Exact inequalities involving power mean, arithmetic mean, and identric mean$^{\ast }$

Yu-Ming Chu$^{\S }$, Ming-Yu Shi$^{\bullet }$ Yue-Ping Jiang$^{\bullet }$

October 10, 2009

$^\S $ Department of Mathematics, Huzhou Teachers College, Huzhou 313000, Zhejiang, China, e-mail: chuyuming@hutc.zj.cn.

$^\bullet $College of Mathematics and Econometrics, Hunan University, Changsha, 410082, Hunan, China, e-mail: mingyulj08@163.com, ypjiang731@163.com.

$^\ast $The work of the first author has been supported by the Natural Science Foundation (Grant no. 11071069) and the Innovation Team Foundation of the Department of Education of Zhejiang Province (Grant no. T200924).

For $p\in \mathbb {R}$, the power mean $M_{p}(a,b)$ of order $p$, identric mean $I(a,b)$ and arithmetic mean $A(a,b)$ of two positive real numbers $a$ and $b$ are defined by

\begin{equation*} M_{p}(a,b)= \begin{cases} \displaystyle \left(\tfrac {a^{p}+b^{p}}{2}\right)^{1/p}, & p\neq 0,\\ \sqrt{ab}, & p=0, \end{cases} \quad I(a,b)= \begin{cases} \displaystyle \tfrac {1}{\rm {e}}\left(b^{b}/a^{a}\right)^{1/(b-a)}, & a\neq b,\\ \displaystyle a, & a=b, \end{cases}\end{equation*}

and $A(a,b)=(a+b)/2$, respectively.

In the article, we answer the questions: What are the least values $p$, $q$ and $r$, such that inequalities $A^{1/2}(a,b)I^{1/2}(a,b)\leq M_{p}(a,b)$, $A(a,b)^{1/3}I^{2/3}(a,b)\leq M_{q}(a,b)$ and $A^{2/3}(a,b)I^{1/3}(a,b)\leq M_{r}(a,b)$ hold for all $a,b{\gt}0$?

MSC. 26E60.

Keywords. Power mean, identric mean, arithmetic mean.

1 Introduction

For $p\in \mathbb {R}$, the power mean $M_{p}(a,b)$ of order $p$ and the identric mean $I(a,b)$ of two positive real numbers $a$ and $b$ are defined by

\begin{equation} M_{p}(a,b)= \begin{cases} \displaystyle \left(\tfrac {a^{p}+b^{p}}{2}\right)^{\tfrac {1}{p}}, & p\neq 0,\\ \sqrt{ab}, & p=0 \end{cases} \end{equation}

1.1

and

\begin{equation} I(a,b)= \begin{cases} \displaystyle \tfrac {1}{\rm {e}}\left(\tfrac {b^{b}}{a^{a}}\right)^{\tfrac {1}{b-a}}, & a\neq b,\\ a, & a=b, \end{cases} \end{equation}

1.4

respectively.

It is well-known that $M_{p}(a,b)$ is continuous and strictly increasing with respect to $p\in \mathbb {R}$ for fixed $a,b{\gt}0$ with $a\neq b$. Recently, both mean values have been the subject of intensive research. In particular, many remarkable inequalities for $M_{p}(a,b)$ and $I(a,b)$ can be found in literature [1-24].

Let $A(a,b)=(a+b)/2$, $L(a,b)=(b-a)/(\log b-\log a)(b\neq a)$ and $L(a,a)=a$, $G(a,b)=\sqrt{ab}$ and $H(a,b)=2ab/(a+b)$ be the arithmetic, logarithmic, geometric and harmonic means of two positive numbers $a$ and $b$, respectively. Then

\begin{align} \min \{ a,b\} & \leq H(a,b)=M_{-1}(a,b)\leq G(a,b)=M_{0}(a,b)\leq L(a,b)\\ & \leq I(a,b) \leq A(a,b)=M_{1}(a,b) \leq \max \{ a,b\} .\nonumber \end{align}

In [25], Alzer and Janous established the following sharp double inequality

\[ M_{\log 2/\log 3}(a,b)\leq \tfrac {2}{3}A(a,b)+\tfrac {1}{3}G(a,b)\leq M_{2/3}(a,b) \]

for all $a,b{\gt}0$.

In [26-28], the authors presented the bounds for $L$ and $I$ in terms of $A$ and $G$ as follows

\[ G^{2/3}(a,b)A^{1/3}(a,b)\leq L(a,b)\leq \tfrac {2}{3}G(a,b)+\tfrac {1}{3}A(a,b) \]

and

\[ \tfrac {1}{3}G(a,b)+\tfrac {2}{3}A(a,b)\leq I(a,b) \]

for all $a,b{\gt}0$.

The following companion of (1.3) provides inequalities for the geometric and arithmetic means of $L$ and $I$, the proof can be found in [29].

\begin{align*} G^{1/2}(a,b)A^{1/2}(a,b)& \leq L^{1/2}(a,b)I^{1/2}(a,b)\leq \tfrac {1}{2}L(a,b)+\tfrac {1}{2}I(a,b)\\ & \leq \tfrac {1}{2} G(a,b)+ \tfrac {1}{2} A(a,b) \end{align*}

for all $a,b{\gt}0$.

The following sharp bounds for $L$, $I$, $(LI)^{1/2}$, and $(L+I)/2$ in terms of power means $M_p(a,b)$ are proved in [29-35].

\begin{eqnarray*} L(a,b)\leq M_{1/3}(a,b), \quad M_{2/3}(a,b)\leq I(a,b)\leq M_{\log 2}(a,b), \end{eqnarray*}

\[ M_{0}(a,b)\leq \sqrt{ L(a,b)I(a,b)}\leq M_{1/2}(a,b), \]

and

\[ \tfrac {1}{2}\left(L(a,b)+I(a,b)\right){\lt}M_{1/2}(a,b) \]

for all $a,b{\gt}0$.

Alzer and Qiu [36] proved

\[ M_c(a,b)\leq \tfrac {1}{2}L(a,b)+\tfrac {1}{2}I(a,b) \]

for all $a$, $b{\gt}0$ with the best possible parameter $c=\log 2/(1+\log 2)$.

The main purpose of this paper is to answer the questions: What are the least values $p$, $q$ and $r$, such that inequalities $A^{1/2}(a,b)I^{1/2}(a,b)\leq M_{p}(a,b)$, $A(a,b)^{1/3}I^{2/3}(a,b)\leq M_{q}(a,b)$ and $A^{2/3}(a,b)I^{1/3}(a,b)\leq M_{r}(a,b)$ hold for all $a$, $b{\gt}0$?

2 Lemmas

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

Lemma 2.1

Let

\begin{align*} g(t)& =(1-r)(t^{p+1}+t^{p}+t+1)\log t+(2r-1)t^{p+1}-\\ & \quad -2rt^{p}+t^{p-1}-t^{2}+2rt+1-2r. \end{align*}

If $(r,p)=\left\{ (\tfrac {1}{3},\tfrac {7}{9}),(\tfrac {2}{3},\tfrac {8}{9}),(\tfrac {1}{2},\tfrac {5}{6})\right\} $, then there exists $\lambda \in (1,+\infty )$, such that $g(t){\gt}0$ for $t\in (1,\lambda )$ and $g(t){\lt}0$ for $t\in (\lambda ,+\infty )$.

Proof â–¼

Let $g_1(t)=t^{1-p}g'(t)$, $g_2(t)=t^{p}g_1'(t)$, $g_3(t)=t^{1-p}g_2'(t)$, $g_4(t)=t^{3}g_3'(t)$, $g_5(t)=t^{p-2}g_4'(t)$, $g_6(t)=t^{3}g_5'(t)$, $g_7(t)=t^{1-p}g_6'(t)$, and $g_8(t)=t^{p}g_7'(t)$. Then simple computations lead to

\begin{eqnarray} & & g(1)=0,\\ & & \lim _{t\rightarrow +\infty } g(t)=-\infty ,\\ & & g_1(t)=(1-r)[t^{1-p}+(1+p)t+p]\log t-2t^{2-p}+(1+r)t^{1-p}\nonumber \\ & & \quad \qquad +(1-r)t^{-p}+(2pr-p+r)t-(1-p)t^{-1}-2pr-r+1,\nonumber \\ & & g_1(1)=0,\\ & & \lim _{t\rightarrow +\infty }g_1(t)=-\infty ,\\ & & g_2(t)=(1-r)[(1+p)t^{p}+1-p]\log t+(pr+1)t^{p}+p(1-r)t^{p-1}\nonumber \\ & & \quad \qquad +(1-p)t^{p-2}-2(2-p)t-p(1-r)t^{-1}-pr-p+2,\nonumber \\ & & g_2(1)=0,\\ & & \lim _{t\rightarrow +\infty }g_2(t)=-\infty .\\ & & g_3(t)=p(1+p)(1-r)\log t-2(2-p)t^{1-p}+(1+p)(1-r)t^{-p}\nonumber \\ & & \quad \qquad +p(1-r)t^{-1-p}-p(1-p)(1-r)t^{-1}-(1-p)(2-p)t^{-2}\nonumber \\ & & \quad \qquad +p^{2}r-pr+2p-r+1,\nonumber \\ & & g_3(1)=6p-4-2r=0,\\ & & \lim _{t\rightarrow +\infty }g_3(t)=-\infty ,\\ & & g_4(t)=p(1-r)[(1+p)t^{2}+(1-p)t-(1-p)t^{2-p}-(1+p)t^{1-p}]\nonumber \\ & & \quad \qquad -2(1-p)(2-p)(t^{3-p}-1),\nonumber \\ & & g_4(1)=0,\\ & & \lim _{t\rightarrow +\infty }g_4(t)=-\infty ,\\ & & g_5(t)=p(1-r)[2(1+p)t^{p-1}+(1-p)t^{p-2}-(1-p)(2-p)t^{-1}\nonumber \\ & & \quad \qquad -(1+p)(1-p)t^{-2}]-2(1-p)(2-p)(3-p),\nonumber \\ & & g_5(1)=4(1-r)p^{2}-2(1-p)(2-p)(3-p), \end{eqnarray}

\begin{eqnarray} & & \lim _{t\rightarrow +\infty }g_5(t)=-2(1-p)(2-p)(3-p){\lt}0,\\ & & g_6(t)=p(1-r)[-2(1+p)(1-p)t^{p+1}-(1-p)(2-p)t^{p}\nonumber \\ & & \quad \qquad +(1-p)(2-p)t+2(1+p)(1-p)],\nonumber \\ & & g_6(1)=0,\\ & & g_7(t)=p(1-p)(1-r)[(2-p)t^{1-p}-2(1+p)^{2}t-p(2-p)],\nonumber \\ & & g_7(1)=-p^{2}(1-p)(7+p)(1-r){\lt}0,\\ & & g_8(t)=p(1-p)(1-r)[-2(1+p)^{2}t^{p}+(1-p)(2-p)], \end{eqnarray}

and

\begin{eqnarray} & & \hspace{-38mm}g_8(1)=-p^{2}(1-p)(7+p)(1-r){\lt}0. \end{eqnarray}

From (2.11) we know that $g_5(1)=\tfrac {296}{729}$ if $(r,p)=(\tfrac {1}{3},\tfrac {7}{9})$, $g_5(1)=\tfrac {388}{729}$ if $(r,p)=(\tfrac {2}{3},\tfrac {8}{9})$, and $g_5(1)=\tfrac {119}{729}$ if $(r,p)\in (\tfrac {1}{2},\tfrac {5}{6})$. Therefore

\begin{eqnarray} g_5(1){\gt}0 \end{eqnarray}

for $(r,p)=\left\{ (\tfrac {1}{3},\tfrac {7}{9}),(\tfrac {2}{3},\tfrac {8}{9}), (\tfrac {1}{2},\tfrac {5}{6})\right\} $.

From (2.15) we clearly see that $g_8(t)$ is strictly decreasing in $[1,+\infty )$, then (2.16) implies that $g_8(t){\lt}0$ for $t\in [1,+\infty )$. Hence that $g_7(t)$ is strictly decreasing in $[1,+\infty )$.

From (2.14) and the monotonicity of $g_7(t)$ we know that $g_7(t){\lt}0$ for $t\in [1,+\infty )$. Hence $g_6(t)$ is strictly decreasing in $[1,+\infty )$.

From (2.13) and the monotonicity of $g_6(t)$ we know that $g_6(t){\lt}0$ for $t\in [1,+\infty )$. Hence that $g_5(t)$ is strictly decreasing in $[1,+\infty )$.

Inequalities (2.12) and (2.17) together the monotonicity of $g_5(t)$ imply that there exists $t_0\in (1,+\infty )$, such that $g_5(t){\gt}0$ for $t\in (1,t_0)$ and $g_5(t){\lt}0$ for $t\in (t_0,+\infty )$. Hence $g_4(t)$ is strictly increasing in $[1,t_0]$ and strictly decreasing in $[t_0,+\infty )$.

From equations (2.9) and (2.10) together with the monotonicity of $g_4(t)$ we clearly see that there exists $t_1\in (1,+\infty )$, such that $g_4(t){\gt}0$ for $t\in (1,t_1)$ and $g_4(t){\lt}0$ for $t\in (t_1,+\infty )$. Hence $g_3(t)$ is strictly increasing in $[1,t_1]$ and strictly decreasing in $[t_1,+\infty )$.

Equations (2.7) and (2.8) together with the monotonicity of $g_3(t)$ imply that there exists $t_2\in (1,+\infty )$, such that $g_3(t){\gt}0$ for $t\in (1,t_2)$ and $g_3(t){\lt}0$ for $t\in (t_2,+\infty )$. Hence $g_2(t)$ is strictly increasing in $[1,t_2]$ and strictly decreasing in $[t_2,+\infty )$.

It follows from equations (2.5) and (2.6) together with the monotonicity of $g_2(t)$ that there exists $t_3\in (1,+\infty )$, such that $g_2(t){\gt}0$ for $t\in (1,t_3)$ and $g_2(t){\lt}0$ for $t\in (t_3,+\infty )$. Hence $g_1(t)$ is strictly increasing in $[1,t_3]$ and strictly decreasing in $[t_3,+\infty )$.

Equations (2.3) and (2.4) together with the monotonicity of $g_1(t)$ lead to the conclusion that there exists $\lambda _5\in (1,+\infty )$, such that $g_1(t){\gt}0$ for $t\in (1,t_4)$ and $g_1(t){\lt}0$ for $t\in (t_4,+\infty )$. Hence $g(t)$ is strictly increasing in $[1,t_4]$ and strictly decreasing in $[t_4,+\infty )$.

Therefore, Lemma 2.1 follows from equations (2.1) and (2.2) together with the monotonicity of $g(t)$.

3 Main Results

Theorem 3.1

For all $a,b{\gt}0$ one has

\[ A^{1/3}(a,b)I^{2/3}(a,b)\leq M_{7/9}(a,b), \]

\[ A^{2/3}(a,b)I^{1/3}(a,b)\leq M_{8/9}(a,b), \]

and

\[ A^{1/2}(a,b)I^{1/2}(a,b)\leq M_{5/6}(a,b), \]

each inequality holds equality if and only if $a=b$, and the parameters $7/9$, $8/9$ and $5/6$ in each inequality cannot be improved.

Proof â–¼

If $a=b$, then we clearly see that

\begin{align*} A^{1/2}(a,b)I^{1/2}(a,b)& =M_{5/6}(a,b)=A^{1/3}(a,b)I^{2/3}(a,b)=M_{7/9}(a,b)\\ & = A^{2/3}(a,b)I^{1/3}(a,b)=M_{8/9}(a,b)=a. \end{align*}

If $a\neq b$, without loss of generality, we assume that $a{\gt}b$. Let $(\alpha ,p)=\left\{ (1/3,7/9),(2/3,8/9),(1/2,5/6)\right\} $ and $t=a/b{\gt}1$, then (1.1) and (1.2) lead to

\begin{equation} M_p(a,b) \! -\! A^{\alpha }(a,b)I^{1-\alpha }(a,b) =b\left[\left( \tfrac {t^{p}+1}{2}\right)^{1/p}\! -\! \left(\tfrac {t+1}{2}\right)^{\alpha }\left(\tfrac {1}{\rm {e}}\cdot t^{\tfrac {t}{t-1}}\right)^{1-\alpha }\right]. \end{equation}

3.1

Let

\[ f(t)=\displaystyle \tfrac {1}{p}\log \tfrac {1+t^{p}}{2}-\alpha \log \tfrac {t+1}{2} -(1-\alpha )\tfrac {t}{t-1}\log t+(1-\alpha ), \]

then

\begin{eqnarray} & & \lim _{t\rightarrow 1^{+}}f(t)=0, \\ & & \lim _{t\rightarrow \infty }f(t)=(1-\alpha )+(\alpha -\tfrac {1}{p})\log 2 \end{eqnarray}

and

\begin{eqnarray} & & f’(t)=\tfrac {g(t)}{(t+1)(t-1)^{2}(t^{p}+1)}, \end{eqnarray}

where

\begin{align*} g(t)& =(1-\alpha )(t^{p+1}+t^{p}+t+1)\log t\\ & \quad +(2\alpha -1)t^{p+1}-2\alpha t^{p}+t^{p-1}-t^{2}+2\alpha t+1-2\alpha . \end{align*}

From (3.3) we know that $\displaystyle \lim _{t\rightarrow \infty }f(t)=2(7-10\log 2)/21{\gt}0$ if $(\alpha ,p)=(\tfrac {1}{3},\tfrac {7}{9})$, $\displaystyle \lim _{t\rightarrow \infty }f(t)=(8-11\log 2)/24{\gt}0$ if $(\alpha ,p)=(\tfrac {2}{3},\tfrac {8}{9})$, and $\displaystyle \lim _{t\rightarrow \infty }f(t)=(5-7\log 2)/10{\gt}0$ if $(\alpha ,p)=(\tfrac {1}{2},\tfrac {5}{6})$. Hence we get

\begin{eqnarray} \lim _{t\rightarrow \infty }f(t){\gt}0. \end{eqnarray}

Equation (3.4) and Lemma 2.1 lead to the conclusion that there exists $\lambda \in (1,+\infty )$, such that $f(t)$ is strictly increasing in $[1,\lambda ]$ and strictly decreasing in $[\lambda ,+\infty )$. Therefore, $A^{1/3}(a,b)I^{2/3}(a,b){\lt} M_{7/9}(a,b)$, $A^{2/3}(a,b)I^{1/3}(a,b){\lt}M_{8/9}(a,b)$, and $A^{1/2}(a,b)I^{1/2}(a,b){\lt}M_{5/6}(a,b)$ for all $a,b{\gt}0$ with $a\neq b$ follow from (3.1), (3.2), (3.5) and the monotonicity of $f(t)$.

Next, we prove that the parameters $7/9$, $8/9$ and $5/6$ in each inequality cannot be improved.

For any $0{\lt}\varepsilon {\lt}7/9$, $0{\lt}x{\lt}1$ and $x\rightarrow 0$, making use of Taylor expansion one has

\begin{align} & \log \left[A^{1/2}(1,1+x)I^{1/2}(1,1+x)\right]-\log M_{5/6-\varepsilon }(1,1+x)\\ & =\tfrac {1}{2}\log (1+\tfrac {x}{2})+\tfrac {1+x}{2x}\log (1+x)-\tfrac {1}{2}-\tfrac {6}{5-6\varepsilon }\log \tfrac {1+(1+x)^{5/6-\varepsilon }}{2}\nonumber \\ & =\tfrac {\varepsilon }{8}x^{2}+o(x^{2}),\nonumber \end{align}

\begin{align} & \log \left[A^{1/3}(1,1+x)I^{2/3}(1,1+x)\right]-\log M_{7/9-\varepsilon }(1,1+x)\\ & =\tfrac {1}{3}\log (1+\tfrac {x}{2})+\tfrac {2(1+x)}{3x}\log (1+x)-\tfrac {2}{3}-\tfrac {9}{7-9\varepsilon }\log \tfrac {1+(1+x)^{7/9-\varepsilon }}{2}\nonumber \\ & =\tfrac {\varepsilon }{8}x^{2}+o(x^{2})\nonumber \end{align}

and

\begin{align} & \log \left[A^{2/3}(1,1+x)I^{1/3}(1,1+x)\right]-\log M_{8/9-\varepsilon }(1,1+x)\\ & =\tfrac {2}{3}\log (1+\tfrac {x}{2})+\tfrac {1+x}{3x}\log (1+x)-\tfrac {1}{3}-\tfrac {9}{8-9\varepsilon }\log \tfrac {1+(1+x)^{8/9-\varepsilon }}{2}\nonumber \\ & =\tfrac {\varepsilon }{8}x^{2}+o(x^{2}).\nonumber \end{align}

Equations (3.6)–(3.8) imply that for any $0{\lt}\varepsilon {\lt}7/9$, there exists $0{\lt}\delta =\delta (\varepsilon ){\lt}1$, such that

\[ A^{1/2}(1,1+x)I^{1/2}(1,1+x){\gt} M_{5/6-\varepsilon }(1,1+x), \]

\[ A^{1/3}(1,1+x)I^{2/3}(1,1+x){\gt} M_{7/9-\varepsilon }(1,1+x) \]

and

\[ A^{2/3}(1,1+x)I^{1/3}(1,1+x){\gt} M_{8/9-\varepsilon }(1,1+x) \]

for $x\in (0,\delta )$.

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Exact inequalities involving power mean, arithmetic mean, and identric mean\(^{\ast }\)

1 Introduction

2 Lemmas

3 Main Results

Bibliography