Volume 2010, Article ID 698012,13pages doi:10.1155/2010/698012
Research Article
A New Method to Study Analytic Inequalities
Xiao-Ming Zhang and Yu-Ming Chu
Department of Mathematics, Huzhou Teachers College, Huzhou 313000, China
Correspondence should be addressed to Yu-Ming Chu,[email protected]
Received 16 October 2009; Accepted 24 December 2009
Academic Editor: Kunquan Lan
Copyrightq2010 X.-M. Zhang and Y.-M. Chu. 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.
We present a new method to study analytic inequalities involving nvariables. Regarding its applications, we proved some well-known inequalities and improved Carleman’s inequality.
1. Monotonicity Theorems
Throughout this paper, we denoteRthe set of real numbers andRthe set of strictly positive
real numbers,n∈N,n≥2.
In this section, we present the main results of this paper.
Theorem 1.1. Suppose thata, b ∈ Rwitha < bandc ∈ a, b,f : a, bn → Rhas continuous partial derivatives and
Dm
x1, x2, . . . , xn−1, c| min
1≤k≤n−1{xk} ≥c, xm1≤maxk≤n−1{xk}/c
, m1,2, . . . , n−1.
1.1
If∂fx/∂xm>0for allx∈Dmm1,2, . . . , n−1, then
fy1, y2, . . . , yn−1, c
≥fc, c, . . . c, c, 1.2
Proof. Without loss of generality, since we assume thatn3 andy1> y2 > c.
Forx1∈y2, y1, we clearly see thatx1, y2, c∈D1, then
∂fx
∂x1
xx1,y2,c
>0. 1.3
From the continuity of the partial derivatives offand
∂fx
∂x1
xy2,y2,c
>0, 1.4
we know that there existsε >0 such thaty2−ε≥cand
∂fx
∂x1
xx1,y2,c
>0, 1.5
for anyx1 ∈ y2−ε, y2. Hence, since f·, y2, c : x1 ∈ y2−ε, y1 → fx1, y2, cis strictly
monotone increasing, then we have
fy1, y2, c
> fy2, y2, c
> fy2−ε, y2, c
. 1.6
Next, forx2∈y2−ε, y2, theny2−ε, x2, c∈D2and
∂fx ∂x2
x
y2−ε,x2,c
>0. 1.7
Hence, we get
fy1, y2, c
> fy2, y2, c
> fy2−ε, y2, c
> fy2−ε, y2−ε, c
. 1.8
Ify2−εc, thenTheorem 1.1is true. Otherwise, we repeat the above process and we clearly
see that the first and second variables in f are decreasing and no less than c. Let s, t be their limit values, respectively, thenfy1, y2, c > fs, t, cands, t ≥ c. Ifs c, t c, then Theorem 1.1is also true; otherwise, we repeat the above process again and denote pandq
the greatest lower bounds for the first and the second variables , respectively. We clearly see thatpqc; therefore,fy1, y2, c> fc, c, candTheorem 1.1is true.
Theorem 1.2. Suppose thata, b ∈ Rwitha < bandc ∈ a, b,f : a, bn → Rhas continuous partial derivatives and
Em
x1, x2, . . . , xn−1, c| max
1≤k≤n−1{xk} ≤c, xm1≤mink≤n−1{xk}/c
, m1,2, . . . , n–1. 1.9
If∂fx/∂xm<0for allx∈Em m1,2, . . . , n−1, then
fy1, y2, . . . , yn−1, c
≥fc, c, . . . c, c, 1.10
for allym∈a, c m1,2, . . . , n−1.
It follows from Theorems1.1and1.2that we get the following Corollaries1.3–1.6.
Corollary 1.3. Suppose thata, b∈Rwitha < b,f:a, bn → Rhas continuous partial derivatives and
Dm
x x1, x2, . . . , xn|a≤ min
1≤k≤n{xk}< xm1max≤k≤n{xk} ≤b
, m1,2, . . . , n. 1.11
If∂fx/∂xm>0for allx∈Dmandm1,2, . . . , n, then
fx1, x2, . . . , xn≥fxmin, xmin, . . . , xmin, 1.12
for allxm∈a, b m1,2, . . . , nwithxminmin1≤k≤n{xk}.
Corollary 1.4. Supposea, b∈Rwitha < b, then
D1
x x1, x2, . . . , xn|a≤ min
1≤k≤n{xk}< x11max≤k≤n{xk} ≤b
, 1.13
and f : a, bn → Ris symmetric with continuous partial derivatives. If∂fx/∂x1 > 0 for all
x x1, x2, . . . , xn∈D1, then
fx1, x2, . . . , xn≥fxmin, xmin, . . . , xmin, 1.14
wherexminmin1≤k≤n{xk}. Equality holds if and only ifx1x2· · ·xn.
Corollary 1.5. Supposea, b∈Rwitha < b,f :a, bn → Rhas continuous partial derivatives and
Em
x x1, x2, . . . , xn|a≤xm min
1≤k≤n{xk}<1max≤k≤n{xk} ≤b
If∂fx/∂xm<0for allx∈Emandm1,2, . . . , n, then
fx1, x2, . . . , xn≥fxmax, xmax, . . . , xmax, 1.16
wherexmaxmax1≤k≤n{xk}. Equality holds if and only ifx1x2· · ·xn.
Corollary 1.6. Supposea, b∈Rwitha < b, then
En
x x1, x2, . . . , xn|a≤xn min
1≤k≤n{xk}<1max≤k≤n{xk} ≤b
, 1.17
andf : a, bn → Ris symmetric with continuous partial derivatives . If∂fx/∂xn < 0for all
x x1, x2, . . . , xn∈En, then
fx1, x2, . . . , xn≥fxmax, xmax, . . . , xmax, 1.18
wherexmaxmax1≤k≤n{xk}. Equality holds if and only ifx1x2· · ·xn.
2. Unifying Proof of Some Well-Known Inequality
In this section, we denotea a1, a2, . . . , an,aminmin1≤k≤n{ak},amaxmax1≤k≤n{ak},and
Dm{a|amamax> amin>0}, m1,2, . . . , n. 2.1
Proposition 2.1 Power Mean Inequality. If the power mean Mra of order r is defined by
Mra 1/n
n
i1ari
1/r for
r /0 and M0a
n
i1a1i/n, thenMra ≥ Msa forr > s;
equality holds if and only ifa1a2 · · ·an.
Proof. It is well known thatMrais symmetric with respect toa1, a2, . . . , anandr →Mra
is continuous. Without loss of generality, we assume thatr, s /0. Then
fa 1
rln
n i1ari
n
−1
sln
n i1asi
n
, a∈Rn,
∂fa
∂a1
ar−1 1
n
i1ari
−as1−1
n i1asi
n
i2
ar1−1asi −as1−1ari
n
i1ari ·
n
i1asi
n
i2as1−1ari
a1/air−s−1
n
i1ari ·
n
i1asi
.
Ifa∈D1, then∂fa/∂a1>0. It follows fromCorollary 1.4that we get
fa1, a2, . . . , an≥famin, amin, . . . , amin,
n i1ari
n
1/r ≥
n i1asi
n
1/s
, Mra≥Msa.
2.3
Equality holds if and only ifa1a2· · ·an.
Proposition 2.2Holder Inequality. Suppose thatx1, x2, . . . , xn,y1, y2, . . . , yn∈Rn p, q >
1. If1/p1/q1, then
n
k1 xpk
1/p n
k1 yqk
1/q ≥n
k1
xkyk. 2.4
Proof. Letb b1, b2, . . . , bn∈Rnand
f:a∈Rn−→
n
k1 bk
1/p n
k1 bkak
1/q −n
k1
bka1k/q, a∈Rn. 2.5
Ifa∈D1, then
∂fa
∂a1
1
qb1
n
k1 bk
1/p n
k1 bkak
1/q−1
−1
qb1a 1/q−1 1
1
qb1a −1/p
1
n
k1 bkak
−1/p⎡
⎣ n
k1 bk
1/p
a11/p−
n
k1 bkak
1/p⎤
⎦
> 1 qb1a
−1/p
1
n
k1 bkak
−1/p⎡
⎣ n
k1 bk
1/p
a11/p−
n
k1 bka1
1/p⎤
⎦
0.
2.6
Similarly, ifa∈Dm m2,3, . . . , n, then∂fa/∂am>0. FromTheorem 1.1, we get
fa1, a2, . . . , an≥famin, amin, . . . , amin,
n
k1
bk
1/p n
k1
bkak
1/q ≥n
k1
bka1k/q.
2.7
Proposition 2.3Minkowski Inequality. Suppose thatx1, x2, . . . , xn,y1, y2, . . . , yn ∈Rn. If
p >1, then
n
k1 xpk
1/p
n
k1 ykp
1/p
≥ n k1
xkyk
p
1/p
. 2.8
Proof. Letb b1, b2, . . . , bn∈Rnand
f:a∈Rn
−→
n
k1 bkak
1/p − n
k1 bk
a1k/p1p
1/p
, a∈Rn
. 2.9
Ifa∈D1, then
∂fa
∂a1
1
pb1
n
k1 bkak
1/p−1
−1
pb1a 1/p−1 1
a11/p1p−1
n
k1 bk
a1k/p1p
1/p−1
1
pb1
n
k1 bkak
1/p−1 n
k1 bk
a1k/p1p
1/p−1
·
⎡
⎣ n
k1 bk
a1k/p1p
1−1/p
−1a−11/pp−1
n
k1 bkak
1−1/p⎤
⎦
1
pb1
n
k1 bkak
1/p−1 n
k1 bk
a1k/p1p
1/p−1
·
⎡
⎣ n
k1 bk
a1k/p1p
1−1/p − n
k1 bk
a1k/pa1k/pa−11/pp
1−1/p⎤
⎦
> 1 pb1
n
k1 bkak
1/p−1 n
k1 bk
a1k/p1p
1/p−1
·
⎡
⎣ n
k1 bk
a1k/p1p
1−1/p − n
k1 bk
a1k/pa11/pa−11/pp
1−1/p⎤
⎦
0.
2.10
Similarly, Ifa ∈Dm m 2,3, . . . , n, then∂fa/∂am > 0. It follows fromTheorem 1.1that
we get
fa1, a2, . . . , an≥famin, amin, . . . , amin,
n
k1 bkak
1/p
≥ n k1
bk
a1k/p1p
1/p − n
k1 bk
1/p
.
2.11
3. A Brief Proof for Hardy’s Inequality
Ifan ≥0 n∈N, n≥1with∞n1a p
n<∞, then the well-known Hardy’s inequalitysee1,
Theorem 326is
p p−1
p∞
n1 apn≥
∞ n1 1 n n k1 ak p
. 3.1
In this section, we establish the following result involving Hardy’s inequality.
Theorem 3.1. Letn∈N,n≥1, andak≥0 k∈N, k≥1. If
Bn min
1≤k≤n
k−1
2
1/p
ak
, 3.2
then
p p−1
p n
k1 apk−
n k1 ⎛ ⎝1 k k j1 aj ⎞ ⎠ p
≥Bn
⎡
⎣ p
p−1
p n
k1
1
k−1/2−
n k1 ⎛ ⎝1 k k j1 1
j−1/21/p
⎞ ⎠
p⎤
⎦.
3.3
Proof. Letbk k−1/21/pak, then inequality3.3is equivalent to
p p−1
p n
k1 bpk k−1/2 −
n k1 ⎛ ⎝1 k k j1 bj
j−1/21/p
⎞ ⎠
p
≥Bn
⎡
⎣ p
p−1
p n
k1
1
k−1/2 −
n k1 ⎛ ⎝1 k k j1 1
j−1/21/p
⎞ ⎠
p⎤
⎦,
3.4
andBnmin1≤k≤n{bk}. Let
Dm
b|bmmax
1≤k≤n{bk}>1min≤k≤n{bk}>0
, m1,2, . . . , n,
f:b∈0,∞n−→
p p−1
p n
k1 bpk k−1/2 −
n k1 ⎛ ⎝1 k k j1 bj
j−1/21/p
⎞ ⎠
p
.
Ifb∈Dm m1,2, . . . , n, then
∂fb
∂bm p
p p−1
p
bpm−1
m−1/2 −
n
km
⎡ ⎢
⎣ p
kpm−1/21/p
⎛
⎝k
j1 bj
j−1/21/p
⎞ ⎠
p−1⎤
⎥ ⎦
> pb
p−1
m
m−1/21/p ·
⎡ ⎢ ⎣
p p−1
p 1
m−1/21−1/p −
∞
km
1
kp
⎛
⎝k
j1
1
j−1/21/p
⎞ ⎠
p−1⎤
⎥ ⎦.
3.6
Making use of the well-known Hadamard’s inequality of convex functions, we get
∂fb
∂bm >
pbpm−1
m−1/21/p
⎡
⎣ p
p−1
p 1
m−1/21−1/p −
∞
km
1
kp
k1/2
1/2
1
x−1/21/pdx
p−1⎤
⎦
pb
p−1
m
m−1/21/p
p
p−1
p 1
m−1/21−1/p −
p p−1
p−1∞
km
k−2p1/p
> pb
p−1
m
m−1/21/p
p
p−1
p 1
m−1/21−1/p −
p p−1
p−1∞
m−1/2
x−2p1/pdx
0.
3.7
ThenTheorem 1.1leads to
fb1, b2, . . . , bn≥fBn, Bn, . . . , Bn, 3.8
and we clearly see that inequalities3.4and3.3are true.
Corollary 3.2. Letn∈N,n≥1, andak≥0k∈N, k≥1. If
Bn min
1≤k≤n
k−1
2
1/p
ak
, 3.9
then
p p−1
p n
k1 apk−
n k1 ⎛ ⎝1 k k j1 aj ⎞ ⎠ p
> Bn
p p−1
p n
k1
1
k2k−1 ≥Bn
p p−1
p
.
Proof. From inequality3.3, we clearly see that
p p−1
p n
k1 bpk k−1/2 −
n
k1
⎛
⎝1
k
k
j1 bj
j−1/21/p
⎞ ⎠
p
> Bn
p
p−1
p n
k1
1
k−1/2 −
n
k1
1
k
k1/2
1/2
1
x−1/21/pdx
p
Bn
p p−1
p n
k1
1
k−1/2 − 1
k
Bn
p p−1
p n
k1
1
k2k−1
≥Bn
p p−1
p
.
3.11
Remark 3.3. Ifn → ∞, then inequality3.1follows from inequality3.10.
4. A Refinement of Carleman’s Inequality
Ifan≥0n∈N, n≥1with 0<∞n1an<∞, then the well-known Carleman’s inequality is
∞
n1
n
!
k1 ak
1/n
< e ∞
n1
an, 4.1
with the best possible constant factoresee2.
Recently, Yang and Debnath3gave a strengthened version of4.1as follows:
∞
n1
n
!
k1 ak
1/n
< e ∞
n1
1− 1 2n2
an. 4.2
Some other strengthened versions of 4.1 were given in 4–9. In this section, we give a refinement for Carleman’s inequalityseeCorollary 4.4.
Lemma 4.1. Ifm∈Nandm≥1, then
e
1− 2 3m7
1
m > ∞
km
1
kk!1/k, 4.3
e
1− 2 3m10
1
m1 >
1
Proof. Letψm e1−2/3m71/m−k∞m1/kk!1/k, then inequalityψm> ψm1
is equivalent to inequality
1−2m2 3m7
2m
3m10 >
m1
em!1/m. 4.5
If 1≤m≤16, then simple computation leads to inequality4.5. Ifm≥17, then it is not difficult to verify that√2πm≥e7/3and
√
2πm≥e21m271m70/9m239m50. 4.6
Ifx >0, thene >11/xx; this implies that
e > 1 21m
271m70
9m239m50m
9m239m50m/21m271m70
. 4.7
From inequalities4.6and4.7, we get
√
2πm > 1 21m
271m70
9m239m50m
m
,
2πm1/2m> m13m73m10 m9m239m50 ,
m5 3m7
2m
3m10 >
m1
m2πm1/2m.
4.8
From the well-known Stirling Formulam! √2πmm/emexpθm/12m 0 < θm < 1, we
get
m!>√2πm
m e
m
. 4.9
Therefore, inequality4.5follows from inequalities4.8and4.9.
From the monotonicity of sequence{ψm}∞m1and limm→∞ψm 0, we getψm>
Meanwhile, we have
"
2πm1> e2/3,
"
2πm1> e2m2/3m8,
"
2πm1>
1 2 3m8
3m8/2·2m2/3m8 ,
2πm11/2m2> 3m10
3m8 ,
e
1− 2 3m10
1
m1 >
e
m12πm11/2m2.
4.10
Therefore, inequality4.4follows from inequalities4.10and
m1!>
#
2πm1
m1
e
m1
. 4.11
Theorem 4.2. Letn∈N,n≥1, andak>0 k1,2, . . . , n. IfBnmin1≤k≤n{kak}, then
e
n
k1
1− 2 3k7
ak− n
k1
⎛
⎝!k
j1
aj
⎞ ⎠
1/k ≥Bn
e n k1
1− 2 3k7
1 k − n k1 1
k!1/k . 4.12
Proof. Letbkkak,k1,2, . . . , n, andb b1, b2, . . . , bn,
Dm
b|bmmax
1≤k≤n{bk}>1min≤k≤n{bk}>0
, m1,2, . . . , n,
f :b∈Rn
−→e
n
k1
1− 2 3k7
bk k − n k1 ⎛ ⎝1 k! k ! j1 bj ⎞ ⎠ 1/k
, b∈Rn
.
4.13
Then inequality4.12is equivalent to the following inequality:
e
n
k1
1− 2 3k7
bk k − n k1 ⎛ ⎝1 k! k ! j1 bj ⎞ ⎠ 1/k ≥Bn
e n k1
1− 2 3k7
1 k − n k1 1
k!1/k , 4.14
Ifb∈Dm m1,2, . . . , n, then
∂fb
∂bm e
1− 2 3m7
1
m−
n
km
1
kbm
⎛
⎝1
k!
k
!
j1
bj
⎞ ⎠
1/k
> e
1− 2 3m7
1
m−
n
km
1
kk!1/k
> e
1− 2 3m7
1
m−
∞
km
1
kk!1/k.
4.15
From inequality4.3and∂fb/∂bm>0 together withTheorem 1.1, we clearly see that
fb1, b2, . . . , bn≥fBn, Bn, . . . , Bn. 4.16
Therefore, inequality4.14is proved.
Corollary 4.3. Letn∈N,n≥1, andak>0k1,2, . . . , n. IfBnmin1≤k≤n{kak}, then
e
n
k1
1− 2 3k7
ak− n
k1
⎛
⎝!k
j1
aj
⎞ ⎠
1/k ≥Bn
4 5e−1
. 4.17
Proof. Let Tm emk11 − 2/3k 71/k −mk11/k!1/k m 1,2, . . . , n, then inequality4.4implies that{Tm}nm1is a strictly increasing sequence. Then from inequality
4.12we get
e
n
k1
1− 2 3k7
ak− n
k1
⎛
⎝!k
j1 aj
⎞ ⎠
1/k
≥BnTn≥BnT1 Bn
4 5e−1
. 4.18
Letn → ∞; thus, we know thatCorollary 4.4is true.
Corollary 4.4. Ifan≥0 n∈N, n≥1with0<∞n1an<∞, then
∞
n1
⎛
⎝!n
j1 aj
⎞ ⎠
1/n ≤e
∞
n1
1− 2 3n7
an. 4.19
Remark 4.5. Many other applications forTheorem 1.1appeared in10.
Acknowledgments
National Nature Science Foundation of China under Grant no. 60850005, the Nature Science Foundation of Zhejiang Province under Grant no. Y607128, the Nature Science Foundation of China Central Radio & TV University under Grant no. GEQ1633, and the Nature Science Foundation of Zhejiang Broadcast & TV University under Grant no. XKT-07G19.
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