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<title>Optimal inequalities related to the logarithmic, identric, arithmetic and harmonic means\(^\bullet \): Optimal inequalities related to the logarithmic, identric, arithmetic and harmonic means\(^\bullet \)</title>
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<h1>Optimal inequalities related to the logarithmic, identric, arithmetic and harmonic means\(^\bullet \)</h1>
<p class="authors">
<span class="author">Wei-feng Xia\(^\ast \) Yu-ming Chu\(^\S \)</span>
</p>
<p class="date">June 2, 2010.</p>
</div>
<p>\(^\ast \) School of Teacher Education, Huzhou Teachers College, Huzhou 313000, Zhejiang, China, e-mail: <span class="tt">xwf212@hutc.zj.cn</span>. </p>
<p>\(^\S \)Department of Mathematics, Huzhou Teachers College, Huzhou 313000, Zhejiang, China, e-mail: <span class="tt">chuyuming@hutc.zj.cn</span>. </p>
<p>\(^\bullet \) The work of the first author has been supported by the Natural Science Foundation of Zhejiang Province (Grant no. Y7080185) and the work of the second author has been supported by the Natural Science Foundation (Grant no. 11071069) and the Innovation Team Foundation of the Department of Education of Zhejiang Province (Grant no. T200924). </p>

<div class="abstract"><p> The logarithmic mean \(L(a,b)\), identric mean \(I(a,b)\), arithmetic mean \(A(a,b)\) and harmonic mean \(H(a,b)\) of two positive real values \(a\) and \(b\) are defined by </p>
<div class="displaymath" id="a0000000002">
  \begin{align*} & L(a,b)= \begin{cases}  \tfrac {b-a}{\log b-\log a},&  a\neq b,\\ a,& a=b, \end{cases}\\ & I(a,b)= \begin{cases}  \tfrac {1}{{\rm e}}\left(\tfrac {b^b}{a^a}\right)^{\tfrac {1}{b-a}},&  a\neq b,\\ a,& a=b, \end{cases}\end{align*}
</div>
<p> \(A(a,b)=\tfrac {a+b}{2}\) and \(H(a,b)=\tfrac {2ab}{a+b}\), respectively. </p>
<p>In this article, we answer the questions: What are the best possible parameters \(\alpha _{1},\alpha _{2},\beta _{1}\) and \(\beta _{2}\), such that \(\alpha _{1}A(a,b)+(1-\alpha _{1})H(a,b)\leq L(a,b)\leq \beta _{1}A(a,b)+(1-\beta _{1})H(a,b)\) and \(\alpha _{2}A(a,b)+(1-\alpha _{2})H(a,b)\leq I(a,b)\leq \beta _2A(a,b)+(1-\beta _{2})H(a,b)\) hold for all \(a,b{\gt}0\)? </p>
<p><b class="bf">MSC.</b> 26E60. </p>
<p><b class="bf">Keywords.</b> Logarithmic mean, identric mean, arithmetic mean, harmonic mean. </p>
</div>
<h1 id="a0000000003">1 Introduction</h1>
<p>The logarithmic mean \(L(a,b)\) and identric mean \(I(a,b)\) of two positive real values \(a\) and \(b\) are defined by </p>
<div class="equation" id="a0000000004">
<p>
  <div class="equation_content">
    \begin{equation} L(a,b)= \begin{cases}  \tfrac {b-a}{\log b-\log a},&  a\neq b,\\ a,& a=b \end{cases} \end{equation}
  </div>
  <span class="equation_label">1.1</span>
</p>
</div>
<p> and </p>
<div class="equation" id="a0000000005">
<p>
  <div class="equation_content">
    \begin{equation} I(a,b)= \begin{cases}  \tfrac {1}{{\rm e}}\left(\tfrac {b^b}{a^a}\right)^{\tfrac {1}{b-a}},&  a\neq b,\\ a,& a=b, \end{cases} \end{equation}
  </div>
  <span class="equation_label">1.4</span>
</p>
</div>
<p> respectively. In the recent past, both mean values have been the subject of intensive research. In particular, many remarkable inequalities for logarithmic mean or identric mean can be found in the literature [1–25]. It might be surprising that the logarithmic mean has applications in physics, economics, and even in meteorology [26–28]. In [26] the authors study a variant of Jensen’s functional equation involving \(L(a,b)\), which appears in a heat conduction problem. </p>
<p>The power mean \(M_{p}(a,b)\) of order \(p\) is defined by </p>
<div class="displaymath" id="a0000000006">
  \begin{equation*} M_{p}(a,b)= \begin{cases}  \left(\tfrac {a^{p}+b^{p}}{2}\right)^{\tfrac {1}{p}},&  p\neq 0,\\ \sqrt{ab},& p=0. \end{cases}\end{equation*}
</div>
<p>If we denote by \(A(a,b)=\tfrac {1}{2}(a+b)\), \(G(a,b)=\sqrt{ab}\) and \(H(a,b)=\tfrac {2ab}{a+b}\) the arithmetic mean, geometric mean and harmonic mean of two positive numbers \(a\) and \(b\), respectively, then it is well-known that </p>
<div class="displaymath" id="a0000000007">
  \begin{align}  \min \{ a,b\} & \leq H(a,b)=M_{-1}(a,b)\leq G(a,b)=M_{0}(a,b)\\ & \leq L(a,b)\leq I(a,b) \leq A(a,b)=M_{1}(a,b) \leq \max \{ a,b\} ,\nonumber \end{align}
</div>
<p> and all inequalities are strict for \(a\neq b\). </p>
<p>In [9, 12, 30] the authors present bounds for \(L(a,b)\) in terms of \(G(a,b)\) and \(A(a,b)\). </p>
<div class="displaymath" id="a0000000008">
  \[ G^{\tfrac {2}{3}}(a,b)A^{\tfrac {1}{3}}(a,b){\lt}L(a,b){\lt}\tfrac {2}{3}G(a,b)+\tfrac {1}{3}A(a,b) \]
</div>
<p> for all \(a,b{\gt}0\) with \(a\neq b\). </p>
<p>The following companion of (1.3) provides inequalities for the geometric and arithmetic means of \(L(a,b)\) and \(I(a,b)\). A proof can be found in [7]. </p>
<div class="displaymath" id="a0000000009">
  \begin{equation*}  G^{\tfrac {1}{2}}(a,b)A^{\tfrac {1}{2}}(a,b){\lt}L^{\tfrac {1}{2}}(a,b)I^{\tfrac {1}{2}}(a,b){\lt}\tfrac {1}{2}L(a,b)+\tfrac {1}{2}I(a,b){\lt}\tfrac {1}{2}G(a,b)+\tfrac {1}{2}A(a,b) \end{equation*}
</div>
<p> for all \(a,b{\gt}0\) with \(a\neq b\). </p>
<p>The following bounds for \(L(a,b)\), \(I(a,b)\), \((L(a,b)I(a,b))^{\tfrac {1}{2}}\), and \(\tfrac {L(a,b)+I(a,b)}{2}\) in terms of power means are proved in [4, 6–8, 10, 24, 30]. </p>
<div class="displaymath" id="a0000000010">
  \begin{eqnarray*} & & M_{0}(a,b){\lt}L(a,b){\lt}M_{\tfrac {1}{3}}(a,b),\\ & & M_{\tfrac {2}{3}}(a,b){\lt}I(a,b){\lt}M_{\log 2}(a,b),\\ & & M_{0}(a,b){\lt}L^{\tfrac {1}{2}}(a,b)I^{\tfrac {1}{2}}(a,b){\lt}M_{\tfrac {1}{2}}(a,b) \end{eqnarray*}
</div>
<p> and </p>
<div class="displaymath" id="a0000000011">
  \[ M_{\tfrac {\log 2}{1+\log 2}}(a,b){\lt}\tfrac {1}{2}L(a,b)+\tfrac {1}{2}I(a,b){\lt}M_{\tfrac {1}{2}}(a,b) \]
</div>
<p> for all \(a,b{\gt}0\) with \(a\neq b\). </p>
<p>The main purpose of this paper is to answer the questions: What are the best possible parameters \(\alpha _{1},\alpha _{2},\beta _{1}\) and \(\beta _{2}\), such that \(\alpha _{1}A(a,b)+(1-\alpha _{1})H(a,b)\leq L(a,b)\leq \beta _{1}A(a,b)+(1-\beta _{1})H(a,b)\) and \(\alpha _{2}A(a,b)+(1-\alpha _{2})H(a,b)\leq I(a,b)\leq \beta _2A(a,b)+(1-\beta _{2})H(a,b)\) hold for all \(a,b{\gt}0\)? </p>
<h1 id="a0000000012">2 Lemmas</h1>
<p><div class="lemma_thmwrapper " id="a0000000013">
  <div class="lemma_thmheading">
    <span class="lemma_thmcaption">
    Lemma
    </span>
    <span class="lemma_thmlabel">2.1</span>
  </div>
  <div class="lemma_thmcontent">
  <p>The function \(g(t)=(t^{2}+4t+1)\log t-3t^{2}+3{\gt}0\) for \(t\in (1,+\infty )\). </p>

  </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>By simple computation we have </p>
<div class="equation" id="a0000000015">
<p>
  <div class="equation_content">
    \begin{equation}  g(1)=0, \end{equation}
  </div>
  <span class="equation_label">2.1</span>
</p>
</div>
<div class="displaymath" id="a0000000016">
  \begin{equation*}  g’(t)=(2t+4)\log t-5t+\tfrac {1}{t}+4, \end{equation*}
</div>
<div class="equation" id="a0000000017">
<p>
  <div class="equation_content">
    \begin{equation}  g'(1)=0, \end{equation}
  </div>
  <span class="equation_label">2.2</span>
</p>
</div>
<div class="displaymath" id="a0000000018">
  \begin{equation*}  g”(t)=2\log t+\tfrac {4}{t}-\tfrac {1}{t^{2}}-3, \end{equation*}
</div>
<div class="equation" id="a0000000019">
<p>
  <div class="equation_content">
    \begin{equation}  g''(1)=0 \end{equation}
  </div>
  <span class="equation_label">2.3</span>
</p>
</div>
<p> and </p>
<div class="equation" id="a0000000020">
<p>
  <div class="equation_content">
    \begin{equation}  g'''(t)=\tfrac {2}{t}-\tfrac {4}{t^{2}}+\tfrac {2}{t^{3}}=\tfrac {2}{t^{3}}(t-1)^2. \end{equation}
  </div>
  <span class="equation_label">2.4</span>
</p>
</div>
<p>Equation (2.4) leads to \(g'''(t){\gt}0\) for \(t\in (1,+\infty )\), then \(g''(t)\) is strictly increasing in \((1,+\infty )\). Hence \(g(t){\gt}0 \) for \(t\in (1,+\infty )\) follows from the monotonicity of \(g''(t)\) and (2.1)–(2.3). </p>
<p><div class="lemma_thmwrapper " id="a0000000021">
  <div class="lemma_thmheading">
    <span class="lemma_thmcaption">
    Lemma
    </span>
    <span class="lemma_thmlabel">2.2</span>
  </div>
  <div class="lemma_thmcontent">
  <p>The function \(g(t)=(5t^{3}+19t^{2}+19t+5)\log t-14t^{3}-6t^{2}+6t+14{\gt}0\) for \(t\in (1,+\infty )\). </p>

  </div>
</div> </p>
<p><div class="proof_wrapper" id="a0000000022">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div>By elementary computation we get </p>
<div class="equation" id="a0000000023">
<p>
  <div class="equation_content">
    \begin{equation}  g(1)=0, \end{equation}
  </div>
  <span class="equation_label">2.5</span>
</p>
</div>
<div class="displaymath" id="a0000000024">
  \begin{equation*}  g’(t)=(15t^{2}+38t+19)\log t-37t^{2}+7t+\tfrac {5}{t}+25, \end{equation*}
</div>
<div class="equation" id="a0000000025">
<p>
  <div class="equation_content">
    \begin{equation}  g'(1)=0, \end{equation}
  </div>
  <span class="equation_label">2.6</span>
</p>
</div>
<div class="displaymath" id="a0000000026">
  \begin{equation*}  g”(t)=(30t+38)\log t-59t+\tfrac {19}{t}-\tfrac {5}{t^{2}}+45, \end{equation*}
</div>
<div class="equation" id="a0000000027">
<p>
  <div class="equation_content">
    \begin{equation}  g''(1)=0, \end{equation}
  </div>
  <span class="equation_label">2.7</span>
</p>
</div>
<div class="displaymath" id="a0000000028">
  \begin{equation*}  g”’(t)=30\log t+\tfrac {38}{t}-\tfrac {19}{t^2}+\tfrac {10}{t^3}-29, \end{equation*}
</div>
<div class="equation" id="a0000000029">
<p>
  <div class="equation_content">
    \begin{equation}  g'''(1)=0 \end{equation}
  </div>
  <span class="equation_label">2.8</span>
</p>
</div>
<p> and </p>
<div class="equation" id="a0000000030">
<p>
  <div class="equation_content">
    \begin{equation}  g^{(4)}(t)=\tfrac {1}{t^4}(t-1)(30t^{2}-8t+30). \end{equation}
  </div>
  <span class="equation_label">2.9</span>
</p>
</div>
<p>Equation (2.9) leads to \(g^{(4)}(t){\gt}0\) for \(t\in (1,+\infty )\), then \(g'''(t)\) is strictly increasing in \((1,+\infty )\). Hence \(g(t){\gt}0\) for \(t\in (1,+\infty )\) follows from the monotonicity of \(g'''(t)\) and (2.5)–(2.8). </p>
<p><div class="lemma_thmwrapper " id="a0000000031">
  <div class="lemma_thmheading">
    <span class="lemma_thmcaption">
    Lemma
    </span>
    <span class="lemma_thmlabel">2.3</span>
  </div>
  <div class="lemma_thmcontent">
  <p>If \(g(t)=-[t^{3}+(2{\rm e}-1)t^{2}+(2{\rm e}-1)t+1]\log t+(2{\rm e}-2)t^{3}+(-2{\rm e}+6)t^{2}+(2{\rm e}-6)t-2{\rm e}+2\), then there exists \(\lambda \in (1,+\infty )\) such that \(g(t){\gt}0\) for \(t\in (1,\lambda )\) and \(g(t){\lt}0\) for \(t\in (\lambda ,+\infty )\). </p>

  </div>
</div> </p>
<p><div class="proof_wrapper" id="a0000000032">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div>Elementary computation yields </p>
<div class="equation" id="a0000000033">
<p>
  <div class="equation_content">
    \begin{equation}  g(1)=0,\quad \lim _{t\to +\infty }g(t)=-\infty , \end{equation}
  </div>
  <span class="equation_label">2.10</span>
</p>
</div>
<div class="displaymath" id="a0000000034">
  \begin{equation*}  g’(t)=-[3t^{2}+(4{\rm e}-2)t+2{\rm e}-1]\log t +(6{\rm e}-7)t^{2}+(-6{\rm e}+13)t-\tfrac {1}{t}-5, \end{equation*}
</div>
<div class="equation" id="a0000000035">
<p>
  <div class="equation_content">
    \begin{equation}  g'(1)=0,\quad \lim _{t\to +\infty }g'(t)=-\infty , \end{equation}
  </div>
  <span class="equation_label">2.11</span>
</p>
</div>
<div class="displaymath" id="a0000000036">
  \begin{equation*}  g”(t)=-(6t+4{\rm e}-2)\log t+(12{\rm e}-17)t-\tfrac {2{\rm e}-1}{t}+\tfrac {1}{t^2}-10{\rm e}+15, \end{equation*}
</div>
<div class="equation" id="a0000000037">
<p>
  <div class="equation_content">
    \begin{equation}  g''(1)=0,\quad \lim _{t\to +\infty }g''(t)=-\infty , \end{equation}
  </div>
  <span class="equation_label">2.12</span>
</p>
</div>
<div class="displaymath" id="a0000000038">
  \begin{equation*}  g”’(t)=-6\log t-\tfrac {4{\rm e}-2}{t}+\tfrac {2{\rm e}-1}{t^2}-\tfrac {2}{t^3}+12{\rm e}-23, \end{equation*}
</div>
<div class="equation" id="a0000000039">
<p>
  <div class="equation_content">
    \begin{equation}  g'''(1)=10{\rm e}-24>0,\quad \lim _{t\to +\infty }g'''(t)=-\infty , \end{equation}
  </div>
  <span class="equation_label">2.13</span>
</p>
</div>
<div class="displaymath" id="a0000000040">
  \begin{eqnarray}  g^{(4)}(t)=-\tfrac {1}{t^4}(t-1)[6t^2+(8-4{\rm e})t+6]. \end{eqnarray}
</div>
<p>Equation (2.14) implies that \(g^{(4)}(t){\lt}0\) for \(t\in (1,+\infty )\), then \(g'''(t)\) is strictly decreasing in \((1,+\infty )\). </p>
<p>From (2.13) and the monotonicity of \(g'''(t)\) we clearly see that there exists \(t_{1}\in (1,+\infty )\), such that \(g'''(t){\gt}0\) for \(t\in (1,t_{1})\) and \(g'''(t){\lt}0\) for \(t\in (t_{1},+\infty )\). Hence we know that \(g''(t)\) is strictly increasing in \([1,t_{1})\) and strictly decreasing in \([t_{1},+\infty )\). </p>
<p>The monotonicity of \(g''(t)\) and (2.12) imply that there exists \(t_{2}\in (1,+\infty )\), such that \(g''(t){\gt}0\) for \(t\in (1,t_{2})\) and \(g''(t){\lt}0\) for \(t\in (t_{2},+\infty )\). Hence we know that \(g'(t)\) is strictly increasing in \([1,t_{2})\) and strictly decreasing in \([t_{2},+\infty )\). </p>
<p>From (2.11) and the monotonicity of \(g'(t)\) we clearly see that there exists \(t_{3}\in (1,+\infty )\), such that \(g'(t){\gt}0\) for \(t\in (1,t_{3})\) and \(g'(t){\lt}0\) for \(t\in (t_{3},+\infty )\). Hence we conclude that \(g(t)\) is strictly increasing in \([1,t_{3})\) and strictly decreasing in \([t_{3},+\infty )\). </p>
<p>The monotonicity of \(g(t)\) and (2.10) imply that there exists \(\lambda \in (1,+\infty )\), such that \(g(t){\gt}0\) for \(t\in (1,\lambda )\) and \(g(t){\lt}0\) for \(t\in (\lambda ,+\infty )\). </p>
<h1 id="a0000000041">3 Main Results</h1>
<p><div class="theorem_thmwrapper " id="a0000000042">
  <div class="theorem_thmheading">
    <span class="theorem_thmcaption">
    Theorem
    </span>
    <span class="theorem_thmlabel">3.1</span>
  </div>
  <div class="theorem_thmcontent">
  <p>The double inequality </p>
<div class="displaymath" id="a0000000043">
  \begin{equation*}  \alpha _{1}A(a,b)+(1-\alpha _{1})H(a,b)\leq L(a,b)\leq \beta _{1}A(a,b)+(1-\beta _{1})H(a,b) \end{equation*}
</div>
<p> holds for all \(a,b{\gt}0\) if and only if \(\alpha _{1}\leq 0\) and \(\beta _{1}\geq \tfrac {2}{3}\). </p>

  </div>
</div> </p>
<p><div class="proof_wrapper" id="a0000000044">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div>If \(a=b\), then \(\alpha _{1}A(a,b)+(1-\alpha _{1})H(a,b)= L(a,b)= \beta _{1}A(a,b)+(1-\beta _{1})H(a,b)=a\) for all \(\alpha _{1},\beta _{1}\in \mathbb {R}\). Next, we assume that \(a\neq b\). </p>
<p>Firstly, we prove that \(H(a,b){\lt}L(a,b){\lt}\tfrac {2}{3}A(a,b)+\tfrac {1}{3}H(a,b)\). From (1.3) we know that \(H(a,b){\lt}L(a,b)\) is true, so we only need to prove that \(L(a,b){\lt}\tfrac {2}{3}A(a,b)+\tfrac {1}{3}H(a,b)\). </p>
<p>Without loss of generality, we assume that \(a{\gt}b\). Let \(t=\tfrac {a}{b}{\gt}1\), then simple computation leads to </p>
<div class="displaymath" id="a0000000045">
  \begin{align}  \tfrac {2}{3}A(a,b)+\tfrac {1}{3}H(a,b)-L(a,b)& =b\left[\tfrac {t+1}{3}+\tfrac {2t}{3(t+1)}-\tfrac {t-1}{\log t}\right]\\ & =\tfrac {b[(t^{2}+4t+1)\log t-3t^{2}+3]}{3(t+1)\log t}.\nonumber \end{align}
</div>
<p>Therefore, \(L(a,b){\lt}\tfrac {2}{3}A(a,b)+\tfrac {1}{3}H(a,b)\) follows from Lemma 2.1 and (3.1). </p>
<p>Secondly, we prove that the parameters \(\alpha _{1}\leq 0\) and \(\beta _{1}\geq \tfrac {2}{3}\) cannot be improved. </p>
<p>For any \(0{\lt}\varepsilon {\lt}1\) and \(0{\lt}x{\lt}1\), from (1.1) we have </p>
<div class="displaymath" id="a0000000046">
  \begin{align} & \lim _{x\to 0}[\varepsilon A(1,x)+(1-\varepsilon )H(1,x)-L(1,x)]=\\ & \quad =\lim _{x\to 0}\left[\varepsilon \cdot \tfrac {1+x}{2}+(1-\varepsilon )\cdot \tfrac {2x}{1+x}-\tfrac {x-1}{\log x}\right] =\tfrac {\varepsilon }{2}.\nonumber \end{align}
</div>
<p>Equation (3.2) implies that for any \(0{\lt}\varepsilon {\lt}1\), there exists \(0{\lt}\delta =\delta (\varepsilon ){\lt}1\), such that \(\varepsilon A(1,x)+(1-\varepsilon )H(1,x){\gt}L(1,x)\) for \(x\in (0,\delta )\). Hence the parameter \(\alpha _{1}\leq 0\) cannot be improved. </p>
<p>Next, for any \(0{\lt}\varepsilon {\lt}1\) and \(0{\lt}x{\lt}1\), from (1.1) we get </p>
<div class="displaymath" id="a0000000047">
  \begin{align} & L(1+x,1)-\left[(\tfrac {2}{3}-\varepsilon )A(1+x,1)+(\tfrac {1}{3}+\varepsilon )H(1+x,1)\right]=\\ & =\tfrac {x}{\log (1+x)}-\tfrac {(\tfrac {2}{3}-\varepsilon )x^{2}+4x+4}{2(x+2)}\nonumber \\ & =\tfrac {h(x)}{2(x+2)\log (1+x)},\nonumber \end{align}
</div>
<p> where \(h(x)=2x(x+2)-[(\tfrac {2}{3}-\varepsilon )x^{2}+4x+4]\log (1+x)\). </p>
<p>Let \(x\to 0\) and using Taylor expansion we obtain </p>
<div class="displaymath" id="a0000000048">
  \begin{eqnarray}  h(x)=\varepsilon x^3+o(x^3). \end{eqnarray}
</div>
<p>Equations (3.3) and (3.4) imply that for any \(0{\lt}\varepsilon {\lt}1\), there exists \(0{\lt}\delta =\delta (\varepsilon ){\lt}1\), such that \(L(1+x,1){\gt}(\tfrac {2}{3}-\varepsilon )A(1+x,1)+(\tfrac {1}{3}+\varepsilon )H(1+x,1) \) for \(x\in (0,\delta )\). Hence the parameter \(\beta _{1}\geq \tfrac {2}{3}\) cannot be improved. </p>
<p><div class="theorem_thmwrapper " id="a0000000049">
  <div class="theorem_thmheading">
    <span class="theorem_thmcaption">
    Theorem
    </span>
    <span class="theorem_thmlabel">3.2</span>
  </div>
  <div class="theorem_thmcontent">
  <p>The double inequality </p>
<div class="displaymath" id="a0000000050">
  \[ \alpha _{2}A(a,b)+(1-\alpha _{2})H(a,b)\leq I(a,b)\leq \beta _{2}A(a,b)+(1-\beta _{2})H(a,b) \]
</div>
<p> holds for all \(a,b{\gt}0\) if and only if \(\alpha _{2}\leq \tfrac {2}{{\rm e}}\) and \(\beta _{2}\geq \tfrac {5}{6}\). </p>

  </div>
</div> </p>
<p><div class="proof_wrapper" id="a0000000051">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div>If \(a=b\), then \(\alpha _{2}A(a,b)+(1-\alpha _{2})H(a,b)=I(a,b)=\beta _{2}A(a,b)+(1-\beta _{2})H(a,b)=a\) for all \(\alpha _{2},\beta _{2}\in \mathbb {R}\). Next, we assume that \(a\neq b\). </p>
<p>Firstly, we prove that \(\tfrac {2}{{\rm e}}A(a,b)+(1-\tfrac {2}{{\rm e}})H(a,b){\lt} I(a,b)\) and the parameter \(\alpha _{2}\leq \tfrac {2}{{\rm e}}\) cannot be improved. </p>
<p>Without loss of generality, we assume that \(a{\gt}b\). Let \(t=\tfrac {a}{b}{\gt}1\), then </p>
<div class="displaymath" id="a0000000052">
  \begin{align} & I(a,b)-\left[\tfrac {2}{{\rm e}}A(a,b)+(1-\tfrac {2}{{\rm e}})H(a,b)\right]=\\ & \quad =\tfrac {b}{{\rm e}}\left[t^{\tfrac {t}{t-1}}-(t+1)-({\rm e}-2)\tfrac {2t}{t+1}\right].\nonumber \end{align}
</div>
<p>Let \(f(t)=\log t^{\tfrac {t}{t-1}}-\log \left[(t+1)+({\rm e}-2)\tfrac {2t}{t+1}\right] \), then elementary computation yields </p>
<div class="equation" id="a0000000053">
<p>
  <div class="equation_content">
    \begin{equation}  \lim _{t\to 1}f(t)=0,\quad \lim _{t\to +\infty }f(t)=0 \end{equation}
  </div>
  <span class="equation_label">3.6</span>
</p>
</div>
<p> and </p>
<div class="equation" id="a0000000054">
<p>
  <div class="equation_content">
    \begin{equation}  f'(t)=\tfrac {g(t)}{(t+1)(t-1)^{2}[t^{2}+(2{\rm e}-2)t+1]}, \end{equation}
  </div>
  <span class="equation_label">3.7</span>
</p>
</div>
<p> where \(g(t)=-[t^{3}+(2{\rm e}-1)t^{2}+(2{\rm e}-1)t+1]\log t+(2{\rm e}-2)t^{3}+(-2{\rm e}+6)t^{2}+(2{\rm e}-6)t-2{\rm e}+2\). </p>
<p>From (3.7) and Lemma 2.3 we know that there exists \(\lambda \in (1,+\infty )\), such that \(f(t)\) is strictly increasing in \((1,\lambda )\) and strictly decreasing in \((\lambda ,+\infty )\). Then (3.6) and the monotonicity of \(f(t)\) imply that \(f(t){\gt}0\) for \(t\in (1,+\infty )\), and from (3.5) we know that \(I(a,b){\gt}\tfrac {2}{{\rm e}}A(a,b)+(1-\tfrac {2}{{\rm e}})H(a,b)\) for \(a,b{\gt}0\) with \(a\neq b\). </p>
<p>Next, we prove that the parameter \(\alpha _{2}\leq \tfrac {2}{{\rm e}}\) cannot be improved. </p>
<p>For any \(0{\lt}\varepsilon {\lt}1\) and \(0{\lt}x{\lt}1\), from (1.2) we have </p>
<div class="displaymath" id="a0000000055">
  \begin{align} & \lim _{x\to 0}\left[(\tfrac {2}{{\rm e}}+\varepsilon )A(1,x)+(1-\tfrac {2}{{\rm e}}-\varepsilon )H(1,x)-I(1,x)\right]=\\ & \quad =\lim _{x\to 0}\left[(\tfrac {2}{{\rm e}}+\varepsilon )\cdot \tfrac {1+x}{2}+(1-\tfrac {2}{{\rm e}}-\varepsilon )\cdot \tfrac {2x}{1+x}-\tfrac {1}{{\rm e}}x^{\tfrac {x}{x-1}}\right]\nonumber \\ & \quad =\tfrac {\varepsilon }{2}.\nonumber \end{align}
</div>
<p>Equation (3.8) implies that for any \(0{\lt}\varepsilon {\lt}1\), there exists \(0{\lt}\delta =\delta (\varepsilon ){\lt}1\), such that \((\tfrac {2}{{\rm e}}+\varepsilon )A(1,x)+(1-\tfrac {2}{{\rm e}}-\varepsilon )H(1,x){\gt}I(1,x)\) for \(x\in (0,\delta )\). Hence the parameter \(\alpha _{2}\leq \tfrac {2}{{\rm e}}\) cannot be improved. </p>
<p>Secondly, we prove that \(I(a,b){\lt}\tfrac {5}{6}A(a,b)+\tfrac {1}{6}H(a,b)\) and the parameter \(\beta _{2}\geq \tfrac {5}{6}\) cannot be improved. </p>
<p>Let \(t=\tfrac {a}{b}{\gt}1\), then from (1.2) we have </p>
<div class="displaymath" id="a0000000056">
  \begin{align} & \tfrac {5}{6}A(a,b)+\tfrac {1}{6}H(a,b)-I(a,b)= b\left[\tfrac {5t^{2}+14t+5}{12(t+1)}-\tfrac {1}{{\rm e}}t^{\tfrac {t}{t-1}}\right].\end{align}
</div>
<p>Let \(f(t)=\log \left[ \tfrac {5t^{2}+14t+5}{12(t+1)}\right]-\log \left(\tfrac {1}{{\rm e}}t^{\tfrac {t}{t-1}}\right)\), then </p>
<div class="equation" id="a0000000057">
<p>
  <div class="equation_content">
    \begin{equation}  f(1)=0 \end{equation}
  </div>
  <span class="equation_label">3.10</span>
</p>
</div>
<p> and </p>
<div class="equation" id="a0000000058">
<p>
  <div class="equation_content">
    \begin{equation}  f'(t)=\tfrac {g(t)}{(t-1)(t^{2}-1)(5t^{2}+14t+5)}, \end{equation}
  </div>
  <span class="equation_label">3.11</span>
</p>
</div>
<p> where \(g(t)=(5t^{3}+19t^{2}+19t+5)\log t-14t^{3}-6t^{2}+6t+14\). </p>
<p>From Lemma 2.2 and (3.11) together with (3.10) we clearly see that \(f(t){\gt}0\) for \(t\in (1,+\infty )\). Hence from (3.9) we know that \(\tfrac {5}{6}A(a,b)+\tfrac {1}{6}H(a,b){\gt}I(a,b)\) for \(a,b{\gt}0\) with \(a\neq b\). </p>
<p>Next, we prove that the parameter \(\beta _{2}\geq \tfrac {5}{6}\) cannot be improved. </p>
<p>For any \(0{\lt}\varepsilon {\lt}1\) and \(0{\lt}x{\lt}1\), from (1.2) we get </p>
<div class="displaymath" id="a0000000059">
  \begin{align} & I(1+x,1)-\left[(\tfrac {5}{6}-\varepsilon )A(1+x,1)+(\tfrac {1}{6}+\varepsilon )H(1+x,1)\right]=\\ & \quad =\tfrac {1}{{\rm e}}(1+x)^{\tfrac {1+x}{x}}-\tfrac {(\tfrac {5}{6}-\varepsilon )x^{2}+4x+4}{2(2+x)}\nonumber \\ & \quad =\tfrac {h(x)}{2(2+x)},\nonumber \end{align}
</div>
<p> where \(h(x)=\tfrac {2}{{\rm e}}(2+x)(1+x)^{\tfrac {1+x}{x}}-(\tfrac {5}{6}-\varepsilon )x^{2}-4x-4\). </p>
<p>Let \(x\to 0\) and using Taylor expansion we obtain </p>
<div class="displaymath" id="a0000000060">
  \begin{align}  h(x)& =2(2+x)\left[1+\tfrac {1}{2}x-\tfrac {1}{24}x^{2}+o(x^{2})\right]-(\tfrac {5}{6}-\varepsilon )x^{2}-4x-4\\ & =\varepsilon x^{2}+o(x^{2}).\nonumber \end{align}
</div>
<p>Equations (3.12) and (3.13) imply that for any \(0{\lt}\varepsilon {\lt}1\), there exists \(0{\lt}\delta =\delta (\varepsilon ){\lt}1\), such that \(I(1+x,1){\gt}(\tfrac {5}{6}-\varepsilon )A(1+x,1)+(\tfrac {1}{6}+\varepsilon )H(1+x,1)\) for \(x\in (0,\delta )\). Hence the parameter \(\beta _{2}\geq \tfrac {5}{6}\) cannot be improved. </p>
<p><small class="footnotesize">  </small></p>
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