Hindawi Publishing Corporation
Abstract and Applied Analysis
Volume 2010, Article ID 108920, 12 pages
doi:10.1155/2010/108920
Research Article
Sharp Power Mean Bounds for the Combination of
Seiffert and Geometric Means
Yu-Ming Chu,1 Ye-Fang Qiu,2 and Miao-Kun Wang2
1
2
Department of Mathematics, Huzhou Teachers College, Huzhou 313000, China
Department of Mathematics, Zhejiang Sci-Tech University, Hangzhou 310018, China
Correspondence should be addressed to Yu-Ming Chu, chuyuming2005@yahoo.com.cn
Received 29 March 2010; Revised 23 July 2010; Accepted 29 July 2010
Academic Editor: Lance Littlejohn
Copyright q 2010 Yu-Ming Chu et al. 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 answer the question: for α ∈ 0, 1, what are the greatest value p and the least value q such
that the double inequality Mp a, b < P α a, bG1−α a, b < Mq a, b holds for all a, b > 0 with a / b.
Here, Mp a, b, P a, b, and Ga, b denote the power of order p, Seiffert, and geometric means of
two positive numbers a and b, respectively.
1. Introduction
For p ∈ R, the power mean Mp a, b of order p and the Seiffert mean P a, b of two positive
numbers a and b are defined by
⎧
⎪
ap bp 1/p
⎪
⎨
, p/
0,
2
Mp a, b ⎪
⎪
⎩√ab,
p 0,
⎧
a−b
⎪
⎪
, a/
b,
⎨
a/b − π
P a, b 4 arctan
⎪
⎪
⎩a,
a b.
1.1
The main properties of the power mean are given in 1. It is well known that Mp a, b is
strictly increasing with respect to p for fixed a, b > 0 with a /
b. Recently, the power mean
has been the subject of intensive research. In particular, many remarkable inequalities for the
power mean can be found in the literature 2–16.
2
Abstract and Applied Analysis
The Seiffert mean P was introduced by Seiffert in 17, it can be rewritten in the
following symmetric form see 18, 2.4:
P a, b ⎧
⎪
⎨
a−b
,
2 arcsina − b/a b
⎪
⎩a,
a
/ b,
a b.
1.2
√
b−log a, b /
a,
Let Aa, b 1/2a b, Ga, b ab, La, b b−a/log
Ha, b a,
ba,
b
a 1/b−a
, b/
a,
be the arithmetic, geometric, logarithmic,
2ab/a b and Ia, b 1/eb /aa, ba
harmonic, and identric means of two positive numbers a and b, respectively. Then it is well
known that
min{a, b} < Ha, b M−1 a, b < Ga, b M0 a, b
< La, b < Ia, b < Aa, b M1 a, b < max{a, b}
1.3
for all a, b > 0 with a /
b.
In 9, Alzer and Janous presented the sharp power mean bounds for the sum
2/3Aa, b 1/3Ga, b as follows:
Mlog 2/ log 3 a, b <
1
2
Aa, b Ga, b < M2/3 a, b
3
3
1.4
for all a, b > 0 with a /
b.
In 17, Seiffert proved that
La, b < P a, b < Ia, b
1.5
for all a, b > 0 with a / b.
The following power mean bounds for the Seiffert mean was given by Jagers 19:
M1/2 a, b < P a, b < M2/3 a, b
1.6
for all a, b > 0 with a /
b.
In 20, 21, the authors presented the bounds for the Seiffert mean P in terms of A and
G as follows:
P a, b >
3Aa, bGa, b
,
Aa, b 2Ga, b
1
1
2
1
Aa, b Ga, b < P a, b < Aa, b Ga, b,
2
2
3
3
P a, b > A1/3 a, bG2/3 a, b
for all a, b > 0 with a /
b.
1.7
Abstract and Applied Analysis
3
The following sharp lower power mean bounds for 1/3Ga, b 2/3Ha, b,
2/3Ga, b 1/3Ha, b, and P a, b can be found in 4, 6:
2
1
Ga, b Ha, b > M−2/3 a, b,
3
3
1
2
Ga, b Ha, b > M−1/3 a, b,
3
3
1.8
P a, b > Mlog 2/ log π a, b
for all a, b > 0 with a /
b.
The purpose of this paper is to answer the question: for α ∈ 0, 1, what are the greatest
value p and the least value q such that the double inequality Mp a, b < P α a, bG1−α a, b <
b.
Mq a, b holds for all a, b > 0 with a /
2. Lemmas
In order to prove our main result, we need several lemmas which we present in this section.
Lemma 2.1. Let λ ∈ 0, 1/3, x ∈ 1, ∞ and hx 1−3λx2λ1 −13λx2λ −13λx1−3λ.
Then there exists x0 > 1 such that hx < 0 for x ∈ 1, x0 , hx > 0 for x ∈ x0 , ∞ and hx0 0.
Proof. Simple computations lead to
h1 −12λ < 0,
2.1
lim hx ∞,
2.2
h x 1 − 3λ2λ 1x2λ − 2λ1 3λx2λ−1 − 1 3λ,
h 1 −6λ1 2λ < 0,
2.3
lim h x ∞,
2.4
h x 2λx2λ−2 1 2λ1 − 3λx 1 − 2λ1 3λ > 0.
2.5
x → ∞
x → ∞
Inequality 2.5 implies that h x is strictly increasing in 1, ∞. Then 2.3 and 2.4
lead to that there exists x1 > 1 such that h x < 0 for x ∈ 1, x1 and h x > 0 for x ∈ x1 , ∞.
Hence, hx is strictly decreasing in 1, x1 and strictly increasing in x1 , ∞.
Therefore, Lemma 2.1 follows from 2.1 and 2.2 together with the monotonicity of
hx.
√
Lemma 2.2. If λ ∈ 1/ 10, 1/3, then the following statements are true:
1 300λ4 324λ3 29λ2 − 45λ − 8 < 0;
2 −1176λ5 − 24λ4 114λ3 10λ2 − 3λ − 1 < 0;
3 −24λ5 72λ4 178λ3 6λ2 − 25λ − 3 < 0.
4
Abstract and Applied Analysis
Proof. Simple computations lead to
√
3
29λ2 − 45λ − 8 < 300 × 1/34 324 × 1/33 29 × 1/32 − 45/ 10 − 8
1 300λ4 324λ√
590 − 243 10/54 < 0;
√ 5
√ 4
2 −1176λ5 − 24λ4 114λ3 √10λ2 − 3λ − 1 < −1176 √
× 1/ 10 − 24 × 1/ 10 114 ×
1/33 10 × 1/32 − 3/ 10 − 1 3070 − 1107 10/750 < 0;
√ 5
10 72 × 1/34 3 −24λ5 72λ4 178λ3 6λ2 √− 25λ − 3 < −24 × 1/
√
3
2
178 × 1/3 6 × 1/3 − 25/ 10 − 3 34750 − 17037 10/6750 < 0.
√
Lemma 2.3. If α ∈ 0, 3/ 10, then
I α a, bG1−α a, b < M2α/3 a, b
2.6
holds for all a, b > 0 with a /
b.
Proof.√Without loss of generality, we assume that a > b. Let t a/b > 1 and β α/3 ∈
0, 1/ 10. Then
logM2α/3 a, b − log I α a, bG1−α a, b
log M2β a, b − log I 3β a, bG1−3β a, b
3βt
1 − 3β
1 t2β
1
log
−
log t −
log t 3β.
2β
2
t−1
2
ft 3βt
1 − 3β
1 t2β
1
log
−
log t −
log t 3β.
2β
2
t−1
2
2.7
Let
2.8
Then simple computations lead to
limft 0,
2.9
t→1
f t gt
t − 12
,
2.10
where gt t2β1 − 2t2β t2β−1 / 1 t2β − 3βt − 1 3β log t − 1 − 3βt − 2 1/t/2,
g1 0,
g t g1 t
2 ,
2
2t 1 t2β
2.11
Abstract and Applied Analysis
5
where g1 t 1−3βt4β2 6βt4β1 −13βt4β −2βt2β2 4βt2β1 −2βt2β −13βt2 6βt1−3β,
g1 1 0,
2.12
g1 1 0,
2.13
g1 1 0.
2.14
Let g2 t g1 t/8βt2β−3 and g3 t g2 t/2βt2β−3 . Then
g2 t 1 − 3β 2β 1 4β 1 t2β2 3β 4β 1 4β − 1 t2β1
− 1 3β 4β − 1 2β − 1 t2β − β β 1 2β 1 t2
β 2β 1 2β − 1 t − β β − 1 2β − 1 ,
2.15
g2 1 0,
√
√
g2 1 2 1 − 10β 1 10β ≥ 0,
2.17
√
√
g2 1 2 6β 1 1 − 10β 1 10β ≥ 0,
2.18
2 g3 t 2 1 − 3β 2β 1 4β 1 β 1 t2 3β 4β 1 4β − 1
2 × 2β 1 2β − 1 t − 2 1 3β 4β − 1 2β − 1 β − 1 ,
g3 1 9β 3 − 28β2 > 0,
g3 t 2β 1 4β 1 4 1 − 3β 2β 1 β 1 t 3β 4β − 1 2β − 1 ,
g3 1 2β 1 4β 1 4 3β − 46β2 ,
min
β∈0,1/10
4 3β − 46β
2
√
3 10 − 6
> 0.
10
2.16
2.19
2.20
2.21
2.22
2.23
√
From 2.21−2.23 we clearly see that g3 t > g3 1 ≥ 3 10 − 6/102β 14β 1 > 0
for t ∈ 1, ∞, hence g3 t is strictly increasing in 1, ∞. Then 2.20 implies that g3 t > 0 for
t ∈ 1, ∞, hence g2 t is strictly increasing in 1, ∞.
It follows from 2.18 and the monotonicity of g2 t that g2 t > 0 for t ∈ 1, ∞, hence
g2 t is strictly increasing in 1, ∞. Then 2.17 implies that g2 t > 0 for t ∈ 1, ∞, therefore
g2 t is strictly increasing in 1, ∞.
Equation 2.16 and the monotonicity of g2 t lead to that g2 t > 0 for t ∈ 1, ∞, so
g1 t is strictly increasing in 1, ∞.
From 2.9−2.14 and the monotonicity of g1 t we can deduce that
ft > 0
for t ∈ 1, ∞.
2.24
6
Abstract and Applied Analysis
Therefore, Lemma 2.3 follows from 2.7 and 2.8 together with 2.24.
Remark 2.4. In 22, Theorem 3.1, the authors proved that
I 1/2 a, bG1/2 a, b < M1/3 a, b,
2.25
I 2/3 a, bG1/3 a, b < M4/9 a, b,
2.26
I 1/3 a, bG2/3 a, b < M2/9 a, b
2.27
for all a, b > 0 with a /
b.
Obviously, 2.6 is the generalization of 2.25−2.27.
√
√
Remark 2.5. If α ∈ 3/ 10, 1, then β α/3 ∈ 1/ 10, 1/3 and 2.17 leads to
√
√
g2 1 2 1 − 10β 1 10β < 0.
2.28
Inequality 2.28 and the continuity of g2 t imply that there exists δ > 0 such that
g 2 t < 0
2.29
for t ∈ 1, 1 δ.
From 2.29 and 2.9−2.16 we can deduce that
ft < 0
2.30
for t ∈ 1, 1 δ.
Equations 2.7 and 2.8 together with 2.30 lead to
I α a, bG1−α a, b > M2α/3 a, b
2.31
for all a, b > 0 with a/b ∈√
1, 1 δ ∪ 1/1 δ, 1.
Therefore, α0 3/ 10 is the largest value in 0, 1 such that inequality 2.6 holds for
α ∈ 0, α0 .
3. Main Result
Theorem 3.1. If α ∈ 0, 1, then
M0 a, b < P α a, bG1−α a, b < M2α/3 a, b
3.1
holds for all a, b > 0 with a /
b, and M0 a, b and M2α/3 a, b are the best possible lower and upper
power mean bounds for the product Pα a, bG1−α a, b.
Abstract and Applied Analysis
7
Proof. For all a, b > 0 with a /
b, from 1.3, 1.5 and Lemma 2.1 we clearly see that
P α a, bG1−α a, b > M0 a, b
3.2
P α a, bG1−α a, b < M2α/3 a, b
3.3
for all α ∈ 0, 1, and
√
for α ∈ 0, 3/ 10.
√
b.
Next, we prove that 3.3 is also true for α ∈ 3/ 10, 1 and all a, b > 0 with a /
Without
loss
of
generality,
we
assume
that
a
>
b.
Let
t
a/b
>
1
and
β
α/3 ∈
√
1/ 10, 1/3. Then 1.1 leads to
logM2α/3 a, b − log P α a, bG1−α a, b
log M2β a, b − log P 3β a, bG1−3β a, b
1
1 t4β
t2 − 1
log
− 3β log
− 1 − 3β log t.
2β
2
4 arctan t − π
Ft 1
1 t4β
t2 − 1
log
− 3β log
− 1 − 3β log t.
2β
2
4 arctan t − π
3.4
Let
3.5
Then simple computations lead to
limFt 0,
F t t→1
1 − 3β t4β2 − 1 3β t4β − 1 3β t2 1 − 3β
12β
.
2
t 14 arctan t − π
tt2 − 1 t4β 1
3.6
3.7
From Lemma 2.1 we know that there exists λ0 > 1 such that
1 − 3β λ0 4β2 − 1 3β λ0 4β − 1 3β λ0 2 1 − 3β 0,
1 − 3β t4β2 − 1 3β t4β − 1 3β t2 1 − 3β < 0
3.9
1 − 3β t4β2 − 1 3β t4β − 1 3β t2 1 − 3β > 0
3.10
3.8
for t ∈ 1, λ0 , and
for t ∈ λ0 , ∞.
8
Abstract and Applied Analysis
We divide two cases to prove that
F t > 0
3.11
for t > 1.
Case 1. t ∈ λ0 , ∞. Then 3.11 follows from 3.7 and 3.8 together with 3.10.
Case 2. t ∈ 1, λ0 . Then 3.7 can be written as
1 − 3β t4β2 − 1 3β t4β − 1 3β t2 1 − 3β
F1 t,
F t tt2 − 1 t4β 1 4 arctan t − π
3.12
where
12βt t2 − 1 t4β 1
F1 t 4 arctan t 2
− π.
t 1 1 − 3β t4β2 − 1 3β t4β − 1 3β t2 1 − 3β
3.13
Let x t2 ∈ 1, λ0 2 , then 3.13 leads to
F1 1 0,
F1 t t2 1
4F2 x
2 ,
4β2
− 1 3β t4β − 1 3β t2 1 − 3β
1 − 3β t
2 3.14
where
F2 x 18β2 − 9β 1 x4β3 − 18β2 − 3β 1 x4β2
− 18β2 3β 1 x4β1 18β2 9β 1 x4β
2 6β2 − 1 x2β3 − 2 6β2 − 1 x2β2
− 2 6β2 − 1 x2β1 2 6β2 − 1 x2β
18β2 9β 1 x3 − 18β2 3β 1 x2
− 18β2 − 3β 1 x 18β2 − 9β 1 .
3.15
Abstract and Applied Analysis
9
Let F3 x x4−2β /8βF2 4 x, then 3.15 leads to
F2 1 0,
F2 1 0,
3.16
F2 1 0,
F2 1 0,
F3 x 18β2 − 9β 1 4β 1 4β 3 2β 1 x2β3
− 18β2 − 3β 1 2β 1 4β 1 4β − 1 x2β2
− 18β2 3β 1 4β 1 4β − 1 2β − 1 x2β1
18β2 9β 1 4β − 1 2β − 1 4β − 3 x2β
6β − 1 2β 3 β 1 2β 1 x3
3.17
2
− 6β2 − 1 β 1 2β 1 2β − 1 x2
− 6β2 − 1 2β 1 2β − 1 β − 1 x
6β2 − 1 2β − 1 β − 1 2β − 3 .
√
From β ∈ 1/ 10, 1/3 and x > 1, we clearly see that 18β2 − 9β 14β 14β 32β 1x2β3 < 18β2 − 9β 14β 14β 32β 1x3 , −18β2 − 3β 12β 14β 14β − 1x2β2 <
−18β2 − 3β 12β 14β 14β − 1x2 , −18β2 3β 14β 14β − 12β − 1x2β1 <
−18β2 3β 14β 14β − 12β − 1x2 and 18β2 9β 14β − 12β − 14β − 3x2β <
18β2 9β 14β − 12β − 14β − 3x. These inequalities and Lemma 2.2 lead to
F3 x < 18β2 − 9β 1 4β 1 4β 3 2β 1 x3
− 18β2 − 3β 1 2β 1 4β 1 4β − 1 x2
− 18β2 3β 1 4β 1 4β − 1 2β − 1 x2
18β2 9β 1 4β − 1 2β − 1 4β − 3 x
6β2 − 1 2β 3 β 1 2β 1 x3
− 6β2 − 1 β 1 2β 1 2β − 1 x2
− 6β2 − 1 2β 1 2β − 1 β − 1 x
6β2 − 1 2β − 1 β − 1 2β − 3
10
Abstract and Applied Analysis
2β 300β4 324β3 29β2 − 45β − 8 x3
−1176β5 − 24β4 114β3 10β2 − 3β − 1 x2
2 276β5 − 276β4 3β3 43β2 − 3β − 1 x
24β5 − 72β4 62β3 − 6β2 − 11β 3
< 2β 300β4 324β3 29β2 − 45β − 8 x
−1176β5 − 24β4 114β3 10β2 − 3β − 1 x
2 276β5 − 276β4 3β3 43β2 − 3β − 1 x
24β5 − 72β4 62β3 − 6β2 − 11β 3
−24β5 72β4 178β3 6β2 − 25β − 3 x
24β5 − 72β4 62β3 − 6β2 − 11β 3
< −24β5 72β4 178β3 6β2 − 25β − 3
24β5 − 72β4 62β3 − 6β2 − 11β 3
12β 20β2 − 3 < 0.
3.18
From 3.16−3.18 we can deduce that
F2 x < 0
3.19
for x ∈ 1, λ0 2 .
Equation 3.14 together with 3.19 imply that
F1 t < 0
3.20
for t ∈ 1, λ0 .
Therefore, 3.11 follows from 3.9 and 3.12 together with 3.20.
It follows from 3.4−3.6 and 3.11 that
P α a, bG1−α a, b < M2α/3 a, b
3.21
√
for α ∈ 3/ 10, 1 and all a, b > 0 with a /
b.
At last, we prove that M0 a, b and M2α/3 a, b are the best possible lower- and upperpower mean bounds for the product P α a, bG1−α a, b, respectively.
Abstract and Applied Analysis
11
For any 0 < ε < 2/3α and x > 0, from 1.1 one has
Mε 1, x
lim
x → ∞ P α 1, xG1−α 1, x
lim
x → ∞
1 x−ε /21/ε
1−α/2
∞,
α x
√
1 − x−1 / 4 arctan x − π
3.22
2α/3−ε 2α/3−ε
P α 1, 1 xG1−α 1, 1 x
− M2α/3−ε 1, 1 x
Jx
√
4 arctan 1 x − π
α2α−3ε/3 ,
3.23
√
where Jx xα2α−3ε/3 1 x1−α2α−3ε/6 − 1/2 4 arctan 1 x − πα2α−3ε/3 1 1 x2α−3ε/3 .
Let x → 0, making use of the Taylor expansion we get
Jx 1
ε2α − 3εx1/3α2α−3ε2 o x1/3α2α−3ε2 .
24
3.24
Equation 3.22 implies that for any 0 < ε < 2/3α there exists X Xε, α > 1 such
that Mε 1, x > P α 1, xG1−α 1, x for x ∈ X, ∞.
Equations 3.23 and 3.24 imply that for any 0 < ε < 2/3α there exists δ δε, α >
0 such that P α 1, 1 xG1−α 1, 1 x > M2α/3−ε 1, 1 x for x ∈ 0, δ.
Acknowledgments
The authors wish to thank the anonymous referees for their very careful reading of the paper
and fruitful comments and suggestions. This paper is partly supported by N S Foundation of
China under Grant 60850005, N S Foundation of Zhejiang Province under Grants Y7080106
and Y607128, and the Innovation Team Foundation of the Department of Education of
Zhejiang Province under Grant T200924.
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