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<title>Logarithmic mean and weighted sum of geometric and anti-harmonic means: Logarithmic mean and weighted sum of geometric and anti-harmonic means</title>
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<h1>Logarithmic mean and weighted sum of geometric and anti-harmonic means</h1>
<p class="authors">
<span class="author">Mira-Cristiana Anisiu\(^\ast \), Valeriu Anisiu\(^\S \)</span>
</p>
<p class="date">July 12, 2012</p>
</div>
<p>\(^\ast \)“T. Popoviciu” Institute of Numerical Analysis, P.O. Box 68, 400110 Cluj-Napoca, Romania, e-mail: <span class="tt">mira@math.ubbcluj.ro</span>. </p>
<p>\(^\S \)“Babeş-Bolyai” University, Faculty of Mathematics and Computer Science, 1 Kogălniceanu St., 400084 Cluj-Napoca, Romania, e-mail: <span class="tt">anisiu@math.ubbcluj.ro</span>. </p>

<div class="abstract"><p> We consider the problem of finding the optimal values \(\alpha ,\  \beta \in \mathbb {R}\) for which the inequality</p>
<div class="displaymath" id="a0000000002">
  \[  \alpha G(a,b)+(1-\alpha )C(a,b){\lt}L(a,b){\lt}\beta G(a,b)+(1-\beta )C(a,b)  \]
</div>
<p> 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 \(b\). </p>
<p><b class="bf">MSC.</b> 26-04, 26D15, 26E60. </p>
<p><b class="bf">Keywords.</b> Two-variable means, weighted arithmetic mean, inequalities, symbolic computer algebra. </p>
</div>
<h1 id="a0000000003">1 Introduction</h1>
<p>Given \(a,b{\gt}0\), \(a\neq b\), the <em>geometric</em>, <em>logarithmic</em> and <em>anti-harmonic</em> means are defined by</p>
<div class="displaymath" id="a0000000004">
  \[  G=\sqrt{ab},\  \  L=\tfrac {b-a}{\ln b-\ln a},\   C=\tfrac {a^{2}+b^{2}}{a+b}.  \]
</div>
<p> It is well-known that </p>
<div class="equation" id="0">
<p>
  <div class="equation_content">
    \begin{equation}  G < L < C. \label{0}\end{equation}
  </div>
  <span class="equation_label">1</span>
</p>
</div>
<p> In this paper we find the values of the parameters \(\alpha ,\  \beta \in \mathbb {R}\) for which the inequality</p>
<div class="displaymath" id="a0000000005">
  \[  \alpha G(a,b)+(1-\alpha )C(a,b){\lt}L(a,b){\lt}\beta G(a,b)+(1-\beta )C(a,b)  \]
</div>
<p> holds for all positive numbers \(a\neq b\). </p>
<p>Recently, results of this type have been obtained for various triplets of means. Not being exhaustive, we mention Alzer and Qiu <span class="cite">
	[
	<a href="#al-qiu" >1</a>
	]
</span> for geometric, exponential (identric) and arithmetic means, Xia and Chu <span class="cite">
	[
	<a href="#xc" >4</a>
	]
</span> for harmonic, logarithmic respectively identric and arithmetic means, and Chu et al. <span class="cite">
	[
	<a href="#CQWW" >3</a>
	]
</span> for harmonic, Seiffert and arithmetic means. Several theorems concerning three means chosen from</p>
<div class="equation" id="1">
<p>
  <div class="equation_content">
    \begin{equation}  H < G < L < I < A < Q < S < C \label{1}\end{equation}
  </div>
  <span class="equation_label">2</span>
</p>
</div>
<p> are proved in <span class="cite">
	[
	<a href="#aa" >2</a>
	]
</span>. For those means Symbolic Algebra Program <em>Maple</em> was used to find the interval where the parameters \(\alpha \  \)and \(\beta \) can vary, and then the proofs were given. </p>
<p>We can use <em>Maple</em> 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. </p>
<h1 id="a0000000006">2 Main result</h1>
<p><div class="theorem_thmwrapper theorem-style-plain" id="5">
  <div class="theorem_thmheading">
    <span class="theorem_thmcaption">
    <sc element at 0x2453173644752>
    </span>
    <span class="theorem_thmlabel">1</span>
  </div>
  <div class="theorem_thmcontent">
  <p>The inequality<div class="displaymath" id="a0000000007">
  
</div> αG(a,b)+(1-α)C(a,b)&lt;L(a,b) &lt; βG(a,b)+(1-β)C(a,b) <div class="displaymath" id="a0000000008">
  
</div> 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 <span class="rm">(<a href="#h">7</a>)</span> which is greater than \(1\), and with \(g\) defined in <span class="rm">(<a href="#g">5</a>)</span>. </p>

  </div>
</div> </p>
<div class="proof_wrapper" id="a0000000009">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  <p>The double inequality (<a href="#5">1</a>) is equivalent to</p>
<div class="equation" id="8">
<p>
  <div class="equation_content">
    \begin{equation}  \beta <\tfrac {C(a,b)-L(a,b)}{C(a,b)-G(a,b)}<\alpha . \label{8}\end{equation}
  </div>
  <span class="equation_label">3</span>
</p>
</div>
<p> 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 </p>
<div class="equation" id="9">
<p>
  <div class="equation_content">
    \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}
  </div>
  <span class="equation_label">4</span>
</p>
</div>
<p> The function \(f\) is obviously bounded, \(0\leq f(t)\leq 1\). We shall find \(\inf f\) and \(\sup f\) for \(t{\gt}1\). </p>
<p>Define </p>
<div class="equation" id="g">
<p>
  <div class="equation_content">
    \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}
  </div>
  <span class="equation_label">5</span>
</p>
</div>
<p> In order to find \(\inf f=\inf g\) and \(\sup f=\sup g\) we shall show first that </p>
<div class="equation" id="gmin">
<p>
  <div class="equation_content">
    \begin{equation}  \fbox{$g^\prime $ has a unique root in $(1,\infty ).$} \tag {*}\label{gmin}\end{equation}
  </div>
  <span class="equation_label">6</span>
</p>
</div>
<p> 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 \(\beta _{0}=\inf g=g(x_{0}),\) \(\sup g=\max (1,8/9)=1\). </p>
<p>So, it remains to prove (<a href="#gmin">6</a>). </p>
<p>The derivative of \(g\) is given by</p>
<div class="displaymath" id="a0000000010">
  \[  g^{\prime }(x)=\tfrac {h(x)}{2\, x\left( x-1\right) ^{3}\left( x^{2}+x+1\right) ^{2}\left( \ln x\right) ^{2}},  \]
</div>
<p> where</p>
<div class="equation" id="h">
<p>
  <div class="equation_content">
    \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}
  </div>
  <span class="equation_label">7</span>
</p>
</div>
<p> The equation \(g^{\prime }(x)=0\) is equivalent to \(h(x)=0\), hence to </p>
<div class="equation" id="ech">
<p>
  <div class="equation_content">
    \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}
  </div>
  <span class="equation_label">8</span>
</p>
</div>
<p> where</p>
<div class="equation" id="p">
<p>
  <div class="equation_content">
    \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}
  </div>
  <span class="equation_label">9</span>
</p>
</div>
<p>We have considered in (<a href="#ech">8</a>) the positive root of the quadratic in \(\ln x\) equation \(h(x)=0\). Let us denote the left hand side of (<a href="#ech">8</a>) by \(k(x).\) </p>
<p>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 <em>Maple</em> will help us to do and organize the computations. </p>
<p>We are interested in the numerator of \(k^{\prime }(x)\) expressed in terms of \(d=\sqrt{p},\) where \(p\) is the polynomial given in (<a href="#p">9</a>), i.e.</p>
<p>\({\gt}\)  </p>

<p>The numerator of \(k^{\prime }(x)\) is given by </p>

<p><span class="ttfamily"> assuming d &gt; 0:</span> </p>
<p><span class="ttfamily">\({\gt}\) ndk := collect(%,d);</span> </p>
<p><div class="displaymath" id="a0000000011">
  
</div> 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<br />+1d-1-4 x-658 x^8-306 x^5-586 x^7-478 x^6-478 x^10-586 x^9<br />-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 <div class="displaymath" id="a0000000012">
  
</div> </p>
<p>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 \(p_{1}^{2}p-p_{0}^{2}\). </p>
<p>We can factorize this polynomial using <em>Maple</em>: </p>
<p><span class="ttfamily">\({\gt}\) p0:= coeff(ndk,d,0):</span> </p>
<p><span class="ttfamily">\({\gt}\) p1:= coeff(ndk,d,1):</span> </p>

<div class="displaymath" id="a0000000013">
  \[ \begin{array}[c]{c}-\left( x-1\right) ^{4}\left( x+1\right) ^{4}\left( x^{2}+1\right) ^{2}\left( x^{2}+x+1\right) \left( x^{4}+4\, x^{2}+1\right) ^{2}\\ \left( x^{10}-x^{9}-3\, x^{8}-44\, x^{7}-94\, x^{6}-150\, x^{5}-94\, x^{4}-44\, x^{3}-3\, x^{2}-x+1\right) \end{array}  \]
</div>
<p> It follows that any root of \(k^{\prime }(x)\) in \((1,\infty )\) must be a root of the 10th degree polynomial</p>
<div class="displaymath" id="a0000000014">
  \[  P=x^{10}-x^{9}-3\, x^{8}-44\, x^{7}-94\, x^{6}-150\, x^{5}-94\, x^{4}-44\, x^{3}-3\, x^{2}-x+1.  \]
</div>
<p> But the polynomial \(P\) has a unique root in \((1,\infty ).\) This can be verified using the Sturm sequence. </p>
<p>Indeed, <em>Maple</em> gives: </p>
<p><span class="ttfamily">\({\gt}\) sturm(P,x,0,infinity);</span></p>
<div class="displaymath" id="a0000000015">
  \[  1  \]
</div>
<p> 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 (<a href="#gmin">6</a>). </p>
<p>The unique solution \(x_{0}\) of \(g^{\prime }(x)=0\) can be easily approximated by using the command </p>
<p><span class="ttfamily">\({\gt}\) Digits:=30:</span> </p>
<p><span class="ttfamily">\({\gt}\) x0:=fsolve(h(x),x=4..infinity);</span></p>
<div class="displaymath" id="a0000000016">
  \[  x0:=7.27177296398582281915348781959  \]
</div>
<p> giving \(g(x_{0})=0.87002762...\). </p>

  </div>
</div>
<p><small class="footnotesize">  </small></p>
<div class="bibliography">
<h1>Bibliography</h1>
<dl class="bibliography">
  <dt><a name="al-qiu">1</a></dt>
  <dd><p><span class="scshape">H. Alzer</span> and <span class="scshape">S.-L. Qiu</span>, <i class="it">Inequalities for means in two variables</i>, Arch. Math. (Basel), <b class="bfseries">80</b> (2003), pp.&#160;201–215. </p>
</dd>
  <dt><a name="aa">2</a></dt>
  <dd><p><span class="scshape">M.-C. Anisiu</span> and <span class="scshape">V. Anisiu</span>, <i class="it">Bilateral inequalities for means</i>, communication, 9th Joint Conference on Mathematics and Computer Science (MACS 2012), Siófok, Hungary, February 9–12, Abstracts, p. 15, 2012. </p>
</dd>
  <dt><a name="CQWW">3</a></dt>
  <dd><p><span class="scshape">Y.-M.Chu, Y.-F. Qiu, M.-K. Wang</span> and <span class="scshape"> G.-D. Wang</span>, <i class="it">The optimal convex combination bounds of arithmetic and harmonic means for the Seiffert’s mean</i>, J. Inequal. Appl., <b class="bf">2010</b>, Article ID 436457, 2010, 7 pages, doi:10.1155/2010/436457. </p>
</dd>
  <dt><a name="xc">4</a></dt>
  <dd><p><a href ="http://ictp.acad.ro/jnaat/journal/article/view/2010-vol39-no2-art10"> <span class="scshape">W.-F. Xia</span> and <span class="scshape">Y.-M. Chu</span>, <i class="it">Optimal inequalities related to the logarithmic, identric, arithmetic and harmonic means</i>, Rev. Anal. Numér. Théor. Approx., <b class="bfseries">39</b> (2010) no. 2, pp.&#160;176–183. <img src="img-0001.png" alt="\includegraphics[scale=0.1]{ext-link.png}" style="width:12.0px; height:10.700000000000001px" />
</a> </p>
</dd>
</dl>


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