Hindawi Publishing Corporation
Journal of Inequalities and Applications
Volume 2011, Article ID 686834, 9 pages
doi:10.1155/2011/686834
Research Article
The Optimal Convex Combination Bounds for
Seiffert’s Mean
Hong Liu
1
and Xiang-Ju Meng
2
1
College of Mathematics and Computer Science, Hebei University, Baoding 071002, China
2
Department of Mathematics, Baoding College, Baoding 071002, China
Correspondence should be addressed to Hong Liu,
Received 28 November 2010; Accepted 28 February 2011
Academic Editor: P. Y. H. Pang
Copyright q 2011 H. Liu and X J. Meng. 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 derive some optimal convex combination bounds related to Seiffert’s mean. We find the
greatest values α
1
, α
2
and the least values β
1
, β
2
such that the double inequalities α
a −b
4arctan
a/b
− π
.
1.1
Recently, the inequalities for means have been the subject of intensive research. In particular,
many remarkable inequalities for P can be found in the literature 2–6.Seiffert’s mean P can
be rewritten as see 5,equation2.4
P
a, b
a −b
2arcsin
a −b
/
a b
. 1.2
2 Journal of Inequalities and Applications
Let Ca, ba
<C
a, b
< max
{
a, b
}
. 1.3
In 7,Seiffert proved that
P
a, b
>
3A
a, b
G
a, b
A
a, b
2G
a, b
<P
a, b
<βA
a, b
1 −β
H
a, b
1.5
holds for all a, b > 0witha
/
b.
For more results, see 9–23.
The purpose of the present paper is to find the greatest values α
1
,α
2
and the least
values β
1
,β
2
G
a, b
,
α
2
C
a, b
1 − α
2
H
a, b
<P
a, b
<β
2
C
a, b
2/9 and β
1
1/π.
Proof. Firstly, we prove that
P
a, b
<
1
π
C
a, b
1 −
1
π
G
a, b
,
P
a, b
pC
a, b
1 −p
G
a, b
bP
t
2
, 1
− b
pC
t
2
, 1
4arctant − π
f
t
,
2.2
where
f
t
t
4
− 1
pt
4
1 −p
t
3
1 −p
2
t
2
1
pt
4
1 −p
t
3
1 −p
t p
2
g
t
,
2.4
where
t
3
− 3
5p − 1
t
2
− 2
5p − 1
t − 4p
2
− p 1.
2.5
We divide the proof into two cases.
Case 1 p 2/9.Inthiscase,
g
t
1
81
47t
4
76t
3
2 −9p
8
2 −
9
π
< 0, lim
t →∞
g
t
∞, 2.8
g
t
−6
4p
2
p − 1
t
5
− 10
24
2 −9p
24
2 −
9
π
< 0, lim
t →∞
g
t
∞, 2.10
g
t
−30
4p
2
p −1
t
8
17 −
70
π
−
9
π
2
< 0, lim
t →∞
g
t
∞, 2.12
g
t
−120
4p
2
p − 1
25
π
−
9
π
2
< 0, lim
t →∞
g
t
∞, 2.14
g
4
t
−360
4p
2
p − 1
t
2
− 240
∞, 2.16
g
5
t
−720
4p
2
p − 1
t −1200p 240, 2.17
g
5
1
960
1 −2p − 3p
2
960
1 −
2
π
−
3
exists λ
2
> 1suchthatg
t < 0fort ∈ 1,λ
2
and g
t > 0fort ∈ λ
2
, ∞;henceg
t is
strictly decreasing in 1,λ
2
and strictly increasing for λ
2
, ∞. As this goes on, there exists
λ
3
> 1suchthatft is strictly decreasing in 1,λ
3
and strictly increasing in λ
3
, ∞.Note
that if p 1/π, then the second equality in 2.4 becomes
lim
t →∞
f
P
t
2
, 1
<α
1
C
t
2
, 1
1 − α
1
G
t
2
, 1
, 2.21
for t ∈ 1, 1 δ.
Finally, we prove that 1/πCa, b1 −1/πGa, b is the best possible upper convex
combination bound of the contraharmonic and geometric means for Seiffert’s mean.
If β
t →∞
β
1
t
4
1 −β
1
t
3
1 −β
1
t β
1
4arctant − π
t
4
− 1
β
1
π<1.
2.23
for t ∈ X, ∞.
Secondly, we present the optimal convex combination bounds of the contraharmonic
and harmonic means for Seiffert’s mean as follows.
Theorem 2.2. The double inequality α
2
Ca, b1 − α
2
Ha, b <Pa, b <β
2
Ca, b1 −
β
2
Ha, b holds for all a, b > 0 with a
/
b ifandonlyifα
2
1/π and β
2
5/12.
Proof. Firstly, we prove that
P
a, b
<
5
12
C
a, b
,
2.24
for all a, b > 0witha
/
b.
6 Journal of Inequalities and Applications
Without loss of generality, we assume that a>b.Lett
a/b > 1andp ∈
{1/π, 5/12}.Then1.1 leads to
P
a, b
−
pC
a, b
1 −p
H
a, b
4
2
1 −p
t
2
p
t
2
1
4arctant − π
f
t
,
2.25
where
f
t
t
p
− π,
f
t
4
t −1
2
t
2
1
pt
4
2
1 −p
t
2
p
2
4
2
2p
2
− 5p 2
t
3
p
2
− 6p 2
t
2
−2p
2
− p 1
t −p
2
.
2.28
We divide the proof into two cases.
Case 1 p 5/12.Inthiscase,
g
12
5
− π<0.
2.30
Journal of Inequalities and Applications 7
Case 2 p 1/π.From2.28 we have that
g
1
2
5 −12p
2
5 −
12
π
> 0, lim
t →∞
g
t
−∞, 2.31
g
2
p
2
− 6p 2
t −2p
2
− p 1,
2.32
g
t
6
5 −12p
6
5 −
12
π
> 0, lim
t →∞
g
2p
2
− 5p 2
t 2p
2
− 12p 4,
2.34
g
t
4
18 − 41p − 8p
2
4
18 −
41
π
−
8
π
2
> 0, lim
t →∞
2
24p
2
− 60p 24,
2.36
g
1
12
11 − 22p − 16p
2
12
11 −
22
π
−
16
π
2
> 0, lim
t →∞
g
24
7 −
11
π
−
24
π
2
> 0, lim
t →∞
g
t
−∞, 2.39
g
5
t
−720p
2
t −240p
2
− 120p 120, 2.40
g
t > 0fort ∈ 1,λ
4
and g
4
t < 0fort ∈ λ
4
, ∞, and hence g
t is strictly increasing
in 1,λ
4
and strictly decreasing for λ
1
, ∞.From2.37 and the monotonicity of g
t,there
exists λ
5
> 1suchthatg
t > 0fort ∈ 1,λ
5
and g
t < 0fort ∈ λ
5
, ∞;henceg
t is
strictly increasing in 1,λ
g
1
2
5 −12β
2
> 0. 2.43
From this result and the continuity of gt we clearly see that there e xists δ δβ
2
> 0
such that gt > 0fort ∈ 1, 1 δ. Then the last equality in 2.27 implies that f
t > 0for
t ∈ 1 , 1 δ.Thusft is increasing for t ∈ 1, 1 δ.Dueto2.27, ft > 0fort ∈ 1, 1 δ,
which is equivalent to, by 2.25,
P
t
2
, 1
>β
2
C
t
2
1 −α
2
H
t
2
, 1
P
t
2
, 1
lim
t →∞
α
2
t
4
− 2
1 −α
2
t
2
, 1
1 −α
2
H
t
2
, 1
>P
t
2
, 1
2.46
for t ∈ X, ∞.
Acknowledgments
The authors wish to thank the anonymous referees for their very careful reading of the paper
and fruitful comments and suggestions. This research is partly supported by N S Foundation
of Hebei Province Grant A2011201011, and the Youth Foundation of Hebei University
Grant 2010Q24.
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