# New sharp bounds for logarithmic mean and identric mean

- Zhen-Hang Yang
^{1}Email author

**2013**:116

https://doi.org/10.1186/1029-242X-2013-116

© Yang; licensee Springer 2013

**Received: **9 November 2012

**Accepted: **22 February 2013

**Published: **20 March 2013

## Abstract

For $x,y>0$ with $x\ne y$, let $L=L(x,y)$, $I=I(x,y)$, $A=A(x,y)$, $G=G(x,y)$, ${A}_{r}={A}^{1/r}({x}^{r},{y}^{r})$ denote the logarithmic mean, identric mean, arithmetic mean, geometric mean and *r*-order power mean, respectively. We find the best constant $p,q>0$ such that the inequalities

hold, respectively. From them some new inequalities for means are derived. Lastly, our new lower bound for the logarithmic mean is compared with several known ones, which shows that our results are superior to others.

**MSC:** 26D07, 26E60, 05A15, 15A18.

### Keywords

logarithmic mean identric mean power mean inequality## 1 Introduction

*x*and

*y*with $x\ne y$ are defined by

*r*of the positive real numbers

*x*and

*y*is defined by

The main properties of these means are given in [1]. In particular, the function $r\mapsto {A}_{r}(x,y)$ ($x\ne y$) is continuous and strictly increasing on ℝ. As special cases, the arithmetic mean and geometric mean are $A=A(x,y)={A}_{1}(x,y)$ and $G=G(x,y)={A}_{0}(x,y)$, respectively.

This result, or a part of it, has been rediscovered and reproved many times (see, *e.g.*, [4–7]).

*p*. Only when $p\in (0,1)$, however, the inequality ${A}_{p}^{1-p}{G}^{p}>G$ would be true. In 2009, another better lower bound for

*L*was given by Zhu [14], that is,

*I*, Stolarsky [17, 18] and Pittenger [19] presented lower and upper bounds for

*I*as follows:

Other inequalities for *L* and *I* and their applications can be found in the literature [20–33].

hold.

are even on $(-\mathrm{\infty},\mathrm{\infty})$, and therefore we assume that $p,q>0$ in what follows.

Our main results are stated as follows.

**Theorem 1** *Let* $p,q>0$. *Then inequalities* (1.12) *hold for all* $x,y>0$ *with* $x\ne y$ *if and only if* $p\ge {p}_{0}=1/\sqrt{5}$ *and* $0<q\le 1/3$, *and the function* $p\mapsto {A}_{p}^{1/(3p)}{G}^{1-1/(3p)}$ *is decreasing on* $(0,\mathrm{\infty})$.

**Theorem 2** *Let* $p,q>0$. *Then inequalities* (1.13) *hold for all* $x,y>0$ *with* $x\ne y$ *if and only if* $p\ge 2/3$ *and* $0<q\le {q}_{0}=\sqrt{10}/5=0.63246$, *and the function* $p\mapsto {A}_{p}^{2/(3p)}{G}^{1-2/(3p)}$ *is decreasing on* $(0,\mathrm{\infty})$.

We will prove two theorems above by hyperbolic function theory. For this end, we need the following lemma, which tells us an inequality for bivariate homogeneous means can be equivalently changed into the form of hyperbolic functions.

**Lemma 1**

*Let*$M(x,y)$

*be a homogeneous mean of positive arguments*

*x*

*and*

*y*.

*Then*

*where* $t=\frac{1}{2}ln(x/y)$.

where $t=\frac{1}{2}ln(x/y)>0$. And then, due to Lemma 1, Theorem 1 and Theorem 2 can be restated as equivalent ones, respectively.

**Theorem 1′**

*Let*$p,t>0$.

*Then the inequality*

*holds for all* $t>0$ *if and only if* $p\ge {p}_{0}=1/\sqrt{5}$ *and the function* $p\mapsto {(coshpt)}^{1/(3{p}^{2})}$ *is decreasing on* $(0,\mathrm{\infty})$. *Inequality* (1.16) *is reversed if and only if* $0<p\le 1/3$.

**Theorem 2′**

*Let*$p,t>0$.

*Then the inequality*

*holds for all* $t>0$ *if and only if* $p\ge 2/3$ *and the function* $p\mapsto {(coshpt)}^{2/(3{p}^{2})}$ *is decreasing on* $(0,\mathrm{\infty})$. *Inequality* (1.17) *is reversed if and only if* $0<p\le {q}_{0}=\sqrt{10}/5$.

Therefore, we will prove Theorem 1′ and Theorem 2′ instead of Theorem 1 and Theorem 2 in the sequel.

## 2 Proof of Theorem 1′

In order to prove Theorem 1′, we first give the following lemmas.

**Lemma 2**

*For*$t>0$,

*let the function*$U:(0,\mathrm{\infty})\mapsto (0,\mathrm{\infty})$

*be defined by*

*Then*

*U*

*is decreasing on*$(0,\mathrm{\infty})$

*with*

*Proof*

*V*is decreasing on $(0,\mathrm{\infty})$, and so $V(p)<{lim}_{p\to 0}V(p)=0$. Therefore, ${U}^{\mathrm{\prime}}(p)<0$, that is to say,

*U*is decreasing on $(0,\mathrm{\infty})$. And, by L’Hospital’s rule, we have

which proves the lemma. □

**Remark 1**From Lemmas 1 and 2 it follows that the function $p\mapsto {({A}_{p}/G)}^{1/p}$ is decreasing on $(0,\mathrm{\infty})$, and

**Lemma 3**

*Let*$p>0$

*and let*$f:(0,\mathrm{\infty})\mapsto (-\mathrm{\infty},\mathrm{\infty})$

*be the function defined by*

*Proof*Using L’Hospital’s rule gives (2.5). To obtain (2.6), we write $f(t)$ as

from which (2.6) easily follows.

This lemma is proved. □

**Lemma 4** *For* $p>0$, *let* *f* *be defined by* (2.4). *Then* *f* *is increasing if* $p\ge 1/\sqrt{5}$ *and decreasing if* $0<p\le 1/3$.

*Proof*

Now we are ready to prove desired results.

*f*is increasing if $p\ge 1/\sqrt{5}$. To this end, by (2.7) in combination with (2.8) and (2.9), it suffices to show that ${a}_{n}\ge 0$ for $n\in \mathbb{N}$. We easily check that ${a}_{1}={a}_{2}=0$ and

which together with (2.11) yields $v(n,p)>0$, and from (2.10) it is derived that ${a}_{n+1}>0$. By mathematical induction, we conclude that ${a}_{n}\ge 0$ for $n\in \mathbb{N}$, which proves part one of this lemma.

*f*is decreasing if $p\le 1/3$. Likewise, it needs to be shown that ${a}_{n}\le 0$ for $n\in \mathbb{N}$. As mentioned previously, ${a}_{1}={a}_{2}=0$ but

which leads to $v(n,p)<0$, and from (2.10) we have ${a}_{n+1}<0$ for $n\ge 3$. By mathematical induction, it is obtained that ${a}_{n}\le 0$ for $n\in \mathbb{N}$.

This completes the proof. □

Now we prove Theorem 1′.

*Proof of Theorem 1′* It is clear that (1.16) (or its reverse inequality) is equivalent to $f(t)>0$ (or <0), where $f(t)$ is defined by (2.4).

which yields $p\ge 1/\sqrt{5}$.

*f*is increasing on $(0,\mathrm{\infty})$, hence

for all $t>0$.

which leads to $0<p\le 1/3$.

*f*by Lemma 4 we conclude that

for all $t>0$.

(iii) Lastly, from Lemma 2 we easily conclude that the function $p\mapsto {(coshpt)}^{1/(3{p}^{2})}$ is decreasing on $(0,\mathrm{\infty})$.

Thus the proof is accomplished. □

## 3 Proof of Theorem 2′

The following lemmas are useful.

**Lemma 5**

*Let*$p>0$

*and let*$g:(0,\mathrm{\infty})\mapsto (-\mathrm{\infty},\mathrm{\infty})$

*be the function defined by*

*Proof*Since $g(t)\to 0$ as $t\to {0}^{+}$, using L’Hospital’s rule yields (3.2). To obtain (3.3), we have to change $g(t)$ as follows:

from which (3.3) follows.

Thus the lemma is proved. □

**Lemma 6** *For* $p>0$, *let the function* *g* *be defined by* (3.1). *Then* *g* *is increasing if* $p\ge 2/3$ *and decreasing if* $0<p\le \sqrt{10}/5$.

*Proof*

leads to (3.10).

Now we are in a position to prove our results.

(i) We first prove that *g* is increasing if $p\ge 2/3$. For this end, it is enough to show that ${b}_{n}\le 0$ by (3.5) and (3.6). Indeed, we have ${b}_{1}={b}_{2}=0$ and ${b}_{3}=-16{p}^{-4}(5{p}^{2}-2)<0$. Suppose that ${b}_{n}\le 0$ for $n\ge 4$. From (3.8) and (3.9) we have ${b}_{n+1}<0$, which proves part one of this lemma by mathematical induction.

(ii) Next we prove that *g* is decreasing if $0<p\le \sqrt{10}/5$. It suffices to prove that ${b}_{n}\ge 0$. We have seen that ${b}_{1}={b}_{2}=0$, but ${b}_{3}=-16{p}^{-4}(5{p}^{2}-2)\ge 0$. Using (3.8) and (3.10), we conclude that ${b}_{n+1}\ge 0$ if ${b}_{n}\ge 0$ for $n\ge 3$. By mathematical induction, part two of this lemma is proved.

Thus the proof ends. □

Based on the above lemmas, Theorem 2′ can be easily proved.

*Proof of Theorem 2′* It is clear that (1.17) (or its reverse inequality) is equivalent to $g(t)>0$ (or <0), where $g(t)$ is defined by (3.1).

which leads to $p\ge 2/3$.

for all $t>0$.

Solving the inequalities yields $0<p\le \sqrt{10}/5$.

*g*we have

for all $t>0$.

(iii) Lastly, due to Lemma 2, the function $p\mapsto {(coshpt)}^{1/(3{p}^{2})}$ is clearly decreasing on $(0,\mathrm{\infty})$.

This proves the proof. □

## 4 Corollaries

Using Theorem 1 and (2.2), the following corollaries are immediate.

**Corollary 1**

*We have*

*holds for* $x,y>0$ *with* $x\ne y$, *where the constants* ${p}_{0}=1/\sqrt{5}$ *and* $1/3$ *are the best constants*.

By Theorem 2 and (2.3), we obtain the following.

**Corollary 2**

*We have*

*holds for* $x,y>0$ *with* $x\ne y$, *where* $2/3$ *and* ${q}_{0}=\sqrt{10}/5$ *are the best constants*.

**Remark 2**Neuman [9] has derived some bounds for certain differences of bivariate means, one of which is as follows:

where $x,y>0$ with $x\ne y$.

Employing Theorem 1, Theorem 2 and (2.2), we can prove an interesting chain of inequalities involving the logarithmic mean, identric mean, power mean and geometric mean.

**Corollary 3**

*Let*$p\ge 2/3$, $1/\sqrt{5}\le q\le \sqrt{10}/5$, $0<r\le \sqrt{10}/10$.

*Then the inequalities*

*hold for* $x,y>0$ *with* $x\ne y$, *where* ${I}_{1/2}={I}^{2}(\sqrt{x},\sqrt{y})$.

*Proof*By Remark 1, we see that the function $p\mapsto {A}_{p}^{1/(3p)}{G}^{1-1/(3p)}$ is decreasing on $(0,\mathrm{\infty})$, and by (2.2) it is deduced that

if $p\ge 2/3$ and $0<r\le \sqrt{10}/10$.

The second and third inequalities are equivalent to (1.17), which hold if and only if $p\ge 2/3$ and $q\le \sqrt{10}/5$ by Theorem 2, respectively.

If $1/\sqrt{5}\le q\le \sqrt{10}/5$, then by Theorem 1 the fourth and fifth inequalities hold.

Squaring both sides of (4.8) and (4.9) yields the sixth and seventh inequality, respectively.

The proof is finished. □

From Lemma 6 with (3.4) another known interesting inequality can be reobtained. It should be noted that the second inequality in (4.10) first appeared in [33] and was reproved by Neuman and Sándor [36].

**Corollary 4**

*For*$x,y>0$

*with*$x\ne y$,

*we have*

*where* $2\sqrt{2}{e}^{-1}$ *is the best constant*.

*Proof*From (3.4) it is obtained that

which is equivalent to (4.10).

Thus the proof is completed. □

## 5 Comparison of some lower bounds for logarithmic mean

*L*such as

and so on, some of which have been proved to be comparable and others remain to be compared further. As applications of our main results, we will discuss them in this section. To this end, we first give a lemma.

**Lemma 7** ([[33], Conclusion 1])

*The function* $r\mapsto {A}_{r}$ *is strictly log*-*concave on* $[0,\mathrm{\infty})$.

Now we compare ${A}_{{p}_{0}}^{1/(3{p}_{0})}{G}^{1-1/(3{p}_{0})}$ with ${((7A+8G)/15)}^{5/7}{G}^{2/7}$.

**Lemma 8**

*Let*$p>0$.

*Then the inequalities*

*hold for all* $x,y>0$ *with* $x\ne y$, *where* ${p}_{0}=1/\sqrt{5}$ *and* ${p}_{1}=7/15$ *cannot be improved*.

*Proof*By Lemma 1, to prove (5.1), it suffices to show that

Clearly, if we prove that for all $n\in \mathbb{N}$, ${v}_{n}\ge 0$ if $p={p}_{0}=1/\sqrt{5}$ and ${v}_{n}\le 0$ if $p={p}_{1}=7/15$, then inequalities (5.1) are valid.

which together with (5.3) yields ${v}_{n+1}>{v}_{n}{(\frac{1}{{p}_{0}}-1)}^{2}>0$ under the inductive assumption ${v}_{n}>0$ for $n\ge 3$. By mathematical induction, it is acquired that ${v}_{n}\ge 0$ for all $n\in \mathbb{N}$.

for all $n\in \mathbb{N}$.

which yields $0<p\le {p}_{0}=1/\sqrt{5}$. On the other hand, by Lemma 2 the function $p\mapsto \frac{1}{3{p}^{2}}ln(coshpt)$ is decreasing on $(0,\mathrm{\infty})$. Therefore, ${p}_{0}=1/\sqrt{5}$ is the largest constant such that ${D}_{1}(t)>0$ for all $t>0$.

In the same way, we can prove ${p}_{1}=7/15$ is the smallest constant such that ${D}_{1}(t)<0$ for all $t>0$.

This completes the proof. □

Next let us compare $\sqrt{\frac{2A+G}{3}G}$ and ${A}_{2/3}^{2/3}{A}_{1/3}^{-1/3}{G}^{2/3}$.

**Lemma 9**

*For*$x,y>0$

*with*$x\ne y$,

*we have*

*Proof*Suppose that $x>y>0$ and let $x/y={t}^{6}$. Then inequality (5.4) can be equivalently changed into

which completes the proof. □

Using Lemmas 7-9, we can easily prove the following.

**Proposition 1**

*For*$x,y>0$

*with*$x\ne y$

*and*${p}_{0}=\sqrt{5}/5$, ${p}_{2}=\sqrt{10}/5=0.63246$,

*we have*

*Proof* The first, second and third inequalities follow from Theorem 1 and Lemma 8. Since $7/15<1/2<{p}_{2}=\sqrt{10}/5$, by Theorem 1 the fourth and fifth ones follow. The sixth and seventh ones are obtained from the second and third ones of (4.7).

By the known inequality ${A}_{2/3}>(2A+G)/3$ (see [[33], (5.13)]), we easily get the eighth one.

Lemma 9 shows that the ninth one holds.

The last one is equivalent to ${A}_{2/3}>{A}^{1/2}{A}_{1/3}^{1/2}$, which easily follows from Lemma 7.

This completes the proof. □

Lastly, we compare ${A}_{{p}_{0}}^{1/(3{p}_{0})}{G}^{1-1/(3{p}_{0})}$ with ${A}_{p}^{1-p}{G}^{p}$ ($0<p<1$).

**Proposition 2**

*Let*${p}_{0}=1/\sqrt{5}$

*and*$0<p<1$.

*Then*

*holds for all* $x,y>0$ *with* $x\ne y$ *if* $1-1/(3{p}_{0})\le p<1$. *While* ${A}_{{p}_{0}}^{1/(3{p}_{0})}{G}^{1-1/(3{p}_{0})}$ *and* ${A}_{p}^{1-p}{G}^{p}$ *are not comparable if* $0<p<1-1/(3{p}_{0})$.

*Proof*(i) By a simple equivalent transformation, inequality (5.6) can be changed into

which proves inequality (5.6).

which implies that there are numbers ${t}_{2}>{t}_{1}>0$ such that ${D}_{3}(t)>0$ when $t\in (0,{t}_{1})$ and ${D}_{3}(t)<0$ when $t\in ({t}_{2},\mathrm{\infty})$. Consequently, $\frac{1}{3{p}_{0}^{2}}ln(cosh{p}_{0}t)$ and $\frac{1-p}{p}lncoshpt$ are not comparable on $(0,\mathrm{\infty})$ if $0<p<1-1/(3{p}_{0})$, which is the desired result.

Thus the proof is finished. □

**Remark 3** From the above two propositions, as far as the lower bounds for the logarithmic mean are concerned, our new lower bound ${A}_{{p}_{0}}^{1/(3{p}_{0})}{G}^{1-1/(3{p}_{0})}$ seems to be superior to most known ones.

## Declarations

### Acknowledgements

The author would like to thank Mr. Pi for her help. The author also wishes to thank the reviewer(s) who gave some important and valuable advises.

## Authors’ Affiliations

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