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<title>An optimal double inequality among the one-parameter, arithmetic and harmonic means\(^\ast \): An optimal double inequality among the one-parameter, arithmetic and harmonic means\(^\ast \)</title>
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<h1>An optimal double inequality among the one-parameter, arithmetic and harmonic means\(^\ast \)</h1>
<p class="authors">
<span class="author">Wang Miao-Kun\(^\S \), Qiu Ye-Fang\(^\S \) Chu Yu-Ming\(^\S \)</span>
</p>
<p class="date">October 14, 2010.</p>
</div>
<p>\(^\S \)Department of Mathematics, Huzhou Teachers College, Xueshi Str. no. 1, 313000, Huzhou, China, e-mail: <span class="tt">wmk000@126.com, qiuyefang861013@126.com, <br />chuyuming2005@yahoo.com.cn</span>. </p>
<p>\(^\ast \)The work of the last author has been supported by the Natural Science Foundation of China (Grant No. 11071069), the Natural Science Foundation of Zhejiang Province (Grant No. Y7080106) and the Innovation Team Foundation of the Department of Education of Zhejiang Province (Grant No. T200924) </p>

<div class="abstract"><p> For \(p\in \mathbb {R}\), the one-parameter mean \(J_{p}(a,b)\), arithmetic mean \(A(a,b)\), and harmonic mean \(H(a,b)\) of two positive real numbers \(a\) and \(b\) are defined by </p>
<div class="displaymath" id="a0000000002">
  \begin{equation*}  J_{p}(a,b)= \begin{cases}  \tfrac {p(a^{p+1}-b^{p+1})}{(p+1)(a^p-b^p)}, &  a\neq b,p\neq 0,-1,\\ \tfrac {a-b}{\log {a}-\log {b}}, &  a\neq b,p=0,\\ \tfrac {ab(\log {a}-\log {b})}{a-b}, &  a\neq b,p=-1,\\ a, &  a=b, \end{cases}\end{equation*}
</div>
<p> \(A(a,b)=\tfrac {a+b}{2}\), and \(H(a,b)=\tfrac {2ab}{a+b}\), respectively. </p>
<p>In this paper, we answer the question: For \(\alpha \in (0,1)\), what are the greatest value \(r_{1}\) and the least value \(r_{2}\) such that the double inequality \(J_{r_{1}}(a,b){\lt}\alpha A(a,b)+ (1-\alpha )H(a,b){\lt}J_{r_{2}}(a,b)\) holds for all \(a,b{\gt}0\) with \(a\neq b\)? </p>
<p><b class="bf">MSC.</b> 33E05, 26E60. </p>
<p><b class="bf">Keywords.</b> One-parameter mean, arithmetic mean, harmonic mean. </p>
</div>
<h1 id="a0000000003">1 Introduction</h1>
<p>  For \(p\in \mathbb {R}\), the one-parameter mean \(J_{p}(a,b)\), arithmetic mean \(A(a,b)\), and harmonic mean \(H(a,b)\) of two positive real numbers \(a\) and \(b\) are defined by </p>
<div class="equation" id="a0000000004">
<p>
  <div class="equation_content">
    \begin{equation}  J_{p}(a,b)= \begin{cases}  \tfrac {p(a^{p+1}-b^{p+1})}{(p+1)(a^p-b^p)},& a\neq b,p\neq 0,-1,\\ \tfrac {a-b}{\log {a}-\log {b}},& a\neq b,p=0,\\ \tfrac {ab(\log {a}-\log {b})}{a-b},& a\neq b,p=-1,\\ a,& a=b, \end{cases} \end{equation}
  </div>
  <span class="equation_label">1</span>
</p>
</div>
<p> \(A(a,b)=\tfrac {a+b}{2}\), and \(H(a,b)=\tfrac {2ab}{a+b}\), respectively. </p>
<p>Recently, the one-parameter mean \(J_{p}(a,b)\) has been the subject of intensive research. In particular, many remarkable inequalities and properties for the one-parameter mean \(J_{p}\) can be found in the literature [1–7]. </p>
<p>It is well-known that the one-parameter mean \(J_{p}(a,b)\) is continuous and strictly increasing with respect to \(p\in \mathbb {R}\) for fixed \(a,b{\gt}0\) with \(a\neq b\) [5]. Many mean values are special case of the one-parameter mean, for example </p>
<div class="displaymath" id="a0000000005">
  \[ \begin{array}{lll}& J_{1}(a,b)=\tfrac {a+b}{2}=A(a,b),& \mbox{the arithmetic mean,}\\ & J_{\tfrac {1}{2}}(a,b)=\tfrac {a+\sqrt{ab}+b}{3}=He(a,b),& \mbox{the Heronian mean,}\\ & J_{-\tfrac {1}{2}}(a,b)=\sqrt{ab}=G(a,b),& \mbox{the geometric mean} \end{array} \]
</div>
<p> and </p>
<div class="displaymath" id="a0000000006">
  \[ J_{-2}(a,b)=\tfrac {2ab}{a+b}=H(a,b),\quad \quad \mbox{the harmonic mean.} \]
</div>
<p>For \(r\in \mathbb {R}\), the power mean \(M_{r}(a,b)\) of order \(r\) of two positive numbers \(a\) and \(b\) is defined by </p>
<div class="equation" id="a0000000007">
<p>
  <div class="equation_content">
    \begin{equation}  M_{r}(a,b)= \begin{cases}  (\tfrac {a^{r}+b^{r}}{2})^{\tfrac {1}{r}},& r\neq 0,\\ \sqrt{ab},& r=0. \end{cases} \end{equation}
  </div>
  <span class="equation_label">6</span>
</p>
</div>
<p>The main properties of the power mean are given in [8]. In particular, \(M_{r}(a,b)\) is continuous and strictly increasing with respect to \(r\in \mathbb {R}\) for fixed \(a,b{\gt}0\) with \(a\neq b\). </p>
<p>In [9], Alzer and Janous established the following sharp double inequality (see also [9, p. 350]) </p>
<div class="displaymath" id="a0000000008">
  \begin{eqnarray*}  M_{\tfrac {\log 2}{\log 3}}(a,b){\lt}\tfrac {2}{3}J_{1}(a,b)+\tfrac {1}{3}J_{-\tfrac {1}{2}}(a,b){\lt}M_{\tfrac {2}{3}}(a,b) \end{eqnarray*}
</div>
<p> for all \(a,b{\gt}0\) with \(a\neq b\). </p>
<p>In [10], Mao proved </p>
<div class="displaymath" id="a0000000009">
  \begin{eqnarray*}  M_{\tfrac {1}{3}}(a,b)\leq \tfrac {1}{3}J_{1}(a,b)+\tfrac {2}{3}J_{-\tfrac {1}{2}}(a,b) \leq M_{\tfrac {1}{2}}(a,b) \end{eqnarray*}
</div>
<p> for all \(a,b{\gt}0\), and \(M_{\tfrac {1}{3}}(a,b)\) is the best possible lower power mean bound for the sum \(\tfrac {1}{3}J_{1}(a,b)+\tfrac {2}{3}J_{-\tfrac {1}{2}}(a,b)\). </p>
<p>The purpose of this paper is to answer the question: For \(\alpha \in (0,1)\), what are the greatest value \(r_{1}\) and the least value \(r_{2}\) such that the double inequality </p>
<div class="displaymath" id="a0000000010">
  \[ J_{r_{1}}(a,b){\lt}\alpha A(a,b)+(1-\alpha )H(a,b){\lt}J_{r_{2}}(a,b) \]
</div>
<p> holds for all \(a,b{\gt}0\) with \(a\neq b\)? </p>
<h1 id="a0000000011">2 Lemmas</h1>
<p> In order to establish our main result we need two lemmas, which we present in this section. </p>
<p><div class="lemma_thmwrapper " id="a0000000012">
  <div class="lemma_thmheading">
    <span class="lemma_thmcaption">
    Lemma
    </span>
    <span class="lemma_thmlabel">1</span>
  </div>
  <div class="lemma_thmcontent">
  <p>If \(t{\gt}1\), then </p>
<div class="displaymath" id="a0000000013">
  \begin{eqnarray} -\log {t}+\tfrac {(t-1)(t^2+10t+1)}{6t(t+1)}{\gt}0.\end{eqnarray}
</div>

  </div>
</div> </p>
<p><div class="proof_wrapper" id="a0000000014">
  <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="displaymath" id="a0000000015">
  \begin{eqnarray} h(t)=-\log {t}+\tfrac {(t-1)(t^2+10t+1)}{6t(t+1)}.\end{eqnarray}
</div>
<p> Then simple computations lead to </p>
<div class="displaymath" id="a0000000016">
  \begin{eqnarray} h(1)=0,\end{eqnarray}
</div>
<div class="displaymath" id="a0000000017">
  \begin{eqnarray} h’(t)=\tfrac {(t-1)^4}{6t^2(t+1)^2}{\gt}0\end{eqnarray}
</div>
<p> for \(t{\gt}1\). </p>
<p>Therefore, Lemma 1 follows from (4)–(6). </p>
<p><div class="lemma_thmwrapper " id="a0000000018">
  <div class="lemma_thmheading">
    <span class="lemma_thmcaption">
    Lemma
    </span>
    <span class="lemma_thmlabel">2</span>
  </div>
  <div class="lemma_thmcontent">
  <p>If \(t{\gt}1\), then </p>
<div class="displaymath" id="a0000000019">
  \begin{eqnarray} \log {t}-\tfrac {3(t^2-1)}{t^2+4t+1}{\gt}0.\end{eqnarray}
</div>

  </div>
</div> </p>
<p><div class="proof_wrapper" id="a0000000020">
  <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="displaymath" id="a0000000021">
  \begin{eqnarray} g(t)=\log {t}-\tfrac {3(t^2-1)}{t^2+4t+1}.\end{eqnarray}
</div>
<p> Then simple computations lead to </p>
<div class="displaymath" id="a0000000022">
  \begin{eqnarray} g(1)=0,\end{eqnarray}
</div>
<div class="displaymath" id="a0000000023">
  \begin{eqnarray} g’(t)=\tfrac {(t-1)^4}{t(t^2+4t+1)^2}{\gt}0.\end{eqnarray}
</div>
<p> for \(t{\gt}1\). </p>
<p>Therefore, Lemma 2 follows from (8)–(10).</p>
<h1 id="a0000000024">3 Main Result</h1>
<p> <div class="theorem_thmwrapper " id="a0000000025">
  <div class="theorem_thmheading">
    <span class="theorem_thmcaption">
    Theorem
    </span>
    <span class="theorem_thmlabel">3</span>
  </div>
  <div class="theorem_thmcontent">
  <p>Inequality </p>
<div class="displaymath" id="a0000000026">
  \begin{eqnarray*}  J_{3\alpha -2}(a,b){\lt}\alpha A(a,b)+(1-\alpha )H(a,b){\lt}J_{\tfrac {\alpha }{2-\alpha }}(a,b) \end{eqnarray*}
</div>
<p> holds for all \(a,b{\gt}0\) with \(a\neq b\), and \(J_{3\alpha -2}(a,b)\) and \(J_{\tfrac {\alpha }{2-\alpha }}(a,b)\) are the best possible lower and upper one-parameter mean bounds for the sum \(\alpha A(a,b)+(1-\alpha )H(a,b)\), respectively. </p>

  </div>
</div> </p>
<p><div class="proof_wrapper" id="a0000000027">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div>We first prove that </p>
<div class="displaymath" id="a0000000028">
  \begin{eqnarray}  \alpha A(a,b)+(1-\alpha )H(a,b){\gt}J_{3\alpha -2}(a,b) \end{eqnarray}
</div>
<p> for \(\alpha \in (0,1)\) and all \(a,b{\gt}0\) with \(a\neq b\). </p>
<p>Without loss of generality, we assume that \(a{\gt}b\) and take \(t={\tfrac {a}{b}}{\gt}1\). We divide the proof into three cases. </p>
<p><i class="it">Case 1.</i> If \(\alpha =\tfrac {1}{3}\), then from (1) we have </p>
<div class="displaymath" id="a0000000029">
  \begin{align} & \alpha A(a,b)+(1-\alpha )H(a,b)-J_{3\alpha -2}(a,b)=\nonumber \\ & \quad =b[\tfrac {1+t}{6}+\tfrac {4t}{3(1+t)}-\tfrac {t\log {t}}{t-1}]\\ & \quad =\tfrac {b t}{t-1}[-\log {t}+\tfrac {(t-1)(t^2+10t+1)}{6t(1+t)}].\nonumber \end{align}
</div>
<p>Therefore, inequality (11) follows from (12) and Lemma 1. </p>
<p><i class="it">Case 2.</i> If \(\alpha =\tfrac {2}{3}\), then (1) leads to </p>
<div class="displaymath" id="a0000000030">
  \begin{align} & \alpha A(a,b)+(1-\alpha )H(a,b)-J_{3\alpha -2}(a,b)=\nonumber \\ & \quad =b[\tfrac {1+t}{3}+\tfrac {2t}{3(1+t)}-\tfrac {t-1}{\log {t}}]\\ & \quad =\tfrac {b(t^2+4t+1)}{3(1+t)\log {t}}[\log {t}-\tfrac {3(t^2-1)}{t^2+4t+1}].\nonumber \end{align}
</div>
<p>Therefore, inequality (11) follows from (13) and Lemma 2. </p>
<p><i class="it">Case 3.</i> If \(\alpha \in (0,\tfrac {1}{3})\cup (\tfrac {1}{3},\tfrac {2}{3})\cup (\tfrac {2}{3},1)\), then (1) implies that </p>
<div class="displaymath" id="a0000000031">
  \begin{align} & \alpha A(a,b)+(1-\alpha )H(a,b)-J_{3\alpha -2}(a,b)=\nonumber \\ & \quad =b[\alpha \tfrac {1+t}{2}+(1-\alpha )\tfrac {2t}{1+t}-\tfrac {(3\alpha -2)(t^{3\alpha -1}-1)}{(3\alpha -1)(t^{3\alpha -2}-1)}]. \end{align}
</div>
<p> Let </p>
<div class="displaymath" id="a0000000032">
  \begin{eqnarray}  f(t)=\alpha \tfrac {1+t}{2}+(1-\alpha )\tfrac {2t}{1+t}-\tfrac {(3\alpha -2)(t^{3\alpha -1}-1)}{(3\alpha -1)(t^{3\alpha -2}-1)}, \end{eqnarray}
</div>
<p> then \(f(t)\) can be rewritten as </p>
<div class="displaymath" id="a0000000033">
  \begin{eqnarray}  f(t)& =& \tfrac {f_{1}(t)}{2(3\alpha -1)(t+1)(t^{3\alpha -2}-1)}, \end{eqnarray}
</div>
<p> where </p>
<div class="displaymath" id="a0000000034">
  \begin{eqnarray}  f_{1}(t)& =& (1-\alpha )(4-3\alpha )t^{3\alpha }+2\alpha (4-3\alpha )t^{3\alpha -1}+\alpha (3\alpha -1)t^{3\alpha -2}\nonumber \\ & & -\alpha (3\alpha -1)t^2-2\alpha (4-3\alpha )t-(1-\alpha )(4-3\alpha ). \end{eqnarray}
</div>
<p> Note that </p>
<div class="displaymath" id="a0000000035">
  \begin{eqnarray}  f_{1}(1)=0, \end{eqnarray}
</div>
<div class="displaymath" id="a0000000036">
  \begin{eqnarray}  {f_{1}}’(t)& =& 3\alpha (1-\alpha )(4-3\alpha )t^{3\alpha -1}+2\alpha (4-3\alpha )(3\alpha -1)t^{3\alpha -2}\nonumber \\ & & +\alpha (3\alpha -1)(3\alpha -2)t^{3\alpha -3}-2\alpha (3\alpha -1)t\\ & & -2\alpha (4-3\alpha ),\nonumber \end{eqnarray}
</div>
<div class="displaymath" id="a0000000037">
  \begin{eqnarray}  {f_{1}}’(1)=0, \end{eqnarray}
</div>
<div class="displaymath" id="a0000000038">
  \begin{eqnarray}  {f_{1}}”(t)& =& 3\alpha (1-\alpha )(3\alpha -1)(4-3\alpha )t^{3\alpha -2}+2\alpha (4-3\alpha )(3\alpha -1)\nonumber \\ & & \times (3\alpha -2)t^{3\alpha -3}+3\alpha (3\alpha -1)(3\alpha -2)(\alpha -1)t^{3\alpha -4}\\ & & -2\alpha (3\alpha -1),\nonumber \end{eqnarray}
</div>
<div class="displaymath" id="a0000000039">
  \begin{eqnarray}  {f_{1}}”(1)& =& 0 \end{eqnarray}
</div>
<p> and </p>
<div class="displaymath" id="a0000000040">
  \begin{eqnarray}  {f_{1}}”’(t)& =& 3\alpha (1-\alpha )(4-3\alpha )(3\alpha -1)(3\alpha -2)t^{3\alpha -5}(t-1)^2. \end{eqnarray}
</div>
<p>We divide the proof into two subcases. </p>
<p><i class="it">Subcase 1.</i> If \(\alpha \in (0,\tfrac {1}{3})\cup (\tfrac {2}{3},1)\), then from (23) we clearly see that \({f_{1}}'''(t){\gt}0\) for \(t\in (1,\infty )\). Then (18) and (20) together with (22) imply that \(f_{1}(t){\gt}0\) for \(t\in (1,\infty )\). Note that </p>
<div class="displaymath" id="a0000000041">
  \begin{eqnarray}  (3\alpha -1)(t^{3\alpha -2}-1){\gt}0 \end{eqnarray}
</div>
<p> for \(t{\gt}1\). </p>
<p>Therefore, inequality (11) follows from (14)–(16) and (24) together with the fact that \(f_{1}(t){\gt}0\) for \(t\in (1,\infty )\). </p>
<p><i class="it">Subcase 2.</i> If \(\alpha \in (\tfrac {1}{3},\tfrac {2}{3})\), then from (23) we clearly see that \({f_{1}}'''(t){\lt}0\) for \(t\in (1,\infty )\). Then (18) and (20) together with (22) imply that \(f_{1}(t){\lt}0\) for \(t\in (1,\infty )\). Note that </p>
<div class="displaymath" id="a0000000042">
  \begin{eqnarray}  (3\alpha -1)(t^{3\alpha -2}-1){\lt}0 \end{eqnarray}
</div>
<p> for \(t{\gt}1\). </p>
<p>Therefore, inequality (11) follows from (14)–(16) and (25) together with the fact that \(f_{1}(t){\lt} 0\) for \(t\in (1,\infty )\). </p>
<p>Next, we prove that </p>
<div class="displaymath" id="a0000000043">
  \begin{eqnarray}  \alpha A(a,b)+(1-\alpha )H(a,b){\lt}J_{\tfrac {\alpha }{2-\alpha }}(a,b) \end{eqnarray}
</div>
<p> for \(\alpha \in (0,1)\) and all \(a,b{\gt}0\) with \(a\neq b\). </p>
<p>Without loss of generality, we assume that \(a{\gt}b\). Let \(t={\tfrac {a}{b}}{\gt}1\), then from (1) we have </p>
<div class="displaymath" id="a0000000044">
  \begin{align} & \alpha A(a,b)+(1-\alpha )H(a,b)-J_{\tfrac {\alpha }{2-\alpha }}(a,b)=\\ & \quad =b[\alpha \tfrac {1+t}{2}+(1-\alpha )\tfrac {2t}{1+t}-\tfrac {\alpha (t^{\tfrac {2}{2-\alpha }}-1)}{2(t^{\tfrac {\alpha }{2-\alpha }}-1)}].\nonumber \end{align}
</div>
<p>Let </p>
<div class="displaymath" id="a0000000045">
  \begin{eqnarray}  F(t)=\alpha \tfrac {1+t}{2}+(1-\alpha )\tfrac {2t}{1+t}-\tfrac {\alpha (t^{\tfrac {2}{2-\alpha }}-1)}{2(t^{\tfrac {\alpha }{2-\alpha }}-1)}, \end{eqnarray}
</div>
<p> then \(F(t)\) can be rewritten as </p>
<div class="displaymath" id="a0000000046">
  \begin{eqnarray}  F(t)=\tfrac {tF_{1}(t)}{2(t+1)(t^{\tfrac {\alpha }{2-\alpha }}-1)}, \end{eqnarray}
</div>
<p> where </p>
<div class="displaymath" id="a0000000047">
  \begin{eqnarray}  F_{1}(t)=(4-3\alpha )t^{\tfrac {\alpha }{2-\alpha }}+\alpha t^{\tfrac {2(\alpha -1)}{2-\alpha }}-\alpha t-4+3\alpha . \end{eqnarray}
</div>
<p> Note that </p>
<div class="displaymath" id="a0000000048">
  \begin{eqnarray}  F_{1}(1)=0, \end{eqnarray}
</div>
<div class="displaymath" id="a0000000049">
  \begin{eqnarray}  {F_{1}}’(t)=\tfrac {\alpha }{2-\alpha }[(4-3\alpha )t^{\tfrac {2(\alpha -1)}{2-\alpha }}+2(\alpha -1)t^{\tfrac {3\alpha -4}{2-\alpha }}-2+\alpha ], \end{eqnarray}
</div>
<div class="displaymath" id="a0000000050">
  \begin{eqnarray}  {F_{1}}’(1)=0 \end{eqnarray}
</div>
<p> and </p>
<div class="displaymath" id="a0000000051">
  \begin{eqnarray}  {F_{1}}”(t)=\tfrac {2\alpha (4-3\alpha )(1-\alpha )}{(2-\alpha )^2}t^{\tfrac {4\alpha -6}{2-\alpha }}(1-t){\lt}0 \end{eqnarray}
</div>
<p> for \(t\in (1,\infty )\) and \(\alpha \in (0,1)\). </p>
<p>Therefore, inequality (26) follows from (27)–(29), (31), (33) and (34). </p>
<p>At last, we prove that \(J_{3\alpha -2}(a,b)\) and \(J_{\tfrac {\alpha }{2-\alpha }}(a,b)\) are the best possible lower and upper one-parameter mean bounds for the sum \(\alpha A(a,b)+(1-\alpha )H(a,b)\), respectively. </p>
<p>For any \(0{\lt}\varepsilon {\lt}\tfrac {\alpha }{2-\alpha }\) and \(x{\gt}0\), from (1) one has </p>
<div class="displaymath" id="a0000000052">
  \begin{eqnarray}  \lim _{x\rightarrow \infty }\tfrac {\alpha A(x,1)+(1-\alpha ) H(x,1)}{J_{\tfrac {\alpha }{2-\alpha }-\varepsilon }(x,1)} =\tfrac {2\alpha -\alpha (2-\alpha )\varepsilon }{2\alpha -2(2-\alpha )\varepsilon }{\gt}1. \end{eqnarray}
</div>
<p>For any \(\varepsilon {\gt}0\), \(\varepsilon \neq 1-3\alpha \), \(\varepsilon \neq 2-3\alpha \) and \(x{\gt}0\), let \(x\rightarrow 0\), making use of (1) and the Taylor expansion one has </p>
<div class="displaymath" id="a0000000053">
  \begin{align} & {J_{3\alpha -2+\varepsilon }(1+x,1)-\alpha A(1+x,1)-(1-\alpha )H(1+x,1)}=\nonumber \\ & \quad =1+\tfrac {x}{2}+\tfrac {3\alpha -3+\varepsilon }{12}x^2+o(x^2)-\alpha (1+\tfrac {x}{2})-(1-\alpha )\nonumber \\ & \qquad \times (1+\tfrac {x}{2}-\tfrac {1}{4}x^2+o(x^2))\\ & \quad =\tfrac {\varepsilon }{12}x^2+o(x^2).\nonumber \end{align}
</div>
<p>Inequality (35) implies that for any \(0{\lt}\varepsilon {\lt}\tfrac {\alpha }{2-\alpha }\) there exists \(X=X(\alpha ,\varepsilon ){\gt}1\) such that \({\alpha A(x,1)+(1-\alpha )H(x,1)}{\gt}J_{\tfrac {\alpha }{2-\alpha }-\varepsilon }(x,1)\) for \(x\in (X,\infty )\), and inequality (36) implies that for any \(\varepsilon {\gt}0\), \(\varepsilon \neq 1-3\alpha \) and \(\varepsilon \neq 2-3\alpha \) there exists \(\delta =\delta (\alpha ,\varepsilon ){\gt}0\) such that \(J_{3\alpha -2+\varepsilon }(1+x,1){\gt}\alpha A(1+x,1)+(1-\alpha )H(1+x,1)\) for \(x\in (0,\delta )\).</p>
<p><div class="acknowledgement_thmwrapper " id="a0000000054">
  <div class="acknowledgement_thmheading">
    <span class="acknowledgement_thmcaption">
    Acknowledgement
    </span>
  </div>
  <div class="acknowledgement_thmcontent">
  <p>The authors wish to thank the anonymous referees for their very careful reading of the manuscript and fruitful comments and suggestions. </p>

  </div>
</div> </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">H. Alzer</i>, <i class="it">On Stolarsky’s mean value family</i>, Internat. J. Math. Ed. Sci. Tech., <b class="bf">20</b>(1), pp.&#160;186–189, 1987. </p>
</dd>
  <dt><a name="2">2</a></dt>
  <dd><p><i class="sc">H. Alzer</i>, <i class="it">Über eine einparametrige Familie Von Mittelwerten</i>, Bayer. Akad. Wiss. Math.-Natur. Kl. Sitzungsber., <b class="bf">1987</b>, pp.&#160;1–9, 1988. </p>
</dd>
  <dt><a name="3">3</a></dt>
  <dd><p><i class="sc">H. Alzer</i>, <i class="it">Über eine einparametrige Familie Von Mittelwerten II</i>, Bayer. Akad. Wiss. Math.-Natur. Kl. Sitzungsber., <b class="bf">1988</b>, pp.&#160;23–29, 1989. </p>
</dd>
  <dt><a name="4">4</a></dt>
  <dd><p><i class="sc">F. Qi</i>, <i class="it">The extended mean values: definition, properties, monotonicities, comparison, convexities, generalizations, and applications</i>, Cubo Math. Educ., <b class="bf">5</b>(3), pp.&#160;63–90, 2003. </p>
</dd>
  <dt><a name="5">5</a></dt>
  <dd><p><i class="sc">W.-S. Cheung</i> and <i class="sc">F. Qi</i>, <i class="it">Logarithmic convexity of the one-parameter mean values</i>, Taiwanese J. Math., <b class="bf">11</b>(1), pp.&#160;231–237, 2007. </p>
</dd>
  <dt><a name="6">6</a></dt>
  <dd><p><i class="sc">F. Qi</i>, <i class="sc">P. Cerone</i>, <i class="sc">S.S. Dragomir</i> and <i class="sc">H.M. Srivastava</i>, <i class="it">Alternative proofs for monotonic and logarithmically convex properties of one-parameter mean values</i>, Appl. Math. Comput., <b class="bf">208</b>(1), pp.&#160;129–133, 2009. </p>
</dd>
  <dt><a name="7">7</a></dt>
  <dd><p><i class="sc">N.-G. Zheng</i>, <i class="sc">Z.-H. Zhang</i> and <i class="sc">X.-M. Zhang</i>, <i class="it">Schur-convexity of two types of one-parameter mean values in \(n\) variables</i>, J. Inequal. Appl., Art. ID 78175, 10 pages, 2007. </p>
</dd>
  <dt><a name="8">8</a></dt>
  <dd><p><i class="sc">P.S. Bullen</i>, <i class="sc">D.S. Mitrinović</i> and <i class="sc">P.M. Vasić</i>, <i class="it">Means and Their Inequalities</i>, D. Reidel Pubishing Co., Dordrecht, 1988. </p>
</dd>
  <dt><a name="9">9</a></dt>
  <dd><p><i class="sc">H. Alzer</i> and <i class="sc">W. Janous</i>, <i class="it">Solution of problem 8\(^*\)</i>, Crux Math., <b class="bf">13</b>, pp.&#160;173–178, 1987. </p>
</dd>
  <dt><a name="10">10</a></dt>
  <dd><p><i class="sc">Q.-J. Mao</i>, <i class="it">Power mean, logarithmic mean and Heronian dual mean of two positive number</i>, J. Suzhou Coll. Edu., <b class="bf">16</b>(1–2), pp.&#160;82–85, 1999. </p>
</dd>
</dl>


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