Hindawi Publishing Corporation
Journal of Applied Mathematics
Volume 2011, Article ID 618929, 9 pages
doi:10.1155/2011/618929
Research Article
Optimal Inequalities between
Harmonic, Geometric, Logarithmic, and
Arithmetic-Geometric Means
Yu-Ming Chu and Miao-Kun Wang
Department of Mathematics, Huzhou Teachers College, Huzhou 313000, China
Correspondence should be addressed to Yu-Ming Chu, chuyuming2005@yahoo.com.cn
Received 19 July 2011; Accepted 4 September 2011
Academic Editor: Laurent Gosse
Copyright q 2011 Y.-M. Chu and M.-K. Wang. 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 find the least values p, q, and s in 0, 1/2 such that the inequalities Hpa 1 − pb,
pb 1 − pa > AGa, b, Gqa1−qb, qb 1−qa > AGa, b, and Lsa1−sb, sb 1−sa >
AGa, b hold for all a, b > 0 with a /
b, respectively. Here AGa, b, Ha, b, Ga, b, and La, b
denote the arithmetic-geometric, harmonic, geometric, and logarithmic means of two positive
numbers a and b, respectively.
1. Introduction
The classical arithmetic-geometric mean AGa, b of two positive real numbers a and b is
defined as the common limit of sequences {an } and {bn }, which are given by
an1 a0 a,
b0 b,
an bn
,
2
bn1 1.1
an bn .
√
ab, La, b a − b/log a − log b,
Let Ha, b 2ab/a b, Ga, b Ia, b 1/ebb /aa 1/b−a , Aa, b a b/2, and Mp a, b ap bp /21/p p / 0
√
and M0 a, b ab be the harmonic, geometric, logarithmic, identric, arithmetic, and p-th
2
Journal of Applied Mathematics
power means of two positive numbers a and b with a /
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 M−1 a, b < max{a, b}
1.2
for all a, b > 0 with a /
b.
Recently, the inequalities for means have been the subject of intensive research. In
particular, many remarkable inequalities for arithmetic-geometric mean can be found in the
literature 1–9.
Carlson and Vuorinen 2, and Bracken 9 proved that
1.3
La, b < AGa, b
for all a, b > 0 with a /
b.
In 3, Vamanamurthy and Vuorinen established the following inequalities:
AGa, b < Ia, b < Aa, b,
AGa, b < M1/2 a, b,
π
La, b,
2
AG a2 , b2
M1 a, b <
< M2 a, b
AGa, b
1.4
AGa, b <
for all a, b > 0 with a / b.
We recall the Gauss identity 6, 7
π
AG 1, r Kr 2
for r ∈ 0, 1 and r given by
Kr √
1 − r 2 . As usual, K and E denote the complete elliptic integrals 8
π/2 1 − r 2 sin2 θ
−1/2
dθ 0
Er π/2 0
1.5
1 − r 2 sin2 θ
1/2
∞
π
1/2, n2 2n
r ,
2 n0 n!2
K r K r ,
∞
π
−1/2, n1/2, n 2n
dθ r ,
2 n0
n!2
where a, 0 1 for a /
0, and a, n n−1
k0 a
k.
E r E r ,
1.6
Journal of Applied Mathematics
3
For fixed a, b > 0 with a /
b and x ∈ 0, 1/2, let
f1 x Hxa 1 − xb, xb 1 − xa,
1.7
f2 x Gxa 1 − xb, xb 1 − xa,
1.8
f3 x Lxa 1 − xb, xb 1 − xa.
1.9
Then it is not difficult to verify that f1 x, f2 x, and f3 x are continuous and strictly
increasing in 0, 1/2, respectively. Note that f1 0 Ha, b < AGa, b < f1 1/2 Aa, b,
f2 0 Ga, b < AGa, b < f2 1/2 Aa, b and f3 0 La, b < AGa, b < f3 1/2 Aa, b.
Therefore, it is natural to ask what are the least values p, q, and s in 0, 1/2 such that
the inequalities Hpa 1 − pb, pb 1 − pa > AGa, b, Gqa 1 − qb, qb 1 − qa >
AGa, b, and Lsa1−sb, sb1−sa > AGa, b hold for all a, b > 0 with a /
b, respectively.
The main purpose of this paper is to answer these questions. Our main results are Theorems
1.1–1.3.
Theorem 1.1. If p ∈ 0, 1/2, then inequality
H pa 1 − p b, pb 1 − p a > AGa, b
1.10
holds for all a, b > 0 with a /
b if and only if p ≥ 1/4.
Theorem 1.2. If q ∈ 0, 1/2, then inequality
G qa 1 − q b, qb 1 − q a > AGa, b
holds for all a, b > 0 with a /
b if and only if q ≥ 1/2 −
√
1.11
2/4.
Theorem 1.3. If s ∈ 0, 1/2, then inequality
Lsa 1 − sb, sb 1 − sa > AGa, b
holds for all a, b > 0 with a /
b if and only if s ≥ 1/2 −
√
1.12
3/4.
2. Lemmas
In order to establish our main results we need several formulas and lemmas, which we
present in this section.
4
Journal of Applied Mathematics
For 0 < r < 1, the following derivative formulas were presented in 6, Appendix E, pp.
474-475:
dE E − K
dK E − r 2 K
,
,
dr
dr
r
rr 2
d E − r2K
dK − E rE
rK,
2,
dr
dr
r
√ 2 r
K
1 rKr.
1r
2.1
2.2
The following Lemma 2.1 can be found in 6, Theorem 3.217 and Exercise 3.434.
Lemma 2.1. 1 1 r 2 Er − 2r 2 Kr is strictly increasing from 0, 1 onto 0, 1;
2 Er/r 1/2 is strictly increasing from 0, 1 onto π/2, ∞.
Lemma 2.2. Inequality
2
1
Kr 1 − r 2 > 1
π
2
2.3
holds for all r ∈ 0, 1.
Proof. Let
⎡
⎤
2
1
fr log⎣ Kr 1 − r 2 ⎦.
π
2
2.4
Then simple computations lead to
f0 0,
2
1
r
Er − 2r 2 Kr
2
r
Er − r Kr
f r −
.
2 − r2
rr 2 Kr
rr 2 2 − r 2 Kr
2.5
2.6
It follows from Lemma 2.1 1 and 2.6 that f r > 0 for r ∈ 0, 1, which implies that
fr is strictly increasing in 0, 1.
Therefore, inequality 2.3 follows from 2.4 and 2.5 together with the monotonicity
of fr.
Lemma 2.3. Inequality
√
√ 2 3
2 3r
rKr > log
√
π
2 − 3r
holds for all r ∈ 0, 1.
2.7
Journal of Applied Mathematics
5
Proof. Let
√ √
2 3r
2 3
.
rKr − log
gr √
π
2 − 3r
2.8
Then simple computations lead to
g0 0,
2.9
√
√
√
1 3r 2 Er
4 3
Er − r 2 Kr
2 3
2 3
g r −
Kr r
− 2π .
2
2
π
4 − 3r
rr r 3/2 r 1/2
π 1 3r 2
2.10
Clearly the function r → 1 3r 2 /r 3/2 is strictly decreasing from 0, 1 onto 4, ∞.
Then 2.10 and Lemma 2.1 2 lead to the conclusion that g r > 0 for r ∈ 0, 1. Thus, gr
is strictly increasing in 0, 1.
Therefore, inequality 2.7 follows from 2.8 and 2.9 together with the monotonicity
of gr.
3. Proof of Theorems 1.1–1.3
Proof of Theorem 1.1. Let λ 1/4, then from the monotonicity of the function f1 x Hxa 1 − xb, xb 1 − xa in 0, 1/2 we know that to prove inequality 1.10 we only need to
prove that
AGa, b < Hλa 1 − λb, λb 1 − λa
3.1
for all a, b > 0 with a /
b.
From 1.1 and 1.7 we clearly see that both AGa, b and Hλa1−λb, λb 1−λa
are symmetric and homogeneous of degree 1. Without loss of generality, we can assume that
a 1 > b. Let t b ∈ 0, 1 and r 1 − t/1 t, then from 1.5 we have
Hλa 1 − λb, λb 1 − λa − AGa, b π
t 33t 1
−
.
81 t
2Kt 3.2
Let
π
t 33t 1
−
.
81 t
2Kt 3.3
π
π
2 r2 − r
−
F1 r,
41 r
21 rKr 81 rKr
3.4
Ft Then making use of 2.2 we get
Ft 6
Journal of Applied Mathematics
where F1 r 2/π4 − r 2 Kr − 4. Note that
F1 r ∞
1/2, n2
n0
4r 2
n!2
∞
1/2, n 12
n 1!2
n0
r 2n 4 − r 2 − 4
r 2n − r 2
∞
1/2, n2
n0
n!2
r 2n
3.5
∞
1/2, n2 n0 n
1!2
3n2 2n r 2n1 > 0.
Therefore, inequality 3.1 follows from 3.2–3.5.
Next, we prove that the parameter p λ 1/4 is the best possible parameter in 0, 1/2
such that inequality 1.10 holds for all a, b > 0 with a /
b.
Since for 0 < p < 1/2 and small x > 0,
AG1, 1 − x 2K
√
1
1
1 − x − x2 o x3 ,
2
16
2x − x2
π
1
1 2
H p1 − x 1 − p, 1 − p 1 − x p 1 − x −p2 p −
x o x3 .
2
4
3.6
3.7
It follows from 3.6 and 3.7 that inequality AG1, 1 − x ≤ Hp1 − x 1 − p, 1 −
p1 − x p holds for small x only p ≥ 1/4.
Remark 3.1. For 0 < p < 1/2 and x > 0, one has
H px 1 − p, 1 − p x p
4 px 1 − p 1 − p x p
lim
K x ∞.
lim
x→0
x→0
AG1, x
1 xπ
3.8
Equation 3.8 implies that there does not exist p ∈ 0, 1/2 such that AG1, x >
Hpx 1 − p, 1 − px p for all x ∈ 0, 1.
√
Proof of Theorem 1.2. Let μ 1/2 − 2/4, then from the monotonicity of the function f2 x Gxa 1 − xb, xb 1 − xa in 0, 1/2 we know that to prove inequality 1.11 we only
need to prove that
AGa, b < G μa 1 − μ b, μb 1 − μ a
for all a, b > 0 with a /
b.
3.9
Journal of Applied Mathematics
7
From 1.1 and 1.8 we clearly see that both AGa, b and Gμa1−μb, μb 1−μa
are symmetric and homogeneous of degree 1. Without loss of generality, we can assume that
a 1 > b. Let t b ∈ 0, 1 and r 1 − t/1 t, then from 1.5 we have
μ 1 − μ t μt 1 − μ −
G μa 1 − μ b, μb 1 − μ a − AGa, b π
.
2Kt 3.10
Let
Gt μ 1 − μ t μt 1 − μ −
π
.
2Kt 3.11
Then making use of 2.2 we have
⎡
⎤
π
2
1
⎣ Kr 1 − r 2 − 1⎦.
Gt 21 rKr π
2
3.12
Therefore, inequality 3.9 follows from 3.10–3.12 together with Lemma 2.2.
√
Next, we prove that the parameter q μ 1/2 − 2/4 is the best possible parameter
in 0, 1/2 such that inequality 1.11 holds for all a, b > 0 with a /
b.
Since for 0 < q < 1/2 and small x > 0,
1
1
−4q2 4q − 1 x2 o x3 .
G q1 − x 1 − q, 1 − q 1 − x q 1 − x 2
8
3.13
It follows from 3.6 and 3.13 that inequality
AG1, 1 − x ≤ Gq1 − x 1 − q, 1 −
√
q1 − x q holds for small x only q ≥ 1/2 − 2/4.
Remark 3.2. For 0 < q < 1/2 and x > 0, one has
G qx 1 − q, 1 − q x q
2 lim
qx 1 − q 1 − q x q K x ∞.
x→0
x→0π
AG1, x
lim
3.14
Equation 3.14 implies that there does not exist q ∈ 0, 1/2 such that AG1, x >
Gqx 1 − q, 1 − qx q for all x ∈ 0, 1.
√
Proof of Theorem 1.3. Let β 1/2 − 3/4, then from the monotonicity of f3 x Lxa 1 −
xb, xb 1 − xa in 0, 1/2 we know that to prove inequality 1.12 we only need to prove
that
AGa, b < L βa 1 − β b, βb 1 − β a
for all a, b > 0 with a /
b.
3.15
8
Journal of Applied Mathematics
From 1.1 and 1.9 we clearly see that both AGa, b and Lβa 1 − βb, βb 1 − βa
are symmetric and homogeneous of degree 1. Without loss of generality, we can assume that
a 1 > b. Let t b ∈ 0, 1 and r 1 − t/1 t, then from 1.5 one has
L βa 1 − β b, βb 1 − β a − AGa, b
√
31 − t
π
√ √ √ √ − 2Kt .
2 log
2− 3 t2 3
2 3 t2− 3
3.16
Let
√
Jt 2 log
31 − t
π
√ √ √ √ − 2Kt .
2− 3 t2 3
2 3 t2− 3
3.17
Then from 2.2 we get
Jt 21 rKr log
π
2
√ √ gr,
3r
2 − 3r
3.18
where gr is defined as in Lemma 2.3.
Therefore, inequality 3.15 follows from 3.16–3.18 together with Lemma 2.3.
√
Next, we prove that the parameter s β 1/2 − 3/4 is the best possible parameter
in 0, 1/2 such that inequality 1.12 holds for all a, b > 0 with a /
b.
Since for 0 < s < 1/2 and small x > 0,
1
1
−4s2 4s − 1 x2 o x3 .
Ls1 − x 1 − s, 1 − s1 − x s 1 − x 2
12
3.19
It follows from 3.6 and 3.19 that inequality
AG1, 1 − x ≤ Ls1 − x 1 − s, 1 −
√
s1 − x s holds for small x only s ≥ 1/2 − 3/4.
Remark 3.3. For 0 < s < 1/2 and x > 0, one has
lim
x→0
Lsx 1 − s, 1 − sx s
2 1 − 2s1 − x
lim K x
∞.
x→0π
AG1, x
logsx 1 − s/1 − sx s
3.20
Equation 3.20 implies that there exist no values s ∈ 0, 1/2 such that AG1, x >
Lsx 1 − s, 1 − sx s for all x ∈ 0, 1.
Acknowledgments
The authors wish to thank the anonymous referees for their careful reading of the paper
and their fruitful comments and suggestions. This research was supported by the Natural
Science Foundation of China under Grant 11071069, and Innovation Team Foundation of the
Department of Education of Zhejiang Province under Grant T200924.
Journal of Applied Mathematics
9
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