# Monotonicity of the Ratio of the Power and Second Seiffert Means with Applications

Authors: Zhen-Hang Yang; Ying-Qing Song; Yu-Ming Chu
Journal: Abstract and Applied Analysis
 (2014)

---

## Abstract

We present the necessary and sufficient condition for the monotonicity of the ratio of the power and second Seiffert means. As applications, we get the sharp upper and lower bounds for the second Seiffert mean in terms of the power mean.


---

## Body

## 1. Introduction

Throughout this paper, we assume thata
,
b
>
0 with a
≠
b. The second Seiffert mean T
(
a
,
b
) and rth power mean M
r
(
a
,
b
) of a and b are defined by
(1)
T
(
a
,
b
)
=
a
-
b
2
arctan
(
(
a
-
b
)
/
(
a
+
b
)
)
,
(2)
M
r
(
a
,
b
)
=
(
a
r
+
b
r
2
)
1
/
r
(
r
≠
0
)
,
M
0
(
a
,
b
)
=
a
b
,
respectively.It is well-known that the power meanM
r
(
a
,
b
) is strictly increasing with respect to r
∈
R for fixed a
,
b
>
0 with a
≠
b. In the recent past, both mean values have been the subject of intensive research. In particular, many remarkable inequalities for T and M
r can be found in the literature [1–5].Seiffert [6] proved that the double inequality
(3)
M
1
(
a
,
b
)
<
T
(
a
,
b
)
<
M
2
(
a
,
b
)
holds for all a
,
b
>
0 with a
≠
b.In [7], Hästö proved that the function T
(
1
,
x
)
/
M
p
(
1
,
x
) is strictly increasing on [
1
,
∞
) if p
≤
1 and presented an improvement for the first inequality in (3).Costin and Toader [8] proved that the inequality
(4)
T
(
a
,
b
)
>
M
3
/
2
(
a
,
b
)
holds for all a
,
b
>
0 with a
≠
b.In [9], Witkowski proved that the double inequality
(5)
2
2
π
M
2
(
a
,
b
)
<
T
(
a
,
b
)
<
4
π
M
1
(
a
,
b
)
holds for all a
,
b
>
0 with a
≠
b.Recently, the following optimal estimations for the second Seiffert mean by power means were obtained independently in [10, 11]:
(6)
M
log
⁡
2
/
(
log
⁡
π
-
log
⁡
2
)
(
a
,
b
)
<
T
(
a
,
b
)
<
M
5
/
3
(
a
,
b
)
for all a
,
b
>
0 with a
≠
b.The main purpose of this paper is to give the necessary and sufficient condition for the monotonicity of the functionT
(
1
,
x
)
/
M
p
(
1
,
x
) on (
0,1
) and present the best possible parameters α and β such that the double inequality
(7)
α
M
5
/
3
(
a
,
b
)
<
T
(
a
,
b
)
≤
β
M
log
⁡
2
/
(
log
⁡
π
-
log
⁡
2
)
(
a
,
b
)
holds for all a
,
b
>
0 with a
≠
b.


## 2. Main Results

In order to prove our main results we first establish a lemma.Lemma 1.
Letf
(
p
,
x
) be defined on R
×
(
0,1
) by
(8)
f
(
p
,
x
)
=
(
1
-
x
)
(
1
+
x
p
)
(
1
+
x
2
)
(
1
+
x
p
-
1
)
-
arctan
1
-
x
1
+
x
.
Then there exists λ
∈
(
0,1
) such that f
(
p
,
x
) is strictly decreasing with respect to x on (
0
,
λ
] and strictly increasing with respect to x on [
λ
,
1
) if p
∈
(
1,5
/
3
).Proof.
Let(9)
f
1
(
p
,
x
)
=
(
1
-
p
)
x
p
+
(
1
+
p
)
x
p
-
1
-
(
1
+
p
)
x
p
-
2
+
(
p
-
1
)
x
p
-
3
-
2
x
2
p
-
3
+
2
.
Then,
(10)
f
1
(
p
,
1
)
=
0
,
f
1
(
p
,
0
+
)
=
∞
,
(11)
∂
f
(
p
,
x
)
∂
x
=
-
x
(
1
-
x
)
(
1
+
x
2
)
2
(
1
+
x
p
-
1
)
2
f
1
(
p
,
x
)
,
(12)
x
4
-
p
∂
f
1
(
p
,
x
)
∂
x
=
-
2
(
2
p
-
3
)
x
p
-
p
(
p
-
1
)
x
3
+
(
p
-
1
)
(
p
+
1
)
x
2
-
(
p
+
1
)
(
p
-
2
)
x
+
(
p
-
1
)
(
p
-
3
)
∶
=
f
2
(
p
,
x
)
,
(13)
f
2
(
p
,
0
)
=
(
p
-
1
)
(
p
-
3
)
<
0
,
f
2
(
p
,
1
)
=
2
(
5
-
3
p
)
>
0
,
(14)
∂
f
2
(
p
,
x
)
∂
x
=
-
2
p
(
2
p
-
3
)
x
p
-
1
-
3
p
(
p
-
1
)
x
2
+
2
(
p
-
1
)
(
p
+
1
)
x
-
(
p
+
1
)
(
p
-
2
)
.
We divide two cases to prove that∂
f
2
(
p
,
x
)
/
∂
x
>
0 for all x
∈
(
0,1
) and p
∈
(
1,5
/
3
).Case 1.
Consider thatp
∈
(
1,3
/
2
]. From (14) we clearly see that
(15)
∂
3
f
2
(
p
,
x
)
∂
x
3
=
-
2
p
(
p
-
1
)
[
3
+
(
2
-
p
)
(
3
-
2
p
)
x
p
-
3
]
<
0
,
(16)
∂
f
2
∂
x
(
p
,
0
)
=
(
p
+
1
)
(
2
-
p
)
>
0
,
∂
f
2
∂
x
(
p
,
1
)
=
2
p
(
5
-
3
p
)
>
0
.
Equation (15) implies that ∂
f
2
(
p
,
x
)
/
∂
x is strictly concave with respect to x on the interval (
0,1
). Then (16) and the basic properties of concave function lead to the conclusion that
(17)
∂
f
2
∂
x
(
p
,
x
)
>
(
1
-
x
)
∂
f
2
∂
x
(
p
,
0
)
+
x
∂
f
2
∂
x
(
p
,
1
)
>
0
.Case 2.
Consider thatp
∈
(
3
/
2,5
/
3
). Making use of the weighted arithmetic-geometric inequality λ
a
+
(
1
-
λ
)
b
≥
a
λ
b
1
-
λ
(
0
≤
λ
≤
1
) we get
(18)
x
p
-
1
≤
(
p
-
1
)
x
+
(
2
-
p
)
.
Equations (14) and (18) lead to
(19)
∂
f
2
(
p
,
x
)
∂
x
≥
-
2
p
(
2
p
-
3
)
[
(
p
-
1
)
x
+
(
2
-
p
)
]
-
3
p
(
p
-
1
)
x
2
+
2
(
p
-
1
)
(
p
+
1
)
x
-
(
p
+
1
)
(
p
-
2
)
=
-
3
p
(
p
-
1
)
x
2
-
2
(
p
-
1
)
(
2
p
2
-
4
p
-
1
)
x
+
(
p
-
2
)
(
4
p
2
-
7
p
-
1
)
∶
=
f
3
(
p
,
x
)
.
Note that(20)
∂
2
f
3
(
p
,
x
)
∂
x
2
=
-
6
p
(
p
-
1
)
<
0
,
f
3
(
p
,
1
)
=
2
p
(
5
-
3
p
)
>
0
,
f
3
(
p
,
0
)
=
4
(
p
-
2
)
(
p
-
65
+
7
8
)
(
p
+
65
+
7
8
)
>
0
.
It follows from (20) and the concavity of the function f
3
(
p
,
x
) with respect to x on the interval (
0,1
) that
(21)
f
3
(
p
,
x
)
>
(
1
-
x
)
f
3
(
p
,
0
+
)
+
x
f
3
(
p
,
1
)
>
0
.
Therefore, ∂
f
2
(
p
,
x
)
/
∂
x
>
0 follows from (19) and (21).
Next we prove the desired result. From (12) and (13) together with the fact that ∂
f
2
(
p
,
x
)
/
∂
x
>
0 we clearly see that there exists λ
1
∈
(
0,1
) such that f
1
(
p
,
x
) is strictly decreasing with respect to x on (
0
,
λ
1
] and strictly increasing with respect to x on [
λ
1
,
1
). Therefore, Lemma 1 follows easily from (10) and (11) together with the piecewise monotonicity of f
1
(
p
,
x
) with respect to x on the interval (
0,1
).Theorem 2.
LetF
(
p
,
x
) be defined on R
×
(
0,1
) by
(22)
F
(
p
,
x
)
=
log
⁡
T
(
1
,
x
)
M
p
(
1
,
x
)
=
log
⁡
1
-
x
2
arctan
(
(
1
-
x
)
/
(
1
+
x
)
)
-
1
p
log
⁡
1
+
x
p
2
(
p
≠
0
)
,
(23)
F
(
0
,
x
)
=
lim
⁡
p
→
0
F
(
p
,
x
)
=
log
⁡
1
-
x
2
arctan
(
(
1
-
x
)
/
(
1
+
x
)
)
-
1
2
log
⁡
x
.
Then the following statements are true.(1)
F
(
p
,
x
) is strictly increasing with respect to x on (
0,1
) if and only if p
≥
5
/
3.
(2)
F
(
p
,
x
) is strictly decreasing with respect to x on (
0,1
) if and only if p
≤
1.
(3)
Ifp
∈
(
1,5
/
3
), then there exists μ
∈
(
0,1
) such that F
(
p
,
x
) is strictly increasing with respect to x on (
0
,
μ
] and strictly decreasing with respect to x on [
μ
,
1
).Proof.
It follows from (22) and (23) that
(24)
∂
F
(
p
,
x
)
∂
x
=
1
+
x
p
-
1
x
(
1
-
x
)
(
1
+
x
p
)
arctan
(
(
1
-
x
)
/
(
1
+
x
)
)
f
(
p
,
x
)
,
where f
(
p
,
x
) is defined by (8). And
(25)
∂
f
(
p
,
x
)
∂
x
=
-
x
(
1
-
x
)
(
1
+
x
2
)
2
(
1
+
x
p
-
1
)
g
(
p
,
x
)
,
where
(26)
g
(
p
,
x
)
=
(
1
-
p
)
x
p
+
(
1
+
p
)
x
p
-
1
-
2
x
2
p
-
3
-
(
1
+
p
)
x
p
-
2
+
(
p
-
1
)
x
p
-
3
+
2
.
(1) IfF
(
p
,
x
) is strictly increasing with respect to x on (
0,1
), then (24) leads to f
(
p
,
x
)
>
0 for all x
∈
(
0,1
). Making use of L’Höspital’s rule and (8) we get
(27)
lim
⁡
x
→
1
-
f
(
p
,
x
)
(
1
-
x
)
3
=
1
24
(
3
p
-
5
)
≥
0
,
which implies that p
≥
5
/
3.
Ifp
≥
5
/
3, then from (8) and (26) together with the fact that the function p
→
(
1
+
x
p
)
/
(
1
+
x
p
-
1
) is strictly increasing on R we get
(28)
f
(
p
,
x
)
≥
f
(
5
3
,
x
)
,
(29)
g
(
5
3
,
x
)
=
2
3
x
-
4
/
3
(
1
-
x
1
/
3
)
3
(
1
+
x
2
/
3
)
×
(
1
+
3
x
1
/
3
+
5
x
2
/
3
+
3
x
+
x
4
/
3
)
>
0
for all x
∈
(
0,1
).
Equations (8) and (25) together with inequality (29) lead to the conclusion that
(30)
f
(
5
3
,
x
)
>
f
(
5
3
,
1
)
=
0
for all x
∈
(
0,1
).
Therefore,F
(
p
,
x
) is strictly increasing with respect to x on (
0,1
) which follows easily from (24), (28), and (30).
(2) IfF
(
p
,
x
) is strictly decreasing with respect to x on (
0,1
), then (24) implies that f
(
p
,
x
)
<
0 for all x
∈
(
0,1
). In particular, we have f
(
p
,
0
+
)
≤
0 and p
≤
1. Indeed, if p
>
1, then (8) leads to the conclusion that f
(
p
,
0
+
)
=
1
-
π
/
4
>
0.
Ifp
≤
1, then from (8) and (26) together with the fact that the function p
→
(
1
+
x
p
)
/
(
1
+
x
p
-
1
) is strictly increasing on R we get
(31)
f
(
p
,
x
)
≤
f
(
1
,
x
)
,
(32)
g
(
1
,
x
)
=
4
(
1
-
1
x
)
<
0
for all x
∈
(
0,1
).
Equations (8) and (25) together with inequality (32) lead to the conclusion that
(33)
f
(
1
,
x
)
<
f
(
1,1
)
=
0
for all x
∈
(
0,1
).
Therefore,F
(
p
,
x
) is strictly decreasing with respect to x on (
0,1
) which follows easily from (24), (31), and (33).
(3) Ifp
∈
(
1,5
/
3
), then (8) leads to
(34)
f
(
p
,
0
)
=
1
-
π
4
>
0
,
f
(
p
,
1
)
=
0
.
It follows from Lemma1 and (34) that we clearly see that there exists μ
∈
(
0,1
) such that f
(
p
,
x
)
>
0 for x
∈
(
0
,
μ
) and f
(
p
,
x
)
<
0 for x
∈
(
μ
,
1
). Then from (24) we get Theorem 2(3) immediately.Theorem 3.
For alla
,
b
>
0 with a
≠
b, the double inequality
(35)
α
M
5
/
3
(
a
,
b
)
<
T
(
a
,
b
)
≤
β
M
log
⁡
2
/
(
log
⁡
π
-
log
⁡
2
)
(
a
,
b
)
holds with the best possible constants β
=
e
F
(
log
⁡
2
/
(
log
⁡
π
-
log
⁡
2
)
,
μ
)
=
1.0136
… and α
=
2
8
/
5
/
π
=
0.9649
…, where μ is the solution of the equation f
(
log
⁡
2
/
(
log
⁡
π
-
log
⁡
2
)
,
x
)
=
0 on (
0,1
)
and
f
(
p
,
x
) and F
(
p
,
x
) are defined by (8) and (22), respectively.Proof.
Without loss of generality, we assume thata
>
b
>
0. Let x
=
b
/
a
∈
(
0,1
); then from (1) and (2) we get
(36)
log
⁡
T
(
a
,
b
)
-
log
⁡
M
p
(
a
,
b
)
=
F
(
p
,
x
)
.
Ifp
=
5
/
3, then from (22) and Theorem 2(1) we get
(37)
F
(
5
3
,
x
)
>
F
(
5
3
,
0
)
=
log
⁡
2
8
/
5
π
.
Therefore, the first inequality in (35) with the best possible constant α
=
2
8
/
5
/
π follows from (36) and (37) together with the monotonicity of F
(
5
/
3
,
x
) given in Theorem 2(1).
Ifp
=
log
⁡
2
/
(
log
⁡
π
-
log
⁡
2
)
∈
(
1,5
/
3
), then Lemma 1 and (24) together with (34) imply that there exists μ
∈
(
0,1
) such that f
(
log
⁡
2
/
(
log
⁡
π
-
log
⁡
2
)
,
x
)
=
0, and F
(
log
⁡
2
/
(
log
⁡
π
-
log
⁡
2
)
,
x
) is strictly increasing on (
0
,
μ
] and strictly decreasing on [
μ
,
1
). Therefore, we have
(38)
F
(
log
⁡
2
log
⁡
π
-
log
⁡
2
,
x
)
≤
F
(
log
⁡
2
log
⁡
π
-
log
⁡
2
,
μ
)
.
Making use of MATHEMATICA software, numerical computations show that(39)
0.186930110570624
<
μ
<
0.186930110570625
,
e
F
(
log
⁡
2
/
(
log
⁡
π
-
log
⁡
2
)
,
μ
)
=
1.0136
…
.
Therefore, the second inequality in (35) with the best possible constant β
=
e
F
(
log
⁡
2
/
(
log
⁡
π
-
log
⁡
2
)
,
μ
)
=
1.0136
… follows from (36) and (38) together with the piecewise monotonicity of F
(
log
⁡
2
/
(
log
⁡
π
-
log
⁡
2
)
,
x
).Corollary 4.
The double inequality(40)
Q
2
(
a
,
b
)
L
p
-
1
(
a
,
b
)
<
T
(
a
,
b
)
<
Q
2
(
a
,
b
)
L
q
-
1
(
a
,
b
)
holds for all a
,
b
>
0 with a
≠
b if and only if p
≥
5
/
3 and q
≤
1, where Q
(
a
,
b
)
=
(
a
2
+
b
2
)
/
2 and L
p
(
a
,
b
)
=
(
a
p
+
1
+
b
p
+
1
)
/
(
a
p
+
b
p
) are, respectively, the quadratic and pth Lehmer means of a and b.Proof.
Without loss of generality, we assume thata
>
b
>
0. Let x
=
b
/
a
∈
(
0,1
). Then from Theorem 2 and (24) we clearly see that the f
(
p
,
x
)
>
0 if and only if p
≥
5
/
3 and f
(
p
,
x
)
<
0 if and only if p
≤
1. Then (8) leads to the conclusion that the inequalities
(41)
(
1
-
x
)
(
1
+
x
p
)
(
1
+
x
2
)
(
1
+
x
p
-
1
)
>
arctan
1
-
x
1
+
x
,
(42)
(
1
-
x
)
(
1
+
x
q
)
(
1
+
x
2
)
(
1
+
x
q
-
1
)
<
arctan
1
-
x
1
+
x
hold for all x
∈
(
0,1
) if and only p
≥
5
/
3 and q
≤
1.
Therefore, Corollary4 follows easily from inequalities (41) and (42) together with (1).Corollary 5.
Leta
1
,
b
1
,
a
2
,
b
2
>
0 with a
1
/
b
1
<
a
2
/
b
2. Then Theorem 2 leads to the following Ky Fan type inequality:
(43)
T
(
a
1
,
b
1
)
T
(
a
2
,
b
2
)
<
(
>
)
M
p
(
a
1
,
b
1
)
M
p
(
a
2
,
b
2
)
if p
≥
5
/
3
(
p
≤
1
).



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*Source: 840130-2014-07-20.xml*