Optimal generalized Heronian mean bounds for the logarithmic mean

Hong-Xing Shi; Bo-Yong Long; Yu-Ming Chu

doi:10.1186/1029-242X-2012-63

Research
Open Access

Optimal generalized Heronian mean bounds for the logarithmic mean

Hong-Xing Shi¹,
Bo-Yong Long² and
Yu-Ming Chu³Email author

Journal of Inequalities and Applications20122012:63

https://doi.org/10.1186/1029-242X-2012-63

Received: 22 December 2011
Accepted: 13 March 2012
Published: 13 March 2012

Abstract

In this article, we establish a double inequality between the generalized Heronian and logarithmic means. The achieved result is inspired by the articles of Lin and Shi et al., and the methods from Janous. The inequalities we obtained improve the existing corresponding results and, in some sense, are optimal.

2010 Mathematics Subject Classification: 26E60.

Keywords

generalized Heronian mean
logarithmic mean
power mean

1 Introduction

The logarithmic mean L(a, b) of two positive numbers a and b is defined by

L (a, b) = \{\begin{matrix} \frac{a - b}{log a - log b}, & a \neq b, \\ a, & a = b . \end{matrix}

(1.1)

Recently, the logarithmic mean has been^: the subject of intensive research. In particular, many remarkable inequalities for the logarithmic mean can be found in the literature [1–30]. It might be surprising that the logarithmic mean has applications in physics, economics, and even in meteorology [31–33]. In [31], Kahlig and Matkowski study a variant of Jensen's functional equation involving the logarithmic mean, which appears in a heat conduction problem. A representation of the logarithmic mean as an infinite product and an iterative algorithm for computing the logarithmic mean as the common limit of two sequences of special geometric and arithmetic means are given in [17]. In [34, 35] it is shown that the logarithmic mean can be expressed in terms of Gauss's hypergeometric function ₂F₁. And, in [35], Carlson and Gastafson prove that the reciprocal of the logarithmic mean is strictly totally positive, that is, every n × n determinant with elements $\frac{1}{L (x_{i}, y_{i})}$ , where 0 < x₁< x₂< ... < x_n and 0 < y₁< y₂< ... < y_n are positive for all n ≥ 1.

The power mean of order r of two positive numbers a and b is defined by

M_{r} (a, b) = \{\begin{matrix} {(\frac{a^{r} + b^{r}}{2})}^{\frac{1}{r}} & r \neq 0, \\ \sqrt{a b}, & r = 0 . \end{matrix}

(1.2)

It is well-known that M_r (a, b) is continuous and strictly increasing with respect to $r \in ℝ$ for fixed a, b > 0 with a ≠ b.

Lin [18] presents the sharp power mean bounds for logarithmic mean as follows:

M_{0} (a, b) < L (a, b) < M_{\frac{1}{3}} (a, b)

(1.3)

for all a, b > 0 with a ≠ b.

For ω ≥ 0 and

p \in ℝ

the generalized Heronian mean H_p,ω (a, b) of two positive numbers a and b is introduced by Shi et al. [36] as follows:

H_{p, ω} (a, b) = \{\begin{matrix} {[\frac{a^{p} + ω {(a b)}^{\frac{p}{2}} + b^{p}}{ω + 2}]}^{\frac{1}{p}}, & p \neq 0, \\ \sqrt{a b}, & p = 0 . \end{matrix}

(1.4)

From (1.2) and (1.4) we clearly see that H_p_,0(a, b) = M_p (a, b) for all $p \in ℝ$ and a, b > 0. It easily follows from (1.4) that H_p,ω (a, b) is continuous with respect to $p \in ℝ$ for fixed a, b > 0 and ω ≥ 0, strictly increasing with respect to $p \in ℝ$ for fixed a, b > 0 with a ≠ b and ω ≥ 0, strictly decreasing with respect to ω ≥ 0 for fixed a, b > 0 with a ≠ b and p > 0, and strictly increasing with respect to ω ≥ 0 for fixed a, b > 0 with a ≠ b and p < 0.

In [37], Janous prove that

L (a, b) < H_{1,4} (a, b)

(1.5)

for all a, b > 0 with a ≠ b.

The purpose of this article is to find the greatest value p = p(ω) and the least value q = q(ω) such that the double inequality H_p,ω (a, b) < L(a, b) < H_q,ω (a, b) holds for fixed ω ≥ 0 and all a, b > 0 with a ≠ b.

2 Main result

Theorem 2.1. For fixed ω ≥ 0 and all a, b > 0 with a ≠ b we have

H_{0, ω} (a, b) < L (a, b) < H_{\frac{ω + 2}{6}, ω} (a, b),

(2.1)

and $H_{\frac{ω + 2}{6}, ω} (a, b)$ and H_0,ω(a, b) are the best possible upper and lower generalized Heronian mean bounds of the logarithmic mean L(a, b), respectively.

Proof. Without loss of generality, we assume a > b and put

t = \frac{a}{b} > 1

. Then from (1.1) and (1.4) we get

\begin{gathered} log [L (a, b)] - log [H_{\frac{ω + 2}{6}, ω} (a, b)] \\ = log (\frac{t - 1}{log t}) - \frac{6}{ω + 2} log (\frac{1 + ω t^{{\frac{ω + 2}{12}}_{}} + t^{\frac{ω + 2}{6}}}{ω + 2}) . \end{gathered}

(2.2)

Let

f (t) = log (\frac{t - 1}{log t}) - \frac{6}{ω + 2} log (\frac{1 + ω t^{{\frac{ω + 2}{12}}_{}} + t^{\frac{ω + 2}{6}}}{ω + 2}) .

(2.3)

Then simple computations lead to

lim_{t \to 1 +} f (t) = 0,

(2.4)

f^{'} (t) = \frac{f_{1} (t)}{t (t - 1) (1 + ω t^{\frac{ω + 2}{12}} + t^{\frac{ω + 2}{6}}) log t},

(2.5)

where

f_{1} (t) = (\frac{ω}{2} t^{\frac{ω + 14}{12}} + t^{\frac{ω + 2}{6}} + \frac{ω}{2} t^{\frac{ω + 2}{12}} + t) log t - t^{\frac{ω + 8}{6}} + t^{\frac{ω + 2}{6}} - ω t^{\frac{ω + 14}{12}} + ω t^{\frac{ω + 2}{12}} - t + 1

.

f_{1} (1) = 0,

(2.6)

\begin{gathered} {f^{'}}_{1} (t) = [\frac{ω (ω + 14)}{24} t^{\frac{ω + 2}{12}} + \frac{ω (ω + 2)}{24} t^{\frac{ω + 10}{12}} + \frac{ω + 2}{6} t^{\frac{ω + 4}{6}} + 1]] log t \\ - \frac{ω + 8}{6} t^{\frac{ω + 2}{6}} + \frac{ω + 8}{6} t^{\frac{ω + 4}{6}} - \frac{ω (ω + 8)}{12} t^{\frac{ω + 2}{12}} + \frac{ω (ω + 8)}{12} t^{\frac{ω - 10}{12}}, \end{gathered}

(2.7)

f_{1}^{'} (1) = 0,

(2.8)

f_{1}^{″} (t) = \frac{1}{288} t^{\frac{ω - 10}{6}} f_{2} (t),

(2.9)

where

\begin{gathered} f_{2} (t) = [ω (ω + 2) (ω + 14) t^{\frac{10 - ω}{12}} + ω (ω + 2) (ω - 10) t^{\frac{ω + 2}{12}} + 8 (ω + 2) (ω - 4)] log t - 2 ω (ω^{2} + \\ 4 ω - 68) t^{\frac{10 - ω}{12}} + 2 ω (ω^{2} + 4 ω - 68) t^{- \frac{ω + 2}{12}} + 288 t^{\frac{4 - ω}{6}} - 8 (ω + 2) (ω + 8) t + 8 (ω^{2} + 10 ω - 20) \end{gathered}

.

f_{2} (1) = 0,

(2.10)

f_{2}^{'} (t) = \frac{1}{12} t^{- \frac{ω + 14}{12}} f_{3} (t),

(2.11)

where

\begin{gathered} f_{3} (t) = - [ω (ω + 2) (ω + 14) (ω - 10) t - ω {(ω + 2)}^{2} (ω - 10)] log t + 2 ω (ω^{3} - 12 ω + 848) t + \\ 2 ω {(ω + 2)}^{2} (4 - ω) + 96 (ω + 2) (ω - 4) t^{\frac{ω + 12}{12}} - 576 (ω - 4) t^{\frac{10 - ω}{12}} - 96 (ω + 2) (ω + 8) t^{\frac{ω + 14}{12}} . \end{gathered}

f_{3} (1) = 0,

(2.12)

\begin{array}{l} {f^{'}}_{3} (t) & = - ω (ω + 2) (ω + 14) (ω - 10) log t - ω {(ω + 2)}^{2} (ω - 10) t^{- 1} \\ + 8 {(ω + 2)}^{2} (ω - 4) t^{\frac{ω + 10}{12}} - 8 (ω + 2) (ω + 8) (ω + 14) t^{\frac{ω + 2}{12}} \\ + 48 (ω - 4) (ω - 10) t^{\frac{ω + 2}{12}} + ω (ω^{3} - 6 ω^{2} + 108 ω + 1976), \end{array}

(2.13)

f_{3}^{'} (1) = 0,

(2.14)

f_{3}^{″} (t) = t^{- 2} f_{4} (t),

(2.15)

where

\begin{gathered} f_{4} (t) = - ω (ω + 2) (ω + 14) (ω - 10) t + ω {(ω + 2)}^{2} (ω - 10) + \frac{2}{3} {(ω + 2)}^{2} (ω - 4) (ω - \\ 10) t^{\frac{ω + 2}{12}} - \frac{2}{3} {(ω + 2)}^{2} (ω + 8) (ω + 14) t^{\frac{ω + 14}{12}} - 4 (ω - 2) (ω - 4) (ω - 10) t^{\frac{10 - ω}{12}} . \end{gathered}

f_{4} (1) = - 8 (ω + 2) (5 ω^{2} - 10 ω + 32) < 0,

(2.16)

\begin{array}{l} {f^{'}}_{4} (t) & = - ω (ω + 2) (ω + 14) (ω - 10) + \frac{1}{18} {(ω + 2)}^{3} (ω - 4) (ω - 10) t^{\frac{ω - 10}{12}} \\ - \frac{1}{18} {(ω + 2)}^{2} (ω + 8) {(ω + 14)}^{2} t^{\frac{ω + 2}{12}} \\ + \frac{1}{3} (ω + 2) (ω - 4) {(ω - 10)}^{2} t^{- \frac{ω + 2}{12}}, \end{array}

(2.17)

f_{4}^{'} (1) = - \frac{2}{3} (ω + 2) (5 ω^{3} + 60 ω^{2} - 108 ω + 448) < 0,

(2.18)

f_{4}^{″} (t) = \frac{1}{216} {(ω + 2)}^{2} t^{- \frac{ω + 14}{12}} f_{5} (t),

(2.19)

where

f_{5} (t) = (ω + 2) (ω - 4) {(ω - 10)}^{2} t^{\frac{ω - 4}{6}} - (ω + 2) (ω + 8) {(ω + 14)}^{2} t^{\frac{ω - 2}{6}} - 6 (ω - 4) {(ω - 10)}^{2}

.

f_{5} (1) = - 6 (11 ω^{3} + 36 ω^{2} + 588 ω + 256) < 0,

(2.20)

f_{5}^{'} (t) = \frac{ω + 2}{6} t^{\frac{ω - 10}{6}} f_{6} (t),

(2.21)

where

f_{6} (t) = {(ω - 4)}^{2} {(ω - 10)}^{2} - (ω + 2) (ω + 8) {(ω + 14)}^{2} t,

(2.22)

f_{6} (1) = - 6 (11 ω^{3} + 36 ω^{2} + 588 ω + 256) < 0 .

(2.23)

From (2.22) we clearly see that f₆(t) is strictly decreasing in [1, +∞), then (2.23) leads to that

f_{6} (t) < 0

(2.24)

for t ∈ (1, + ∞).

It easily follows from (2.5)-(2.21) and (2.24) that

f^{'} (t) < 0

(2.25)

for t ∈ (1, + ∞).

Therefore, $L (a, b) < H_{\frac{ω + 2}{6}, ω} (a, b)$ follows from (2.2)-(2.4) and (2.25).

On the other hand, H_0,ω(a, b) = M₀(a, b) < L(a, b) follows from (1.3).

Next, we prove that H_0,ω(a, b) and $H_{\frac{ω + 2}{6}, ω} (a, b)$ are the optimal lower and upper generalized Heronian mean bounds of the logarithmic mean L(a, b).

For any

0 < ε < \frac{ω + 2}{6}

, ω ≥ 0 and x > 0, from (1.1) and (1.4) we have

\begin{gathered} lim_{x \to + \infty} \frac{H_{ε, ω} (1, x)}{L (1, x)} \\ = {(ω + 2)}^{- \frac{1}{ε}} lim_{x \to + \infty} [\frac{{(1 + ω x^{- \frac{ε}{2}} + x^{- ε})}^{\frac{1}{ε}}}{1 - \frac{1}{x}} log x] \\ = + \infty, \end{gathered}

(2.26)

\begin{gathered} log [L (1, 1 + x)] - log [H_{\frac{ω + 2}{6} - ε, ω} (1, 1 + x)] \\ = \frac{ε}{4 (ω + 2)} x^{2} + o (x^{3}) (x \to 0) . \end{gathered}

(2.27)

Equations (2.26) and (2.27) imply that for any ω ≥ 0 and $0 < ε < \frac{ω + 2}{6}$ there exist sufficiently large X = X(ε, ω) > 1 and sufficiently small δ = δ (ε, ω) > 0; such that H_ε,ω (1, x) > L(1, x) for x ∈ (X, +∞) and $L (1, 1 + x) > H_{\frac{ω + 2}{6} - ε, ω} (1, 1 + x)$ for x ∈ (0, δ).

Remark 2.1. If we take ω = 0, then inequality (2.1) reduce to inequality (1.3).

Remark 2.2. If we take ω = 4, then the upper bound in inequality (2.1) becomes the upper bound in inequality (1.5).

Declarations

Acknowledgements

This study was partly supported by the Natural Science Foundation of China (Grant nos. 11071069, 11171307, 11171105), the Social Science Foundation of China (Grant no. 10BTJ001), the Natural Science Foundation of Hunan Province (Grant no. 09JJ6003), and the Innovation Team Foundation of the Department of Education of Zhejiang Province (Grant no. T200924).

Authors’ Affiliations

(1)

School of Primary Education, Chuxiong Normal University, Chuxiong, 675000, China

(2)

School of Mathematical Science, Anhui University, Hefei, 230039, China

(3)

School of Mathematics and Computational Science, Hunan City University, Yiyang, 413000, China

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