Hindawi Publishing Corporation
Journal of Applied Mathematics
Volume 2012, Article ID 379785, 12 pages
doi:10.1155/2012/379785
Research Article
A Best Possible Double Inequality for
Power Mean
Yong-Min Li,1 Bo-Yong Long,2 and Yu-Ming Chu1
1
2
Department of Mathematics, Huzhou Teachers College, Huzhou 313000, China
School of Mathematics Science, Anhui University, Hefei 230039, China
Correspondence should be addressed to Yu-Ming Chu, chuyuming2005@yahoo.com.cn
Received 29 February 2012; Accepted 9 September 2012
Academic Editor: Huijun Gao
Copyright q 2012 Yong-Min Li 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 any p, q ∈ R with p /
q and p /
− q, what are the greatest value
λ
λp, q and the least value μ μp, q, such that the double inequality Mλ a, b <
Mp a, bMq a, b < Mμ a, b holds for all a, b > 0 with a /
b? Where Mp a, b is the pth power
mean of two positive numbers a and b.
1. Introduction
For p ∈ R, the pth power mean Mp a, b of two positive numbers a and b is defined by
⎧
p
p 1/p
⎪
⎪
⎨ a b
,
2
Mp a, b ⎪
⎪
⎩√ab,
p/
0,
1.1
p 0.
It is well known that Mp a, b is continuous and strictly increasing with respect to
p ∈ R for fixed a, b > 0 with a / b. Many classical means are special case of the
√ power
mean, for example, M−1 a, b Ha, b 2ab/a b, M0 a, b Ga, b ab, and
M1 a, b Aa, b a b/2 are the harmonic, geometric, and arithmetic means of a
and b, respectively. Recently, the power mean has been the subject of intensive research. In
particular, many remarkable inequalities and properties for the power mean can be found in
literature 1–15.
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Let La, b a − b/log a − log b and Ia, b 1/eaa /bb 1/a−b be the logarithmic
and identric means of two positive numbers a and b with a /
b, respectively. Then it is well
known that
Ha, b M−1 a, b < Ga, b M0 a, b < La, b < Ia, b < Aa, b M1 a, b
1.2
for all a, b > 0 with a /
b.
In 16–22, the authors presented the sharp power mean bounds for L, I, IL1/2 , and
L I/2 as follows:
M0 a, b < La, b < M1/3 a, b,
M2/3 a, b < Ia, b < Mlog 2 a, b,
1
M0 a, b < La, bIa, b < M1/2 a, b,
La, b Ia, b < M1/2 a, b
2
1.3
for all a, b > 0 with a / b.
Alzer and Qiu 12 proved that the inequality
1
La, b Ia, b > Mp a, b
2
1.4
holds for all a, b > 0 with a /
b if and only if p ≤ log 2/1 log 2 0.40938 . . ..
The following sharp bounds for the sum αAa, b 1 − αLa, b, and the products
Aα a, bL1−α a, b and Gα a, bL1−α a, b in terms of power means were proved in 5, 8 as
follows:
Mlog 2/log 2−log α a, b < αAa, b 1 − αLa, b < M12α/3 a, b,
M0 a, b < Aα a, bL1−α a, b < M12α/3 a, b,
1.5
M0 a, b < Gα a, bL1−α a, b < M1−α/3 a, b
for any α ∈ 0, 1 and all a, b > 0 with a /
b.
In 2, 7, the authors answered the question: for any α ∈ 0, 1, what are the greatest
values p1 p1 α, p2 p2 α, p3 p3 α, and p4 p4 α, and the least values q1 q1 α,
q2 q2 α, q3 q3 α, and q4 q4 α, such that the inequalities
Mp1 a, b < P α a, bL1−α a, b < Mq1 a, b,
Mp2 a, b < Aα a, bG1−α a, b < Mq2 a, b,
Mp3 a, b < Gα a, bH 1−α a, b < Mq3 a, b,
Mp4 a, b < Aα a, bH 1−α a, b < Mq4 a, b
hold for all a, b > 0 with a /
b?
1.6
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3
In 4, the authors presented the greatest value p pα, β and the least value q qα, β such that the double inequality
Mp a, b < Aα a, bGβ a, bH 1−α−β a, b < Mq a, b
1.7
holds for all a, b > 0 with a /
b and α, β > 0 with α β < 1.
It is the aim of this paper to answer the question: for any p, q ∈ R with p /
q and p /
−q,
what are the greatest value λ λp, q and the least value μ μp, q, such that the double
inequality
Mλ a, b <
Mp a, bMq a, b < Mμ a, b
1.8
holds for all a, b > 0 with a / b?
2. Main Result
In order to establish our main result, we need a lemma which we present in this section.
Lemma 2.1. Let p, q /
0, p /
q and x > 1. Then
2
Mp x, 1Mq x, 1 < Mpq/2
x, 1
2.1
2
Mp x, 1Mq x, 1 > Mpq/2
x, 1
2.2
for p q > 0, and
for p q < 0.
Proof. From 1.1, we have
2
log Mp x, 1Mq x, 1 − log Mpq/2
x, 1
1 xp 1
1 xq
4
1 xpq/2
1
log
log
−
log
.
p
2
q
2
pq
2
fx 1
1 xp 1
1 xq
4
1 xpq/2
log
log
−
log
,
p
2
q
2
pq
2
2.3
Let
2.4
then simple computations lead to
f1 0,
2
1 − xpq/2 xp/2 − xq/2
f x .
x1 xp 1 xq 1 xpq/2
2.5
2.6
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Equation 2.6 implies that
f x < 0
2.7
f x > 0
2.8
for p q > 0, and
for p q < 0.
Therefore, inequality 2.1 follows from 2.3–2.5 and inequality 2.7, and inequality
2.2 follows from 2.3–2.5 and inequality 2.8.
Let
E0 E1 E2 p, q ∈ R2 : p, q > 0, p > q ,
p, q ∈ R2 : p, q < 0, p > q ,
E0 E1 E2 p, q ∈ R2 : p > 0, q 0 ,
E3 p, q ∈ R2 : p > 0, q < 0, p q > 0 ,
E4 p, q ∈ R2 : p 0, q < 0 ,
E5 p, q ∈ R2 : p > 0, q < 0, p q < 0 ,
E6 E5 E6 p, q ∈ R2 : p q ,
E3 E4 p, q ∈ R2 : p −q ,
p, q ∈ R2 : p, q > 0, p < q ,
p, q ∈ R2 : p, q < 0, p < q ,
p, q ∈ R2 : p 0, q > 0 ,
p, q ∈ R2 : p < 0, q > 0, p q > 0 ,
p, q ∈ R2 : p < 0, q 0 ,
p, q ∈ R2 : p < 0, q > 0, p q < 0 .
2.9
Then we clearly see that R2 6i0 Ei 6i0 Ei , and it is not difficult to verify that the
identity Mp a, bMq a, b Mpq/2 a, b holds for all a, b > 0 if p, q ∈ E0 E0 . Let
⎧
2pq
⎪
⎪
,
⎪
⎪
⎪
⎨ p q
λ
pq
⎪
,
⎪
⎪
⎪
⎪
⎩ 2
0,
⎧
2pq
⎪
⎪
,
⎪
⎪
⎪
⎨ p q
μ
pq
⎪
,
⎪
⎪
⎪
⎪
⎩ 2
0,
p, q ∈ E1 E1 ,
p, q ∈ E2 E2 E5 E5 E6 E6 ,
p, q ∈ E3 E3 E4 E4 ,
p, q ∈ E2
E2 ,
p, q ∈ E1 E1 E3 E3 E4 E4 ,
p, q ∈ E5 E5 E6 E6 .
Then we have Theorem 2.2 as follows.
2.10
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5
Theorem 2.2. The double inequality
Mλ a, b <
Mp a, bMq a, b < Mμ a, b
2.11
holds for all a, b > 0 with a /
b, and Mλ a, b and Mμ a, b are the best possible lower and upper
power mean bounds for the geometric mean of Mp a, b and Mq a, b.
Proof. From 1.1, we clearly see that Mp a, b is symmetric and homogenous of degree 1.
Without loss of generality, we assume that b 1, a x > 1 and p > q. We divide the proof of
inequality 2.11 into three cases.
Case 1. p, q ∈ E1 E2 . Then from Lemma 2.1, we clearly see that
Mp x, 1Mq x, 1 < Mpq/2 x, 1
2.12
Mp x, 1Mq x, 1 > Mpq/2 x, 1
2.13
for p, q ∈ E1 , and
for p, q ∈ E2 .
From 1.1, we get
2
log Mp x, 1Mq x, 1 − log M2pq/pq
x, 1
1 xp 1
1 xq p q
1 x2pq/pq
1
log
log
−
log
.
p
2
q
2
pq
2
Fx 1
1 xp 1
1 xq p q
1 x2pq/pq
log
log
−
log
,
p
2
q
2
pq
2
2.14
Let
2.15
then simple computations lead to
F1 0,
F x x1 xp 1
2.16
xq Gx
,
xq 1 x2pq/pq
2.17
where
Gx xp−q − x2pqp
2
−q2 /pq
2xp − x2pq/pq − 2xqp−q/pq 1,
G1 0,
G x xpq−q
2
−p−q/pq
2.18
2.19
Hx,
2.20
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Journal of Applied Mathematics
where
2pq p2 − q2 p
2
2
x 2pxp q /pq
Hx p − q xpp−q/pq −
pq
2pq q 2q p − q
x −
,
−
pq
pq
2
2 p−q
,
H1 pq
H x p
xp−1 Ix,
pq
2.21
2.22
2.23
where
2
Ix p − q x−2pq/pq 2 p2 q2 x−qp−q/pq − 2q2 x−p−q − 2pq − p2 q2 ,
2
I1 2 p − q ,
2q p − q q2 −pq−p−q/pq
x
I x Jx,
pq
2.24
2.25
2.26
where
Jx −p p − q x−q q p q x−pp−q/pq − p2 − q2 ,
J1 −2p p − q ,
2
2
J x pq p − q x−q−1 1 − xq −p 2pq/pq .
2.27
2.28
2.29
If p, q ∈ E1 , then 2.15, 2.18, 2.21, 2.22, 2.24, 2.25, 2.27, and 2.28 lead to
x → ∞
lim Fx 0,
2.30
lim Gx −∞,
2.31
lim Hx −∞,
2.32
H1 > 0,
2.33
lim Ix −2pq − p2 q2 < 0,
2.34
I1 > 0,
lim Jx − p2 q2 < 0,
2.35
2.36
J1 < 0.
2.37
x → ∞
x → ∞
x → ∞
x → ∞
Journal of Applied Mathematics
7
We divide the discussion into two subcases.
Subcase 1.1. p, q ∈ E1 . Then 2.26 and 2.29 together with inequalities 2.36 and 2.37
imply that Ix is strictly decreasing in 1, ∞. In fact, if q2 − p2 2pq/p q ≥ 0, then
2.29 and inequality 2.37 imply that Jx < 0 for x ∈ 1, ∞. If q2 − p2 2pq/p q < 0,
then 2.29 and inequality 2.36 lead to the conclusion that Jx < 0 for x ∈ 1, ∞.
From inequalities 2.34 and 2.35 together with the monotonicity of Ix, we know
that there exists λ1 > 1 such that Ix > 0 for x ∈ 1, λ1 and Ix < 0 for x ∈ λ1 , ∞. Then
2.23 leads to the conclusion that Hx is strictly increasing in 1, λ1 and strictly decreasing
in λ1 , ∞.
It follows from 2.32 and 2.33 together with the piecewise monotonicity of Hx
that there exists λ2 > λ1 > 1 such that Hx > 0 for 1, λ2 and Hx < 0 for λ2 , ∞. Then
2.20 leads to the conclusion that Gx is strictly increasing in 1, λ2 and strictly decreasing
in λ2 , ∞.
From 2.17, 2.19 and 2.31 together with the piecewise monotonicity of Gx, we
clearly see that there exists λ3 > λ2 > 1 such that Fx is strictly increasing in 1, λ3 and
strictly decreasingin λ3 , ∞.
Therefore, Mp x, 1Mq x, 1 > M2pq/pq x, 1 follows from 2.14–2.16 and 2.30
together with the piecewise monotonicity of Fx.
Subcase 1.2. p, q ∈ E2 . Then 2.30 and 2.35 again hold, and 2.18, 2.21, 2.22, and 2.28
lead to
lim Gx ∞,
2.38
lim Hx ∞,
2.39
H1 < 0,
2.40
J1 > 0.
2.41
x → ∞
x → ∞
It follows from 2.29 and inequalities q2 −p2 2pq/pq < 0 and 2.41 that Jx > 0
for x ∈ 1, ∞. Then 2.26 and inequality 2.35 lead to the conclusion that Ix > 0 for
x ∈ 1, ∞. Therefore, Hx is strictly increasing in 1, ∞ follows from 2.23.
It follows from 2.20 and 2.39 together with inequality 2.40 and the monotonicity
of Hx that there exists μ1 > 1 such that Gx is strictly decreasing in 1, μ1 and strictly
increasing in μ1 , ∞.
From 2.17, 2.19 and 2.38 together with the piecewise monotonicity of Gx, we
clearly see that there exists μ2 > μ1 > 1 such that Fx is strictly decreasing in 1, μ2 and
strictly increasing
in μ2 , ∞.
Therefore, Mp x, 1Mq x, 1 < M2pq/pq x, 1 follows from 2.14–2.16 and 2.30
together with the piecewise monotonicity of Fx.
Case 2. p, q ∈ E3 E5 . Clearly, we have M0 x, 1 < Mp x, 1Mq x, 1 for p, q ∈ E3 and
M0 x, 1 > Mp x, 1Mq x, 1 for p, q ∈ E5 . Therefore, we need only to prove that
M0 x, 1Mr x, 1 < Mr/2 x, 1
2.42
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for r > 0, and
M0 x, 1Mr x, 1 > Mr/2 x, 1
2.43
for r < 0.
From 1.1, one has
2
logM0 x, 1Mr x, 1 − log Mr/2
x, 1 1
1 xr 4
1 xr/2
1
log x log
− log
.
2
r
2
r
2
2.44
Let
fx 1
1 xr 4
1 xr/2
1
log x log
− log
,
2
r
2
r
2
2.45
then simple computations lead to
f1 0,
3
xr/2 − 1
.
2x1 xr 1 xr/2
2.46
f x −
2.47
If r > 0 or r < 0, resp., then 2.47 leads to the conclusion that fx is strictly
decreasing or increasing, resp. in 1, ∞. Therefore, inequalities 2.42 and 2.43 follow
from 2.44–2.46 and the monotonicity of fx.
Case 3. p, q ∈ E4 E6 . Then from Lemma 2.1, we clearly see that Mpq/2 x, 1 >
Mp x, 1Mq x, 1 for p, q ∈ E4 and Mp x, 1Mq x, 1 > Mpq/2 x, 1 for p, q ∈ E6 .
Therefore, we need only to prove that
Mp x, 1Mq x, 1 > M0 x, 1
2.48
Mp x, 1Mq x, 1 < M0 x, 1
2.49
for p, q ∈ E4 , and
for p, q ∈ E6 .
From 1.1, we get
1 xp 1
1 xq
1
log
− log x.
log Mp x, 1Mq x, 1 − log M02 x, 1 log
p
2
q
2
2.50
Let
fx 1
1 xp 1
1 xq
log
log
− log x,
p
2
q
2
2.51
Journal of Applied Mathematics
9
then simple computations lead to
f x f1 0,
2.52
xpq − 1
.
x1 xp 1 xq 2.53
If p, q ∈ E4 or E6 , resp., then 2.53 implies that fx is strictly increasing or
decreasing, resp. in 1, ∞. Therefore, inequalities 2.48 and 2.49 follow from 2.50–
2.52 and the monotonicity of fx.
Next, we prove that Mλ a, b and Mμ a, b are the best possible lower and upper
power mean bounds for the geometric mean of Mp a, b and Mq a, b. We divide the proof
into six cases.
Case A. p, q ∈ E1 . Then for any ∈ 0, p q/2 and x > 0, from 1.1, one has
2
Mp 1 x, 1Mq 1 x, 1 − Mpq/2−
1 x, 1
1 1 xp
2
1/p 1 1 xq
2
1/q
2
M2pq/pq
x, 1
lim
x → ∞ Mp x, 1Mq x, 1
1 1 xpq/2−
−
2
4/pq−2
,
2
2pq /pq2pqpq > 1.
2.54
2.55
Letting x → 0 and making use of Taylor expansion, we get
1 1 xp
2
1/p 1 1 xq
2
x2 o x2 .
4
1/q
1 1 xpq/2−
−
2
4/pq−2
2.56
Equations 2.54 and 2.56 together with inequality 2.55 imply that for any
∈ 0, p q/2 , there exist δ1 δ1 > 0 and X1 X1 p, q, > 1 such that
Mp 1 x, 1Mq 1 x, 1 > Mpq/2− 1 x, 1 for x ∈ 0, δ1 and Mp x, 1Mq x, 1 <
M2pq/pq x, 1 for x ∈ X1 , ∞.
Case B. p, q ∈ E2 . Then for ∈ 0, −p q/2 and x > 0, making use of 1.1 and Taylor
expansion, we have
2
Mpq/2
1 x, 1 − Mp 1 x, 1Mq 1 x, 1 lim
x → ∞
Mp x, 1Mq x, 1
2
M2pq/pq−
x, 1
2
2
x o x2
4
x −→ 0,
2pq /pq2pq−pq > 1.
2.57
2.58
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Journal of Applied Mathematics
Equation 2.57 and inequality 2.58 imply that for any ∈ 0, −p q/2, there
exist δ2 δ2 > 0 and X2 X2 p, q, > 1 such that Mpq/2 1 x, 1 >
Mp 1 x, 1Mq 1 x, 1 for x ∈ 0, δ2 and Mp x, 1Mq x, 1 > M2pq/pq− x, 1 for
x ∈ X2 , ∞.
Case C. p, q ∈ E3 . Then for ∈ 0, p/2 and x > 0, making use of 1.1 and Taylor expansion,
we have
2
Mp 1 x, 1M0 1 x, 1 − Mp/2−
1 x, 1 2
x o x2
4
x −→ 0,
2.59
M2 x, 1
∞.
x → ∞ Mp x, 1M0 x, 1
lim
Equation 2.59 leads to theconclusion that for any ∈ 0, p/2, there exist δ3 δ3 >
0 and X3 X3 p, > 1 such that Mp 1 x, 1M0 1 x, 1 > Mp/2− 1 x, 1 for x ∈ 0, δ3 and M x, 1 > Mp x, 1M0 x, 1 for x ∈ X3 , ∞.
Case D. p, q ∈ E4 . Then for ∈ 0, p q/2 and x > 0, making use of 1.1 and Taylor
expansion, we have
2
Mp 1 x, 1Mq 1 x, 1 − Mpq/2−
1 x, 1 2
x o x2
4
x −→ 0,
2.60
M2 x, 1
∞.
x → ∞ Mp x, 1Mq x, 1
lim
Equation 2.60 implies that for any ∈ 0, p
q/2, there exist δ4 δ4 > 0 and
X4 X4 p, q, > 1 such that Mpq/2− 1 x, 1 < Mp 1 x, 1Mq 1 x, 1 for x ∈ 0, δ4 and M x, 1 > Mp x, 1Mq x, 1 for x ∈ X4 , ∞.
Case E. p, q ∈ E5 . Then for any ∈ 0, −q/2 and x > 0, making use of 1.1 and
Taylor expansion, one has
2
Mq/2
1 x, 1 − M0 1 x, 1Mq 1 x, 1 lim
x → ∞
M0 x, 1Mq x, 1
2
M−
x, 1
2
x o x2
4
x −→ 0,
2.61
∞.
Equation 2.61 leads to the conclusion that for any ∈ 0, −q/2, there exist δ5 δ5 > 0 and X5 X5 q, > 1 such that Mq/2 1 x, 1 > M0 1 x, 1Mq 1 x, 1 for
x ∈ 0, δ5 and M− x, 1 < M0 x, 1Mq x, 1 for x ∈ X5 , ∞.
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11
Case F. p, q ∈ E6 . Then for any ∈ 0, −p q/2 and x > 0, making use of 1.1 and Taylor
expansion, one has
2
Mpq/2
1 x, 1 − Mp 1 x, 1Mq 1 x, 1 lim
x → ∞
Mp x, 1Mq x, 1
2
M−
x, 1
2
x o x2
4
x −→ 0,
2.62
∞.
Equation 2.62 shows that for any ∈ 0, −p
q/2, there exist δ6 δ6 > 0 and
X6 X6 p, q, > 1 such that Mpq/2 1 x, 1 > Mp 1 x, 1Mq 1 x, 1 for x ∈ 0, δ6 2
and Mp x, 1Mq x, 1 > M−
x, 1 for x ∈ X6 , ∞.
Acknowledgments
This research was supported by the Natural Science Foundation of China under Grants
11071069 and 11171307 and the Innovation Team Foundation of the Department of Education
of Zhejiang Province under Grant T200924.
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21 A. O. Pittenger, “The symmetric, logarithmic and power means,” Univerzitet u Beogradu. Publikacije
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