R E S E A R C H
Open Access
Reverse Beckenbach-Dresher’s inequality
Chang-Jian Zhao
1*and Wing-Sum Cheung
2*Correspondence: [email protected]
1Department of Mathematics, China Jiliang University, Hangzhou, 310018, P.R. China
Full list of author information is available at the end of the article
Abstract
In the paper, we establish an inverse of Beckenbach-Dresher’s integral inequality, which provides new estimates on inequality of this type.
MSC: 26D15
Keywords: Beckenbach’s inequality; Radon’s inequality; Beckenbach-Dresher’s inequality
1 Introduction
The well-known inequality due to Beckenbach can be stated as follows (see [], also see [], p.).
Theorem A If≤p≤,and xi,yi> for i= , , . . . ,n,then
n
i=(xi+yi)p
n
i=(xi+yi)p–≤
n i=x
p i
n i=x
p–
i
+
n i=y
p i
n i=y
p–
i
. (.)
An integral analogue of Beckenbach’s inequality easily follows.
Theorem B Let≤p≤.If f and g are positive and continuous functions on[a,b],then
b
a(f(x) +g(x))pdx
b
a(f(x) +g(x))p–dx ≤
b a f(x)pdx
b
a f(x)p–dx
+
b a g(x)pdx
b
a g(x)p–dx
. (.)
An extension of Beckenbach’s inequality was obtained by Dresher [] by an ingenious method using moment-space theory.
Theorem C Let f and g be positive and continuous functions on[a,b].If p≥≥r≥,
then
b
a(f(x) +g(x))pdx
b
a(f(x) +g(x))rdx
/(p–r)
≤
b a fp(x)dx
b a fr(x)dx
/(p–r)
+
b a gp(x)dx
b a gr(x)dx
/(p–r)
. (.)
The inequality which we shall call Beckenbach-Dresher’s inequality. In fact, this re-sult was also established by Danskin [], who employed a combination of Hölder’s and Minkowski’s inequalities.
Beckenbach-Dresher’s inequality was studied extensively and numerous variants, gen-eralizations, and extensions appeared in the literature (see [–] and the references cited therein). Research of reverse Beckenbach-Dresher’s integral inequality is rare (see [] and []). The aim of this paper is to discuss reverse Beckenbach-Dresher’s integral inequality and establish the following reversed Beckenbach-Dresher integral inequality by deriving reverse Hölder’s, Minkowski’s and Radon’s integral inequalities.
Theorem Let f and g be continuous functions on[a,b], <m≤f(x)≤Mand <m≤
g(x)≤M.If p≥≥r≥,then
· b
a(f(x) +g(x))pdx
b
a(f(x) +g(x))rdx
/(p–r)
≥
b a fp(x)dx
b a fr(x)dx
/(p–r)
+
b a gp(x)dx
b a gr(x)dx
/(p–r)
, (.)
where
= Lα,β(s,t,S,T)
α,β(m,m,M,M)
,
α +
β = ,α> , (.)
Lα,β(s,t,S,T) =
ϒα,β
sT–m+m ,stS– m
m+,St–m+m ,s–STm+m m+, m> , (.) s=minm(b–a)/p,m(b–a)/p , S=max
M(b–a)/p,M(b–a)/p ,
t=minm(b–a)/r,m(b–a)/r , T=max
M(b–a)/r,M(b–a)/r ,
ϒα,β(m,m,M,M) = max
Cα,β
Mα
mα
(b–a)
, m
β
Mβ(b–a)
,
Cα,β
mα
Mα
(b–a)
, M
β
mβ(b–a)
, (.)
Cα,β(ξ,η) =
ξ/α+η/β
ξ/αη/β , (.)
and
α,β(m,m,M,M)
=maxϒα,β
m, (m+m)α–,M, (M+M)α–
,
ϒα,β
m, (m+m)α–,M, (M+M)α– . (.)
2 Proof of theorem
Lemma .[] If <m≤a≤M, <m≤b≤M, α+
β= andα> ,then
maxCα,β(M,m),Cα,β(m,M) ·αβa/αb/β≥aβ+bα, (.)
with equality if and only if either(a,b) = (m,M)or(a,b) = (M,m),where Cα,β(ξ,η)is
as in(.).
Lemma . Let f and g be positive continuous functions on[a,b], α +β = ,α> and fα
and gβbe integrable on[a,b].If <m
≤f(x)≤Mand <m≤g(x)≤M,then
b
a
fα(x)dx
/α b a
gβ(x)dx /β
≤ϒα,β(m,m,M,M)·
b
a
f(x)g(x)dx, (.)
with equality if and only if fαand gβare proportional,whereϒ
α,β(m,m,M,M)is as in
(.).
The inequality is reversed if <α< orα< .
Proof If we set successively
¯
a=f
α(x)
X , X= b
a
fα(x)dx,
¯
b=g
β(x)
Y , Y= b
a
gβ(x)dx.
Notice that
mα
Mα(b–a)
≤ ¯a≤ M
α
mα(b–a)
,
and
mβ
Mβ(b–a)≤ ¯b≤
Mβ mβ(b–a). By using Lemma ., we have
max
Cα,β
Mα
mα
(b–a)
, m
β
Mβ(b–a)
,Cα,β
mα
Mα
(b–a)
, M
β
mβ(b–a)
· f(x)g(x)
X/αY/β
≥
α fα(x)
X +
β gβ(x)
Y ,
with equality if and only if either
(a¯,b¯) =
mα
Mα
(b–a)
, M
β
mβ(b–a)
or
(a¯,b¯) =
Mα
mα
(b–a)
, m
β
Mβ(b–a)
.
Therefore
ϒα,β(m,m,M,M)·
b
a f(x)g(x)dx
X/αY/β ≥
α b
a f
α(x)dx
X +
β b
ag
β(x)dx
Y = . (.)
In the following, we discuss the equality condition of (.). In view of the equality con-ditions of Lemma ., the equality in (.) holds if and only if
fα(x)
b
a fα(x)dx
, g
β(x)
b
a gβ(x)dx
=
mα
Mα
(b–a)
, M
β
mβ(b–a)
,
or
fα(x)
b
a fα(x)dx
, g
β(x)
b
a gβ(x)dx
=
Mα
mα
(b–a)
, m
β
Mβ(b–a)
.
Hencefα(x) =μgβ(x), where
μ= m
α
m
β
MβMα
fα α
gββ,
or
μ=M
α
M
β
mβmα
fα α
gββ
is a constant. It follows that the equality in (.) holds if and only iffαandgβare
propor-tional.
This proof is completed.
Lemma . Let f and g be non-negative continuous functions on[a,b].If <m≤f(x)≤
M, <m≤g(x)≤Mandα> ,then
b
a
f(x) +g(x)αdx /α
≥α,β(m,m,M,M)
b
a
fα(x)dx /α
+
b
a
gα(x)dx /α
, (.)
with equality if and only if f and g are proportional,whereα,β(m,m,M,M)is as in
(.).
The inequality is reversed if <α< orα< .
Proof From the hypotheses, we have
f(x) +g(x)αα=f(x)f(x) +g(x)α–+g(x)f(x) +g(x)α–. (.)
By using Lemma ., we obtain
f(x)f(x) +g(x)α–≥ϒα,β
m, (m+m)α–,M, (M+M)α–
–
×f(x)α·f(x) +g(x)αα/β, (.)
with equality if and only iffα(x) and (f(x) +g(x))α are proportional. It follows that the
g(x)f(x) +g(x)α–≥ϒα,β
m, (m+m)α–,M, (M+M)α–
–
×g(x)α·f(x) +g(x)αα/β, (.)
with equality if and only ifgα(x) and (f(x) +g(x))α are proportional. It follows that the
equality holds if and only iff(x) andg(x) are proportional. Hence
f(x) +g(x)αα≥α,β(m,m,M,M)·f(x) +g(x)
α/β
α f(x)α+g(x)α
, (.)
whereα,β(m,m,M,M) =max{M,N},
M=ϒα,β
m, (m+m)α–,M, (M+M)α–
,
and
N=ϒα,β
m, (m+m)α–,M, (M+M)α–
.
Dividing both sides of (.) byf(x) +g(x)α/β
α , we have
f(x) +g(x)α≥α,β(m,m,M,M)·f(x)α+g(x)α
. (.)
Moreover, in view of the equality conditions of (.) and (.), it follows that the equality in (.) holds if and only iff(x) andg(x) are proportional.
This proof is completed.
Lemma . Let f and g be continuous functions on[a,b], <m≤f(x)≤M and <
m≤g(x)≤M.If m> ,then
b
a
fm+(x)
gm(x) dx≤Lα,β(m,m,M,M)
(abf(x)dx)m+
(abg(x)dx)m , (.)
where Lα,β(m,m,M,M)is as in(.).
Proof Letα=m+ ,β= (m+ )/mand replacingf(x) andg(x) byu(x) andv(x) in (.), respectively, we have
b
a
u(x)m+dx
/(m+) b a
v(x)(m+)/mdx m/(m+)
≤ϒα,β(m,m,M,M)·
b
a
u(x)v(x)dx. (.)
Taking for
u(x) =
f(x)
g(x)
/(m+)
, v(x) =fm/(m+)(x)g/(m+)(x)
in (.), and in view of
m
M
m+
≤u(x)≤
M
m
and
m m m+
m
m+
≤v(x)≤M
m m+
M
m+
,
we obtain
ϒα,β
mM–
m+,mm+m
m
m+
,
Mm–
m+,Mm+m
M
m+
b
a
f(x)dx
≥
b
a
f(x)
g(x)dx
/(m+) b a
f(x)g/m(x)dx m/(m+)
.
Hence
b
a
f(x)
g(x)dx
≤[ϒα,β((mM–)
m+,m
m m+
m
m+
, (Mm– )
m+,M
m m+
M
m+
)
b af(x)dx]
m+
(abf(x)g/m(x)dx)m . (.)
On the other hand, in (.), replacing f(x) andg(x) byu(x) andv(x), respectively, and lettingu(x) =f(x) andv(x) = (gf((xx)))m, and in view of
m≤u(x)≤M
and
m
M
m
≤v(x)≤
M
m
m
,
we have
b
a
fm+(x)
gm(x) dx
≤[ϒα,β(mM –m+m
, (mmM– )
m m+,Mm–
m m+
, (m– MM)
m m+)b
af(x)dx]m+
(abg(x)dx)m
=Lα,β(m,m,M,M)(
b
a f(x)dx)m+
(abg(x)dx)m .
This proof is completed.
Letf(x) andg(x) reduce to positive real sequencesai andbi(i= , . . . ,n), respectively,
and with appropriate changes in the proof of (.), we have the following.
Lemma . Let aiand bibe positive real sequences and <m≤ai≤M, <m≤bi≤
M,i= , . . . ,n.If m> ,then
n
i=
am+
i
bmi ≤Lα,β(m,m,M,M)
(ni=ai)m+
(ni=bi)m
, (.)
This is just an inverse of the following well-known Radon’s inequality[],p.
n
i=
ami+ bm
i ≥(
n i=ai)m+
(ni=bi)m
,
where m> ,ai≥and bi> ,i= , , . . . ,n.
Proof of Theorem Let
α=
b
a
fp(x)dx /p
, β=
b
a
fr(x)dx /r
,
α=
b
a
gp(x)dx /p
, β=
b
a
gr(x)dx /r
,
then
<m(b–a)/p≤α≤M(b–a)/p,
<m(b–a)/p≤α≤M(b–a)/p,
<m(b–a)/r≤β≤M(b–a)/r,
and
<m(b–a)/r≤β≤M(b–a)/r.
Let
s=minm(b–a)/p,m(b–a)/p , S=max
M(b–a)/p,M(b–a)/p
and
t=minm(b–a)/r,m(b–a)/r , T=max
M(b–a)/r,M(b–a)/r .
From reverse Radon’s inequality (.) in Lemma ., we have, form> ,
αm+ βm +
αm+
βm ≤Lα,β(s,t,S,T)
(α+α)m+
(β+β)m
. (.)
Ifm=p–rr, then
fp(x)dx
fr(x)dx
/(p–r)
+
gp(x)dx
gr(x)dx
/(p–r)
≤Lα,β(s,t,S,T)
[(fp(x)dx)/p+ (gp(x)dx)/p]p/(p–r)
[(fr(x)dx)/r+ (gr(x)dx)/r]r/(p–r) . (.)
On the other hand, by using the Minkowski inequality (.) and its reverse form, with
p≥ and <r≤, respectively,
α,β(m,m,M,M)p
fp(x)dx /p
+
gp(x)dx /pp
≤ f(x) +g(x)pdx, (.)
with equality if and only iff andgare proportional, and
α,β(m,m,M,M)r
fr(x)dx /r
+
gr(x)dx /rr
≥ f(x) +g(x)r(x)dx, (.)
with equality if and only iff andgare proportional.
From (.), (.) and (.), (.) follows. This proof is completed.
Competing interests
The authors declare that they have no competing interests. Authors’ contributions
C-JZ and W-SC jointly contributed to the main results. All authors read and approved the final manuscript. Author details
1Department of Mathematics, China Jiliang University, Hangzhou, 310018, P.R. China.2Department of Mathematics, The University of Hong Kong, Pokfulam Road, Hong Kong, P.R. China.
Acknowledgements
The first author’s research is supported by the Natural Science Foundation of China (11371334). The second author’s research is partially supported by a HKU Seed Grant for Basic Research. The authors express their grateful thanks to the two referees for their excellent suggestions and comments.
Received: 11 September 2014 Accepted: 27 April 2015 References
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