# Sharp Power Mean Bounds for the One-Parameter Harmonic Mean

Authors: Yu-Ming Chu; Li-Min Wu; Ying-Qing Song
Journal: Journal of Function Spaces
 (2015)

---

## Abstract

We present the best possible parametersα
=
α
(
r
) and β
=
β
(
r
) such that the double inequality M
α
(
a
,
b
)
<
H
r
(
a
,
b
)
<
M
β
(
a
,
b
) holds for all r
∈
(
0
,
1
/
2
) and a
,
b
>
0 with a
≠
b, where M
p
(
a
,
b
)
=
[
(
a
p
+
b
p
)
/
2
]
1
/
p  (
p
≠
0
) and M
0
(
a
,
b
)
=
a
b and H
r
(
a
,
b
)
=
2
[
r
a
+
(
1
-
r
)
b
]
[
r
b
+
(
1
-
r
)
a
]
/
(
a
+
b
) are the power and one-parameter harmonic means of a and b, respectively.


---

## Body

## 1. Introduction

Forp
∈
R and a
,
b
>
0, the pth power mean M
p
(
a
,
b
) of a and b is defined by (1)
M
p
a
,
b
=
a
p
+
b
p
2
1
/
p
p
≠
0
,
M
0
a
,
b
=
a
b
.It is well known thatM
p
(
a
,
b
) is strictly increasing with respect to p
∈
R for fixed a
,
b
>
0 with a
≠
b, symmetric and homogeneous of degree 1. Many classical means are special cases of the power mean: for example, M
-
1
(
a
,
b
)
=
2
a
b
/
(
a
+
b
)
=
H
(
a
,
b
) is the harmonic mean, M
0
(
a
,
b
)
=
a
b
=
G
(
a
,
b
) is the geometric mean, M
1
(
a
,
b
)
=
(
a
+
b
)
/
2
=
A
(
a
,
b
) is the arithmetic mean, and M
2
(
a
,
b
)
=
(
a
2
+
b
2
)
/
2
=
Q
(
a
,
b
) is the quadratic mean. The main properties of the power mean are given in [1]. Recently, the power mean has attracted the attention of many researchers. In particular, many remarkable inequalities for the power mean can be found in the literature [2–10].LetL
(
a
,
b
)
=
(
b
-
a
)
/
(
log
⁡
b
-
log
⁡
a
), P
(
a
,
b
)
=
(
a
-
b
)
/
[
2
arcsin
(
(
a
-
b
)
/
(
a
+
b
)
)
], I
(
a
,
b
)
=
1
/
e
(
b
b
/
a
a
)
1
/
(
b
-
a
), T
(
a
,
b
)
=
(
a
-
b
)
/
[
2
arctan
(
(
a
-
b
)
/
(
a
+
b
)
)
], and C
(
a
,
b
)
=
(
a
2
+
b
2
)
/
(
a
+
b
) be the logarithmic, first Seiffert, identric, second Seiffert, and contraharmonic means of two distinct positive real numbers a and b, respectively. Then it is well known that the inequalities (2)
H
a
,
b
<
G
a
,
b
<
L
a
,
b
<
P
a
,
b
<
I
a
,
b
<
A
a
,
b
<
T
a
,
b
<
Q
a
,
b
<
C
a
,
bhold for all a
,
b
>
0 with a
≠
b.Lin [11] proved that the double inequality (3)
M
p
a
,
b
<
L
a
,
b
<
M
q
a
,
bholds for all a
,
b
>
0 with a
≠
b if and only if p
≤
0 and q
≥
1
/
3.In [12], Pittenger presented the best possible parameters λ
=
λ
(
p
) and μ
=
μ
(
p
) such that the double inequality (4)
M
λ
a
,
b
<
L
p
a
,
b
<
M
μ
a
,
bholds for all a
,
b
>
0 with a
≠
b, where L
p
(
a
,
b
)
=
[
(
b
p
+
1
-
a
p
+
1
)
/
(
(
p
+
1
)
(
b
-
a
)
)
]
1
/
p
(
p
≠
0
,
-
1
), L
0
(
a
,
b
)
=
I
(
a
,
b
), and L
-
1
(
a
,
b
)
=
L
(
a
,
b
) is the generalized logarithmic mean of a and b.Jagers [13] and Seiffert [14] proved that the double inequalities (5)
M
1
/
2
a
,
b
<
P
a
,
b
<
M
2
/
3
a
,
b
,
M
1
a
,
b
<
T
a
,
b
<
M
2
a
,
bhold for all a
,
b
>
0 with a
≠
b.In [15, 16], the authors proved that the double inequalities (6)
M
log
⁡
⁡
2
/
1
+
log
⁡
⁡
2
a
,
b
<
I
a
,
b
+
L
a
,
b
2
<
M
1
/
2
a
,
b
,
M
0
a
,
b
<
I
a
,
b
L
a
,
b
<
M
1
/
2
a
,
bhold for alla
,
b
>
0 with a
≠
b.Costin and Toader [17] proved that the double inequality (7)
M
4
/
3
a
,
b
<
T
a
,
b
<
M
5
/
3
a
,
bholds for all a
,
b
>
0 with a
≠
b.In [18–20], the authors proved that the double inequalities (8)
M
p
a
,
b
<
P
a
,
b
<
M
q
a
,
b
,
M
r
a
,
b
<
T
a
,
b
<
M
s
a
,
bhold for all a
,
b
>
0 with a
≠
b if and only if p
≤
log
⁡
π
/
log
⁡
2, q
≥
2
/
3, r
≤
log
⁡
2
/
(
log
⁡
π
-
log
⁡
2
), and s
≥
5
/
3.Čizmesija [21] proved that p
=
-
α
/
2 and q
=
log
⁡
2
/
[
log
⁡
4
-
log
⁡
(
1
-
α
)
] are the best possible parameters such that the double inequality M
p
(
a
,
b
)
<
[
H
(
a
,
b
)
+
α
G
(
a
,
b
)
+
(
1
-
α
)
A
(
a
,
b
)
]
/
2
<
M
q
(
a
,
b
) holds for all α
∈
(
0,1
) and a
,
b
>
0 with a
≠
b.In [22, 23], the authors proved that the inequalities(9)
M
p
K
r
,
E
r
>
π
2
,
M
λ
K
r
,
K
1
-
r
2
≤
K
2
2
≤
M
μ
K
r
,
K
1
-
r
2hold for all r
∈
(
0,1
) if and only if p
≥
-
1
/
2, λ
≤
log
⁡
2
/
(
log
⁡
π
-
log
⁡
2
)
-
log
⁡
[
K
(
2
/
2
)
]
=
-
4.18
,
…, and μ
≥
1
-
4
K
4
(
2
/
2
)
/
π
2
=
-
3.789
,
…, where K
(
r
)
=
∫
0
π
/
2
‍
(
1
-
r
2
sin
2
θ
)
-
1
/
2
d
θ and E
(
r
)
=
∫
0
π
/
2
‍
(
1
-
r
2
sin
2
θ
)
1
/
2
d
θ are, respectively, the complete elliptic integrals of the first and second kinds.Letp
∈
[
0,1
] and N be the bivariate symmetric mean. Then, the one-parameter mean N
p
(
a
,
b
) was defined by Neuman [24] as follows: (10)
N
p
a
,
b
=
N
p
a
+
1
-
p
b
,
p
b
+
1
-
p
a
.Letp
1
,
q
1
,
p
2
,
q
2
,
λ
1
,
λ
2
,
λ
3
,
λ
,
μ
∈
(
0,1
/
2
) and p
3
,
q
3
∈
(
1
/
2,1
). Then, the authors in [25–28] proved that the inequalities (11)
H
p
1
a
,
b
<
P
a
,
b
<
H
q
1
a
,
b
,
L
p
2
a
,
b
<
P
a
,
b
<
L
q
2
a
,
b
,
H
λ
1
a
,
b
>
G
a
,
b
,
H
λ
2
a
,
b
>
L
a
,
b
,
G
λ
3
a
,
b
>
L
a
,
b
,
C
p
3
a
,
b
<
T
a
,
b
<
C
q
3
a
,
b
,
G
λ
a
,
b
<
I
a
,
b
<
G
μ
a
,
bhold for all a
,
b
>
0 with a
≠
b if and only if p
1
≤
1
/
2
-
1
-
2
/
π
/
2, q
1
≥
1
/
2
-
6
/
12, p
2
≤
λ
0, q
2
≥
1
/
2
-
2
/
4, λ
1
≥
1
/
2
-
2
/
4, λ
2
≥
1
/
2
-
3
/
6, λ
3
≥
1
/
2
-
6
/
6, p
3
≤
1
/
2
+
4
/
π
-
1
/
2, q
3
≥
1
/
2
+
3
/
6, λ
≤
1
/
2
-
1
-
4
/
e
2
/
2, and μ
≥
1
/
2
-
3
/
6, where λ
0 is the unique solution of the equation log
⁡
[
(
1
-
λ
)
/
λ
]
=
π
(
1
-
2
λ
).The main purpose of this paper is to present the best possible parametersα
=
α
(
r
) and β
=
β
(
r
) such that the double inequality M
α
(
a
,
b
)
<
H
r
(
a
,
b
)
<
M
β
(
a
,
b
) holds for all r
∈
(
0,1
/
2
) and a
,
b
>
0 with a
≠
b.


## 2. Lemmas

In order to prove our main result we need three lemmas, which we present in this section.Lemma 1.
The inequality(12)
-
8
r
2
+
8
r
-
1
+
log
⁡
⁡
2
log
⁡
⁡
2
r
1
-
r
<
0holds for all r
∈
(
0,1
/
2
).Proof.
It is not difficult to verify thatlog
⁡
[
2
r
(
1
-
r
)
]
<
0 for all r
∈
(
0,1
/
2
). Therefore, we only need to prove that f
(
r
)
>
0 for r
∈
(
0,1
/
2
), where f
(
r
)
=
(
-
8
r
2
+
8
r
-
1
)
log
⁡
[
2
r
(
1
-
r
)
]
+
log
⁡
2. Simple computations lead to (13)
f
1
2
=
0
,
f
′
r
=
1
-
2
r
g
r
,where (14)
g
r
=
8
log
⁡
⁡
2
r
1
-
r
-
1
r
1
-
r
+
8
,
(15)
g
1
2
=
4
1
-
log
⁡
⁡
4
<
0
,
g
′
r
=
1
-
2
r
8
r
1
-
r
+
1
r
2
1
-
r
2
>
0for all r
∈
(
0,1
/
2
).
Inequality (15) implies that g
(
r
)
<
0 for all r
∈
(
0,1
/
2
). Then, from (13) we clearly see that f
(
r
)
>
0 for all r
∈
(
0,1
/
2
).Lemma 2.
The inequality(16)
-
16
r
2
+
16
r
-
2
-
8
r
2
-
8
r
-
1
log
⁡
⁡
2
log
⁡
⁡
2
r
1
-
r
+
log
⁡
⁡
2
log
⁡
⁡
2
r
1
-
r
2
<
0holds for all r
∈
(
0,1
/
2
).Proof.
Letp
=
-
log
⁡
2
/
log
⁡
[
2
r
(
1
-
r
)
] and q
=
-
8
r
2
+
8
r
-
1. Then, it is not difficult to verify that (17)
0
<
p
<
1for all r
∈
(
0,1
/
2
).
It follows from Lemma1 that(18)
p
>
qfor all r
∈
(
0,1
/
2
).
Inequalities (17) and (18) lead to (19)
-
16
r
2
+
16
r
-
2
-
8
r
2
-
8
r
-
1
log
⁡
⁡
2
log
⁡
⁡
2
r
1
-
r
+
log
⁡
⁡
2
log
⁡
⁡
2
r
1
-
r
2
=
2
q
-
q
+
2
p
+
p
2
=
-
p
-
q
2
-
p
<
0for all r
∈
(
0,1
/
2
).Lemma 3.
The inequality(20)
-
6
r
2
+
6
r
-
12
r
2
-
12
r
+
1
log
⁡
⁡
2
log
⁡
⁡
2
r
1
-
r
+
log
⁡
⁡
2
log
⁡
⁡
2
r
1
-
r
2
>
0holds for all r
∈
(
0,1
/
2
).Proof.
Letp
=
-
log
⁡
2
/
log
⁡
[
2
r
(
1
-
r
)
] and q
=
-
8
r
2
+
8
r
-
1. Then, it follows from (17) and (18) that (21)
-
6
r
2
+
6
r
-
12
r
2
-
12
r
+
1
log
⁡
⁡
2
log
⁡
⁡
2
r
1
-
r
+
log
⁡
⁡
2
log
⁡
⁡
2
r
1
-
r
2
=
3
4
q
+
1
+
1
-
3
2
q
+
1
p
+
p
2
=
3
4
q
+
1
1
-
2
p
+
p
p
+
1
>
3
4
q
+
1
1
-
2
p
+
p
q
+
1
=
1
4
q
+
1
3
-
2
p
=
2
r
1
-
r
3
-
2
p
>
0for all r
∈
(
0,1
/
2
).


## 3. Main Results

Theorem 4.
The double inequality(22)
M
α
a
,
b
<
H
r
a
,
b
<
M
β
a
,
bholds for all r
∈
(
0,1
/
2
) and a
,
b
>
0 with a
≠
b if and only if α
≤
-
8
r
2
+
8
r
-
1 and β
≥
-
log
⁡
2
/
log
⁡
[
2
r
(
1
-
r
)
].Proof.
Without loss of generality, we assume thata
=
x
∈
(
1
,
∞
) and b
=
1. Let log
⁡
H
r
(
x
,
1
)
-
log
⁡
M
p
(
x
,
1
)
=
f
(
x
), where (23)
f
x
=
log
⁡
⁡
r
1
-
r
x
2
+
1
-
2
r
+
2
r
2
x
+
r
1
-
r
1
+
x
+
1
+
p
log
⁡
⁡
2
-
log
⁡
⁡
1
+
x
p
p
.Then, simple computations lead to(24)
f
1
=
0
,
(25)
f
′
x
=
g
x
r
1
-
r
x
2
+
1
-
2
r
+
2
r
2
x
+
r
1
-
r
1
+
x
1
+
x
p
x
1
-
p
,where (26)
g
x
=
-
r
2
+
r
x
3
-
p
+
-
2
r
2
+
2
r
x
2
-
p
+
3
r
2
-
3
r
+
1
x
1
-
p
+
-
3
r
2
+
3
r
-
1
x
2
+
2
r
2
-
2
r
x
+
r
2
-
r
,
(27)
g
1
=
0
,
(28)
g
′
1
=
-
8
r
2
+
8
r
-
1
-
p
,
(29)
g
′′
1
=
-
16
r
2
+
16
r
-
2
+
p
8
r
2
-
8
r
-
1
+
p
2
,
(30)
g
′′′
x
=
1
-
p
x
-
p
-
2
h
x
,where(31)
h
x
=
-
r
2
+
r
3
-
p
2
-
p
x
2
-
-
2
r
2
+
2
r
p
2
-
p
x
+
3
r
2
-
3
r
+
1
p
p
+
1
,
(32)
h
1
=
6
-
r
2
+
r
+
p
12
r
2
-
12
r
+
1
+
p
2
,
(33)
h
′
x
=
2
r
-
r
2
2
-
p
3
-
p
x
-
p
.
We divide the proof into two cases.
Case 1 (
p
=
-
8
r
2
+
8
r
-
1
). We divide the discussion into two subcases.
Subcase 1.1 (
p
=
-
8
r
2
+
8
r
-
1
and
r
∈
(
0
,
(
2
-
2
)
/
4
)
∪
(
2
-
2
)
/
4,1
/
2
)
). Then, we clearly see that p
∈
(
-
1,0
)
∪
(
0,1
), and (28), (29), (32), and (33) lead to (34)
g
′
1
=
g
′′
1
=
0
,
(35)
h
1
=
1
4
p
+
1
3
-
2
p
>
0
,
(36)
h
′
x
=
1
4
p
+
1
2
-
p
3
-
p
x
-
p
>
1
4
p
+
1
2
-
p
3
-
2
p
>
0for x
∈
(
1
,
∞
).
It follows easily from (24), (25), (27), (30), and (34)–(36) that (37)
H
r
x
,
1
>
M
-
8
r
2
+
8
r
-
1
x
,
1for all r
∈
(
0,1
/
2
).
Subcase 1.2 (
p
=
-
8
r
2
+
8
r
-
1
and
r
=
(
2
-
2
)
/
4
). Then, p
=
0, and (37) follows from (38)
H
r
x
,
1
-
M
-
8
r
2
+
8
r
-
1
x
,
1
=
x
2
+
6
x
+
1
4
1
+
x
-
x
=
x
-
1
2
4
1
+
x
.
Case 2 (
p
=
-
log
⁡
2
/
log
⁡
[
2
r
(
1
-
r
)
]
). Then, we clearly see that p
∈
(
0,1
), and Lemmas 1–3 and (23), (26), (28), (29), and (32) lead to (39)
lim
x
→
+
∞
⁡
f
x
=
0
,
(40)
lim
x
→
+
∞
⁡
g
x
=
+
∞
,
(41)
lim
x
→
+
∞
⁡
g
′
x
=
+
∞
,
(42)
g
′
1
<
0
,
(43)
lim
x
→
+
∞
⁡
g
′′
x
=
+
∞
,
(44)
g
′′
1
<
0
,
(45)
h
1
>
0and (36) again holds.
It follows from (30), (36), and (45) that g
′′ is strictly increasing on (
1
,
∞
). Then, (43) and (44) lead to the conclusion that there exists μ
0
>
1 such that g
′ is strictly decreasing on (
1
,
μ
0
] and strictly increasing on [
μ
0
,
∞
).
From (41) and (42) together with the piecewise monotonicity of g
′ we clearly see that there exists μ
1
>
1 such that g is strictly decreasing on (
1
,
μ
1
] and strictly increasing on [
μ
1
,
∞
). Then, (25), (27), and (40) lead to the conclusion that there exists μ
2
>
1 such that f is strictly decreasing on (
1
,
μ
2
] and strictly increasing on [
μ
2
,
∞
). Therefore, (46)
H
r
x
,
1
<
M
-
log
⁡
⁡
2
/
log
⁡
⁡
2
r
1
-
r
x
,
1for all r
∈
(
0,1
/
2
) follows from (24) and (39) together with the piecewise monotonicity of f.
Next, we prove thatα
=
-
8
r
2
+
8
r
-
1 and β
=
-
log
⁡
2
/
log
⁡
[
2
r
(
1
-
r
)
] are the best possible parameters such that the double inequality (47)
M
α
a
,
b
<
H
r
a
,
b
<
M
β
a
,
bholds for all r
∈
(
0,1
/
2
) and a
,
b
>
0 with a
≠
b.
Letr
∈
(
0,1
/
2
), α
0
=
-
8
r
2
+
8
r
-
1, β
0
=
-
log
⁡
2
/
log
⁡
[
2
r
(
1
-
r
)
], ε
∈
(
0
,
β
0
), and t
>
0. Then, we have (48)
H
r
1,1
+
t
-
M
α
0
+
ε
1,1
+
t
=
2
1
-
r
1
+
t
+
r
r
1
+
t
+
1
-
r
t
+
2
-
1
+
1
+
t
α
0
+
ε
2
1
/
α
0
+
ε
,
(49)
lim
t
→
+
∞
⁡
H
r
1
,
t
M
β
0
-
ε
1
,
t
=
2
ε
/
β
0
β
0
-
ε
>
1
.
Lett
→
0
+ and make use of the Taylor expansion; then, (48) leads to (50)
H
r
1,1
+
t
-
M
α
0
+
ε
1,1
+
t
=
-
ε
8
x
2
+
o
x
2
.
Inequality (49) and equation (50) imply that for any r
∈
(
0,1
/
2
) and ε
∈
(
0
,
β
0
) there exist T
=
T
(
ε
)
>
1 and δ
=
δ
(
ε
)
∈
(
0,1
) such that H
r
(
1
,
t
)
>
M
β
0
-
ε
(
1
,
t
) for t
>
T and H
r
(
1,1
+
t
)
<
M
α
0
+
ε
(
1,1
+
t
) for t
∈
(
0
,
δ
).



---

*Source: 517647-2015-04-30.xml*