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Journal of Inequalities and Applications

Erschienen in:

Open Access 01.12.2014 | Research

Sharp bounds for the arithmetic-geometric mean

verfasst von: Zhen-Hang Yang, Ying-Qing Song, Yu-Ming Chu

Erschienen in: Journal of Inequalities and Applications | Ausgabe 1/2014

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Abstract

In this article, we establish some new inequality chains for the ratio of certain bivariate means, and we present several sharp bounds for the arithmetic-geometric mean.

MSC:26E60, 26D07, 33E05.

Competing interests

The authors declare that they have no competing interests.

Authors’ contributions

All authors contributed equally to the writing of this paper. All authors read and approved the final manuscript.

1 Introduction

Let

R_{+}

be the set of positive real numbers. Then a two-variable continuous function

M : R_{+}^{2} \to R_{+}

is said to be a mean on

R_{+}

if the double inequality

min (a, b) \leq M (a, b) \leq max (a, b)

holds for all

a, b \in R_{+}

The classical arithmetic-geometric mean

A G M (a, b)

of two positive real numbers a and b is defined as the common limit of sequences

{a_{n}}

and

{b_{n}}

, which are given by

A G M (a, b) = lim_{n \to \infty} a_{n} = lim_{n \to \infty} b_{n},

where

a_{0} = a

b_{0} = b

, and for

n \in N

a_{n + 1} = \frac{a_{n} + b_{n}}{2}, b_{n + 1} = \sqrt{a_{n} b_{n}} .

(1.1)

The well-known Gauss identity shows that

A G M (1, \sqrt{1 - r^{2}}) = \frac{π}{2 K (r)}

for

r \in (0, 1)

, where

K (r) = \int_{0}^{π / 2} {(1 - r^{2} {sin}^{2} t)}^{- 1 / 2} d t

[1] is the complete elliptic integral of the first kind.

Let

a, b > 0

with

a \neq b

. Then the well-known Stolarsky mean [2]

S_{p, q} (a, b)

can be expressed as

S_{p, q} (a, b) = {\begin{cases} {(\frac{q}{p} \frac{a^{p} - b^{p}}{a^{q} - b^{q}})}^{1 / (p - q)} & if p q (p - q) \neq 0, \\ {(\frac{b^{p} - a^{p}}{p (ln b - ln a)})}^{1 / p} & if p \neq 0, q = 0, \\ {(\frac{b^{q} - a^{q}}{q (ln b - ln a)})}^{1 / q} & if p = 0, q \neq 0, \\ exp (\frac{b^{p} ln b - a^{p} ln a}{b^{p} - a^{p}} - \frac{1}{p}) & if p = q \neq 0, \\ \sqrt{a b} & if p = q = 0 . \end{cases}

(1.2)

Many bivariate means are the special case of the Stolarsky mean, for example,

S_{2, 1} (a, b) = (a + b) / 2 = A (a, b)

is the arithmetic mean,

S_{0, 0} (a, b) = \sqrt{a b} = G (a, b)

is the geometric mean,

S_{3 / 2, 1 / 2} (a, b) = (a + \sqrt{a b} + b) / 3 = H e (a, b)

is the Heronian mean,

S_{1, 0} (a, b) = (b - a) / (ln b - ln a) = L (a, b)

is the logarithmic mean,

S_{1, 1} (a, b) = 1 / e {(b^{b} / a^{a})}^{1 / (b - a)} = I (a, b)

is the identric (exponential) mean,

S_{2 p, p} (a, b) = A^{1 / p} (a^{p}, b^{p}) = A_{p} (a, b)

the p-order arithmetic (power, Hölder) mean,

S_{3 p / 2, p / 2} (a, b) = H e^{1 / p} (a^{p}, b^{p}) = H e_{p} (a, b)

is the p-order Heronian mean,

S_{p, 0} (a, b) = L^{1 / p} (a^{p}, b^{p}) = L_{p} (a, b)

is the p-order logarithmic mean,

S_{p, p} (a, b) = I^{1 / p} (a^{p}, b^{p}) = I_{p} (a, b)

is the p-order identric (exponential) mean and

S_{p + 1, p} (a, b) = [p (a^{p + 1} - b^{p + 1})] / [(p + 1) (a^{p} - b^{p})] = J_{p} (a, b)

is the one-parameter mean.

Another important family of means is the Gini means [3] defined by

G_{p, q} (a, b) = {\begin{cases} {(\frac{a^{p} + b^{p}}{a^{q} + b^{q}})}^{1 / (p - q)} & if p \neq q, \\ exp (\frac{a^{p} ln a + b^{p} ln b}{a^{p} + b^{p}}) & if p = q, \end{cases}

(1.3)

it also contains many special means, for instance,

G_{1, 0} (a, b) = A (a, b)

is the arithmetic mean,

G_{1, 1} (a, b) = a^{a / (a + b)} b^{b / (a + b)} = I (a^{2}, b^{2}) / I (a, b) = Z (a, b)

is the power-exponential mean,

G_{p, 0} (a, b) = A^{1 / p} (a^{p}, b^{p}) = A_{p} (a, b)

is the p-order arithmetic (power, Hölder) mean,

G_{p, p} (a, b) = Z^{1 / p} (a^{p}, b^{p}) = Z_{p} (a, b)

is the p-order power-exponential mean and

G_{p + 1, p} (a, b) = (a^{p + 1} + b^{p + 1}) / (a^{p} + b^{p}) = L_{p} (a, b)

is the Lehmer mean.

Recently, the inequalities for the bivariate means have been the subject of intensive research. In particular, the bounds for the arithmetic-geometric mean

A G M

have attracted the attention of many mathematicians. It is well known that the double inequality

L (a, b) < A G M (a, b) < L^{2 / 3} (a^{3 / 2}, b^{3 / 2})

(1.4)

holds for all

a, b > 0

with

a \neq b

. The first inequality of (1.4) was first proposed by Carlson and Vuorinen [4], it was proved in the literature [5‐8] by different methods. Vamanamurthy and Vuorinen [9] (also see [5, 6]) proved that

A G M (a, b) < (π / 2) L (a, b)

for all

a, b > 0

with

a \neq b

. The second inequality of (1.4) is due to Borwein and Borwein [10], and Yang [8] presented a simple proof by use of the ‘Comparison Lemma’ [[10], Lemma 2.1].

In [9] Vamanmurthy and Vuorinen presented the upper bounds for the arithmetic-geometric mean

A G M

in terms of the arithmetic mean A, geometric mean G and logarithmic mean L as follows:

\begin{matrix} A G M (a, b) < L_{2} (a, b) = {(A (a, b) L (a, b))}^{1 / 2}, \\ A G M (a, b) < I (a, b) < A (a, b), \\ A G M (a, b) < A_{1 / 2} (a, b) \end{matrix}

for all

a, b > 0

with

a \neq b

In 1995, Sándor [5] proved that the double inequality

\frac{1}{\frac{12 / (5 π)}{L (a, b)} + \frac{1 - 12 / (5 π)}{A (a, b)}} < A G M (a, b) < \frac{1}{\frac{2 / π}{L (a, b)} + \frac{1 - 2 / π}{A (a, b)}},

(1.5)

holds for all

a, b > 0

with

a \neq b

, and it was improved by Alzer and Qiu [[11], Theorem 19] as

\frac{1}{\frac{β_{2}}{L (a, b)} + \frac{1 - β_{2}}{A (a, b)}} < A G M (a, b) < \frac{1}{\frac{α_{2}}{L (a, b)} + \frac{1 - α_{2}}{A (a, b)}},

(1.6)

with the best possible parameters

β_{2} = 3 / 4

and

α_{2} = 2 / π

Other inequalities involving

A G M

can be found in the literature [12‐20].

The aim of this paper is to establish the new inequality chains for the ratio of certain bivariate means, and we present the sharp bounds for the arithmetic-geometric mean

A G M

2 Lemmas

In order to establish our main results we need several lemmas, which we present in this section.

Lemma 1 ([[21], Corollary 1.1])

Let

a, b > 0

with

a \neq b

. Then both

S_{p, 2 m - p} (a, b)

and

G_{p, 2 m - p} (a, b)

are strictly increasing (decreasing) with respect to

p \in (- \infty, m)

(

p \in (m, \infty)

) for fixed

m > 0

Lemma 2 ([[22], Theorem 5], [[23], Theorem 3.4])

Let

a, b, c, d > 0

with

b / a > d / c \geq 1

. Then the ratio of Stolarsky means

R (p, 2 m - p; a, b; c, d) = S_{p, 2 m - p} (a, b) / S_{p, 2 m - p} (c, d)

is strictly increasing (decreasing) with respect to

p \in (- \infty, m)

(

p \in (m, \infty)

) for fixed

m > 0

Lemma 3 ([[24], Theorem 4.1])

Let

a, b, c, d > 0

with

b / a > d / c \geq 1

. Then the ratio of Stolarsky means

R (p, q; a, b; c, d) = S_{p, q} (a, b) / S_{p, q} (c, d)

is strictly log-concave (log-convex) with respect to

p \in ((| q | - q) / 2, \infty)

(

p \in (- \infty, - (| q | + q) / 2)

) for fixed

q \in R

From Lemma 3, we have Corollary 1.

Corollary 1 Let

λ > 0

α \in (0, 1)

and

a, b, c, d > 0

with

b / a > d / c \geq 1

. Then the function

p \mapsto R^{α} (p, 0; a, b; c, d) R^{1 - α} (\frac{λ - α p}{1 - α}, 0; a, b; c, d) = : r (p)

is strictly increasing in

(0, λ)

and strictly decreasing in

(λ, λ / α)

Proof Let

p_{1} = (λ - α p) / (1 - α)

. Then

\begin{array}{rcl} \frac{r^{'} (p)}{r (p)} & = & α {(ln R (p, 0; a, b; c, d))}^{'} + (1 - α) {(ln R (p_{1}, 0; a, b; c, d))}^{'} \\ \times \frac{- α}{1 - α} \\ = & α ({(ln R (p, 0; a, b; c, d))}^{'} - {(ln R (p_{1}, 0; a, b; c, d))}^{'}) \\ = & α (p - p_{1}) {(ln R (ξ, 0; a, b; c, d))}^{″} \\ = & \frac{α}{1 - α} (p - λ) {(ln R (ξ, 0; a, b; c, d))}^{″}, \end{array}

where ξ is between p and

p_{1}

It follows from Lemma 3 that

R (p, 0; a, b; c, d)

is strictly log-concave with respect to

p \in (0, \infty)

and strictly log-convex with respect to

p \in (- \infty, 0)

. Therefore,

r^{'} (p) > 0

for

p \in (0, λ)

and

r^{'} (p) < 0

for

p \in (λ, λ / α)

. □

Lemma 4 ([[25], Corollary 3.1])

Let

a, b, c, d > 0

with

b / a > d / c \geq 1

. Then the function

Q (p) = \frac{\sqrt{S_{p, q} (a, b) S_{2 k - p, q} (a, b)}}{\sqrt{S_{p, q} (c, d) S_{2 k - p, q} (c, d)}}

is strictly decreasing (increasing) in

(k, \infty)

and strictly increasing (decreasing) in

(- \infty, k)

for fixed

q \geq (\leq) 0

k \geq (\leq) 0

with

q^{2} + k^{2} \neq 0

Let

(k, q) = (3 / 2, 0)

(1 / 2, 1)

, respectively. Then Lemma 4 leads to the following.

Corollary 2 Let

a, b, c, d > 0

with

b / a > d / c \geq 1

. Then

(i)

the function

p \mapsto \frac{\sqrt{S_{p, 0} (a, b) S_{3 - p, 0} (a, b)}}{\sqrt{S_{p, 0} (c, d) S_{3 - p, 0} (c, d)}}

is strictly decreasing in

(3 / 2, \infty)

and strictly increasing in

(- \infty, 3 / 2)

;

(ii)

the function

p \mapsto \frac{\sqrt{S_{p, 1} (a, b) S_{1 - p, 1} (a, b)}}{\sqrt{S_{p, 1} (c, d) S_{1 - p, 1} (c, d)}}

is strictly decreasing in

(1 / 2, \infty)

and increasing in

(- \infty, 1 / 2)

Lemma 5 Let

a, b > 0

with

b > a

. Then

b / a > A (a, b) / G (a, b) > 1

Proof Simple computations lead to

\begin{aligned} \frac{b}{a} - \frac{A (a, b)}{G (a, b)} & = \frac{b}{a} - \frac{\sqrt{\frac{b}{a}} + \sqrt{\frac{a}{b}}}{2} \\ = \frac{1}{2} (\sqrt{\frac{b}{a}} - 1) (1 + 2 \sqrt{\frac{b}{a}} + \sqrt{\frac{a}{b}}) > 0 . \end{aligned}

□

Lemma 6 ([26])

Let

x \in (0, 1)

. Then

A G M (1, x) \sim \frac{π / 2}{ln (1 / x)}, x \to 0^{+},

(2.1)

\frac{1}{A G M (1, 1 - x)} = 1 + \frac{1}{2} x + \frac{5}{16} x^{2} + \frac{7}{32} x^{3} + \frac{169}{1, 024} x^{4} + o (x^{4}) .

(2.2)

Lemma 7 is a consequence of the ‘Comparison Lemma’ in [[10], Lemma 2.1].

Lemma 7 Let Φ be a bivariate mean such that

Φ (G (x, y), A (x, y)) < (>) Φ (x, y)

for all

x, y > 0

with

x \neq y

. Then

A G M (a, b) < (>) Φ (a, b)

for all

a, b > 0

with

a \neq b

3 Inequality chains for the ratio of means

In this section, we give some inequality chains for the ratio of certain bivariate means, which will be used to prove our main results in next section.

Proposition 1 Let

a, b, c, d > 0

with

b / a > d / c \geq 1

. Then we have

\frac{\sqrt{A (a, b) G (a, b)}}{\sqrt{A (c, d) G (c, d)}} < \frac{G_{3 / 4, - 1 / 4} (a, b)}{G_{3 / 4, - 1 / 4} (c, d)} < \frac{S_{7 / 4, - 1 / 4} (a, b)}{S_{7 / 4, - 1 / 4} (c, d)} .

(3.1)

Proof The second inequality of (3.1) can be rewritten as

\frac{S_{7 / 4, - 1 / 4} (1, b / a)}{G_{3 / 4, - 1 / 4} (1, b / a)} > \frac{S_{7 / 4, - 1 / 4} (1, d / c)}{G_{3 / 4, - 1 / 4} (1, d / c)} .

Therefore, it suffices to prove that the function

f_{1} (x) = ln \frac{S_{7 / 4, - 1 / 4} (1, x)}{G_{3 / 4, - 1 / 4} (1, x)} = \frac{1}{2} ln \frac{x^{7 / 4} - 1}{7 (1 - x^{- 1 / 4})} - ln \frac{x^{3 / 4} + 1}{x^{- 1 / 4} + 1}

is strictly increasing in

(1, \infty)

. Replacing x by

x^{4}

and differentiating

f_{1}

give

\begin{array}{rcl} 4 x^{3} f_{1}^{'} (x^{4}) & = & \frac{7}{2} \frac{x^{6}}{(x^{7} - 1)} - \frac{1}{2 (x - 1)} - \frac{1}{2 x} - \frac{3 x^{2}}{x^{3} + 1} + \frac{1}{x + 1} \\ = & \frac{(x + 1) (x + x^{2} + 1) {(x - 1)}^{5}}{2 x (x^{2} - x + 1) (x^{6} + x^{5} + x^{4} + x^{3} + x^{2} + x + 1)} > 0 \end{array}

for

x \in (1, \infty)

Similarly, to prove the first inequality of (3.1), it suffices to prove that the function

f_{2} (x) = ln \frac{G_{3 / 4, - 1 / 4} (1, x)}{\sqrt{A (1, x) G (1, x)}} = ln \frac{x^{3 / 4} + 1}{(x^{- 1 / 4} + 1) \sqrt{\frac{x + 1}{2} \sqrt{x}}} = ln \frac{\sqrt{x} - \sqrt[4]{x} + 1}{\sqrt{\frac{x + 1}{2}}}

is strictly increasing on

(1, \infty)

. Replacing x by

x^{4}

and differentiating

f_{2}

yield

4 x^{3} f_{2}^{'} (x^{4}) = \frac{(x + 1) {(x - 1)}^{3}}{(x^{2} - x + 1) (x^{4} + 1)} > 0

for

x \in (1, \infty)

, which completes the proof. □

Proposition 2 Let

a, b > 0

with

a \neq b

. Then we have

\frac{\sqrt{A (a, b) G (a, b)}}{\sqrt{A (G, A) G (G, A)}} < \frac{G_{3 / 4, - 1 / 4} (a, b)}{G_{3 / 4, - 1 / 4} (G, A)} < \frac{S_{7 / 4, - 1 / 4} (a, b)}{S_{7 / 4, - 1 / 4} (G, A)} < 1,

(3.2)

where

G = G (a, b)

and

A = A (a, b)

Proof By symmetry, without loss of generality, we assume that

a < b

. Then from Lemma 5 and Proposition 1 we clearly see that the first and second inequalities of (3.2) hold. Next we prove the last inequality of (3.2). Let

t = ln \sqrt{b / a} > 0

, then the last inequality of (3.2) can be rewritten as

{(\frac{sinh \frac{7 t}{4}}{7 sinh \frac{t}{4}})}^{1 / 2} < {(\frac{1}{7} \frac{{(cosh t)}^{7 / 4} - 1}{1 - {(cosh t)}^{- 1 / 4}})}^{1 / 2} = {(\frac{1}{7} \frac{{(cosh t)}^{2} - {(cosh t)}^{1 / 4}}{{(cosh t)}^{1 / 4} - 1})}^{1 / 2} .

It suffices to prove that the function

g (t) = cosh t - {(\frac{sinh \frac{7 t}{4} + (sinh \frac{t}{4}) {(cosh t)}^{2}}{sinh \frac{7 t}{4} + sinh \frac{t}{4}})}^{4} < 0

for

t > 0

Simple computations lead to

\begin{matrix} sinh 7 t + (sinh t) {(cosh 4 t)}^{2} = 8 (sinh t {cosh}^{2} t) (2 {cosh}^{2} t - 1) (4 {cosh}^{4} t - 2 {cosh}^{2} t - 1), \\ sinh 7 t + sinh t = 8 (sinh t {cosh}^{2} t) (4 {cosh}^{2} t - 3) (2 {cosh}^{2} t - 1), \\ cosh 4 t = 8 {cosh}^{4} t - 8 {cosh}^{2} t + 1, \\ \begin{aligned} g (4 t) & = cosh 4 t - {(\frac{sinh 7 t + (sinh t) {(cosh 4 t)}^{2}}{sinh 7 t + sinh t})}^{4} \\ = (8 {cosh}^{4} t - 8 {cosh}^{2} t + 1) - {(\frac{4 {cosh}^{4} t - 2 {cosh}^{2} t - 1}{4 {cosh}^{2} t - 3})}^{4} : = g_{1} ({cosh}^{2} t), \end{aligned} \end{matrix}

where

g_{1} (x) = (8 x^{2} - 8 x + 1) - {(\frac{4 x^{2} - 2 x - 1}{4 x - 3})}^{4}, x = {cosh}^{2} t > 1 .

g_{1} (x)

can be rewritten as

\begin{array}{rcl} g_{1} (x) & = & - 16 {(x - 1)}^{4} \frac{16 x^{4} + 32 x^{3} - 88 x^{2} + 48 x - 5}{{(4 x - 3)}^{4}} \\ = & - 16 {(x - 1)}^{4} \frac{16 x^{2} {(x - 1)}^{2} + 64 x {(x - 1)}^{2} + 16 x (x - 1) + 5 (x^{2} - 1) + 3 x^{2}}{{(4 x - 3)}^{4}} < 0 \end{array}

for

x > 1

. Therefore,

g (t) < 0

for

t > 0

Thus we complete the proof. □

Proposition 3 Let

a, b, c, d > 0

with

b / a > d / c \geq 1

and

p \in (3 / 2, 2)

. Then

\begin{array}{rcl} \frac{A^{1 / 4} (a, b) L^{3 / 4} (a, b)}{A^{1 / 4} (c, d) L^{3 / 4} (c, d)} & < & \frac{\sqrt{L_{p} (a, b) L_{3 - p} (a, b)}}{\sqrt{L_{p} (c, d) L_{3 - p} (c, d)}} < \frac{L_{3 / 2} (a, b)}{L_{3 / 2} (c, d)} \\ < & \frac{\sqrt{L (a, b) A_{2 / 3} (a, b)}}{\sqrt{L (c, d) A_{2 / 3} (c, d)}} < \frac{\sqrt{L (a, b) I (a, b)}}{\sqrt{L (c, d) I (c, d)}} < \frac{S_{5 / 4, 1 / 4} (a, b)}{S_{5 / 4, 1 / 4} (c, d)} \\ < & \frac{H e_{3 / 4} (a, b)}{H e_{3 / 4} (c, d)} < \frac{A_{1 / 2} (a, b)}{A_{1 / 2} (c, d)} < \frac{I_{3 / 4} (a, b)}{I_{3 / 4} (c, d)} . \end{array}

Proof (i) From part one of Corollary 2 we see that

\frac{\sqrt{S_{2, 0} (a, b) S_{1, 0} (a, b)}}{\sqrt{S_{2, 0} (c, d) S_{1, 0} (c, d)}} < \frac{\sqrt{S_{p, 0} (a, b) S_{3 - p, 0} (a, b)}}{\sqrt{S_{p, 0} (c, d) S_{3 - p, 0} (c, d)}} < \frac{\sqrt{S_{3 / 2, 0} (a, b) S_{3 / 2, 0} (a, b)}}{\sqrt{S_{3 / 2, 0} (c, d) S_{3 / 2, 0} (c, d)}}

for

p \in (3 / 2, 2)

Therefore, the first and second inequalities of Proposition 3 follow from the above inequalities and

\sqrt{S_{1, 0} (a, b) S_{2, 0} (a, b)} = A^{1 / 4} (a, b) L^{3 / 4} (a, b)

together with

S_{3 / 2, 0} (a, b) = L_{3 / 2} (a, b)

(ii)

For the third inequality of Proposition 3. From

A_{2 / 3} (a, b) = S_{4 / 3, 2 / 3} (a, b) = \frac{S_{4 / 3, 0}^{2} (a, b)}{S_{2 / 3, 0} (a, b)}

we clearly see that it suffices to prove

\frac{S_{3 / 2, 0} (a, b) \sqrt{S_{2 / 3, 0} (a, b)}}{S_{3 / 2, 0} (c, d) \sqrt{S_{2 / 3, 0} (c, d)}} < \frac{S_{4 / 3, 0} (a, b) \sqrt{S_{1, 0} (a, b)}}{S_{4 / 3, 0} (c, d) \sqrt{S_{1, 0} (c, d)}} .

Let

(α, λ) = (1 / 3, 11 / 9)

. Then Corollary 1 leads to the conclusion that the function

p \mapsto R^{1 / 3} (p, 0; a, b; c, d) R^{2 / 3} (\frac{11 - 3 p}{6}, 0; a, b; c, d) = : r (p)

is increasing in

(0, 11 / 9)

. Therefore,

r (2 / 3) < r (1)

, that is,

\frac{S_{3 / 2, 0} (a, b) \sqrt{S_{2 / 3, 0} (a, b)}}{S_{3 / 2, 0} (c, d) \sqrt{S_{2 / 3, 0} (c, d)}} < \frac{S_{4 / 3, 0} (a, b) \sqrt{S_{1, 0} (a, b)}}{S_{4 / 3, 0} (c, d) \sqrt{S_{1, 0} (c, d)}} .

(iii)

The fourth inequality of Proposition 3 can be written as

A_{2 / 3} (a, b) / A_{2 / 3} (c, d) < I (a, b) / I (c, d)

, that is,

R (4 / 3, 2 / 3; a, b; c, d) < R (1, 1; a, b; c, d)

. Let

m = 1

, then from Lemma 2 we know that

R (p, 2 - p; a, b; c, d)

is strictly decreasing with respect to

p \in (1, \infty)

(iv)

For the sixth, seventh, and eighth inequalities, let

m = 3 / 4

, then Lemma 2 leads to the conclusion that

R (p, 3 / 2 - p; a, b; c, d)

is strictly decreasing with respect to

p \in (3 / 4, \infty)

. Consequently,

\begin{matrix} R (\frac{5}{4}, \frac{3}{2} - \frac{5}{4}; a, b; c, d) < R (\frac{9}{8}, \frac{3}{2} - \frac{9}{8}; a, b; c, d) \\ < R (1, \frac{3}{2} - 1; a, b; c, d) < R (\frac{3}{4}, \frac{3}{2} - \frac{3}{4}; a, b; c, d), \end{matrix}

which gives the desired results.

(v)

Finally, we prove the fifth inequality. It can be written as

\frac{\sqrt{L (1, b / a) I (1, b / a)}}{S_{5 / 4, 1 / 4} (1, b / a)} < \frac{\sqrt{L (1, d / c) I (1, d / c)}}{S_{5 / 4, 1 / 4} (1, d / c)} .

Thus we need only to prove that the function

h (x) = ln \frac{\sqrt{L (1, x) I (1, x)}}{S_{5 / 4, 1 / 4} (1, x)}

is strictly decreasing in

(1, \infty)

. Let

t = ln \sqrt{x} \in (0, \infty)

. Then

h (x) = \frac{1}{2} ln \frac{sinh t}{t} + \frac{1}{2} (\frac{t cosh t}{sinh t} - 1) - ln \frac{sinh \frac{5 t}{4}}{5 sinh \frac{t}{4}} : = h_{1} (t) .

Differentiating

h_{1} (t)

yields

h_{1}^{'} (t) = - \frac{h_{2} (t)}{4 t sinh \frac{1}{4} t sinh \frac{5}{4} t {sinh}^{2} t},

where

\begin{array}{rcl} h_{2} (t) & = & 2 t^{2} sinh \frac{t}{4} sinh \frac{5 t}{4} + 2 sinh \frac{t}{4} sinh \frac{5 t}{4} {sinh}^{2} t - t cosh \frac{t}{4} sinh \frac{5 t}{4} {sinh}^{2} t \\ + 5 t cosh \frac{5 t}{4} sinh \frac{t}{4} {sinh}^{2} t - 4 t sinh \frac{t}{4} sinh \frac{5 t}{4} cosh t sinh t . \end{array}

We clearly see that it is enough to prove

h_{2} (t) > 0

for

t > 0

Making use of ‘product to sum’ and power series formulas we get

\begin{array}{rcl} h_{2} (t) & = & - t sinh \frac{t}{2} + \frac{1}{4} cosh \frac{t}{2} - \frac{1}{2} cosh \frac{3 t}{2} + t^{2} cosh \frac{3 t}{2} - t sinh \frac{3 t}{2} + \frac{1}{4} cosh \frac{7 t}{2} \\ + \frac{1}{4} cosh t + \frac{11}{4} t sinh t - t^{2} cosh t - \frac{1}{4} cosh 3 t - \frac{1}{4} t sinh 3 t \\ = & \sum_{n = 1}^{\infty} \frac{s (n)}{4^{n + 1} (2 n)!} t^{2 n}, \end{array}

where

s (n) = 7^{2 n} - (\frac{2}{3} n + 1) 6^{2 n} + (64 n^{2} - 80 n - 18) 3^{2 n - 2} - (16 n^{2} - 30 n - 1) 2^{2 n} - 16 n + 1 .

It is easy to verify that

s (1) = s (2) = s (3) = 0

s (4) = 71, 680

. Next we show that

s (n) > 0

for

n \geq 5

. To this end, we rewrite

s (n)

s (n) = 6^{2 n} s_{1} (n) + \frac{1}{9} (16 n^{2} - 30 n - 1) 2^{2 n} s_{2} (n) + s_{3} (n),

where

\begin{matrix} s_{1} (n) = {(\frac{7}{6})}^{2 n} - (\frac{2}{3} n + 1), \\ s_{2} (n) = 3^{2 n} - 9 \times 2^{2 n} = {(\frac{3}{2})}^{2 n} - 9, \\ s_{3} (n) = (48 n^{2} - 50 n - 17) 3^{2 n - 2} - 16 n + 1 . \end{matrix}

Due to

(16 n^{2} - 30 n - 1) = 16 n (n - 2) + (2 n - 1) > 0

for

n \geq 2

, it suffices to prove

s_{i} (n) > 0

for

n \geq 5

i = 1, 2, 3

. Indeed,

s_{1}^{'} (x) = 2 {(\frac{7}{6})}^{2 x} ln \frac{7}{6} - \frac{2}{3} \geq 2 {(\frac{7}{6})}^{10} ln \frac{7}{6} - \frac{2}{3} = 0.77 \dots > 0,

therefore,

s_{1} (n) \geq s_{1} (5) = 20, 455, 153 / 60, 466, 176 > 0

;

s_{2} (n) > s_{2} (3) = 153 / 64 > 0

;

s_{3} (n) > (48 n^{2} - 50 n - 17) - 16 n + 1 = 48 {(n - 2)}^{2} + 126 (n - 2) + 44 > 0

This completes the proof. □

Proposition 4 Let

a, b > 0

with

a \neq b

. Then for

p \in (3 / 2, 2)

we have

\begin{array}{rcl} 1 & < & \frac{A^{1 / 4} (a, b) L^{3 / 4} (a, b)}{A^{1 / 4} (G, A) L^{3 / 4} (G, A)} < \frac{\sqrt{L_{p} (a, b) L_{3 - p} (a, b)}}{\sqrt{L_{p} (G, A) L_{3 - p} (G, A)}} \\ < & \frac{L_{3 / 2} (a, b)}{L_{3 / 2} (G, A)} < \frac{\sqrt{L (a, b) A_{2 / 3} (a, b)}}{\sqrt{L (G, A) A_{2 / 3} (G, A)}} < \frac{\sqrt{L (a, b) I (a, b)}}{\sqrt{L (G, A) I (G, A)}} \\ < & \frac{S_{5 / 4, 1 / 4} (a, b)}{S_{5 / 4, 1 / 4} (G, A)} < \frac{H e_{3 / 4} (a, b)}{H e_{3 / 4} (G, A)} < \frac{A_{1 / 2} (a, b)}{A_{1 / 2} (G, A)} < \frac{I_{3 / 4} (a, b)}{I_{3 / 4} (G, A)}, \end{array}

(3.3)

where

G = G (a, b)

and

A = A (a, b)

Proof Without loss of generality, we assume that

a < b

. Then the second inequality to the last inequality in (3.3) follows easily from Proposition 3 and Lemma 5.

Next we prove the first inequality of (3.3). Let

t = ln \sqrt{b / a} > 0

. Then it equivalent to the inequality

u (t) = \frac{1}{4} ln cosh t + \frac{3}{4} ln \frac{sinh t}{t} - \frac{1}{4} ln \frac{cosh t + 1}{2} - \frac{3}{4} ln \frac{cosh t - 1}{ln cosh t} > 0 .

Differentiating

u (t)

gives

\begin{array}{rcl} u^{'} (t) & = & \frac{3 t {sinh}^{2} t - (t + 3 sinh t cosh t + 2 t cosh t) ln (cosh t)}{2 t (sinh 2 t) ln (cosh t)} \\ = & \frac{t + 3 sinh t cosh t + 2 t cosh t}{2 t (sinh 2 t) ln (cosh t)} \times u_{1} (t), \end{array}

where

u_{1} (t) = \frac{3 t {sinh}^{2} t}{t + 3 sinh t cosh t + 2 t cosh t} - ln (cosh t) .

Differentiating

u_{1} (t)

leads to

u_{1}^{'} (t) = \frac{t sinh t}{{(t + 2 t cosh t + 3 cosh t sinh t)}^{2} cosh t} \times u_{2} (t),

where

u_{2} (t) = \frac{13}{2} t cosh t - 3 sinh t + t cosh 2 t + \frac{3}{2} sinh 2 t + \frac{3}{2} t cosh 3 t - 3 sinh 3 t .

Making use of the power series we get

\begin{aligned} u_{2} (t) = & \frac{13}{2} \sum_{n = 1}^{\infty} \frac{t^{2 n - 1}}{(2 n - 2)!} - 3 \sum_{n = 1}^{\infty} \frac{t^{2 n - 1}}{(2 n - 1)!} + \sum_{n = 1}^{\infty} \frac{2^{2 n - 2} t^{2 n - 1}}{(2 n - 2)!} \\ + \frac{3}{2} \sum_{n = 1}^{\infty} \frac{2^{2 n - 1} t^{2 n - 1}}{(2 n - 1)!} + \frac{3}{2} \sum_{n = 1}^{\infty} \frac{3^{2 n - 2} t^{2 n - 1}}{(2 n - 2)!} - 3 \sum_{n = 1}^{\infty} \frac{3^{2 n - 1} t^{2 n - 1}}{(2 n - 1)!} \\ = & \sum_{n = 1}^{\infty} \frac{v (n)}{(2 n - 1)!} t^{2 n - 1}, \end{aligned}

where

v (n) = (n - \frac{7}{2}) 3^{2 n - 1} + (n + 1) 2^{2 n - 1} + (13 n - \frac{19}{2}) .

Clearly,

v (1) = v (2) = 0

v (3) = 36

and

v (n) > 0

for

n \geq 4

. Therefore,

u_{2} (t) > 0

u_{1} (t)

is strictly increasing in

(0, \infty)

u_{1} (t) > u_{1} (0^{+}) = 0

u^{'} (t) > 0

, and

u (t) > u (0^{+}) = 0

for

t > 0

Thus the proof is finished. □

4 Sharp bounds for $A G M$

In this section, we present several sharp bounds for the arithmetic-geometric mean

A G M

Theorem 1 can be derived from Propositions 1-4 and Lemma 7.

Theorem 1 Let

a, b > 0

with

a \neq b

. Then the inequalities

\begin{array}{rcl} \sqrt{A (a, b) G (a, b)} & < & G_{3 / 4, - 1 / 4} (a, b) < S_{7 / 4, - 1 / 4} (a, b) < A G M (a, b) \\ < & A^{1 / 4} (a, b) L^{3 / 4} (a, b) < \sqrt{L_{p} (a, b) L_{3 - p} (a, b)} \\ < & L_{3 / 2} (a, b) < \sqrt{L (a, b) A_{2 / 3} (a, b)} < \sqrt{L (a, b) I (a, b)} \\ < & S_{5 / 4, 1 / 4} (a, b) < H e_{3 / 4} (a, b) < A_{1 / 2} (a, b) < I_{3 / 4} (a, b) \end{array}

hold for

p \in (3 / 2, 2)

Remark 1 We clearly see that the upper bound

A^{1 / 4} L^{3 / 4}

for

A G M

is better than

L_{3 / 2}

. Moreover, we have

A G M (a, b) < A^{1 / 4} (G, A) L^{3 / 4} (G, A) = A_{1 / 2}^{1 / 4} (a, b) L^{3 / 4} (G, A) .

Theorem 2 The inequality

A G M (a, b) < A^{p} (a, b) L^{1 - p} (a, b)

(4.1)

holds for all

a, b > 0

with

a \neq b

if and only if

p \geq 1 / 4

Proof Let

x > 0

and

x \to 0^{+}

. Then (2.2) and the power series

\frac{1}{A^{p} (1, 1 - x) L^{1 - p} (1, 1 - x)} = 1 + \frac{1}{2} x + \frac{4 - p}{12} x^{2} + o (x^{2})

lead to

lim_{x \to 0^{+}} \frac{\frac{1}{A G M (1, 1 - x)} - \frac{1}{A^{p} (1, 1 - x) L^{1 - p} (1, 1 - x)}}{x^{2}} = \frac{1}{12} (p - \frac{1}{4}) .

Therefore,

p \geq 1 / 4

is the necessary condition for the inequality

A G M (a, b) < A^{p} (a, b) L^{1 - p} (a, b)

to hold for all

a, b > 0

with

a \neq b

. The sufficiency follows easily from the function

p \mapsto A^{p} (a, b) L^{1 - p} (a, b)

is strictly increasing and Theorem 1. □

Let

a, b > 0

with

a \neq b

r \in R

and

α \in (0, 1)

. Then we define

M_{r} (A (a, b), L (a, b); \frac{1}{4})

M_{r} (A, L; \frac{1}{4}) = {(\frac{1}{4} A^{r} + \frac{3}{4} L^{r})}^{1 / r} if r \neq 0 and M_{0} (A, L; \frac{1}{4}) = A^{1 / 4} L^{3 / 4} .

(4.2)

Remark 2 From Theorem 1 and (1.6) we clearly see that the double inequality

M_{- 1} (A (a, b), L (a, b); 1 / 4) < A G M (a, b) < M_{0} (A (a, b), L (a, b); 1 / 4)

holds for all

a, b > 0

with

a \neq b

Moreover, making use of (2.1) and (2.2) we get

\begin{matrix} lim_{x \to 0^{+}} \frac{A G M (1, x) - {(\frac{1}{4} {(\frac{x + 1}{2})}^{r} + \frac{3}{4} {(\frac{x - 1}{ln x})}^{r})}^{1 / r}}{1 / ln (1 / x)} = {\begin{cases} \frac{π}{2} - \infty, & r \geq 0, \\ \frac{π}{2} - {(\frac{3}{4})}^{1 / r}, & r < 0, \end{cases} \\ lim_{x \to 0^{+}} \frac{\frac{1}{A G M (1, 1 - x)} - {(\frac{1}{4} {(\frac{2 - x}{2})}^{r} + \frac{3}{4} {(\frac{- x}{ln (1 - x)})}^{r})}^{1 / r}}{x^{4}} = \frac{r + 1 / 10}{1, 536} . \end{matrix}

Therefore,

p \leq - (ln 4 - ln 3) / (ln π - ln 2) = - 0.63 \dots

and

q \geq - 1 / 10

are the necessary conditions such that the double inequality

M_{p} (A (a, b), L (a, b); \frac{1}{4}) < A G M (a, b) < M_{q} (A (a, b), L (a, b); \frac{1}{4})

(4.3)

holds for all

a, b > 0

with

a \neq b

Conjecture 1 The double inequality

M_{p} (A (a, b), L (a, b); \frac{1}{4}) < A G M (a, b) < M_{q} (A (a, b), L (a, b); \frac{1}{4})

holds for all

a, b > 0

with

a \neq b

if and only if

p \leq - (ln 4 - ln 3) / (ln π - ln 2)

and

q \geq - 1 / 10

Theorem 3 Let

p, q > 3 / 4

. Then the double inequality

S_{p, 3 / 2 - p} (a, b) < A G M (a, b) < S_{q, 3 / 2 - q} (a, b)

(4.4)

holds for all

a, b > 0

with

a \neq b

if and only if

p \geq 7 / 4

and

3 / 4 < q \leq 3 / 2

Proof The sufficiency follows from the function

p \mapsto S_{p, 3 / 2 - p} (a, b)

is strictly decreasing in

(3 / 4, \infty)

by Lemma 2 and Theorem 1.

Next we prove the necessity. It follows from (2.1) and (2.2) together with the power series

\frac{1}{S_{p, 3 / 2 - p} (1, 1 - x)} = 1 + \frac{1}{2} x + \frac{5}{16} x^{2} + \frac{7}{32} x^{3} + \frac{2 p^{2} - 3 p + 316}{1, 920} x^{4} + o (x^{4}),

that

\begin{matrix} lim_{x \to 0^{+}} \frac{\frac{1}{A G M (1, 1 - x)} - \frac{1}{S_{p, 3 / 2 - p} (1, 1 - x)}}{x^{4}} = \frac{169}{1, 024} - \frac{2 p^{2} - 3 p + 316}{1, 920}, \\ lim_{x \to 0^{+}} \frac{A G M (1, x) - S_{q, 3 / 2 - q} (1, x)}{1 / (ln (1 / x))} = {\begin{cases} \frac{π}{2} - \infty, & 3 / 4 < q \leq 3 / 2, \\ \frac{π}{2}, & q > 3 / 2 . \end{cases} \end{matrix}

Therefore,

p \geq 7 / 4

and

3 / 4 < q \leq 3 / 2

are the necessary conditions the double inequality (4.4) to hold for all

a, b > 0

with

a \neq b

. □

Theorem 4 Let

p \geq 1 / 2

. Then the inequality

A G M (a, b) < \sqrt{S_{p, 1} (a, b) S_{1 - p, 1} (a, b)}

(4.5)

holds for all

a, b > 0

with

a \neq b

if and only if

1 / 2 \leq p \leq 1

Proof The sufficiency follows from the function

p \mapsto \sqrt{S_{p, 1} (a, b) S_{1 - p, 1} (a, b)}

is strictly decreasing in

(1 / 2, \infty)

by Corollary 2(ii) and the inequality

A G M (a, b) < \sqrt{L (a, b) I (a, b)} = \sqrt{S_{1, 1} (a, b) S_{0, 1} (a, b)}

in Theorem 1, and the necessity can be derived from the inequality

lim_{x \to 0^{+}} \frac{A G M (1, x) - \sqrt{S_{p, 1} (1, 1 - x) S_{1 - p, 1} (1, 1 - x)}}{1 / (ln (1 / x))} = {\begin{cases} \frac{π}{2} - \infty, & 1 / 2 \leq p \leq 1, \\ \frac{π}{2}, & p > 1 \end{cases} \leq 0 .

□

Making use of the similar methods, we can prove Theorems 5-7, we omit the proofs here.

Theorem 5 Let and

p, q > 1 / 4

. Then the double inequality

G_{p, 1 / 2 - p} (a, b) < A G M (a, b) < G_{q, 1 / 2 - q} (a, b)

holds for all

a, b > 0

with

a \neq b

if and only if

p \geq 3 / 4

and

1 / 4 < q \leq 1 / 2

Theorem 6 The double inequality

S_{p + 1, p} (a, b) < A G M (a, b) < S_{q + 1, q} (a, b)

holds for all

a, b > 0

with

a \neq b

if and only if

p \leq 0

and

q \geq 1 / 4

Theorem 7 The double inequality

H e_{p} (a, b) < A G M (a, b) < H e_{q} (a, b)

holds for all

a, b > 0

with

a \neq b

if and only if

p \leq 0

and

q \geq 3 / 4

Acknowledgements

This research was supported by the Natural Science Foundation of China under Grants 11371125, 11171307, and 61374086, and the Natural Science Foundation of Zhejiang Province under Grant LY13A010004.

Open Access This article is distributed under the terms of the Creative Commons Attribution 2.0 International License ( https://creativecommons.org/licenses/by/2.0 ), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Competing interests

The authors declare that they have no competing interests.

Authors’ contributions

All authors contributed equally to the writing of this paper. All authors read and approved the final manuscript.

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Titel: Sharp bounds for the arithmetic-geometric mean
verfasst von: Zhen-Hang Yang
Ying-Qing Song
Yu-Ming Chu
Publikationsdatum: 01.12.2014
Verlag: Springer International Publishing
Erschienen in: Journal of Inequalities and Applications / Ausgabe 1/2014
Elektronische ISSN: 1029-242X
DOI: https://doi.org/10.1186/1029-242X-2014-192

Springer Professional

Sharp bounds for the arithmetic-geometric mean

Abstract

Competing interests

Authors’ contributions

1 Introduction

2 Lemmas

3 Inequality chains for the ratio of means

4 Sharp bounds for $A G M$

Acknowledgements

Competing interests

Authors’ contributions

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Abstract

Competing interests

Authors’ contributions

1 Introduction

2 Lemmas

3 Inequality chains for the ratio of means

4 Sharp bounds for A G M

Acknowledgements

Competing interests

Authors’ contributions

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4 Sharp bounds for $A G M$