Hindawi Publishing Corporation
Journal of Inequalities and Applications
Volume 2010, Article ID 698012, 13 pages
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, chuyuming2005@yahoo.com.cn
Received 16 October 2009; Accepted 24 December 2009
Academic Editor: Kunquan Lan
Copyright q 2010 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 n variables. Regarding its
applications, we proved some well-known inequalities and improved Carleman’s inequality.
1. Monotonicity Theorems
Throughout this paper, we denote R the set of real numbers and R the 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 that a, b ∈ R with a < b and c ∈ a, b, f : a, bn → R has continuous
partial derivatives and
c ,
Dm x1 , x2 , . . . , xn−1 , c | min {xk } ≥ c, xm max {xk } /
1≤k≤n−1
1≤k≤n−1
m 1, 2, . . . , n − 1.
1.1
If ∂fx/∂xm > 0 for all x ∈ Dm m 1, 2, . . . , n − 1, then
f y1 , y2 , . . . , yn−1 , c ≥ fc, c, . . . c, c,
for all ym ∈ c, b m 1, 2, . . . , n − 1.
1.2
2
Journal of Inequalities and Applications
Proof. Without loss of generality, since we assume that n 3 and y1 > y2 > c.
For x1 ∈ y2 , y1 , we clearly see that x1 , y2 , c ∈ D1 , then
∂fx > 0.
∂x1 xx1 ,y2 ,c
1.3
From the continuity of the partial derivatives of f and
∂fx > 0,
∂x1 xy2 ,y2 ,c
1.4
we know that there exists ε > 0 such that y2 − ε ≥ c and
∂fx > 0,
∂x1 xx1 ,y2 ,c
1.5
for any x1 ∈ y2 − ε, y2 . Hence, since f·, y2 , c : x1 ∈ y2 − ε, y1 → fx1 , y2 , c is strictly
monotone increasing, then we have
f y1 , y2 , c > f y2 , y2 , c > f y2 − ε, y2 , c .
1.6
Next, for x2 ∈ y2 − ε, y2 , then y2 − ε, x2 , c ∈ D2 and
∂fx > 0.
∂x2 xy2 −ε,x2 ,c
1.7
f y1 , y2 , c > f y2 , y2 , c > f y2 − ε, y2 , c > f y2 − ε, y2 − ε, c .
1.8
Hence, we get
If y2 − ε c, then Theorem 1.1 is 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, then fy1 , y2 , c > fs, t, c and s, t ≥ c. If s c, t c, then
Theorem 1.1 is also true; otherwise, we repeat the above process again and denote p and q
the greatest lower bounds for the first and the second variables , respectively. We clearly see
that p q c; therefore, fy1 , y2 , c > fc, c, c and Theorem 1.1 is true.
Similarly, we have the following theorem.
Journal of Inequalities and Applications
3
Theorem 1.2. Suppose that a, b ∈ R with a < b and c ∈ a, b, f : a, bn → R has continuous
partial derivatives and
Em c ,
x1 , x2 , . . . , xn−1 , c | max {xk } ≤ c, xm min {xk } /
1≤k≤n−1
1≤k≤n−1
m 1, 2, . . . , n–1. 1.9
If ∂fx/∂xm < 0 for all x ∈ Em m 1, 2, . . . , n − 1, then
f y1 , y2 , . . . , yn−1 , c ≥ fc, c, . . . c, c,
1.10
for all ym ∈ a, c m 1, 2, . . . , n − 1.
It follows from Theorems 1.1 and 1.2 that we get the following Corollaries 1.3–1.6.
Corollary 1.3. Suppose that a, b ∈ R with a < b, f : a, bn → R has continuous partial derivatives
and
Dm x x1 , x2 , . . . , xn | a ≤ min {xk } < xm max {xk } ≤ b ,
1≤k≤n
1≤k≤n
m 1, 2, . . . , n.
1.11
If ∂fx/∂xm > 0 for all x ∈ Dm and m 1, 2, . . . , n, then
fx1 , x2 , . . . , xn ≥ fxmin , xmin , . . . , xmin ,
1.12
for all xm ∈ a, b m 1, 2, . . . , n with xmin min1≤k≤n {xk }.
Corollary 1.4. Suppose a, b ∈ R with a < b, then
D1 x x1 , x2 , . . . , xn | a ≤ min {xk } < x1 max {xk } ≤ b ,
1≤k≤n
1≤k≤n
1.13
and f : a, bn → R is 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
where xmin min1≤k≤n {xk }. Equality holds if and only if x1 x2 · · · xn .
Corollary 1.5. Suppose a, b ∈ R with a < b, f : a, bn → R has continuous partial derivatives and
Em x x1 , x2 , . . . , xn | a ≤ xm min {xk } < max {xk } ≤ b .
1≤k≤n
1≤k≤n
1.15
4
Journal of Inequalities and Applications
If ∂fx/∂xm < 0 for all x ∈ Em and m 1, 2, . . . , n, then
fx1 , x2 , . . . , xn ≥ fxmax , xmax , . . . , xmax ,
1.16
where xmax max1≤k≤n {xk }. Equality holds if and only if x1 x2 · · · xn .
Corollary 1.6. Suppose a, b ∈ R with a < b, then
En x x1 , x2 , . . . , xn | a ≤ xn min {xk } < max {xk } ≤ b ,
1≤k≤n
1≤k≤n
1.17
and f : a, bn → R is symmetric with continuous partial derivatives . If ∂fx/∂xn < 0 for all
x x1 , x2 , . . . , xn ∈ En , then
fx1 , x2 , . . . , xn ≥ fxmax , xmax , . . . , xmax ,
1.18
where xmax max1≤k≤n {xk }. Equality holds if and only if x1 x2 · · · xn .
2. Unifying Proof of Some Well-Known Inequality
In this section, we denote a a1 , a2 , . . . , an , amin min1≤k≤n {ak }, amax max1≤k≤n {ak }, and
Dm {a | am amax > amin > 0},
m 1, 2, . . . , n.
2.1
Proposition 2.1 Power Mean Inequality. If the power mean Mr a of order r is defined by
n 1/n
1/r
Mr a 1/n ni1 ari for r /
0 and M0 a i1 ai , then Mr a ≥ Ms a for r > s;
equality holds if and only if a1 a2 · · · an .
Proof. It is well known that Mr a is symmetric with respect to a1 , a2 , . . . , an and r → Mr a
is continuous. Without loss of generality, we assume that r, s /
0. Then
1
fa ln
r
n
i1
n
ari
1
− ln
s
n
i1
n
asi
,
a ∈ Rn ,
ar−1
as−1
∂fa
n1 r − n1 s
∂a1
i1 ai
i1 ai
n s−1 r n r−1 s
r−s
s−1 r
−1
i2 a1 ai − a1 ai
i2 a1 ai a1 /ai .
n r n s
n r n s
i1 ai ·
i1 ai
i1 ai ·
i1 ai
2.2
Journal of Inequalities and Applications
5
If a ∈ D1 , then ∂fa/∂a1 > 0. It follows from Corollary 1.4 that we get
fa1 , a2 , . . . , an ≥ famin , amin , . . . , amin ,
n
i1
ari
n
1/r
≥
n
i1
asi
1/s
Mr a ≥ Ms a.
,
n
2.3
Equality holds if and only if a1 a2 · · · an .
Proposition 2.2 Holder Inequality. Suppose that x1 , x2 , . . . , xn , y1 , y2 , . . . , yn ∈ Rn p, q >
1. If 1/p 1/q 1, then
n
k1
p
xk
1/p n
k1
1/q
q
yk
≥
n
xk yk .
2.4
k1
Proof. Let b b1 , b2 , . . . , bn ∈ Rn and
f :a∈
Rn
1/p 1/q
n
n
n
1/q
−→
bk
bk ak
− bk ak ,
k1
k1
a ∈ Rn .
2.5
k1
If a ∈ D1 , then
1/p 1/q−1
n
n
∂fa 1
1
1/q−1
b1
bk
bk ak
− b1 a1
∂a1
q
q
k1
k1
−1/p ⎡
1/p
1/p ⎤
n
n
n
1
−1/p 1/p
⎣
⎦
b1 a1
bk ak
bk
a1 −
bk ak
q
k1
k1
k1
−1/p ⎡
1/p
1/p ⎤
n
n
n
1
−1/p 1/p
⎣
⎦
> b1 a1
bk ak
bk
a1 −
bk a1
q
k1
k1
k1
2.6
0.
Similarly, if a ∈ Dm m 2, 3, . . . , n, then ∂fa/∂am > 0. From Theorem 1.1, we get
fa1 , a2 , . . . , an ≥ famin , amin , . . . , amin ,
1/p 1/q
n
n
n
1/q
bk
bk ak
≥
bk ak .
k1
k1
k1
q
p
p
Therefore, Proposition 2.2 follows from ak yk /xk and bk xk .
2.7
6
Journal of Inequalities and Applications
Proposition 2.3 Minkowski Inequality. Suppose that x1 , x2 , . . . , xn ,y1 , y2 , . . . , yn ∈ Rn . If
p > 1, then
1/p 1/p 1/p
n
n
n
p
p
p
xk
yk
≥
.
xk yk
k1
k1
2.8
k1
Proof. Let b b1 , b2 , . . . , bn ∈ Rn and
f :a∈
Rn
−→
n
1/p
bk ak
k1
n
1/p
p 1/p
−
bk ak 1
,
a ∈ Rn .
2.9
k1
If a ∈ D1 , then
1/p−1
n
n
1/p
1/p
p−1 p 1/p−1
∂fa 1
1
1/p−1
a1 1
b1
bk ak
− b 1 a1
bk ak 1
∂a1
p
p
k1
k1
1/p−1 n
n
1/p
p 1/p−1
1
b1
bk ak
bk ak 1
p
k1
k1
⎡
1−1/p ⎤
n
n
1/p
p 1−1/p p−1
−1/p
⎦
·⎣
bk ak 1
− 1 a1
bk ak
k1
k1
1/p−1 n
n
1/p
p 1/p−1
1
bk ak
bk ak 1
b1
p
k1
k1
⎤
⎡
n
n
1/p
1/p
1−1/p
p 1−1/p
1/p −1/p p
⎦
·⎣
bk ak 1
−
bk ak ak a1
k1
2.10
k1
1/p−1 n
n
1/p
p 1/p−1
1
> b1
bk ak
bk ak 1
p
k1
k1
⎤
⎡
n
n
1/p
1/p
1−1/p
p 1−1/p
1/p −1/p p
⎦
·⎣
bk ak 1
−
bk ak a1 a1
k1
k1
0.
Similarly, If a ∈ Dm m 2, 3, . . . , n, then ∂fa/∂am > 0. It follows from Theorem 1.1 that
we get
fa1 , a2 , . . . , an ≥ famin , amin , . . . , amin ,
1/p 1/p
n
n
n
1/p
p 1/p
bk ak
≥
bk ak 1
−
bk
.
k1
k1
k1
p
p
p
Therefore, Proposition 2.3 follows from ak yk /xk and bk xk .
2.11
Journal of Inequalities and Applications
7
3. A Brief Proof for Hardy’s Inequality
If an ≥ 0 n ∈ N, n ≥ 1 with
Theorem 326 is
∞
p
n1
an < ∞, then the well-known Hardy’s inequality see 1,
p
p−1
p ∞
p
an
≥
n1
∞
n
1
n1
n k1
p
ak
.
3.1
In this section, we establish the following result involving Hardy’s inequality.
Theorem 3.1. Let n ∈ N, n ≥ 1, and ak ≥ 0 k ∈ N, k ≥ 1. If
Bn min
1≤k≤n
1
k−
2
1/p
3.2
ak ,
then
p
p−1
⎛
⎞p
p n
n
k
p
⎝ 1 aj ⎠
ak −
k j1
k1
k1
⎛
⎞p ⎤
⎡
p n
n
k
p
1
1
1
⎝
⎠ ⎦.
−
≥ Bn ⎣
p − 1 k1 k − 1/2 k1 k j1 j − 1/21/p
3.3
Proof. Let bk k − 1/21/p ak , then inequality 3.3 is equivalent to
p
p−1
⎞p
k
bj
1
⎠
− ⎝
k − 1/2 k1 k j1 j − 1/21/p
k1
p n
p
n
bk
⎛
⎛
⎞p ⎤
⎡
p n
n
k
p
1
1
1
⎠ ⎦,
− ⎝
≥ Bn ⎣
p − 1 k1 k − 1/2 k1 k j1 j − 1/21/p
3.4
and Bn min1≤k≤n {bk }. Let
Dm b | bm max {bk } > min {bk } > 0 ,
1≤k≤n
n
f : b ∈ 0, ∞ −→
p
p−1
1≤k≤n
m 1, 2, . . . , n,
⎛
⎞p
n
k
b
1
j
− ⎝
1/p ⎠ .
k
−
1/2
k
j1 j − 1/2
k1
k1
p n
p
bk
3.5
8
Journal of Inequalities and Applications
If b ∈ Dm m 1, 2, . . . , n, then
⎡
⎛
⎞p−1 ⎤
p
p−1
n
k
b
∂fb
p
p
bm
j
⎢
⎝
⎠ ⎥
−
p
⎣
⎦
∂bm
p − 1 m − 1/2 km kp m − 1/21/p j1 j − 1/21/p
>
p−1
pbm
m − 1/21/p
3.6
⎡
⎛
⎞p−1 ⎤
p
∞
k
p
1
1 ⎝
1
⎥
⎢
·⎣
−
1/p ⎠ ⎦.
p
p − 1 m − 1/21−1/p km k
j1 j − 1/2
Making use of the well-known Hadamard’s inequality of convex functions, we get
⎡
k1/2
p−1 ⎤
p
p−1
∞
∂fb
p
pbm
1
1
1
⎣
⎦
>
−
dx
1/p
∂bm
p − 1 m − 1/21−1/p km kp
m − 1/21/p
1/2 x − 1/2
p−1
pbm
m − 1/2
1/p
p−1
>
pbm
m − 1/21/p
p
p−1
p
p−1
p
1
m − 1/2
p
1−1/p
−
1
m − 1/21−1/p
−
p
p−1
p
p−1
p−1 ∞
k−2p1/p
km
p−1 ∞
x−2p1/p dx
m−1/2
0.
3.7
Then Theorem 1.1 leads to
fb1 , b2 , . . . , bn ≥ fBn , Bn , . . . , Bn ,
3.8
and we clearly see that inequalities 3.4 and 3.3 are true.
Corollary 3.2. Let n ∈ N, n ≥ 1, and ak ≥ 0 k ∈ N, k ≥ 1. If
Bn min
1≤k≤n
1
k−
2
1/p
ak ,
3.9
then
p
p−1
⎛
⎞p
p p p
n
n
n
k
p
p
1
1
p
≥ Bn
ak − ⎝
aj ⎠ > Bn
.
k j1
p − 1 k1 k2k − 1
p−1
k1
k1
3.10
Journal of Inequalities and Applications
9
Proof. From inequality 3.3, we clearly see that
p
p−1
⎞p
k
bj
1
⎝
−
1/p ⎠
k
−
1/2
k
j
−
1/2
j1
k1
k1
p n
> Bn
Bn
Bn
≥ Bn
p
n
bk
p
p−1
p
p−1
p
p−1
p
p−1
⎛
p n
n
1
−
k − 1/2 k1
k1
p n k1
1
1
−
k − 1/2 k
k1/2
p
1
1
dx
k 1/2 x − 1/21/p
3.11
p n
p
1
k2k
− 1
k1
.
Remark 3.3. If n → ∞, then inequality 3.1 follows from inequality 3.10.
4. A Refinement of Carleman’s Inequality
If an ≥ 0 n ∈ N, n ≥ 1 with 0 <
∞
n1
an < ∞, then the well-known Carleman’s inequality is
n
∞
!
n1
1/n
ak
<e
∞
an ,
4.1
n1
k1
with the best possible constant factor e see 2.
Recently, Yang and Debnath 3 gave a strengthened version of 4.1 as follows:
n
∞
!
n1
1/n
ak
k1
<e
∞ n1
1−
1
an .
2n 2
4.2
Some other strengthened versions of 4.1 were given in 4–9. In this section, we give a
refinement for Carleman’s inequality see Corollary 4.4.
Lemma 4.1. If m ∈ N and m ≥ 1, then
∞
1
2
1
>
,
3m 7 m km kk!1/k
2
1
1
e 1−
>
.
3m 10 m 1 m 1!1/m1
e 1−
4.3
4.4
10
Journal of Inequalities and Applications
Proof. Let ψm e1−2/3m71/m−
is equivalent to inequality
1−
∞
km 1/kk!
1/k
, then inequality ψm > ψm1
2m
m1
2m 2
>
.
3m 7 3m 10 em!1/m
4.5
If 1 ≤ m ≤ 16, then simple computation leads √
to inequality 4.5.
If m ≥ 17, then it is not difficult to verify that 2πm ≥ e7/3 and
√
2
2
2πm ≥ e21m 71m70/9m 39m50 .
4.6
If x > 0, then e > 1 1/xx ; this implies that
e>
21m2 71m 70
1
9m2 39m 50m
9m2 39m50m/21m2 71m70
.
4.7
From inequalities 4.6 and 4.7, we get
√
2πm >
21m2 71m 70
1
9m2 39m 50m
2πm1/2m >
m
,
m 13m 73m 10
,
m9m2 39m 50
4.8
m5
2m
m1
>
.
3m 7 3m 10 m2πm1/2m
From the well-known Stirling Formula m! get
m! >
√
2πmm/em expθm /12m 0 < θm < 1, we
m
√
m
2πm
.
e
4.9
Therefore, inequality 4.5 follows from inequalities 4.8 and 4.9.
From the monotonicity of sequence {ψm}∞
m1 and limm → ∞ ψm 0, we get ψm >
0; therefore, inequality 4.3 is proved.
Journal of Inequalities and Applications
11
Meanwhile, we have
"
2πm 1 > e2/3 ,
"
"
2πm 1 > e2m2/3m8 ,
2πm 1 >
1
2
3m 8
3m8/2·2m2/3m8
,
4.10
3m 10
,
2πm 11/2m2 >
3m 8
2
1
e
e 1−
>
.
3m 10 m 1 m 12πm 11/2m2
Therefore, inequality 4.4 follows from inequalities 4.10 and
m 1! >
#
m1
2πm 1
e
m1
4.11
.
Theorem 4.2. Let n ∈ N, n ≥ 1, and ak > 0 k 1, 2, . . . , n. If Bn min1≤k≤n {kak }, then
e
n k1
⎛
⎞1/k
k
n
n n
!
1
2
2
1 ⎝
⎠
.
−
aj
≥ Bn e
1−
1−
ak −
1/k
3k 7
3k
7
k
j1
k1
k1
k1 k!
4.12
Proof. Let bk kak , k 1, 2, . . . , n, and b b1 , b2 , . . . , bn ,
Dm f :b∈
Rn
b | bm max {bk } > min {bk } > 0 ,
1≤k≤n
−→ e
n k1
1≤k≤n
n
⎛
k
1!
bk
2
− ⎝
1−
3k 7 k k1 k!
m 1, 2, . . . , n,
⎞1/k
bj ⎠
4.13
,
b∈
Rn .
j1
Then inequality 4.12 is equivalent to the following inequality:
e
n k1
⎛
⎞1/k
k
n
n n
!
1
bk 1 1
2
2
⎝
⎠
,
−
−
1−
1−
bj
≥ Bn e
3k 7 k k1 k! j1
3k 7 k k1 k!1/k
k1
where Bn min1≤k≤n {bk }.
4.14
12
Journal of Inequalities and Applications
If b ∈ Dm m 1, 2, . . . , n, then
⎛
⎞1/k
k
n
∂fb
1 1 ⎝1!
2
−
e 1−
bj ⎠
∂bm
3m 7 m km kbm k! j1
>e 1−
n
1 1
2
−
3m 7 m km kk!1/k
>e 1−
∞
1
2
1 −
.
3m 7 m km kk!1/k
4.15
From inequality 4.3 and ∂fb/∂bm > 0 together with Theorem 1.1, we clearly see that
fb1 , b2 , . . . , bn ≥ fBn , Bn , . . . , Bn .
4.16
Therefore, inequality 4.14 is proved.
Corollary 4.3. Let n ∈ N, n ≥ 1, and ak > 0 k 1, 2, . . . , n. If Bn min1≤k≤n {kak }, then
e
n k1
⎛
⎞1/k
k
n
!
2
4
e−1 .
ak − ⎝ aj ⎠
1−
≥ Bn
3k 7
5
j1
k1
4.17
m
1/k
m 1, 2, . . . , n, then
Proof. Let T m e m
k1 1 − 2/3k 71/k −
k1 1/k!
n
inequality 4.4 implies that {T m}m1 is a strictly increasing sequence. Then from inequality
4.12 we get
e
n k1
⎛
⎞1/k
k
n
!
2
4
ak − ⎝ aj ⎠ ≥ Bn T n ≥ Bn T 1 Bn
e−1 .
1−
3k 7
5
j1
k1
4.18
Let n → ∞; thus, we know that Corollary 4.4 is true.
Corollary 4.4. If an ≥ 0 n ∈ N, n ≥ 1 with 0 <
∞
n1
an < ∞, then
⎛
⎞1/n
n
∞
∞ !
⎝ aj ⎠
≤e
1−
n1
j1
n1
2
an .
3n 7
4.19
Remark 4.5. Many other applications for Theorem 1.1 appeared in 10.
Acknowledgments
The authors wish to thank the anonymous referees for their very careful reading of the
manuscript and fruitful comments and suggestions. This work was partly supported by the
Journal of Inequalities and Applications
13
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.
References
1 G. H. Hardy, J. E. Littlewood, and G. Pólya, Inequalities, Cambridge University Press, Cambridge, UK,
1952.
2 T. Carleman, “Sur les fonctions quasi-analytiques,” in Comptes rendus du Ve Congres des Mathematiciens,
Scandinaves, pp. 181–196, Helsinki, Finland, 1922.
3 B. Yang and L. Debnath, “Some inequalities involving the constant e, and an application to
Carleman’s inequality,” Journal of Mathematical Analysis and Applications, vol. 223, no. 1, pp. 347–353,
1998.
4 H. Alzer, “On Carleman’s inequality,” Portugaliae Mathematica, vol. 50, no. 3, pp. 331–334, 1993.
5 H. Alzer, “A refinement of Carleman’s inequality,” Journal of Approximation Theory, vol. 95, no. 3, pp.
497–499, 1998.
6 M. Johansson, L.-E. Persson, and A. Wedestig, “Carleman’s inequality-history, proofs and some new
generalizations,” JIPAM Journal of Inequalities in Pure and Applied Mathematics, vol. 4, no. 3, article 53,
19 pages, 2003.
7 H. P. Liu and L. Zhu, “New strengthened Carleman’s inequality and Hardy’s inequality,” Journal of
Inequalities and Applications, vol. 2007, Article ID 84104, 7 pages, 2007.
8 J. Pečarić and K. B. Stolarsky, “Carleman’s inequality: history and new generalizations,” Aequationes
Mathematicae, vol. 61, no. 1-2, pp. 49–62, 2001.
9 G. I. Sunouchi and N. Takagi, “A generalization of the Carleman’s inequality theorem,” Proceedings of
the Physico-Mathematical Society of Japan, vol. 16, pp. 164–166, 1934.
10 X. M. Zhang and Y. M. Chu, New Discussion to Analytic Inequality, Harbin Institute of Technology
Press, Harbin, China, 2009.