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<title>Exact inequalities involving power mean, arithmetic mean, and identric mean\(^{\ast }\): Exact inequalities involving power mean, arithmetic mean, and identric mean\(^{\ast }\)</title>
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<h1>Exact inequalities involving power mean, arithmetic mean, and identric mean\(^{\ast }\)</h1>
<p class="authors">
<span class="author">Yu-Ming Chu\(^{\S }\), Ming-Yu Shi\(^{\bullet }\) Yue-Ping Jiang\(^{\bullet }\)</span>
</p>
<p class="date">October 10, 2009</p>
</div>
<p>\(^\S \) Department of Mathematics, Huzhou Teachers College, Huzhou 313000, Zhejiang, China, e-mail: <span class="tt">chuyuming@hutc.zj.cn</span>. </p>
<p>\(^\bullet \)College of Mathematics and Econometrics, Hunan University, Changsha, 410082, Hunan, China, e-mail: <span class="tt">mingyulj08@163.com, ypjiang731@163.com</span>. </p>
<p>\(^\ast \)The work of the first 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> For \(p\in \mathbb {R}\), the power mean \(M_{p}(a,b)\) of order \(p\), identric mean \(I(a,b)\) and arithmetic mean \(A(a,b)\) of two positive real numbers \(a\) and \(b\) are defined by </p>
<div class="displaymath" id="a0000000002">
  \begin{equation*}  M_{p}(a,b)= \begin{cases}  \displaystyle \left(\tfrac {a^{p}+b^{p}}{2}\right)^{1/p}, &  p\neq 0,\\ \sqrt{ab}, &  p=0, \end{cases} \quad I(a,b)= \begin{cases}  \displaystyle \tfrac {1}{\rm {e}}\left(b^{b}/a^{a}\right)^{1/(b-a)}, &  a\neq b,\\ \displaystyle a, &  a=b, \end{cases}\end{equation*}
</div>
<p> and \(A(a,b)=(a+b)/2\), respectively. </p>
<p>In the article, we answer the questions: What are the least values \(p\), \(q\) and \(r\), such that inequalities \(A^{1/2}(a,b)I^{1/2}(a,b)\leq M_{p}(a,b)\), \(A(a,b)^{1/3}I^{2/3}(a,b)\leq M_{q}(a,b)\) and \(A^{2/3}(a,b)I^{1/3}(a,b)\leq M_{r}(a,b)\) hold for all \(a,b{\gt}0\)? </p>
<p><b class="bf">MSC.</b> 26E60. </p>
<p><b class="bf">Keywords.</b> Power mean, identric mean, arithmetic mean. </p>
</div>
<h1 id="a0000000003">1 Introduction</h1>
<p>For \(p\in \mathbb {R}\), the power mean \(M_{p}(a,b)\) of order \(p\) and the identric mean \(I(a,b)\) of two positive real numbers \(a\) and \(b\) are defined by </p>
<div class="equation" id="a0000000004">
<p>
  <div class="equation_content">
    \begin{equation}  M_{p}(a,b)= \begin{cases} \displaystyle \left(\tfrac {a^{p}+b^{p}}{2}\right)^{\tfrac {1}{p}}, &  p\neq 0,\\ \sqrt{ab}, &  p=0 \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} \displaystyle \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. </p>
<p>It is well-known that \(M_{p}(a,b)\) is continuous and strictly increasing with respect to \(p\in \mathbb {R}\) for fixed \(a,b{\gt}0\) with \(a\neq b\). Recently, both mean values have been the subject of intensive research. In particular, many remarkable inequalities for \(M_{p}(a,b)\) and \(I(a,b)\) can be found in literature [1-24]. </p>
<p>Let \(A(a,b)=(a+b)/2\), \(L(a,b)=(b-a)/(\log b-\log a)(b\neq a)\) and \(L(a,a)=a\), \(G(a,b)=\sqrt{ab}\) and \(H(a,b)=2ab/(a+b)\) be the arithmetic, logarithmic, geometric and harmonic means of two positive numbers \(a\) and \(b\), respectively. Then </p>
<div class="displaymath" id="a0000000006">
  \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>In [25], Alzer and Janous established the following sharp double inequality </p>
<div class="displaymath" id="a0000000007">
  \[ M_{\log 2/\log 3}(a,b)\leq \tfrac {2}{3}A(a,b)+\tfrac {1}{3}G(a,b)\leq M_{2/3}(a,b) \]
</div>
<p> for all \(a,b{\gt}0\). </p>
<p>In [26-28], the authors presented the bounds for \(L\) and \(I\) in terms of \(A\) and \(G\) as follows </p>
<div class="displaymath" id="a0000000008">
  \[  G^{2/3}(a,b)A^{1/3}(a,b)\leq L(a,b)\leq \tfrac {2}{3}G(a,b)+\tfrac {1}{3}A(a,b)  \]
</div>
<p> and </p>
<div class="displaymath" id="a0000000009">
  \[  \tfrac {1}{3}G(a,b)+\tfrac {2}{3}A(a,b)\leq I(a,b)  \]
</div>
<p> for all \(a,b{\gt}0\). </p>
<p>The following companion of (1.3) provides inequalities for the geometric and arithmetic means of \(L\) and \(I\), the proof can be found in [29]. </p>
<div class="displaymath" id="a0000000010">
  \begin{align*}  G^{1/2}(a,b)A^{1/2}(a,b)& \leq L^{1/2}(a,b)I^{1/2}(a,b)\leq \tfrac {1}{2}L(a,b)+\tfrac {1}{2}I(a,b)\\ & \leq \tfrac {1}{2} G(a,b)+ \tfrac {1}{2} A(a,b) \end{align*}
</div>
<p> for all \(a,b{\gt}0\). </p>
<p>The following sharp bounds for \(L\), \(I\), \((LI)^{1/2}\), and \((L+I)/2\) in terms of power means \(M_p(a,b)\) are proved in [29-35]. </p>
<div class="displaymath" id="a0000000011">
  \begin{eqnarray*}  L(a,b)\leq M_{1/3}(a,b), \quad M_{2/3}(a,b)\leq I(a,b)\leq M_{\log 2}(a,b), \end{eqnarray*}
</div>
<div class="displaymath" id="a0000000012">
  \[  M_{0}(a,b)\leq \sqrt{ L(a,b)I(a,b)}\leq M_{1/2}(a,b),  \]
</div>
<p> and </p>
<div class="displaymath" id="a0000000013">
  \[  \tfrac {1}{2}\left(L(a,b)+I(a,b)\right){\lt}M_{1/2}(a,b)  \]
</div>
<p> for all \(a,b{\gt}0\). </p>
<p>Alzer and Qiu [36] proved </p>
<div class="displaymath" id="a0000000014">
  \[  M_c(a,b)\leq \tfrac {1}{2}L(a,b)+\tfrac {1}{2}I(a,b)  \]
</div>
<p> for all \(a\), \(b{\gt}0\) with the best possible parameter \(c=\log 2/(1+\log 2)\). </p>
<p>The main purpose of this paper is to answer the questions: What are the least values \(p\), \(q\) and \(r\), such that inequalities \(A^{1/2}(a,b)I^{1/2}(a,b)\leq M_{p}(a,b)\), \(A(a,b)^{1/3}I^{2/3}(a,b)\leq M_{q}(a,b)\) and \(A^{2/3}(a,b)I^{1/3}(a,b)\leq M_{r}(a,b)\) hold for all \(a\), \(b{\gt}0\)? </p>
<h1 id="a0000000015">2 Lemmas</h1>
<p>In order to establish our main results, we need a lemma, which we present in this section. </p>
<p><div class="lemma_thmwrapper " id="a0000000016">
  <div class="lemma_thmheading">
    <span class="lemma_thmcaption">
    Lemma
    </span>
    <span class="lemma_thmlabel">2.1</span>
  </div>
  <div class="lemma_thmcontent">
  <p>Let </p>
<div class="displaymath" id="a0000000017">
  \begin{align*}  g(t)& =(1-r)(t^{p+1}+t^{p}+t+1)\log t+(2r-1)t^{p+1}-\\ & \quad -2rt^{p}+t^{p-1}-t^{2}+2rt+1-2r. \end{align*}
</div>
<p> If \((r,p)=\left\{ (\tfrac {1}{3},\tfrac {7}{9}),(\tfrac {2}{3},\tfrac {8}{9}),(\tfrac {1}{2},\tfrac {5}{6})\right\} \), 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="a0000000018">
  <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 \(g_1(t)=t^{1-p}g'(t)\), \(g_2(t)=t^{p}g_1'(t)\), \(g_3(t)=t^{1-p}g_2'(t)\), \(g_4(t)=t^{3}g_3'(t)\), \(g_5(t)=t^{p-2}g_4'(t)\), \(g_6(t)=t^{3}g_5'(t)\), \(g_7(t)=t^{1-p}g_6'(t)\), and \(g_8(t)=t^{p}g_7'(t)\). Then simple computations lead to </p>
<div class="displaymath" id="a0000000019">
  \begin{eqnarray} & & g(1)=0,\\ & & \lim _{t\rightarrow +\infty } g(t)=-\infty ,\\ & & g_1(t)=(1-r)[t^{1-p}+(1+p)t+p]\log t-2t^{2-p}+(1+r)t^{1-p}\nonumber \\ & & \quad \qquad +(1-r)t^{-p}+(2pr-p+r)t-(1-p)t^{-1}-2pr-r+1,\nonumber \\ & & g_1(1)=0,\\ & & \lim _{t\rightarrow +\infty }g_1(t)=-\infty ,\\ & & g_2(t)=(1-r)[(1+p)t^{p}+1-p]\log t+(pr+1)t^{p}+p(1-r)t^{p-1}\nonumber \\ & & \quad \qquad +(1-p)t^{p-2}-2(2-p)t-p(1-r)t^{-1}-pr-p+2,\nonumber \\ & & g_2(1)=0,\\ & & \lim _{t\rightarrow +\infty }g_2(t)=-\infty .\\ & & g_3(t)=p(1+p)(1-r)\log t-2(2-p)t^{1-p}+(1+p)(1-r)t^{-p}\nonumber \\ & & \quad \qquad +p(1-r)t^{-1-p}-p(1-p)(1-r)t^{-1}-(1-p)(2-p)t^{-2}\nonumber \\ & & \quad \qquad +p^{2}r-pr+2p-r+1,\nonumber \\ & & g_3(1)=6p-4-2r=0,\\ & & \lim _{t\rightarrow +\infty }g_3(t)=-\infty ,\\ & & g_4(t)=p(1-r)[(1+p)t^{2}+(1-p)t-(1-p)t^{2-p}-(1+p)t^{1-p}]\nonumber \\ & & \quad \qquad -2(1-p)(2-p)(t^{3-p}-1),\nonumber \\ & & g_4(1)=0,\\ & & \lim _{t\rightarrow +\infty }g_4(t)=-\infty ,\\ & & g_5(t)=p(1-r)[2(1+p)t^{p-1}+(1-p)t^{p-2}-(1-p)(2-p)t^{-1}\nonumber \\ & & \quad \qquad -(1+p)(1-p)t^{-2}]-2(1-p)(2-p)(3-p),\nonumber \\ & & g_5(1)=4(1-r)p^{2}-2(1-p)(2-p)(3-p), \end{eqnarray}
</div>
<div class="displaymath" id="a0000000020">
  \begin{eqnarray} & & \lim _{t\rightarrow +\infty }g_5(t)=-2(1-p)(2-p)(3-p){\lt}0,\\ & & g_6(t)=p(1-r)[-2(1+p)(1-p)t^{p+1}-(1-p)(2-p)t^{p}\nonumber \\ & & \quad \qquad +(1-p)(2-p)t+2(1+p)(1-p)],\nonumber \\ & & g_6(1)=0,\\ & & g_7(t)=p(1-p)(1-r)[(2-p)t^{1-p}-2(1+p)^{2}t-p(2-p)],\nonumber \\ & & g_7(1)=-p^{2}(1-p)(7+p)(1-r){\lt}0,\\ & & g_8(t)=p(1-p)(1-r)[-2(1+p)^{2}t^{p}+(1-p)(2-p)], \end{eqnarray}
</div>
<p> and </p>
<div class="displaymath" id="a0000000021">
  \begin{eqnarray} & & \hspace{-38mm}g_8(1)=-p^{2}(1-p)(7+p)(1-r){\lt}0. \end{eqnarray}
</div>
<p>From (2.11) we know that \(g_5(1)=\tfrac {296}{729}\) if \((r,p)=(\tfrac {1}{3},\tfrac {7}{9})\), \(g_5(1)=\tfrac {388}{729}\) if \((r,p)=(\tfrac {2}{3},\tfrac {8}{9})\), and \(g_5(1)=\tfrac {119}{729}\) if \((r,p)\in (\tfrac {1}{2},\tfrac {5}{6})\). Therefore </p>
<div class="displaymath" id="a0000000022">
  \begin{eqnarray}  g_5(1){\gt}0 \end{eqnarray}
</div>
<p> for \((r,p)=\left\{ (\tfrac {1}{3},\tfrac {7}{9}),(\tfrac {2}{3},\tfrac {8}{9}), (\tfrac {1}{2},\tfrac {5}{6})\right\} \). </p>
<p>From (2.15) we clearly see that \(g_8(t)\) is strictly decreasing in \([1,+\infty )\), then (2.16) implies that \(g_8(t){\lt}0\) for \(t\in [1,+\infty )\). Hence that \(g_7(t)\) is strictly decreasing in \([1,+\infty )\). </p>
<p>From (2.14) and the monotonicity of \(g_7(t)\) we know that \(g_7(t){\lt}0\) for \(t\in [1,+\infty )\). Hence \(g_6(t)\) is strictly decreasing in \([1,+\infty )\). </p>
<p>From (2.13) and the monotonicity of \(g_6(t)\) we know that \(g_6(t){\lt}0\) for \(t\in [1,+\infty )\). Hence that \(g_5(t)\) is strictly decreasing in \([1,+\infty )\). </p>
<p>Inequalities (2.12) and (2.17) together the monotonicity of \(g_5(t)\) imply that there exists \(t_0\in (1,+\infty )\), such that \(g_5(t){\gt}0\) for \(t\in (1,t_0)\) and \(g_5(t){\lt}0\) for \(t\in (t_0,+\infty )\). Hence \(g_4(t)\) is strictly increasing in \([1,t_0]\) and strictly decreasing in \([t_0,+\infty )\). </p>
<p>From equations (2.9) and (2.10) together with the monotonicity of \(g_4(t)\) we clearly see that there exists \(t_1\in (1,+\infty )\), such that \(g_4(t){\gt}0\) for \(t\in (1,t_1)\) and \(g_4(t){\lt}0\) for \(t\in (t_1,+\infty )\). Hence \(g_3(t)\) is strictly increasing in \([1,t_1]\) and strictly decreasing in \([t_1,+\infty )\). </p>
<p>Equations (2.7) and (2.8) together with the monotonicity of \(g_3(t)\) imply that there exists \(t_2\in (1,+\infty )\), such that \(g_3(t){\gt}0\) for \(t\in (1,t_2)\) and \(g_3(t){\lt}0\) for \(t\in (t_2,+\infty )\). Hence \(g_2(t)\) is strictly increasing in \([1,t_2]\) and strictly decreasing in \([t_2,+\infty )\). </p>
<p>It follows from equations (2.5) and (2.6) together with the monotonicity of \(g_2(t)\) that there exists \(t_3\in (1,+\infty )\), such that \(g_2(t){\gt}0\) for \(t\in (1,t_3)\) and \(g_2(t){\lt}0\) for \(t\in (t_3,+\infty )\). Hence \(g_1(t)\) is strictly increasing in \([1,t_3]\) and strictly decreasing in \([t_3,+\infty )\). </p>
<p>Equations (2.3) and (2.4) together with the monotonicity of \(g_1(t)\) lead to the conclusion that there exists \(\lambda _5\in (1,+\infty )\), such that \(g_1(t){\gt}0\) for \(t\in (1,t_4)\) and \(g_1(t){\lt}0\) for \(t\in (t_4,+\infty )\). Hence \(g(t)\) is strictly increasing in \([1,t_4]\) and strictly decreasing in \([t_4,+\infty )\). </p>
<p>Therefore, Lemma 2.1 follows from equations (2.1) and (2.2) together with the monotonicity of \(g(t)\). </p>
<h1 id="a0000000023">3 Main Results</h1>
<p><div class="theorem_thmwrapper " id="a0000000024">
  <div class="theorem_thmheading">
    <span class="theorem_thmcaption">
    Theorem
    </span>
    <span class="theorem_thmlabel">3.1</span>
  </div>
  <div class="theorem_thmcontent">
  <p>For all \(a,b{\gt}0\) one has </p>
<div class="displaymath" id="a0000000025">
  \[ A^{1/3}(a,b)I^{2/3}(a,b)\leq M_{7/9}(a,b), \]
</div>
<div class="displaymath" id="a0000000026">
  \[ A^{2/3}(a,b)I^{1/3}(a,b)\leq M_{8/9}(a,b), \]
</div>
<p> and </p>
<div class="displaymath" id="a0000000027">
  \[ A^{1/2}(a,b)I^{1/2}(a,b)\leq M_{5/6}(a,b), \]
</div>
<p> each inequality holds equality if and only if \(a=b\), and the parameters \(7/9\), \(8/9\) and \(5/6\) in each inequality cannot be improved. </p>

  </div>
</div> </p>
<p><div class="proof_wrapper" id="a0000000028">
  <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 we clearly see that </p>
<div class="displaymath" id="a0000000029">
  \begin{align*}  A^{1/2}(a,b)I^{1/2}(a,b)& =M_{5/6}(a,b)=A^{1/3}(a,b)I^{2/3}(a,b)=M_{7/9}(a,b)\\ & = A^{2/3}(a,b)I^{1/3}(a,b)=M_{8/9}(a,b)=a. \end{align*}
</div>
<p>If \(a\neq b\), without loss of generality, we assume that \(a{\gt}b\). Let \((\alpha ,p)=\left\{ (1/3,7/9),(2/3,8/9),(1/2,5/6)\right\} \) and \(t=a/b{\gt}1\), then (1.1) and (1.2) lead to </p>
<div class="equation" id="a0000000030">
<p>
  <div class="equation_content">
    \begin{equation}  M_p(a,b) \! -\! A^{\alpha }(a,b)I^{1-\alpha }(a,b) =b\left[\left( \tfrac {t^{p}+1}{2}\right)^{1/p}\! -\! \left(\tfrac {t+1}{2}\right)^{\alpha }\left(\tfrac {1}{\rm {e}}\cdot t^{\tfrac {t}{t-1}}\right)^{1-\alpha }\right]. \end{equation}
  </div>
  <span class="equation_label">3.1</span>
</p>
</div>
<p>Let </p>
<div class="displaymath" id="a0000000031">
  \[  f(t)=\displaystyle \tfrac {1}{p}\log \tfrac {1+t^{p}}{2}-\alpha \log \tfrac {t+1}{2} -(1-\alpha )\tfrac {t}{t-1}\log t+(1-\alpha ), \]
</div>
<p> then </p>
<div class="displaymath" id="a0000000032">
  \begin{eqnarray} & & \lim _{t\rightarrow 1^{+}}f(t)=0, \\ & & \lim _{t\rightarrow \infty }f(t)=(1-\alpha )+(\alpha -\tfrac {1}{p})\log 2 \end{eqnarray}
</div>
<p> and </p>
<div class="displaymath" id="a0000000033">
  \begin{eqnarray} & & f’(t)=\tfrac {g(t)}{(t+1)(t-1)^{2}(t^{p}+1)}, \end{eqnarray}
</div>
<p> where </p>
<div class="displaymath" id="a0000000034">
  \begin{align*}  g(t)& =(1-\alpha )(t^{p+1}+t^{p}+t+1)\log t\\ & \quad +(2\alpha -1)t^{p+1}-2\alpha t^{p}+t^{p-1}-t^{2}+2\alpha t+1-2\alpha . \end{align*}
</div>
<p>From (3.3) we know that \(\displaystyle \lim _{t\rightarrow \infty }f(t)=2(7-10\log 2)/21{\gt}0\) if \((\alpha ,p)=(\tfrac {1}{3},\tfrac {7}{9})\), \(\displaystyle \lim _{t\rightarrow \infty }f(t)=(8-11\log 2)/24{\gt}0\) if \((\alpha ,p)=(\tfrac {2}{3},\tfrac {8}{9})\), and \(\displaystyle \lim _{t\rightarrow \infty }f(t)=(5-7\log 2)/10{\gt}0\) if \((\alpha ,p)=(\tfrac {1}{2},\tfrac {5}{6})\). Hence we get </p>
<div class="displaymath" id="a0000000035">
  \begin{eqnarray}  \lim _{t\rightarrow \infty }f(t){\gt}0. \end{eqnarray}
</div>
<p>Equation (3.4) and Lemma 2.1 lead to the conclusion that there exists \(\lambda \in (1,+\infty )\), such that \(f(t)\) is strictly increasing in \([1,\lambda ]\) and strictly decreasing in \([\lambda ,+\infty )\). Therefore, \(A^{1/3}(a,b)I^{2/3}(a,b){\lt} M_{7/9}(a,b)\), \(A^{2/3}(a,b)I^{1/3}(a,b){\lt}M_{8/9}(a,b)\), and \(A^{1/2}(a,b)I^{1/2}(a,b){\lt}M_{5/6}(a,b)\) for all \(a,b{\gt}0\) with \(a\neq b\) follow from (3.1), (3.2), (3.5) and the monotonicity of \(f(t)\). </p>
<p>Next, we prove that the parameters \(7/9\), \(8/9\) and \(5/6\) in each inequality cannot be improved. </p>
<p>For any \(0{\lt}\varepsilon {\lt}7/9\), \(0{\lt}x{\lt}1\) and \(x\rightarrow 0\), making use of Taylor expansion one has </p>
<div class="displaymath" id="a0000000036">
  \begin{align} & \log \left[A^{1/2}(1,1+x)I^{1/2}(1,1+x)\right]-\log M_{5/6-\varepsilon }(1,1+x)\\ & =\tfrac {1}{2}\log (1+\tfrac {x}{2})+\tfrac {1+x}{2x}\log (1+x)-\tfrac {1}{2}-\tfrac {6}{5-6\varepsilon }\log \tfrac {1+(1+x)^{5/6-\varepsilon }}{2}\nonumber \\ & =\tfrac {\varepsilon }{8}x^{2}+o(x^{2}),\nonumber \end{align}
</div>
<div class="displaymath" id="a0000000037">
  \begin{align} & \log \left[A^{1/3}(1,1+x)I^{2/3}(1,1+x)\right]-\log M_{7/9-\varepsilon }(1,1+x)\\ & =\tfrac {1}{3}\log (1+\tfrac {x}{2})+\tfrac {2(1+x)}{3x}\log (1+x)-\tfrac {2}{3}-\tfrac {9}{7-9\varepsilon }\log \tfrac {1+(1+x)^{7/9-\varepsilon }}{2}\nonumber \\ & =\tfrac {\varepsilon }{8}x^{2}+o(x^{2})\nonumber \end{align}
</div>
<p> and </p>
<div class="displaymath" id="a0000000038">
  \begin{align} & \log \left[A^{2/3}(1,1+x)I^{1/3}(1,1+x)\right]-\log M_{8/9-\varepsilon }(1,1+x)\\ & =\tfrac {2}{3}\log (1+\tfrac {x}{2})+\tfrac {1+x}{3x}\log (1+x)-\tfrac {1}{3}-\tfrac {9}{8-9\varepsilon }\log \tfrac {1+(1+x)^{8/9-\varepsilon }}{2}\nonumber \\ & =\tfrac {\varepsilon }{8}x^{2}+o(x^{2}).\nonumber \end{align}
</div>
<p>Equations (3.6)–(3.8) imply that for any \(0{\lt}\varepsilon {\lt}7/9\), there exists \(0{\lt}\delta =\delta (\varepsilon ){\lt}1\), such that </p>
<div class="displaymath" id="a0000000039">
  \[ A^{1/2}(1,1+x)I^{1/2}(1,1+x){\gt} M_{5/6-\varepsilon }(1,1+x), \]
</div>
<div class="displaymath" id="a0000000040">
  \[ A^{1/3}(1,1+x)I^{2/3}(1,1+x){\gt} M_{7/9-\varepsilon }(1,1+x) \]
</div>
<p> and </p>
<div class="displaymath" id="a0000000041">
  \[ A^{2/3}(1,1+x)I^{1/3}(1,1+x){\gt} M_{8/9-\varepsilon }(1,1+x) \]
</div>
<p> for \(x\in (0,\delta )\). </p>
<p><small class="footnotesize">  </small></p>
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