Sharp bounds for the Neuman-Sándor mean in terms of the power and contraharmonic means

Abstract

In the paper, the authors obtain sharp bounds for the Neuman–Sándor mean in terms of the power and contraharmonic means.

Keywords:

• sharp bound
• Neuman–Sándor mean
• power
• contraharmonic mean

AMS Subject Classifications:

• Primary: 26E60
• Secondary: 26D05
• 33B10

Public Interest Statement

In the paper, the authors obtain a sharp lower bound and a sharp upper bound for the Neuman–Sándor mean in terms of the power and contraharmonic means.

1. Introduction

For positive numbers $a,b>0$ with $a\ne b$, the second Seiffert mean $T(a,b)$, quadratic mean $S(a,b)$, Neuman–Sándor mean $M(a,b)$, and contraharmonic mean $C(a,b)$ are respectively defined in Neuman and Sándor (2003), and Seiffert (1995) by(1.1) $$\begin{array}{cccc}\hfill T(a,b)& =\frac{a-b}{2arctan[(a-b)/(a+b)]},\hfill & \hfill S(a,b)& =\sqrt{\frac{a^2+b^2}{2}}\hfill \end{array}$$(1.1) (1.2) $$\begin{array}{cccc}\hfill M(a,b)& =\frac{a-b}{2\mathrm{a}rcsinh[(a-b)/(a+b)]},\hfill & \hfill C(a,b)& =\frac{a^2+b^2}{a+b}\hfill \end{array}$$(1.2)

It is well known Neuman (2012, 2011), and Neuman and Sándor (2006) that the inequalities$$\begin{array}{c}\hfill M(a,b)<T(a,b)<S(a,b)<C(a,b)\end{array}$$

hold for all $a,b>0$ with $a\ne b$.

In Chu and Hou (2012), Chu, Hou, and Shen (2012), the inequalities(1.3) $$\begin{array}{c}\hfill S(\mathit{\alpha }a+(1-\mathit{\alpha })b,\mathit{\alpha }b+(1-\mathit{\alpha })a)<T(a,b)<S(\mathit{\beta }a+(1-\mathit{\beta })b,\mathit{\beta }b+(1-\mathit{\beta })a)\end{array}$$(1.3)

and(1.4) $$\begin{array}{c}\hfill C(\mathit{\lambda }a+(1-\mathit{\lambda })b,\mathit{\lambda }b+(1-\mathit{\lambda })a)<T(a,b)<C(\mathit{\mu }a+(1-\mathit{\mu })b,\mathit{\mu }b+(1-\mathit{\mu })a)\end{array}$$(1.4)

were proved to be valid for $\frac{1}{2}<\mathit{\alpha },\mathit{\beta },\mathit{\lambda },\mathit{\mu }<1$ and for all $a,b>0$ with $a\ne b$ if and only if(1.5) $$\begin{array}{cc}\hfill \mathit{\alpha }& \le \frac{1}{2}(1+\sqrt{\frac{16}{{\mathit{\pi }}^2}-1}),\mathit{\beta }\ge \frac{3+\sqrt{6}}{6}\hfill \\ \hfill \mathit{\lambda }& \le \frac{1}{2}(1+\sqrt{\frac{4}{\mathit{\pi }}-1}),\mathit{\mu }\ge \frac{3+\sqrt{3}}{6}\hfill \end{array}$$(1.5)

respectively. In Jiang and Qi (2014) and its preprint Jiang and Qi (2013a), the double inequality(1.6) $$\begin{array}{c}\hfill S(\mathit{\alpha }a+(1-\mathit{\alpha })b,\mathit{\alpha }b+(1-\mathit{\alpha })a)<M(a,b)<S(\mathit{\beta }a+(1-\mathit{\beta })b,\mathit{\beta }b+(1-\mathit{\beta })a)\end{array}$$(1.6)

was proved to be valid for $\frac{1}{2}<\mathit{\alpha },\mathit{\beta }<1$ and for all $a,b>0$ with $a\ne b$ if and only if(1.7) $$\begin{array}{c}\hfill \mathit{\alpha }\le \frac{1}{2}\left\{1,+,\sqrt{\frac{1}{[ln(1+\sqrt{2}){]}^2}-1},\right\}\text{and}\mathit{\beta }\ge \frac{3+\sqrt{3}}{6}\end{array}$$(1.7)

For more information on this topic, please refer to recently published papers Chu, Wang, and Gong (2011), Jiang and Qi (2012), Li, Long, and Chu (2012), Li and Qi (2013) and references cited therein.

For $t\in (\frac{1}{2},1)$ and $p\ge \frac{1}{2}$, let(1.8) $$\begin{array}{c}\hfill Q_{t,p}(a,b)=C^p(ta+(1-t)b,tb+(1-t)a)A^{1-p}(a,b)\end{array}$$(1.8)

where $A(a,b)=\frac{a+b}{2}$ is the classical arithmetic mean of $a$ and $b$. Then, by definitions in (1.1) and (1.2), it is easy to see that$$\begin{array}{cc}\hfill Q_{t,1/2}(a,b)& =S(ta+(1-t)b,tb+(1-t)a)\hfill \\ \hfill Q_{t,1}(a,b)& =C(ta+(1-t)b,tb+(1-t)a)\hfill \end{array}$$

and $Q_{t,p}(a,b)$ is strictly increasing with respect to $t\in (\frac{1}{2},1)$.

Motivating by results mentioned above, we naturally ask a question: what are the greatest value $t_1=t_1(p)$ and the least value $t_2=t_2(p)$ in $(\frac{1}{2},1)$ such that the double inequality(1.9) $$\begin{array}{c}\hfill Q_{t_1,p}(a,b)<M(a,b)<Q_{t_2,p}(a,b)\end{array}$$(1.9)

holds for all $a,b>0$ with $a\ne b$ and for all $p\ge \frac{1}{2}$?

The aim of this paper is to answer this question. The solution to this question may be stated as the following Theorem 1.1.

Theorem 1.1

Let $t_1,t_2\in (\frac{1}{2},1)$ and $p\in [\frac{1}{2},\infty )$. Then the double inequality (1.9) holds for all $a,b>0$ with $a\ne b$ if and only if(1.10) $$\begin{array}{c}\hfill t_1\le \frac{1}{2}\left[1,+,\sqrt{(\frac{1}{t^{\ast }}{)}^{1/p}-1},\right]\text{and}{t}_2\ge \frac{1}{2}\left(1,+,\frac{1}{\sqrt{6p}}\right)\end{array}$$(1.10)

where(1.11) $$\begin{array}{c}\hfill t^{\ast }=ln\left(1,+,\sqrt{2},\right)=0.88\cdots \end{array}$$(1.11)

Remark 1.1

When $p=\frac{1}{2}$ in Theorem 1.1, the double inequality (1.9) becomes (1.6).

Remark 1.2

If taking $p=1$ in Theorem 1.1, we can conclude that the double inequality(1.12) $$\begin{array}{c}\hfill C(\mathit{\lambda }a+(1-\mathit{\lambda })b,\mathit{\lambda }b+(1-\mathit{\lambda })a)<M(a,b)<C(\mathit{\mu }a+(1-\mathit{\mu })b,\mathit{\mu }b+(1-\mathit{\mu })a)\end{array}$$(1.12)

holds for all $a,b>0$ with $a\ne b$ if and only if(1.13) $$\begin{array}{c}\hfill \frac{1}{2}<\mathit{\lambda }\le \frac{1}{2}\left[1,+,\sqrt{\frac{1}{ln(1+\sqrt{2})}-1},\right]\text{and}1>\mathit{\mu }\ge \frac{1}{2}\left(1,+,\frac{\sqrt{6}}{6}\right)\end{array}$$(1.13)

Remark 1.3

We note that the paper Li and Zheng (2013) is worth to being read.

2. Lemmas

In order to prove Theorem 1.1, we need the following lemmas.

Lemma 2.1

(Anderson, Vamanamurthy, & Vuorinen, 1997, Theorem 1.25)   For $-\infty <a<b<\infty$, let $f,g:[a,b]\to \mathbb{R}$ be continuous on $[a,b]$ and differentiable on $(a,b)$. If $g^{\prime }(x)\ne 0$ and $\frac{f^{\prime }(x)}{g^{\prime }(x)}$ is strictly increasing (or strictly decreasing, respectively) on $(a,b)$, so are the functions(2.1) $$\begin{array}{c}\hfill \frac{f(x)-f(a)}{g(x)-g(a)}\mathrm{and}\frac{f(x)-f(b)}{g(x)-g(b)}\end{array}$$(2.1)

Lemma 2.2

The function(2.2) $$\begin{array}{c}\hfill h(x)=\frac{(1+x^2)\mathrm{a}rcsinhx}{x}\end{array}$$(2.2)

is strictly increasing and convex on $(0,\infty )$.

Proof

This follows from the following arguments:$$\begin{array}{cc}& h^{\prime }(x)=\frac{x\sqrt{1+x^2}-\mathrm{a}rcsinhx+x^2\mathrm{a}rcsinhx}{x^2}\triangleq \frac{h_1(x)}{x^2}\hfill \\ \hfill & h_1^{\prime }(x)=x(\frac{3x}{\sqrt{1+x^2}}+2\mathrm{a}rcsinhx)\triangleq x{h}_2(x)\hfill \\ \hfill & h_2^{\prime }(x)=\frac{5+2x^2}{{(1+x^2)}^{3/2}}>0\hfill \end{array}$$

on $(0,\infty )$ and$$\begin{array}{c}\hfill \underset{x\to 0^{+}}{lim}{h}_1(x)=\underset{x\to 0^{+}}{lim}{h}_2(x)=0\end{array}$$

Lemma 2.3

For $u\in [0,1]$ and $p\ge \frac{1}{2}$, let(2.3) $$\begin{array}{c}\hfill f_{u,p}(x)=pln(1+u{x}^2)-lnx+ln\mathrm{arcsinh}\mathrm{x}\end{array}$$(2.3)

on $(0,1)$. Then the function $f_{u,p}(x)$ is positive if and only if $6pu\ge 1$ and it is negative if and only if $1+u\le (\frac{1}{t^{\ast }}{)}^{1/p}$, where $t^{\ast }$ is defined by (1.11).

Proof

It is ready that(2.4) $$\begin{array}{c}\hfill \underset{x\to 0^{+}}{lim}{f}_{u,p}(x)=0\end{array}$$(2.4)

and(2.5) $$\begin{array}{c}\hfill \underset{x\to 1^{-}}{lim}{f}_{u,p}(x)=pln(1+u)+ln(t^{\ast })\end{array}$$(2.5)

An easy computation yields(2.6) $$\begin{array}{cc}\hfill f_{u,p}^{\prime }(x)& =\frac{2pux}{1+u{x}^2}+\frac{1}{\sqrt{1+x^2}\mathrm{a}rcsinhx}-\frac{1}{x}\hfill \\ \hfill & =\frac{u[(2p-1)x^2\sqrt{1+x^2}\mathrm{a}rcsinhx+x^3]-[\sqrt{1+x^2}\mathrm{a}rcsinhx-x]}{x(1+u{x}^2)\sqrt{1+x^2}\mathrm{a}rcsinhx}\hfill \\ \hfill & =\frac{(2p-1)x^2\sqrt{1+x^2}\mathrm{a}rcsinhx+x^3}{x(1+u{x}^2)\sqrt{1+x^2}\mathrm{a}rcsinhx}[u-\frac{g_1(x)}{g_2(x)}]\hfill \end{array}$$(2.6)

where$$\begin{array}{c}\hfill g_1(x)=\mathrm{a}rcsinhx-\frac{x}{\sqrt{1+x^2}}\text{and}{g}_2(x)=(2p-1)x^2\mathrm{a}rcsinhx+\frac{x^3}{\sqrt{1+x^2}}\end{array}$$

Furthermore, we have(2.7) $$\begin{array}{c}\hfill g_1(0)=g_2(0)=0\end{array}$$(2.7)

and(2.8) $$\begin{array}{c}\hfill \frac{g_1^{\prime }(x)}{g_2^{\prime }(x)}=\frac{1}{2(2p-1)\sqrt{1+x^2}h(x)+(2p+1)x^2+2p+2}\end{array}$$(2.8)

where $h(x)$ is defined by (2.2). From Lemma 2.2, it follows that the quotient $\frac{g_1^{\prime }(x)}{g_2^{\prime }(x)}$ is strictly decreasing on $(0,1)$. Accordingly, from Lemma 2.1 and (2.7), it is deduced that the ratio $\frac{g_1(x)}{g_2(x)}$ is strictly decreasing on $(0,1)$.

Moreover, making use of L’Hôpital’s rule leads to(2.9) $$\begin{array}{c}\hfill \underset{x\to 0}{lim}\frac{g_1(x)}{g_2(x)}=\frac{1}{6p}\end{array}$$(2.9)

and(2.10) $$\begin{array}{c}\hfill \underset{x\to 1}{lim}\frac{g_1(x)}{g_2(x)}=\frac{\sqrt{2}{t}^{\ast }-1}{\sqrt{2}(2p-1)t^{\ast }+1}\end{array}$$(2.10)

When $u\ge \frac{1}{6p}$, combining (2.6) and (2.9) with the monotonicity of $\frac{g_1(x)}{g_2(x)}$ shows that the function $f_{u,p}(x)$ is strictly increasing on $(0,1)$. Therefore, the positivity of $f_{u,p}(x)$ on $(0,1)$ follows from (2.4) and the increasingly monotonicity of $f_{u,p}(x)$.

When $u\le \frac{\sqrt{2}{t}^{\ast }-1}{\sqrt{2}(2p-1)t^{\ast }+1}$, combining (2.6) and (2.10) with the monotonicity of $\frac{g_1(x)}{g_2(x)}$ reveals that the function $f_{u,p}(x)$ is strictly decreasing on $(0,1)$. Hence, the negativity of $f_{u,p}(x)$ on $(0,1)$ follows from (2.4) and the decreasingly monotonicity of $f_{u,p}(x)$.

When $\frac{\sqrt{2}{t}^{\ast }-1}{\sqrt{2}(2p-1)t^{\ast }+1}<u<\frac{1}{6p}$, from (2.6), (2.9), (2.10) and the monotonicity of the ratio $\frac{g_1(x)}{g_2(x)}$, we conclude that there exists a number $x_0\in (0,1)$ such that $f_{u,p}(x)$ is strictly decreasing in $(0,x_0)$ and strictly increasing in $(x_0,1)$. Denote the limit in (2.5) by $h_p(u)$. Then, from the above arguments, it follows that(2.11) $$\begin{array}{c}\hfill h_p(\frac{1}{6p})=pln(1+\frac{1}{6p})+ln(t^{\ast })>0\end{array}$$(2.11)

and(2.12) $$\begin{array}{c}\hfill h_p(\frac{\sqrt{2}{t}^{\ast }-1}{\sqrt{2}(2p-1)t^{\ast }+1})=pln[1+\frac{\sqrt{2}{t}^{\ast }-1}{\sqrt{2}(2p-1)t^{\ast }+1}]+ln(t^{\ast })<0\end{array}$$(2.12)

Since $h_p(u)$ is strictly increasing for $u>-1$, so it is also in $[\frac{\sqrt{2}{t}^{\ast }-1}{\sqrt{2}(2p-1)t^{\ast }+1},\frac{1}{6p}]$. Thus, the inequalities in (2.11) and (2.12) imply that the function $h_p(u)$ has a unique zero point $u_0=(\frac{1}{t^{\ast }}{)}^{1/p}-1\in (\frac{\sqrt{2}{t}^{\ast }-1}{\sqrt{2}(2p-1)t^{\ast }+1},\frac{1}{6p})$ such that $h_p(u)<0$ for $u\in [\frac{\sqrt{2}{t}^{\ast }-1}{\sqrt{2}(2p-1)t^{\ast }+1},u_0)$ and $h_p(u)>0$ for $u\in (u_0,\frac{1}{6p}]$. As a result, combining (2.4) and (2.5) with the piecewise monotonicity of $f_{u,p}(x)$ reveals that $f_{u,p}(x)<0$ for all $x\in (0,1)$ if and only if $\frac{\sqrt{2}{t}^{\ast }-1}{\sqrt{2}(2p-1)t^{\ast }+1}<u<u_0$. The proof of Lemma 2.3 is complete.

3. Proof of Theorem 1.1

Now we are in a position to prove our Theorem 1.1.

Since both $Q_{t,p}(a,b)$ and $M(a,b)$ are symmetric and homogeneous of degree $1$, without loss of generality, we assume that $a>b$. Let $x=\frac{a-b}{a+b}\in (0,1)$. From (1.2) and (1.8), we obtain$$\begin{array}{cc}\hfill ln\frac{Q_{t,p}(a,b)}{T(a,b)}& =ln\frac{Q_{t,p}(a,b)}{A(a,b)}-ln\frac{T(a,b)}{A(a,b)}\hfill \\ \hfill & =pln[1+{(1-2t)}^2{x}^2]-lnx+ln\mathrm{arcsinh}\mathrm{x}\hfill \end{array}$$

Thus, Theorem 1.1 follows from Lemma 2.3.

Remark 3.1

This is a slightly modified version of the preprint Jiang and Qi (2013b).

Cover image

Source: Author.

Additional information

Funding

The first author was partially supported by the Project of Shandong Province Higher Educational Science and Technology Program [grant number J11LA57], China. The second author was partially supported by the Natural Science Foundation [grant number 2014JQ1006] of Shaanxi Province of China and by the National Natural Science Foundation [grant number 11361038] of China.

Notes on contributors

Feng Qi

Feng Qi is a full Professor in Mathematics at Henan Polytechnic University and Tianjin Polytechnic University, China. He was the founder and the former Head of School of Mathematics and Informatics at Henan Polytechnic University. He was ever a visiting professor at Victoria University in Australia, and University of Hong Kong, and ever a part-time professor atHenan University, Henan Normal University, and Inner Mongolia University for Nationalities in China. He received his PhD degree of Science in Mathematics from University of Science and Technology of China. He is the editor of several international journals. He has published over 500 research articles in reputed international journals. His research interests include the classical analysis, analytic combinatorics, special functions, mathematical inequalities, mathematical means, integral transforms, complex functions, analytic number theory, differential geometry, and mathematical education at universities. For more information, please see his home page at http://qifeng618.wordpress.com and related links therein.