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<title>Sharp inequalities for the Neuman-Sándor mean in terms of arithmetic and contra-harmonic means\(^\bullet \): Sharp inequalities for the Neuman-Sándor mean in terms of arithmetic and contra-harmonic means\(^\bullet \)</title>
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<h1>Sharp inequalities for the Neuman-Sándor mean in terms of arithmetic and contra-harmonic means\(^\bullet \)</h1>
<p class="authors">
<span class="author">Yu-ming Chu\(^{\ast }\), Miao-Kun Wang\(^{\ast }\) Bao-Yu Liu\(^{\S }\)</span>
</p>
<p class="date">May 5, 2013.</p>
<p class="thanks">\(^\bullet \)This research was supported by the Natural Science Foundation of China under Grants 61374086 and 11171307, and the Natural Science Foundation of Zhejiang Province under Grant LY13A010004.</p>
</div>




<div class="abstract"><p> In this paper, we find the greatest values \(\alpha \) and \(\lambda \), and the least values \(\beta \) and \(\mu \) such that the double inequalities </p>
<div class="displaymath" id="a0000000002">
  \[ C^{\alpha }(a,b)A^{1-\alpha }(a,b){\lt}M(a,b){\lt}C^{\beta }(a,b)A^{1-\beta }(a,b) \]
</div>
<p> and </p>
<div class="displaymath" id="a0000000003">
  \begin{align*} & [C(a,b)/6+5 A(a,b)/6]^{\lambda }\left[C^{1/6}(a,b)A^{5/6}(a,b)\right]^{1-\lambda }{\lt}M(a,b){\lt}\\ & \qquad {\lt}[C(a,b)/6+5 A(a,b)/6]^{\mu }\left[C^{1/6}(a,b)A^{5/6}(a,b)\right]^{1-\mu } \end{align*}
</div>
<p> hold for all \(a,b{\gt}0\) with \(a\neq b\), where \(M(a,b)\), \(A(a,b)\) and \(C(a,b)\) denote the Neuman-Sándor, arithmetic, and contra-harmonic means of \(a\) and \(b\), respectively. </p>
<p><b class="bf">MSC.</b> 26E60. </p>
<p><b class="bf">Keywords.</b> Neuman-Sándor mean, arithmetic mean, contra-harmonic mean. </p>
</div>
<h1 id="a0000000004">1 Introduction</h1>
<p> For \(a,b{\gt}0\) with \(a\neq b\) the Neuman-Sándor mean \(M(a,b)\) <span class="cite">
	[
	<a href="#1" >1</a>
	]
</span> is defined by </p>
<div class="displaymath" id="a0000000005">
  \begin{equation*}  M(a,b)=\tfrac {a-b}{2{{\rm {arcsinh}}}\left[(a-b)/(a+b)\right]}, \end{equation*}
</div>
<p> where \({\rm {arcsinh}}(x)=\log (x+\sqrt{1+x^2})\) is the inverse hyperbolic sine function. </p>
<p>Recently, the Neuman-Sándor mean has been the subject of intensive research. In particular, many remarkable inequalities for the Neuman-Sándor mean \(M(a,b)\) can be found in the literature <span class="cite">
	[
	<a href="#1" >1</a>
	]
</span>–<span class="cite">
	[
	<a href="#4" >4</a>
	]
</span>. </p>
<p>Let \(A(a,b)=(a+b)/2\) and \(C(a,b)=(a^2+b^2)/(a+b)\) be the arithmetic and contra-harmonic means of \(a\) and \(b\), respectively. Then from <span class="cite">
	[
	<a href="#1" >1</a>
	]
</span>, <span class="cite">
	[
	<a href="#2" >2</a>
	]
</span> we clearly see that the double inequality </p>
<div class="displaymath" id="a0000000006">
  \begin{equation*}  A(a,b){\lt}M(a,b){\lt}C(a,b) \end{equation*}
</div>
<p> holds for all \(a,b{\gt}0\) with \(a\neq b\). </p>
<p>In <span class="cite">
	[
	<a href="#4" >4</a>
	]
</span>, Neuman proved that the double inequality </p>
<div class="equation" id="a0000000007">
<p>
  <div class="equation_content">
    \begin{equation}  \alpha C(a,b)+(1-\alpha ) A(a,b) < M(a,b) < \beta C(a,b)+(1-\beta ) A(a,b) \end{equation}
  </div>
  <span class="equation_label">1.1</span>
</p>
</div>
<p> holds for all \(a,b{\gt}0\) with \(a\neq b\) if and only if \(\alpha \leq \left(1-\log (\sqrt{2}+1)\right)/\log (\sqrt{2}+1)=0.1345\cdots \) and \(\beta \geq 1/6\), and the inequality </p>
<div class="displaymath" id="a0000000008">
  \begin{align}  C^{\lambda }(a,b)A^{1-\lambda }(a,b){\lt}M(a,b) {\lt} C^{\mu }(a,b)A^{1-\mu }(a,b) \end{align}
</div>
<p> holds true for all \(a,b{\gt}0\) with \(a\neq b\) if \(\mu \geq \log \left((\sqrt{2}+2)/3\right)/\log {2}=0.1865\cdots \) and \(\lambda \leq 1/6\). </p>
<p>The main purpose of this paper is to give some refinements and improvements for inequalities (1.1) and (1.2). Our main results are the following Theorems 1.1 and 1.2. </p>
<p><div class="theorem_thmwrapper " id="a0000000009">
  <div class="theorem_thmheading">
    <span class="theorem_thmcaption">
    Theorem
    </span>
    <span class="theorem_thmlabel">1.1</span>
  </div>
  <div class="theorem_thmcontent">
  <p>The double inequality </p>
<div class="displaymath" id="a0000000010">
  \begin{equation*}  C^{\alpha }(a,b)A^{1-\alpha }(a,b){\lt}M(a,b){\lt}C^{\beta }(a,b)A^{1-\beta }(a,b) \end{equation*}
</div>
<p> holds for all \(a,b{\gt}0\) with \(a\neq b\) if and only if \(\alpha \leq 1/6\) and \(\beta \geq -\log (\log (1+\sqrt{2}))/\log {2}=0.1821\cdots \). </p>

  </div>
</div> </p>
<p><div class="theorem_thmwrapper " id="a0000000011">
  <div class="theorem_thmheading">
    <span class="theorem_thmcaption">
    Theorem
    </span>
    <span class="theorem_thmlabel">1.2</span>
  </div>
  <div class="theorem_thmcontent">
  <p>The double inequality </p>
<div class="displaymath" id="a0000000012">
  \begin{align*} & [C(a,b)/6+5 A(a,b)/6]^{\lambda }\left[C^{1/6}(a,b)A^{5/6}(a,b)\right]^{1-\lambda }{\lt}M(a,b)\\ & \qquad {\lt}[C(a,b)/6+5 A(a,b)/6]^{\mu }\left[C^{1/6}(a,b)A^{5/6}(a,b)\right]^{1-\mu } \end{align*}
</div>
<p> holds for all \(a,b{\gt}0\) with \(a\neq b\) if and only if \(\lambda \leq -[6\log (\log (1+\sqrt{2}))+\log {2}]/[6\log (7/6)-\log {2}]=0.27828\cdots \) and \(\mu \geq 8/25\). </p>

  </div>
</div> </p>
<h1 id="a0000000013">2 Lemmas</h1>
<p> In order to prove our main results we need three lemmas, which we present in this section. <div class="lemma_thmwrapper " id="a0000000014">
  <div class="lemma_thmheading">
    <span class="lemma_thmcaption">
    Lemma
    </span>
    <span class="lemma_thmlabel">2.1</span>
  </div>
  <div class="lemma_thmcontent">
  <p>(See <span class="cite">
	[
	<a href="#5" >5</a>
	, 
	Theorem 1.25
	]
</span>). For \(-\infty {\lt}a{\lt}b{\lt}\infty \), let \(f,g:[a,b]\rightarrow {\mathbb {R}}\) be continuous on \([a,b]\), and be differentiable on \((a,b)\), let \(g'(x)\neq 0\) on \((a,b)\). If \(f^{\prime }(x)/g^{\prime }(x)\) is increasing (decreasing) on \((a,b)\), then so are </p>
<div class="displaymath" id="a0000000015">
  \[ \tfrac {f(x)-f(a)}{g(x)-g(a)}\  \  \mbox{and}\  \  \tfrac {f(x)-f(b)}{g(x)-g(b)}. \]
</div>
<p> If \(f^{\prime }(x)/g^{\prime }(x)\) is strictly monotone, then the monotonicity in the conclusion is also strict. </p>

  </div>
</div> <div class="lemma_thmwrapper " id="a0000000016">
  <div class="lemma_thmheading">
    <span class="lemma_thmcaption">
    Lemma
    </span>
    <span class="lemma_thmlabel">2.2</span>
  </div>
  <div class="lemma_thmcontent">
  <p>(See <span class="cite">
	[
	<a href="#6" >6</a>
	, 
	Lemma 1.1
	]
</span>). Suppose that the power series \(f(x)=\sum \limits _{n=0}^{\infty }a_{n}x^{n}\) and \(g(x)=\sum \limits _{n=0}^{\infty }b_{n}x^{n}\) have the radius of convergence \(r{\gt}0\) and \(b_{n}{\gt}0\) for all \(n\in \{ 0,1,2,\cdots \} \). Let \(h(x)={f(x)}/{g(x)}\), then the following statements are true: </p>
<p>(1) If the sequence \(\{ a_{n}/b_{n}\} _{n=0}^{\infty }\) is (strictly) increasing (decreasing), then \(h(x)\) is also (strictly) increasing (decreasing) on \((0,r)\); </p>
<p>(2) If the sequence \(\{ a_{n}/b_{n}\} \) is (strictly) increasing (decreasing) for \(0{\lt}n\leq n_{0}\) and (strictly) decreasing (increasing) for \(n{\gt}n_{0}\), then there exists \(x_{0}\in (0,r)\) such that \(h(x)\) is (strictly) increasing (decreasing) on \((0,x_{0})\) and (strictly) decreasing (increasing) on \((x_{0},r)\). </p>

  </div>
</div> </p>
<p><div class="lemma_thmwrapper " id="a0000000017">
  <div class="lemma_thmheading">
    <span class="lemma_thmcaption">
    Lemma
    </span>
    <span class="lemma_thmlabel">2.3</span>
  </div>
  <div class="lemma_thmcontent">
  <p>The function </p>
<div class="equation" id="a0000000018">
<p>
  <div class="equation_content">
    \begin{equation}  h(t)=\tfrac {90t+52t\cosh (2t)-66\sinh (2t)+2t\cosh (4t)-3\sinh (4t)}{15t-20t\cosh (2t)+5t\cosh (4t)} \end{equation}
  </div>
  <span class="equation_label">2.1</span>
</p>
</div>
<p> is strictly decreasing on \((0,\log (1+\sqrt{2}))\), where \(\sinh (t)=({\rm e}^{t}-{\rm e}^{-t})/2\) and \(\cosh (t)=({\rm e}^{t}+{\rm e}^{-t})/2\) are the hyperbolic sine and cosine functions, respectively. </p>

  </div>
</div> </p>
<p><div class="proof_wrapper" id="a0000000019">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div>Let </p>
<div class="equation" id="a0000000020">
<p>
  <div class="equation_content">
    \begin{equation}  h_{1}(t)=90t+52t\cosh (2t)-66\sinh (2t)+2t\cosh (4t)-3\sinh (4t), \end{equation}
  </div>
  <span class="equation_label">2.2</span>
</p>
</div>
<div class="equation" id="a0000000021">
<p>
  <div class="equation_content">
    \begin{equation}  h_{2}(t)=15t-20t\cosh (2t)+5t\cosh (4t). \end{equation}
  </div>
  <span class="equation_label">2.3</span>
</p>
</div>
<p> Then making use of power series formulas we have </p>
<div class="displaymath" id="a0000000022">
  \begin{align}  h_{1}(t)=& 90t+52t\sum _{n=0}^{\infty }\tfrac {(2t)^{2n}}{(2n)!}-66\sum _{n=0}^{\infty }\tfrac {(2t)^{2n+1}}{(2n+1)!}\\ & +2t\sum _{n=0}^{\infty }\tfrac {(4t)^{2n}}{(2n)!}-3\sum _{n=0}^{\infty }\tfrac {(4t)^{2n+1}}{(2n+1)!}\nonumber \\ =& 52t\sum _{n=2}^{\infty }\tfrac {(2t)^{2n}}{(2n)!}-66\sum _{n=2}^{\infty }\tfrac {(2t)^{2n+1}}{(2n+1)!} +2t\sum _{n=2}^{\infty }\tfrac {(4t)^{2n}}{(2n)!}\nonumber \\ & -3\sum _{n=2}^{\infty }\tfrac {(4t)^{2n+1}}{(2n+1)!}\nonumber \\ =& \sum _{n=0}^{\infty }\tfrac {[16+13n+(2n-1)2^{2n+2}]2^{2n+7}}{(2n+5)!}t^{2n+5}\nonumber \end{align}
</div>
<p> and </p>
<div class="displaymath" id="a0000000023">
  \begin{align}  h_{2}(t)=& 15t-20t\sum _{n=0}^{\infty }\tfrac {(2t)^{2n}}{(2n)!}+5t\sum _{n=0}^{\infty }\tfrac {(4t)^{2n}}{(2n)!}\\ =& -20t\sum _{n=2}^{\infty }\tfrac {(2t)^{2n}}{(2n)!}+5t\sum _{n=2}^{\infty }\tfrac {(4t)^{2n}}{(2n)!}=\sum _{n=0}^{\infty }\tfrac {5 (2^{2n+2}-1)2^{2n+6}}{(2n+4)!}t^{2n+5}.\nonumber \end{align}
</div>
<p>It follows from (2.1)-(2.5) that </p>
<div class="equation" id="a0000000024">
<p>
  <div class="equation_content">
    \begin{equation}  h(t)=\tfrac {\sum \limits _{n=0}^{\infty }a_{n}t^{2n}}{\sum \limits _{n=0}^{\infty }b_{n}t^{2n}}, \end{equation}
  </div>
  <span class="equation_label">2.6</span>
</p>
</div>
<p> where </p>
<div class="equation" id="a0000000025">
<p>
  <div class="equation_content">
    \begin{equation}  a_{n}=\tfrac {[16+13n+(2n-1)2^{2n+2}]2^{2n+7}}{(2n+5)!},\quad b_{n}=\tfrac {5(2^{2n+2}-1) 2^{2n+6}}{(2n+4)!}. \end{equation}
  </div>
  <span class="equation_label">2.7</span>
</p>
</div>
<p>Equation (2.7) leads to </p>
<div class="equation" id="a0000000026">
<p>
  <div class="equation_content">
    \begin{equation}  \tfrac {a_{n+1}}{b_{n+1}}-\tfrac {a_{n}}{b_{n}}=-\tfrac {6c_{n}}{5(2n+5)(2n+7)(2^{2n+2}-1)(2^{2n+4}-1)}, \end{equation}
  </div>
  <span class="equation_label">2.8</span>
</p>
</div>
<p> where </p>
<div class="equation" id="a0000000027">
<p>
  <div class="equation_content">
    \begin{equation}  c_{n}=(30n^2+135n+110-4^{n+3})4^{n+1}+11. \end{equation}
  </div>
  <span class="equation_label">2.9</span>
</p>
</div>
<p>From (2.9) we get </p>
<div class="equation" id="a0000000028">
<p>
  <div class="equation_content">
    \begin{equation}  c_{0}=195,\quad c_{1}=315,\quad c_{2}=-33525 \end{equation}
  </div>
  <span class="equation_label">2.10</span>
</p>
</div>
<p> and </p>
<div class="displaymath" id="a0000000029">
  \begin{align}  c_{n}& {\lt}(30n^2+135n+110-64n^3)4^{n+1}+11\\ & =\left[10n^2(3-n)+15n(9-n^2)+5(22-n^3)-34n^3\right]4^{n+1}+11\nonumber \\ & {\lt}-34n^3\cdot 4^{n+1}+11{\lt}0\nonumber \end{align}
</div>
<p> for \(n\geq 3\). </p>
<p>Equations (2.8) and (2.10) together with inequality (2.11) lead to the conclusion that the sequence \(\{ a_{n}/b_{n}\} \) is strictly decreasing for \(0\leq n \leq 2\) and strictly increasing for \(n\geq 3\). Then from Lemma 2.2(2) and (2.6) we clearly see that there exists \(t_{0}\in (0,\infty )\) such that \(h(t)\) is strictly decreasing on \((0,t_{0})\) and strictly increasing on \((t_{0},\infty )\). </p>
<p>Let \(t^{*}=\log (1+\sqrt{2})\). Then simple computations lead to </p>
<div class="equation" id="a0000000030">
<p>
  <div class="equation_content">
    \begin{equation}  \sinh (2t^{*})=2\sqrt{2},\  \cosh (2t^{*})=3,\  \sinh (4t^{*})=12\sqrt{2}, \cosh (4t^{*})=17. \end{equation}
  </div>
  <span class="equation_label">2.12</span>
</p>
</div>
<p>Differentiating (2.1) yields </p>
<div class="displaymath" id="a0000000031">
  \begin{align}  h’(t)=& \tfrac {90-80\cosh (2t)+104t\sinh (2t)-10\cosh (4t)+8t\sinh (4t)}{h_{2}(t)}\\ & -\tfrac {15-20\cosh (2t)-40t\sin (2t)+5\cosh (4t)+20t\sinh (4t)}{{h_{2}(t)}^2}h_{1}(t).\nonumber \end{align}
</div>
<p>From (2.2) and (2.3) together with (2.12) and (2.13) we get </p>
<div class="equation" id="a0000000032">
<p>
  <div class="equation_content">
    \begin{equation}  h'(t^*)=\tfrac {-102\sqrt{2}{t^{*}}^2+93t^{*}+21\sqrt{2}}{5{t^{*}}^2}=-0.10035\cdots <0. \end{equation}
  </div>
  <span class="equation_label">2.14</span>
</p>
</div>
<p>From the piecewise monotonicity of \(h(t)\) and inequality (2.14) we clearly see that \(t_{0}{\gt}t^{*}=\log (1+\sqrt{2})\), and the proof of Lemma 2.3 is completed. </p>
<h1 id="a0000000033">3 Proofs of Theorems 1.1 and 1.2</h1>
<p><b class="bf">3.1. Proof of Theorem 1.1</b> Since \(M(a,b)\), \(C(a,b)\) and \(A(a,b)\) are symmetric and homogeneous of degree \(1\). Without loss of generality, we assume that \(a{\gt}b\). Let \(x=(a-b)/(a+b)\) and \(t={\rm {arcsinh}}(x)\). Then \(x\in (0,1)\), \(t\in (0,\log (1+\sqrt{2}))\) and </p>
<div class="equation" id="a0000000034">
<p>
  <div class="equation_content">
    \begin{equation}  \tfrac {\log {[M(a,b)]}-\log {[A(a,b)]}}{\log {[C(a,b)]}-\log {[A(a,b)]}}= \tfrac {\log [{x}/{{\rm {arcsinh}}(x)}]}{\log (1+x^2)}=\tfrac {\log [{\sinh (t)}/{t}]}{2\log [\cosh (t)]}. \end{equation}
  </div>
  <span class="equation_label">3.1</span>
</p>
</div>
<p>Let \(f_{1}(t)=\log [{\sinh (t)}/{t}]\), \(f_{2}(t)=\log [\cosh (t)]\) and </p>
<div class="equation" id="a0000000035">
<p>
  <div class="equation_content">
    \begin{equation}  f(t)=\tfrac {\log [{\sinh (t)}/{t}]}{\log [\cosh (t)]}. \end{equation}
  </div>
  <span class="equation_label">3.2</span>
</p>
</div>
<p> Then \(f_{1}(0^+)=f_{2}(0)=0\), \(f(t)=f_{1}(t)/f_{2}(t)\) and </p>
<div class="displaymath" id="a0000000036">
  \begin{align}  \tfrac {{f_{1}}’(t)}{{f_{2}}’(t)}=& \tfrac {t\cosh ^{2}(t)-\sinh (t)\cosh (t)}{t\sinh ^{2}(t)}=\tfrac {t[\cosh (2t)+1]-\sinh (2t)}{t[\cosh (2t)-1]}\\ =& \tfrac {t\left(\sum \limits _{n=0}^{\infty }2^{2n}t^{2n}/(2n)!+1\right)-\sum \limits _{n=0}^{\infty }2^{2n+1}t^{2n+1}/(2n+1)!} {t\sum \limits _{n=1}^{\infty }2^{2n}t^{2n}/(2n)!}\nonumber \\ =& \tfrac {\sum \limits _{n=1}^{\infty }2^{2n}t^{2n+1}/(2n)!-\sum \limits _{n=1}^{\infty }2^{2n+1}t^{2n+1}/(2n+1)!} {t\sum \limits _{n=1}^{\infty }2^{2n}t^{2n}/(2n)!} =\tfrac {\sum \limits _{n=0}^{\infty }A_{n}t^{2n}}{\sum \limits _{n=0}^{\infty }B_{n}t^{2n}},\nonumber \end{align}
</div>
<p> where \(A_{n}=2^{2n+2}(2n+1)/(2n+3)!\) and \(B_{n}={2^{2n+2}}/{(2n+2)!}\). </p>
<p>Note the \(A_{n}/B_{n}=1-2/(2n+3)\) is strictly increasing for all \(n\geq 0\). Then from Lemma 2.2(1) and (3.3) we know that \({f_{1}}'(t)/{f_{2}}'(t)\) is strictly increasing on \((0,\infty )\). Hence, \(f(t)\) is strictly increasing on \((0,\log (1+\sqrt{2}))\) follows from Lemma 2.1 and the monotonicity of \({f_{1}}'(t)/{f_{2}}'(t)\) together with \(f(0^+)=f_{2}(0)=0\). Moreover, </p>
<div class="equation" id="a0000000037">
<p>
  <div class="equation_content">
    \begin{equation}  \lim \limits _{t\rightarrow 0}f(t)=\lim \limits _{t\rightarrow 0}\tfrac {{f_{1}}’(t)}{{f_{2}}’(t)}=\tfrac {A_{0}}{B_{0}}=\tfrac {1}{3}, \end{equation}
  </div>
  <span class="equation_label">3.4</span>
</p>
</div>
<div class="equation" id="a0000000038">
<p>
  <div class="equation_content">
    \begin{equation}  \lim \limits _{t\rightarrow \log (1+\sqrt{2})}f(t)=-\tfrac {2\log (\log (1+\sqrt{2}))}{\log {2}}. \end{equation}
  </div>
  <span class="equation_label">3.5</span>
</p>
</div>
<p>Therefore, Theorem 1.1 follows easily from (3.1), (3.2), (3.4) and (3.5) together with the monotonicity of \(f(t)\). \(\Box \) </p>
<p><b class="bf">3.2. Proof of Theorem 1.2</b> Since \(M(a,b)\), \(C(a,b)\) and \(A(a,b)\) are symmetric and homogeneous of degree \(1\). Without loss of generality, we assume that \(a{\gt}b\). Let \(x=(a-b)/(a+b)\) and \(t={\rm {arcsinh}}(x)\). Then \(x\in (0,1)\), \(t\in (0,\log (1+\sqrt{2}))\) and </p>
<div class="displaymath" id="a0000000039">
  \begin{align} & \tfrac {\log {M(a,b)}-\log {\left[C^{1/6}(a,b)A^{5/6}(a,b)\right]}}{\log {\left[C(a,b)/6+5A(a,b)/6\right]}-\log {\left[C^{1/6}(a,b)A^{5/6}(a,b)\right]}}=\\ & =\tfrac {\log [{x}/{{\rm {arcsinh}}(x)}]-\log (1+x^2)^{1/6}}{\log (1+x^2/6)-\log (1+x^2)^{1/6}}\nonumber \\ & =\tfrac {\log [{\sinh (t)}/{t}]-[\log \cosh (t)]/3}{\log [1+\sinh ^{2}(t)/6]-[\log \cosh (t)]/3}.\nonumber \end{align}
</div>
<p>Let \(g_{1}(t)=\log [{\sinh (t)}/{t}]-[\log \cosh (t)]/3\), \(g_{2}(t)=\log [1+\sinh ^{2}(t)/6]-[\log \cosh (t)]/3\) and </p>
<div class="equation" id="a0000000040">
<p>
  <div class="equation_content">
    \begin{equation}  g(t)=\tfrac {\log [{\sinh (t)}/{t}]-[\log \cosh (t)]/3}{\log [1+\sinh ^{2}(t)/6]-[\log \cosh (t)]/3}. \end{equation}
  </div>
  <span class="equation_label">3.7</span>
</p>
</div>
<p> Then \(g_{1}(0^+)=g_{2}(0)=0\), \(g(t)=g_{1}(t)/g_{2}(t)\) and </p>
<div class="displaymath" id="a0000000041">
  \begin{equation*}  \tfrac {{g_{1}}’(t)}{{g_{2}}’(t)}=\tfrac {[6+\sinh ^{2}(t)][3t\cosh ^{2}(t)-3\cosh (t)\sinh (t)-t\sinh ^{2}(t)]} {t\sinh (t)[6\sinh (t)\cosh ^{2}(t)-\sinh (t)(6+\sinh ^{2}(t))]}. \end{equation*}
</div>
<p>Elementary computations lead to </p>
<div class="displaymath" id="a0000000042">
  \begin{align*} & [6+\sinh ^{2}(t)][3t\cosh ^{2}(t)-3\cosh (t)\sinh (t)-t\sinh ^{2}(t)]=\nonumber \\ & =\tfrac {45}{4}t+\tfrac {13}{2}t\cosh (2t)-\tfrac {33}{4}\sinh (2t)+\tfrac {t}{4}\cosh (4t)-\tfrac {3}{8}\sinh (4t), \end{align*}
</div>
<div class="displaymath" id="a0000000043">
  \begin{align*} & t\sinh (t)[6\sinh (t)\cosh ^{2}(t)-\sinh (t)(6+\sinh ^{2}(t))]=\nonumber \\ & =\tfrac {15}{8}t-\tfrac {5}{2}t\cosh (2t)+\tfrac {5}{8}t\cosh (4t) \end{align*}
</div>
<p> and </p>
<div class="equation" id="a0000000044">
<p>
  <div class="equation_content">
    \begin{equation}  \tfrac {{g_{1}}’(t)}{{g_{2}}’(t)}=h(t), \end{equation}
  </div>
  <span class="equation_label">3.6</span>
</p>
</div>
<p> where \(h(t)\) is defined as in Lemma 2.3. </p>
<p>It follows from Lemmas 2.1 and 2.3 and (3.8) together with \(g_{1}(0^+)=g_{2}(0)=0\) that \(g(t)\) is strictly decreasing on \((0,\log (1+\sqrt{2}))\). Moreover, </p>
<div class="equation" id="a0000000045">
<p>
  <div class="equation_content">
    \begin{equation}  \lim \limits _{t\rightarrow 0}g(t)=\tfrac {8}{25}, \end{equation}
  </div>
  <span class="equation_label">3.7</span>
</p>
</div>
<div class="equation" id="a0000000046">
<p>
  <div class="equation_content">
    \begin{equation}  \lim \limits _{t\rightarrow \log (1+\sqrt{2})}g(t)=-\tfrac {6\log (\log (1+\sqrt{2}))+\log {2}}{6\log (7/6)-\log {2}}. \end{equation}
  </div>
  <span class="equation_label">3.8</span>
</p>
</div>
<p>Therefore, Theorem 1.2 follows easily from (3.6), (3.7), (3.9) and (3.10) together with the monotonicity of \(g(t)\). \(\Box \) </p>
<p><small class="footnotesize">  </small></p>
<div class="bibliography">
<h1>Bibliography</h1>
<dl class="bibliography">
  <dt><a name="1">1</a></dt>
  <dd><p><i class="sc">E. Neuman</i> and <i class="sc">J. Sándor</i>, <i class="it">On the Schwab-Borchardt mean</i>, Math. Pannon., <b class="bf">14</b> (2003) no. 2, pp.&#160; 253–266. </p>
</dd>
  <dt><a name="2">2</a></dt>
  <dd><p><i class="sc">E. Neuman</i> and <i class="sc">J. Sándor</i>, <i class="it">On the Schwab-Borchardt mean II</i>, Math. Pannon., <b class="bf">17</b> (2006) no. 1, pp.&#160; 49–59. </p>
</dd>
  <dt><a name="3">3</a></dt>
  <dd><p><i class="sc">Y. M. Li</i>, <i class="sc">B. Y. Long</i> and <i class="sc">Y. M. Chu</i>, <i class="it">Sharp bounds for the Neuman-Sándor mean in terms of generalized logarithmic mean</i>, J. Math. Inequal., <b class="bf">6</b> (2012) no. 4, pp.&#160; 567–577. </p>
</dd>
  <dt><a name="4">4</a></dt>
  <dd><p><i class="sc">E. Neuman</i> <i class="it">A note on certain bivariate mean</i>, J. Math. Inequal., <b class="bf">6</b> (2012) no. 4, pp.&#160; 637–643. </p>
</dd>
  <dt><a name="5">5</a></dt>
  <dd><p><i class="sc">G. D. Anderson</i>, <i class="sc">M. K. Vamanamurthy</i> and <i class="sc">M. Vuorinen</i>, <i class="it">Conformal Invariants, Inequalities, and Quasiconformal Maps</i>, New York: John Wiley &amp; Sons, 1997. </p>
</dd>
  <dt><a name="6">6</a></dt>
  <dd><p><i class="sc">S. Simić</i> and <i class="sc">M. Vuorinen</i>, <i class="it">Landen inequalities for zero-balanced hypergeometric functions</i>, Abstr. Appl. Anal., Art. ID 932061, 11 pp., 2012. </p>
</dd>
</dl>


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