Logarithmic mean and weighted sum of geometric and anti-harmonic means

Mira-Cristiana Anisiu$^\ast $, Valeriu Anisiu$^\S $

July 12, 2012

$^\ast $“T. Popoviciu” Institute of Numerical Analysis, P.O. Box 68, 400110 Cluj-Napoca, Romania, e-mail: mira@math.ubbcluj.ro.

$^\S $“Babeş-Bolyai” University, Faculty of Mathematics and Computer Science, 1 Kogălniceanu St., 400084 Cluj-Napoca, Romania, e-mail: anisiu@math.ubbcluj.ro.

We consider the problem of finding the optimal values $\alpha ,\ \beta \in \mathbb {R}$ for which the inequality

\[ \alpha G(a,b)+(1-\alpha )C(a,b){\lt}L(a,b){\lt}\beta G(a,b)+(1-\beta )C(a,b) \]

holds for all $a,b{\gt}0$, $a\neq b$, where $G(a,b),L(a,b)$ and $C(a,b)$ are respectively the geometric, logarithmic and anti-harmonic means of $a$ and $b$.

MSC. 26-04, 26D15, 26E60.

Keywords. Two-variable means, weighted arithmetic mean, inequalities, symbolic computer algebra.

1 Introduction

Given $a,b{\gt}0$, $a\neq b$, the geometric, logarithmic and anti-harmonic means are defined by

\[ G=\sqrt{ab},\ \ L=\tfrac {b-a}{\ln b-\ln a},\ C=\tfrac {a^{2}+b^{2}}{a+b}. \]

It is well-known that

\begin{equation} G < L < C. \label{0}\end{equation}

In this paper we find the values of the parameters $\alpha ,\ \beta \in \mathbb {R}$ for which the inequality

\[ \alpha G(a,b)+(1-\alpha )C(a,b){\lt}L(a,b){\lt}\beta G(a,b)+(1-\beta )C(a,b) \]

holds for all positive numbers $a\neq b$.

Recently, results of this type have been obtained for various triplets of means. Not being exhaustive, we mention Alzer and Qiu [ 1 ] for geometric, exponential (identric) and arithmetic means, Xia and Chu [ 4 ] for harmonic, logarithmic respectively identric and arithmetic means, and Chu et al. [ 3 ] for harmonic, Seiffert and arithmetic means. Several theorems concerning three means chosen from

\begin{equation} H < G < L < I < A < Q < S < C \label{1}\end{equation}

are proved in [ 2 ] . For those means Symbolic Algebra Program Maple was used to find the interval where the parameters $\alpha \ $and $\beta $ can vary, and then the proofs were given.

We can use Maple also to understand the expected degree of difficulty of the proof. Doing so, we found that the problem involving the means $G,$ $L$ and $C$ is among the more difficult ones.

2 Main result

The inequality

αG(a,b)+(1-α)C(a,b)<L(a,b) < βG(a,b)+(1-β)C(a,b)

holds for all positive numbers $a\neq b$ if and only if $\alpha \geq 1$ and $\beta {\lt}\beta _{0}$, where $\beta _{0}=g(x_{0})=0.87002762...$, with $x_{0}$ the unique root of (7) which is greater than $1$, and with $g$ defined in (5).

Proof â–¼

The double inequality (1) is equivalent to

\begin{equation} \beta <\tfrac {C(a,b)-L(a,b)}{C(a,b)-G(a,b)}<\alpha . \label{8}\end{equation}

Without loss of generality, we can consider $0{\lt}a{\lt}b.$ Denoting by $t=b/a,$ $t{\gt}1$, due to the homogeneity of the means, the problem reduces to find $\inf f$ and $\sup f$, where

\begin{equation} f(t)=\tfrac {C(1,t)-L(1,t)}{C(1,t)-G(1,t)}=\tfrac {\left( t^{2}+1\right) \ln t-t^{2}+1}{\left( \sqrt{t}-1\right) ^{2}\left( t+\sqrt{t}+1\right) \ln t}. \label{9}\end{equation}

The function $f$ is obviously bounded, $0\leq f(t)\leq 1$. We shall find $\inf f$ and $\sup f$ for $t{\gt}1$.

Define

\begin{equation} g(x)=f(x^{2})=\tfrac {2\left( x^{4}+1\right) \ln x-x^{4}+1}{2\left( x-1\right) ^{2}\left( x^{2}+x+1\right) \ln x},\ \ \ x>1. \label{g}\end{equation}

In order to find $\inf f=\inf g$ and $\sup f=\sup g$ we shall show first that

\begin{equation} \fbox{$g^\prime $ has a unique root in $(1,\infty ).$} \tag {*}\label{gmin}\end{equation}

Suppose for a moment that this is true and denote by $x_{0}$ this root. We have $\lim _{x\rightarrow 1}g(x)=8/9,$ $\lim _{x\rightarrow \infty }g(x)=1$, $g(7)=0.87003995...{\lt}8/9.$ It follows that $g$ has a minimal point in $(1,\infty )$, so this point must be $x_{0}.$ Furthermore, $g$ must be monotonic in $(1,x_{0})$ and $(x_{0},\infty )$ and so $\beta _{0}=\inf g=g(x_{0}),$ $\sup g=\max (1,8/9)=1$.

So, it remains to prove (6).

The derivative of $g$ is given by

\[ g^{\prime }(x)=\tfrac {h(x)}{2\, x\left( x-1\right) ^{3}\left( x^{2}+x+1\right) ^{2}\left( \ln x\right) ^{2}}, \]

where

\begin{equation} \begin{array}[c]{ll}h(x)= & -2x\left( x+1\right) \left( x^{4}+4x^{2}+1\right) \left( \ln x\right) ^{2}\\ & +x\left( x-1\right) \left( x^{4}+2x^{3}+6x^{2}+2x+1\right) \ln x\\ & +\left( x+1\right) \left( x^{2}+x+1\right) \left( x^{2}+1\right) \left( x-1\right) ^{2}. \end{array} \label{h}\end{equation}

The equation $g^{\prime }(x)=0$ is equivalent to $h(x)=0$, hence to

\begin{equation} \ln x-\tfrac {\left( x^{5}+2\, x^{4}+6\, x^{3}+2\, x^{2}+x+\sqrt{p}\right) \left( x-1\right) }{4x\left( x^{5}+x^{4}+4\, x^{3}+4\, x^{2}+x+1\right) }=0, \label{ech}\end{equation}

where

\begin{equation} \begin{array}[c]{ll}p= & 8\, x^{11}+25\, x^{10}+76\, x^{9}+160\, x^{8}+236\, x^{7}+286\, x^{6}\\ & +236\, x^{5}+160\, x^{4}+76\, x^{3}+25\, x^{2}+8\, x. \end{array} \label{p}\end{equation}

We have considered in (8) the positive root of the quadratic in $\ln x$ equation $h(x)=0$. Let us denote the left hand side of (8) by $k(x).$

We have to show that $k$ has a unique root in $(1,\infty )$. To this aim we compute $k^{\prime }(x)$. The Computer Algebra System Maple will help us to do and organize the computations.

We are interested in the numerator of $k^{\prime }(x)$ expressed in terms of $d=\sqrt{p},$ where $p$ is the polynomial given in (9), i.e.

${\gt}$

The numerator of $k^{\prime }(x)$ is given by

assuming d > 0:

${\gt}$ ndk := collect(%,d);

ndk:= 9 x^2+10 x^3+20 x^4+2 x+2 x^9+20 x^6+9 x^8+10 x^7+x^10+24 x^5
+1d-1-4 x-658 x^8-306 x^5-586 x^7-478 x^6-478 x^10-586 x^9
-306 x^11-154 x^12-64 x^13-64 x^3-22 x^2-154 x^4-x^16-22 x^14-4 x^15

Therefore the numerator $ndk$ of $k^{\prime }(x)$ is of the form $p_{1}d+p_{0}$ ($p_{0}$ and $p_{1}$ being polynomials) and a root of $k^{\prime }(x)$ must be a root of the polynomial $p_{1}^{2}p-p_{0}^{2}$.

We can factorize this polynomial using Maple:

${\gt}$ p0:= coeff(ndk,d,0):

${\gt}$ p1:= coeff(ndk,d,1):

\[ \begin{array}[c]{c}-\left( x-1\right) ^{4}\left( x+1\right) ^{4}\left( x^{2}+1\right) ^{2}\left( x^{2}+x+1\right) \left( x^{4}+4\, x^{2}+1\right) ^{2}\\ \left( x^{10}-x^{9}-3\, x^{8}-44\, x^{7}-94\, x^{6}-150\, x^{5}-94\, x^{4}-44\, x^{3}-3\, x^{2}-x+1\right) \end{array} \]

It follows that any root of $k^{\prime }(x)$ in $(1,\infty )$ must be a root of the 10th degree polynomial

\[ P=x^{10}-x^{9}-3\, x^{8}-44\, x^{7}-94\, x^{6}-150\, x^{5}-94\, x^{4}-44\, x^{3}-3\, x^{2}-x+1. \]

But the polynomial $P$ has a unique root in $(1,\infty ).$ This can be verified using the Sturm sequence.

Indeed, Maple gives:

${\gt}$ sturm(P,x,0,infinity);

\[ 1 \]

We conclude that $k^{\prime }$ has a unique root $r\in $ $(1,\infty )$; actually $r\in $ $(4,5)$ because $k^{\prime }(4){\gt}0,$ $k^{\prime }(5){\lt}0$. So, $k^{\prime }{\gt}0$ in $(1,r)$ and $k^{\prime }{\lt}0$ in $(r,\infty )$. Since $k(1)=0$ and $\lim _{x\rightarrow \infty }k(x)=-\infty $ it follows that $k$ has a unique root in $(1,\infty )$, actually in $(r,\infty )$. So, we have proved (6).

The unique solution $x_{0}$ of $g^{\prime }(x)=0$ can be easily approximated by using the command

${\gt}$ Digits:=30:

${\gt}$ x0:=fsolve(h(x),x=4..infinity);

\[ x0:=7.27177296398582281915348781959 \]

giving $g(x_{0})=0.87002762...$.

Bibliography

1: H. Alzer and S.-L. Qiu, Inequalities for means in two variables, Arch. Math. (Basel), 80 (2003), pp. 201–215.
2: M.-C. Anisiu and V. Anisiu, Bilateral inequalities for means, communication, 9th Joint Conference on Mathematics and Computer Science (MACS 2012), Siófok, Hungary, February 9–12, Abstracts, p. 15, 2012.
3: Y.-M.Chu, Y.-F. Qiu, M.-K. Wang and G.-D. Wang, The optimal convex combination bounds of arithmetic and harmonic means for the Seiffert’s mean, J. Inequal. Appl., 2010, Article ID 436457, 2010, 7 pages, doi:10.1155/2010/436457.
4: W.-F. Xia and Y.-M. Chu, Optimal inequalities related to the logarithmic, identric, arithmetic and harmonic means, Rev. Anal. Numér. Théor. Approx., 39 (2010) no. 2, pp. 176–183. $\includegraphics[scale=0.1]{ext-link.png}$