/a1 j = ( ni=2 ai j+1 )/a1 j+1 ( j = 1,2,...,m), that
pj
pj
p
p
is, if and only if a111 /a1 j = ( ni=2 ai11 )/( ni=2 ai j ) ( j = 2,3,...,m).
On the other hand, using Lemma 2.1 gives
m n
j =1
1/ p j
pj
ai j
≥
i =2
n m
ai j ,
(3.5)
i=2 j =1
pj
p
pj
p
p
pj
where equality holds if and only if a211 /a2 j = a311 /a3 j = · · · = an11 /an j ( 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
pj
pj
pj
p
p
p
in (3.1) if and only if a111 /a1 j = a211 /a2 j = · · · = an11 /an j ( j = 2,3,...,m). This completes
the proof of Theorem 3.1.
As a consequence of Theorem 3.1, putting m = 2, p1 = p, p2 = q, ai1 = ai , ai2 = bi (i =
1,2,...,n) in (3.1), we get the following.
Corollary 3.2. Let p ≥ q > 0, 1/ p + 1/q = 1, and let ai , bi (i = 1,2,...,n) be positive num
p
p
q
q
bers such that a1 − ni=2 ai > 0 and b1 − ni=2 bi > 0. Then
p
a1
−
n
i=2
1/ p p
ai
q
b1
−
n
1/q
q
bi
≤ a1 b1 −
i=2
n
i =2
p
q
p
q
ai bi
p
q
n
bi
a1 b1 ai
−
−
p i=2 a1p b1q
p
q
with equality holding if and only if a1 /b1 = a2 /b2 = · · · = an /bn .
2
(3.6)
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, and let ai , bi (i = 1,2,...,n) be positive
p
p
q
q
numbers such that a1 − ni=2 ai > 0 and b1 − ni=2 bi > 0. Then
p
a1
−
n
1/ p p
ai
q
b1
−
i=2
n
1/q
q
bi
≤ a1 b1 −
i =2
n
i=2
p
q
p
ai bi
p
q
n
ai
bi
a1 b1
−
p − q
max{ p, q} i=2 a1 b1
q
p
2
,
(3.7)
q
with equality holding if and only if a1 /b1 = a2 /b2 = · · · = an /bn .
Obviously, inequalities (3.1), (3.6), and (3.7) are the improvement of Aczél’s inequality
and Popoviciu’s inequality.
4. Integral version of Aczél-Popoviciu-type inequality
As application of Theorem 3.1, we establish here an interesting integral inequality of
Aczél-Popoviciu type.
Theorem 4.1. Let p1 ≥ p2 ≥ · · · ≥ pm > 0, 1/ p1 + 1/ p2 + · · · + 1/ pm = 1, B j > 0 ( j =
pj
1,2,...,m), let f j be positive Riemann integrable functions on [a,b] such that B j −
b pj
a f j (x)dx
> 0 for all j = 1,2,...,m, and let Bm+1 = B1 , pm+1 = p1 , fm+1 = f1 . Then one
has the following inequality:
m
pj
Bj
−
j =1
≤
m
b
a
1/ p j
pj
f j (x)dx
Bj −
b m
j =1
2
p j+1
m b f p j (x)
f j+1 (x)
B1 B2 · · · Bm j
f j (x) dx −
dx .
pj −
p j+1
2p1
a
Bj
B j+1
j =1
j =1
(4.1)
a
Proof. For any positive integer n, we choose an equidistant partition of [a,b] as
a < a+
b−a
b−a
b−a
< ··· < a +
i < ··· < a +
(n − 1) < b,
n
n
n
b−a
Δxi =
, i = 1,2,...,n.
n
pj
b pj
a f j (x)dx
Since the hypothesis B j −
pj
B j − lim
n→∞
n
pj
fj
i =1
a+
(4.2)
> 0 ( j = 1,2,...,m) implies that
i(b − a) b − a
> 0 ( j = 1,2,...,m),
n
n
(4.3)
there exists a positive integer N such that
pj
Bj −
n
i=1
pj
fj
a+
i(b − a) b − a
> 0 ∀n > N, j = 1,2,...,m.
n
n
(4.4)
Shanhe Wu 7
Applying Theorem 3.1, one obtains for any n > N the following inequality:
m
pj
Bj
−
j =1
n
pj
fj
i=1
≤
m
Bj −
j =1
i(b − a) b − a
a+
n
n
1/ p j
n m
i(b − a)
fj a +
n
j =1
i =1
1/ p1 +1/ p2 +···+1/ pm
b−a
n
m
n
i(b − a) b − a
B1 B2 · · · Bm 1 p j
−
a+
pj f j
2p1
n
n
j =1 i=1 B j
−
1
p j+1
f j+1 a +
p j+1
B j+1
i(b − a) b − a
n
n
(4.5)
2
.
Note that 1/ p1 + 1/ p2 + · · · + 1/ pm = 1, the above inequality can be transformed to
m
pj
Bj
−
j =1
n
pj
fj
i=1
≤
m
Bj −
j =1
i(b − a) b − a
a+
n
n
n m
i =1
fj a +
j =1
1/ p j
i(b − a)
n
b−a
n
m
n
i(b − a)
B1 B2 · · · Bm 1 p j
a+
−
pj f j
2p1
n
j =1 i=1 B j
−
1
p j+1
B j+1
p j+1
f j+1
i(b − a)
a+
n
pj
pj
(4.6)
b−a
n
2
,
p j+1
p j+1
where equality holds if and only if f j (a + i(b − a)/n)/B j = f j+1 (a + i(b − a)/n)/B j+1 for
all i = 1,2,...,n ( j = 1,2,...,m).
In view of the hypotheses that f j are positive Riemann integrable functions on [a,b]
pj
and p j > 0 ( j = 1,2,...,m), we conclude that mj=1 f j and f j ( j = 1,2,...,m) are also
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 p1 ≥ p2 ≥ · · · ≥ pm > 0, 1/ p1 + 1/ p2 + · · · + 1/ pm = 1, B j >
pj
b
0 ( j = 1,2,...,m), suppose also that f j ∈ L p j [a,b], B j − a | f j (x)| p j dx > 0 for all j =
1,2,...,m, let Bm+1 = B1 , pm+1 = p1 , fm+1 = f1 . Then the following inequality holds true:
m
pj
Bj −
j =1
≤
m
j =1
b
a
f j (x) p j dx
1/ p j
2
b m b m
f j (x) p j f j+1 (x) p j+1
B1 B2 · · · Bm B j−
f j (x) dx −
−
dx
pj
p j+1
a
j =1
2p1
j =1
a
Bj
B j+1
(4.7)
8
Journal of Inequalities and Applications
pj
p j+1
with equality holding if and only if | f j (x)| p j /B j = | f j+1 (x)| p j+1 /B j+1 ( j = 1,2,...,m) almost everywhere on [a,b].
As a consequence of Theorem 4.1, putting m = 2, p1 = p, p2 = q, B1 = A, B2 = B, f1 =
f , f2 = g in (4.1), we obtain the following.
Corollary 4.4. Let p ≥ q > 0, 1/ p + 1/q = 1, A > 0, B > 0, and let f , g be positive Riemann
b
b
integrable functions on [a,b] such that A p − a f p (x)dx > 0 and B q − a g q (x)dx > 0. Then
Ap −
b
a
1/ p f p (x)dx
≤ AB −
b
a
Bq −
b
a
1/q
g q (x)dx
AB
f (x)g(x)dx −
p
b
f p (x) g q (x)
−
dx
Ap
Bq
a
(4.8)
2
.
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, and let f , g be positive
b
b
Riemann integrable functions on [a,b] such that A p − a f p (x)dx > 0 and B q − a g q (x)dx >
0. Then
p
A −
1/ p b
a
p
f (x)dx
≤ AB −
b
a
q
B −
b
a
1/q
q
g (x)dx
AB
f (x)g(x)dx −
max{ p, q}
b
a
f p (x) g q (x)
−
dx
Ap
Bq
(4.9)
2
.
Acknowledgment
The author would like to express hearty thanks to the anonymous referees for valuable
comments on this paper.
References
[1] J. Aczél, “Some general methods in the theory of functional equations in one variable. New
applications of functional equations,” Uspekhi Matematicheskikh Nauk (N.S.), vol. 11, no. 3(69),
pp. 3–68, 1956 (Russian).
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Shanhe Wu: Department of Mathematics, Longyan College, Longyan, Fujian 364012, China
Email address: wushanhe@yahoo.com.cn