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<title>New Inequalities of Hermite-Hadamard Type for \(HA\)-Convex Functions: New Inequalities of Hermite-Hadamard Type for \(HA\)-Convex Functions</title>
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<h1>New Inequalities of Hermite-Hadamard Type for \(HA\)-Convex Functions</h1>
<p class="authors">
<span class="author">Silvestru Sever Dragomir\(^\ast \)</span>
</p>
<p class="date">April 11, 2017. Accepted: September 11, 2017. Published online: August 6, 2018.</p>
</div>
<p>\(^\ast \)Mathematics, College of Engineering &amp; Science Victoria University, PO Box 14428 Melbourne City, MC 8001, Australia., DST-NRF Centre of Excellence in the Mathematical and Statistical Sciences, School of Computer Science and Applied Mathematics,University of the Witwatersrand, Johannesburg, Private Bag 3, Wits 2050, South Africa, e-mail: <span class="tt">sever.dragomir@vu.edu.au</span>. </p>

<div class="abstract"><p> Some new inequalities of Hermite-Hadamard type for \(HA\)-convex functions defined on positive intervals are given. </p>
<p><b class="bf">MSC.</b> 26D15; 25D10. </p>
<p><b class="bf">Keywords.</b> Convex functions, Integral inequalities, \(HA\)-Convex functions. </p>
</div>
<h1 id="a0000000002">1 Introduction</h1>
<p>Following <span class="cite">
	[
	<a href="#AVV" >1</a>
	]
</span> (see also <span class="cite">
	[
	<a href="#II" >41</a>
	]
</span>) we say that the function \(f:I\subset R\setminus \left\{  0\right\}  \rightarrow \mathbb {R}\) is <i class="itshape">HA-convex</i> or <i class="itshape">harmonically convex</i> if</p>
<div class="equation" id="e.1.1">
<p>
  <div class="equation_content">
    \begin{equation}  f\big( \tfrac {xy}{tx+\left( 1-t\right) y}\big) \leq \left( 1-t\right) f\left( x\right) +tf\left( y\right) \label{e.1.1} \end{equation}
  </div>
  <span class="equation_label">1</span>
</p>
</div>
<p>for all \(x,y\in I\) and \(t\in \left[ 0,1\right] \). If the inequality in (<a href="#e.1.1">1</a>) is reversed, then \(f\) is said to be <i class="itshape">HA-concave </i>or<i class="itshape"> harmonically concave</i>. </p>
<p>In order to avoid any confusion with the class of <i class="itshape">AH-convex </i>functions, namely the functions satisfying the condition</p>
<div class="equation" id="e.1.2">
<p>
  <div class="equation_content">
    \begin{equation}  f\left( \left( 1-t\right) x+ty\right) \leq \tfrac {f\left( x\right) f\left( y\right) }{\left( 1-t\right) f\left( y\right) +tf\left( x\right) }, \label{e.1.2} \end{equation}
  </div>
  <span class="equation_label">2</span>
</p>
</div>
<p>we call the class of functions satisfying (<a href="#e.1.1">1</a>) as <i class="itshape">HA-convex functions</i>. </p>
<p>If \(I\subset \left( 0,\infty \right) \) and \(f\) is convex and nondecreasing function then \(f\) is <i class="itshape">HA-</i>convex<i class="itshape"> </i>and<i class="itshape"> </i>if \(f\) is <i class="itshape">HA</i>-convex and nonincreasing function then \(f\) is convex. </p>
<p>The following simple but important fact is as follows: </p>
<p><div class="criterion_thmwrapper " id="c.1">
  <div class="criterion_thmheading">
    <span class="criterion_thmcaption">
    Criterion
    </span>
    <span class="criterion_thmlabel">1</span>
  </div>
  <div class="criterion_thmcontent">
  <p>If \(\left[ a,b\right] \subset I\subset \left( 0,\infty \right) \) and if we consider the function \(g:\big[ \frac{1}{b},\frac{1}{a}\big] \rightarrow \mathbb {R}\), defined by \(g\left( t\right) =f\big( \frac{1}{t}\big) ,\) then \(f\) is <i class="itshape">HA-</i>convex on \(\left[ a,b\right] \) if and only if \(g\) is convex in the usual sense on \(\big[ \frac{1}{b},\frac{1}{a}\big] .\) </p>

  </div>
</div> </p>
<p>For a convex function \(h:\left[ c,d\right] \rightarrow \mathbb {R}\), the following inequality is well known in the literature as the Hermite-Hadamard inequality</p>
<div class="equation" id="HH">
<p>
  <div class="equation_content">
    \begin{equation}  h\left( \tfrac {c+d}{2}\right) \leq \tfrac {1}{d-c}\int _{c}^{d}h\left( t\right) dt\leq \tfrac {h\left( c\right) +h\left( d\right) }{2} \label{HH} \end{equation}
  </div>
  <span class="equation_label">3</span>
</p>
</div>
<p>for any convex function \(h:\left[ c,d\right] \rightarrow \mathbb {R}\). </p>
<p>For related results, see <span class="cite">
	[
	<a href="#AVV" >1</a>
	]
</span>–<span class="cite">
	[
	<a href="#SSDCon" >18</a>
	]
</span>, <span class="cite">
	[
	<a href="#III.d.DR4" >21</a>
	]
</span>–<span class="cite">
	[
	<a href="#III.b.DCRW" >26</a>
	]
</span>, <span class="cite">
	[
	<a href="#DF" >27</a>
	]
</span>–<span class="cite">
	[
	<a href="#III.b.DRW3" >37</a>
	]
</span> and <span class="cite">
	[
	<a href="#F" >38</a>
	]
</span>–<span class="cite">
	[
	<a href="#PD0" >49</a>
	]
</span>. </p>
<p>If we write the Hermite-Hadamard inequality for the convex function \(g\left( t\right) =f\big( \frac{1}{t}\big) \) on the closed interval \(\big[ \frac{1}{b},\frac{1}{a}\big] ,\) then we have</p>
<div class="equation" id="e.1.3">
<p>
  <div class="equation_content">
    \begin{equation}  f\left( \tfrac {2ab}{a+b}\right) \leq \tfrac {ab}{b-a}\int _{\frac{1}{b}}^{\frac{1}{a}}f\big( \tfrac {1}{t}\big) dt\leq \tfrac {f\left( b\right) +f\left( a\right) }{2}. \label{e.1.3} \end{equation}
  </div>
  <span class="equation_label">4</span>
</p>
</div>
<p>Using the change of variable \(s=\frac{1}{t},\) we have</p>
<div class="displaymath" id="a0000000003">
  \begin{equation*}  \int _{\frac{1}{b}}^{\frac{1}{a}}f\big( \tfrac {1}{t}\big) dt=\int _{a}^{b}\tfrac {f\left( s\right) }{s^{2}}ds \end{equation*}
</div>
<p>and by (<a href="#e.1.3">4</a>) we get</p>
<div class="equation" id="e.1.4">
<p>
  <div class="equation_content">
    \begin{equation}  f\big( \tfrac {2ab}{a+b}\big) \leq \tfrac {ab}{b-a}\int _{a}^{b}\tfrac {f\left( s\right) }{s^{2}}ds\leq \tfrac {f\left( b\right) +f\left( a\right) }{2}. \label{e.1.4} \end{equation}
  </div>
  <span class="equation_label">5</span>
</p>
</div>
<p>The inequality (<a href="#e.1.4">5</a>) has been obtained in a different manner in <span class="cite">
	[
	<a href="#II" >41</a>
	]
</span> by I. Işcan. </p>
<p>The <i class="itshape">identric mean</i> \(I\left( a,b\right) \) is defined by</p>
<div class="displaymath" id="a0000000004">
  \begin{equation*}  I\left( a,b\right) :=\tfrac {1}{e}\big( \tfrac {b^{b}}{a^{a}}\big) ^{\frac{1}{b-a}} \end{equation*}
</div>
<p>while the <i class="itshape">logarithmic mean</i> is defined by</p>
<div class="displaymath" id="a0000000005">
  \begin{equation*}  L\left( a,b\right) :=\tfrac {b-a}{\ln b-\ln a}. \end{equation*}
</div>
<p>In the recent paper <span class="cite">
	[
	<a href="#SSDHA" >25</a>
	]
</span> we established the following inequalities for <i class="itshape">HA</i>-convex functions: </p>
<p><div class="theorem_thmwrapper " id="t.A">
  <div class="theorem_thmheading">
    <span class="theorem_thmcaption">
    Theorem
    </span>
    <span class="theorem_thmlabel">2</span>
  </div>
  <div class="theorem_thmcontent">
  <p>Let \(f:\left[ a,b\right] \subset \left( 0,\infty \right) \rightarrow \mathbb {R}\) be an HA-convex function on the interval \(\left[ a,b\right] .\) Then</p>
<div class="equation" id="e.1.5">
<p>
  <div class="equation_content">
    \begin{equation}  f\left( L\left( a,b\right) \right) \leq \tfrac {1}{b-a}\int _{a}^{b}f\left( x\right) dx\leq \tfrac {\left( L\left( a,b\right) -a\right) bf\left( b\right) +\left( b-L\left( a,b\right) \right) af\left( a\right) }{\left( b-a\right) L\left( a,b\right) }, \label{e.1.5} \end{equation}
  </div>
  <span class="equation_label">6</span>
</p>
</div>

  </div>
</div> </p>
<p>and </p>
<p><div class="theorem_thmwrapper " id="t.B">
  <div class="theorem_thmheading">
    <span class="theorem_thmcaption">
    Theorem
    </span>
    <span class="theorem_thmlabel">3</span>
  </div>
  <div class="theorem_thmcontent">
  <p>Let \(f:\left[ a,b\right] \subset \left( 0,\infty \right) \rightarrow \mathbb {R}\) be a HA-convex function on the interval \(\left[ a,b\right] .\) Then</p>
<div class="equation" id="e.1.6">
<p>
  <div class="equation_content">
    \begin{equation}  f\left( \tfrac {a+b}{2}\right) \tfrac {a+b}{2}\leq \tfrac {1}{b-a}\int _{a}^{b}xf\left( x\right) dx\leq \tfrac {bf\left( b\right) +af\left( a\right) }{2}. \label{e.1.6} \end{equation}
  </div>
  <span class="equation_label">7</span>
</p>
</div>

  </div>
</div> </p>
<p>Motivated by the above results, we establish in this paper some new inequalities of Hermite-Hadamard type for <i class="itshape">HA-</i>convex functions. Some applications for special means are also given. </p>
<h1 id="a0000000006">2 Further Results</h1>
<p>We start with the following characterization of <i class="itshape">HA</i>-convex functions. </p>
<p><div class="theorem_thmwrapper " id="t.2.1">
  <div class="theorem_thmheading">
    <span class="theorem_thmcaption">
    Theorem
    </span>
    <span class="theorem_thmlabel">4</span>
  </div>
  <div class="theorem_thmcontent">
  <p>Let \(f,h:\left[ a,b\right] \subset \left( 0,\infty \right) \rightarrow \mathbb {R}\) be so that \(h\left( t\right) =tf\left( t\right) \) for \(t\in \left[ a,b\right] .\) Then \(f\) is HA-convex on the interval \(\left[ a,b\right] \) if and only if \(h\) is convex on \(\left[ a,b\right] .\) </p>

  </div>
</div> </p>
<p><div class="proof_wrapper" id="a0000000007">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div> Assume that \(f\) is <i class="itshape">HA</i>-convex on the interval \(\left[ a,b\right] .\) Then the function \(g:\big[ \frac{1}{b},\frac{1}{a}\big] \rightarrow \mathbb {R}\), \(g\left( t\right) =f\big( \frac{1}{t}\big) \) is convex on \(\big[ \frac{1}{b},\frac{1}{a}\big] .\) By replacing \(t\) with \(\frac{1}{t}\) we have \(f\left( t\right) =g\big( \frac{1}{t}\big) .\) </p>
<p>If \(\lambda \in \left[ 0,1\right] \) and \(x,y\in \left[ a,b\right] \) then, by the convexity of \(g\) on \(\big[ \frac{1}{b},\frac{1}{a}\big] ,\) we have</p>
<div class="displaymath" id="a0000000008">
  \begin{align*}  h\left( \left( 1-\lambda \right) x+\lambda y\right) &  =\left[ \left( 1-\lambda \right) x+\lambda y\right] f\left( \left( 1-\lambda \right) x+\lambda y\right) \\ &  =\left[ \left( 1-\lambda \right) x+\lambda y\right] g\left( \tfrac {1}{\left( 1-\lambda \right) x+\lambda y}\right) \\ &  =\left[ \left( 1-\lambda \right) x+\lambda y\right] g\Big( \tfrac {\left( 1-\lambda \right) x\frac{1}{x}+\lambda y\frac{1}{y}}{\left( 1-\lambda \right) x+\lambda y}\Big) \\ &  \leq \left[ \left( 1-\lambda \right) x+\lambda y\right] \tfrac {\left( 1-\lambda \right) xg\left( \frac{1}{x}\right) +\lambda yg\left( \frac{1}{y}\right) }{\left( 1-\lambda \right) x+\lambda y} \\ &  =\left( 1-\lambda \right) xg\left( \tfrac {1}{x}\right) +\lambda yg\left( \tfrac {1}{y}\right) \\ &  =\left( 1-\lambda \right) xf\left( x\right) +\lambda yf\left( y\right) =\left( 1-\lambda \right) h\left( x\right) +\lambda h\left( y\right) , \end{align*}
</div>
<p>which shows that \(h\) is convex on \(\left[ a,b\right] .\) </p>
<p>We have \(f\left( t\right) =\tfrac {h\left( t\right) }{t}\) for \(t\in \left[ a,b\right] .\) If \(\lambda \in \left[ 0,1\right] \) and \(x,y\in \left[ a,b\right] \) then, by the convexity of \(h\) on \(\left[ a,b\right] ,\) we have</p>
<div class="displaymath" id="a0000000009">
  \begin{align*}  f\left( \tfrac {xy}{\lambda x+\left( 1-\lambda \right) y}\right) &  =\tfrac {h\left( \tfrac {xy}{\lambda x+\left( 1-\lambda \right) y}\right) }{\frac{xy}{\lambda x+\left( 1-\lambda \right) y}} \\ &  =\tfrac {\lambda x+\left( 1-\lambda \right) y}{xy}h\left( \tfrac {xy}{\lambda x+\left( 1-\lambda \right) y}\right) \\ &  =\tfrac {\lambda x+\left( 1-\lambda \right) y}{xy}h\left( \tfrac {1}{\left( 1-\lambda \right) \frac{1}{x}+\lambda \frac{1}{y}}\right) \\ &  =\tfrac {\lambda x+\left( 1-\lambda \right) y}{xy}h\left( \tfrac {\left( 1-\lambda \right) \frac{1}{x}x+\lambda \frac{1}{y}y}{\left( 1-\lambda \right) \frac{1}{x}+\lambda \frac{1}{y}}\right) \\ &  \leq \tfrac {\lambda x+\left( 1-\lambda \right) y}{xy}\tfrac {\left( 1-\lambda \right) \frac{1}{x}h\left( x\right) +\lambda \frac{1}{y}h\left( y\right) }{\left( 1-\lambda \right) \frac{1}{x}+\lambda \frac{1}{y}} \\ &  =\left( 1-\lambda \right) \tfrac {1}{x}h\left( x\right) +\lambda \tfrac {1}{y}h\left( y\right) =\left( 1-\lambda \right) f\left( x\right) +\lambda f\left( y\right) , \end{align*}
</div>
<p>which shows that \(f\) is <i class="itshape">HA</i>-convex on the interval \(\left[ a,b\right] .\) <div class="proof_wrapper" id="a0000000010">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div> </p>
<p><div class="remark_thmwrapper " id="r.2.1">
  <div class="remark_thmheading">
    <span class="remark_thmcaption">
    Remark
    </span>
    <span class="remark_thmlabel">5</span>
  </div>
  <div class="remark_thmcontent">
  <p>If \(f\) is HA-convex on the interval \(\left[ a,b\right] ,\) then by Theorem <a href="#t.2.1">4</a> the function \(h\left( t\right) =tf\left( t\right) \) is convex on \(\left[ a,b\right] \) and by Hermite-Hadamard inequality (<a href="#HH">3</a>) we get the inequality (<a href="#e.1.6">7</a>). This gives a direct proof of (<a href="#e.1.6">7</a>) and it is simpler than in <span class="cite">
	[
	<a href="#SSDHA" >25</a>
	]
</span>. <span class="qed">â–¡</span></p>

  </div>
</div> </p>
<p>In 1994, <span class="cite">
	[
	<a href="#SSD" >11</a>
	]
</span> (see also <span class="cite">
	[
	<a href="#DP" >32</a>
	, 
	p.
	
	22
	]
</span>) we proved the following refinement of Hermite-Hadamard inequality. For a direct proof that is different from the one in <span class="cite">
	[
	<a href="#SSD" >11</a>
	]
</span>, see the recent paper <span class="cite">
	[
	<a href="#SSDGAn" >24</a>
	]
</span>. </p>
<p><div class="lemma_thmwrapper " id="l.2.1">
  <div class="lemma_thmheading">
    <span class="lemma_thmcaption">
    Lemma
    </span>
    <span class="lemma_thmlabel">6</span>
  </div>
  <div class="lemma_thmcontent">
  <p>Let \(p:\left[ c,d\right] \rightarrow \mathbb {R}\) be a convex function on \(\left[ c,d\right] \). Then for any division \(c=y_{0}{\lt}y_{1}{\lt}...{\lt}y_{n-1}{\lt}y_{n}=d\) with \(n\geq 1\) we have the inequalities</p>
<div class="displaymath" id="e.2.1">
  \begin{align}  p\left( \tfrac {c+d}{2}\right) &  \leq \tfrac {1}{d-c}\sum _{i=0}^{n-1}\left( y_{i+1}-y_{i}\right) p\left( \tfrac {y_{i+1}+y_{i}}{2}\right) \label{e.2.1} \\ &  \leq \tfrac {1}{d-c}\int _{c}^{d}p\left( y\right) dy\leq \tfrac {1}{d-c}\sum _{i=0}^{n-1}\left( y_{i+1}-y_{i}\right) \tfrac {p\left( y_{i}\right) +p\left( y_{i+1}\right) }{2} \notag \\ &  \leq \tfrac {1}{2}\left[ p\left( c\right) +p\left( d\right) \right] . \notag \end{align}
</div>

  </div>
</div> </p>
<p>We can state the following result: </p>
<p><div class="theorem_thmwrapper " id="t.2.2">
  <div class="theorem_thmheading">
    <span class="theorem_thmcaption">
    Theorem
    </span>
    <span class="theorem_thmlabel">7</span>
  </div>
  <div class="theorem_thmcontent">
  <p>Let \(f:\left[ a,b\right] \subset \left( 0,\infty \right) \rightarrow \mathbb {R}\) be a HA-convex function on the interval \(\left[ a,b\right] .\) Then for any division \(a=x_{0}{\lt}x_{1}{\lt}...{\lt}x_{n-1}{\lt}x_{n}=b\) with \(n\geq 1\) we have the inequalities</p>
<div class="displaymath" id="e.2.2">
  \begin{align}  \tfrac {a+b}{2}f\left( \tfrac {a+b}{2}\right) &  \leq \tfrac {1}{2\left( b-a\right) }\sum _{i=0}^{n-1}\left( x_{i+1}^{2}-x_{i}^{2}\right) f\left( \tfrac {x_{i+1}+x_{i}}{2}\right)\   \label{e.2.2} \\ &  \leq \tfrac {1}{b-a}\int _{a}^{b}xf\left( x\right) dx \notag \\ &  \leq \tfrac {1}{b-a}\sum _{i=0}^{n-1}\left( x_{i+1}-x_{i}\right) \tfrac {x_{i}f\left( x_{i}\right) +x_{i+1}f\left( x_{i+1}\right) }{2} \notag \\ &  \leq \tfrac {1}{2}\left[ af\left( a\right) +bf\left( b\right) \right] . \notag \end{align}
</div>

  </div>
</div> </p>
<p>Follows by Lemma <a href="#l.2.1">6</a> for the convex function \(p\left( x\right) =xf\left( x\right) ,\) \(x\in \left[ a,b\right] .\) </p>
<p>If we take \(n=2\) and \(x\in \left[ a,b\right] ,\) then by (<a href="#e.2.2">9</a>) we have</p>
<div class="displaymath" id="e.2.3">
  \begin{align}  \tfrac {a+b}{2}f\big( \tfrac {a+b}{2}\big)&  \leq \label{e.2.3} \tfrac {1}{2\left( b-a\right) }\left[ \big( x^{2}-a^{2}\big) f\left( \tfrac {x+a}{2}\right) +\big( b^{2}-x^{2}\big) f\big( \tfrac {x+b}{2}\big) \right] \\ &  \leq \tfrac {1}{b-a}\int _{a}^{b}tf\left( t\right) dt \notag \\ &  \leq \tfrac {1}{2\left( b-a\right) }\left[ \left( b-a\right) xf\left( x\right) +\left( x-a\right) af\left( a\right) +\left( b-x\right) bf\left( b\right) \right] \notag \\ &  \leq \tfrac {1}{2}\left[ af\left( a\right) +bf\left( b\right) \right] . \notag \end{align}
</div>
<p>If in this inequality we choose \(x=\frac{a+b}{2},\) then we get the inequality</p>
<div class="displaymath" id="e.2.4">
  \begin{align}  \tfrac {a+b}{2}f\big( \tfrac {a+b}{2}\big)& \leq \tfrac {1}{2\left( b-a\right) }\Big[ \tfrac {b+3a}{4}f\big( \tfrac {b+3a}{4}\big) +\tfrac {a+3b}{4}f\big( \tfrac {a+3b}{4}\big) \Big] \label{e.2.4}\\ &  \leq \tfrac {1}{b-a}\int _{a}^{b}tf\left( t\right) dt \notag \\ &  \leq \tfrac {1}{2}\left[ \tfrac {a+b}{2}f\big( \tfrac {a+b}{2}\big) +\tfrac {af\big( a\big) +bf\left( b\right) }{2}\right] \leq \tfrac {1}{2}\left[ af\left( a\right) +bf\left( b\right) \right] . \notag \end{align}
</div>
<p>If we take in (<a href="#e.2.3">10</a>) \(x=\frac{2ab}{a+b},\) then we get</p>
<div class="displaymath" id="e.2.4.a">
  \begin{align}  \tfrac {a+b}{2}f\big( \tfrac {a+b}{2}\big)& \leq \tfrac {1}{4\left( a+b\right) ^{2}}\left[ a^{2}\left( a+3b\right) f\left( \tfrac {a\left( a+3b\right) }{2\left( a+b\right) }\right) +b^{2}\left( 3a+b\right) f\left( \tfrac {b\left( 3a+b\right) }{2\left( a+b\right) }\right) \right]\label{e.2.4.a}\\ &  \leq \tfrac {1}{b-a}\int _{a}^{b}tf\left( t\right) dt \notag \\ &  \leq \tfrac {1}{a+b}\left[ abf\big( \tfrac {2ab}{a+b}\big) +\tfrac {a^{2}f\left( a\right) +b^{2}f\left( b\right) }{2}\right] \leq \tfrac {1}{2}\left[ af\left( a\right) +bf\left( b\right) \right] . \notag \end{align}
</div>
<p>We also have: </p>
<p><div class="theorem_thmwrapper " id="t.2.3">
  <div class="theorem_thmheading">
    <span class="theorem_thmcaption">
    Theorem
    </span>
    <span class="theorem_thmlabel">8</span>
  </div>
  <div class="theorem_thmcontent">
  <p>Let \(f:\left[ a,b\right] \subset \left( 0,\infty \right) \rightarrow \mathbb {R}\) be a HA-convex function on the interval \(\left[ a,b\right] .\) Then for any division \(a=x_{0}{\lt}x_{1}{\lt}...{\lt}x_{n-1}{\lt}x_{n}=b\) with \(n\geq 1\) we have the inequalities</p>
<div class="displaymath" id="e.2.5">
  \begin{align}  f\left( \tfrac {2ab}{a+b}\right) &  \leq \tfrac {ab}{b-a}\sum _{j=0}^{n-1}\left( \tfrac {x_{j+1}-x_{j}}{x_{j+1}x_{j}}\right) f\left( \tfrac {2x_{j+1}x_{j}}{x_{j+1}+x_{j}}\right) \label{e.2.5} \\ &  \leq \tfrac {ab}{b-a}\int _{a}^{b}\tfrac {f\left( x\right) }{x^{2}}dx \notag \\ &  \leq \tfrac {ab}{b-a}\sum _{i=0}^{n-1}\left( \tfrac {x_{j+1}-x_{j}}{x_{j+1}x_{j}}\right) \tfrac {f\left( x_{j}\right) +f\left( x_{j+1}\right) }{2}\leq \tfrac {f\left( b\right) +f\left( a\right) }{2}. \notag \end{align}
</div>

  </div>
</div> </p>
<p><div class="proof_wrapper" id="a0000000011">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div> Consider the convex function \(p\left( x\right) =f\big( \frac{1}{x}\big) \) that is convex on the interval \(\big[ \frac{1}{b},\frac{1}{a}\big] .\) The division \(a=x_{0}{\lt}x_{1}{\lt}...{\lt}x_{n-1}{\lt}x_{n}=b\) with \(n\geq 1\) produces the division \(y_{i}=\frac{1}{x_{n-i}},\) \(i\in \left\{  0,...,n\right\}  \) of the interval \(\big[ \frac{1}{b},\frac{1}{a}\big] .\) </p>
<p>Using the inequality (<a href="#e.2.1">8</a>) we get</p>
<div class="displaymath" id="e.2.6">
  \begin{align}  f\left( \tfrac {1}{\frac{\frac{1}{b}+\frac{1}{a}}{2}}\right) &  \leq \tfrac {1}{\frac{1}{a}-\frac{1}{b}}\sum _{i=0}^{n-1}\left( t\tfrac {1}{x_{n-i-1}}-\tfrac {1}{x_{n-i}}\right) f\left( \tfrac {1}{\frac{\frac{1}{x_{n-i-1}}+\frac{1}{x_{n-i}}}{2}}\right) \label{e.2.6} \\ &  \leq \tfrac {1}{\frac{1}{a}-\frac{1}{b}}\int _{\frac{1}{b}}^{\frac{1}{a}}f\big( \tfrac {1}{t}\big) dt \notag \\ &  \leq \tfrac {1}{\frac{1}{a}-\frac{1}{b}}\sum _{i=0}^{n-1}\left( \tfrac {1}{x_{n-i-1}}-\tfrac {1}{x_{n-i}}\right) \frac{f\left( \frac{1}{\frac{1}{x_{n-i-1}}}\right) +f\left( \frac{1}{\frac{1}{x_{n-i}}}\right) }{2} \notag \\ &  \leq \tfrac {1}{2}\Big[ f\Big( \tfrac {1}{\frac{1}{b}}\Big) +f\Big( \tfrac {1}{\frac{1}{a}}\Big) \Big] \notag \end{align}
</div>
<p>that is equivalent to</p>
<div class="displaymath" id="e.2.7">
  \begin{align}  f\left( \tfrac {2ab}{a+b}\right) &  \leq \tfrac {ab}{b-a}\sum _{i=0}^{n-1}\left( \tfrac {x_{n-i}-x_{n-i-1}}{x_{n-i-1}x_{n-i}}\right) f\left( \tfrac {2x_{n-i-1}x_{n-i}}{x_{n-i}+x_{n-i-1}}\right) \label{e.2.7} \\ &  \leq \tfrac {ab}{b-a}\int _{\frac{1}{b}}^{\frac{1}{a}}f\Big( \tfrac {1}{t}\Big) dt \leq \notag \end{align}
</div>
<div class="displaymath" id="a0000000012">
  \begin{align} &  \leq \tfrac {ab}{b-a}\sum _{i=0}^{n-1}\left( \tfrac {x_{n-i}-x_{n-i-1}}{x_{n-i-1}x_{n-i}}\right) \tfrac {f\left( x_{n-i-1}\right) +f\left( x_{n-i}\right) }{2} \notag \\ &  \leq \tfrac {1}{2}\left[ f\left( b\right) +f\left( a\right) \right] . \notag \end{align}
</div>
<p>By re-indexing the sums and taking into account that</p>
<div class="displaymath" id="a0000000013">
  \begin{equation*}  \int _{\frac{1}{b}}^{\frac{1}{a}}f\big( \tfrac {1}{t}\big) dt=\int _{a}^{b}\tfrac {f\left( x\right) }{x^{2}}dx \end{equation*}
</div>
<p>we obtain the desired result (<a href="#e.2.5">13</a>). <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> </p>
<p><div class="remark_thmwrapper " id="r.2.2">
  <div class="remark_thmheading">
    <span class="remark_thmcaption">
    Remark
    </span>
    <span class="remark_thmlabel">9</span>
  </div>
  <div class="remark_thmcontent">
  <p>If we take \(n=2\) and \(x\in \left[ a,b\right] ,\) then by (<a href="#e.2.5">13</a>) we have, after appropriate calculations, that </p>
<div class="displaymath" id="e.2.8">
  \begin{align}  f\left( \tfrac {2ab}{a+b}\right)& \leq \tfrac {1}{x}\left[ \tfrac {\left( x-a\right) bf\left( \frac{2ax}{a+x}\right) +\left( b-x\right) af\left( \frac{2xb}{x+b}\right) }{b-a}\right] \label{e.2.8} \\ &  \leq \tfrac {ab}{b-a}\int _{a}^{b}\tfrac {f\left( x\right) }{x^{2}}dx \notag \\ &  \leq \tfrac {1}{2}\left[ f\left( x\right) +\tfrac {\left( x-a\right) bf\left( a\right) +\left( b-x\right) af\left( b\right) }{x\left( b-a\right) }\right] \notag \\ &  \leq \tfrac {f\left( b\right) +f\left( a\right) }{2}. \notag \end{align}
</div>
<p>If we take in (<a href="#e.2.8">16</a>) \(x=\frac{2ab}{a+b}\in \left[ a,b\right] ,\) then we get </p>
<div class="displaymath" id="e.2.9">
  \begin{align}  f\big( \tfrac {2ab}{a+b}\big) &  \leq \tfrac {1}{2}\left[ f\big( \tfrac {4ab}{a+3b}\big) +f\big( \tfrac {4ab}{3a+b}\big) \right] \label{e.2.9} \\ &  \leq \tfrac {ab}{b-a}\int _{a}^{b}\tfrac {f\left( x\right) }{x^{2}}dx \notag \\ &  \leq \tfrac {1}{2}\left[ f\left( \tfrac {2ab}{a+b}\right) +\tfrac {f\left( a\right) +f\left( b\right) }{2}\right] \leq \tfrac {f\left( a\right) +f\left( b\right) }{2}. \notag \end{align}
</div>
<p>If we take in (<a href="#e.2.8">16</a>) \(x=\frac{a+b}{2}\in \left[ a,b\right] ,\) then we get </p>
<div class="displaymath" id="e.2.10">
  \begin{align}  f\big( \tfrac {2ab}{a+b}\big) &  \leq \frac{bf\left( \frac{a\left( a+b\right) }{3a+b}\right) +af\left( \frac{b\left( a+b\right) }{a+3b}\right) }{a+b} \label{e.2.10} \\ &  \leq \tfrac {ab}{b-a}\int _{a}^{b}\tfrac {f\left( x\right) }{x^{2}}dx \notag \\ &  \leq \tfrac {1}{2}\left[ f\big( \tfrac {a+b}{2}\big) +\tfrac {bf\left( a\right) +af\left( b\right) }{a+b}\right] \leq \tfrac {f\left( b\right) +f\left( a\right) }{2}. \notag \end{align}
</div>
<p> <span class="qed">â–¡</span></p>

  </div>
</div> </p>
<h1 id="a0000000015">3 Related Results</h1>
<p>We recall some facts on the lateral derivatives of a convex function. </p>
<p>Suppose that \(I\) is an interval of real numbers with interior \(\mathring {I}\) and \(f:I\rightarrow \mathbb {R}\) is a convex function on \(I\). Then \(f\) is continuous on \(\mathring {I}\) and has finite left and right derivatives at each point of \(\mathring {I}\). Moreover, if \(x,y\in \mathring {I}\) and \(x{\lt}y,\) then \(f_{-}^{\prime }\left( x\right) \leq f_{+}^{\prime }\left( x\right) \leq f_{-}^{\prime }\left( y\right) \leq f_{+}^{\prime }\left( y\right) \) which shows that both \(f_{-}^{\prime }\) and \(f_{+}^{\prime }\) are nondecreasing function on \(\mathring {I}\). It is also known that a convex function must be differentiable except for at most countably many points. </p>
<p>For a convex function \(f:I\rightarrow \mathbb {R}\), the subdifferential of \(f\) denoted by \(\partial f\) is the set of all functions \(\varphi :I\rightarrow \left[ -\infty ,\infty \right] \) such that \(\varphi \big( \mathring {I}\big) \subset \mathbb {R}\) and </p>
<div class="equation" id="GI">
<p>
  <div class="equation_content">
    \begin{equation}  f\left( x\right) \geq f\left( a\right) +\left( x-a\right) \varphi \left( a\right) \text{ for any }x,a\in I. \label{GI} \end{equation}
  </div>
  <span class="equation_label">19</span>
</p>
</div>
<p>It is also well known that if \(f\) is convex on \(I,\) then \(\partial f\) is nonempty, \(f_{-}^{\prime }\), \(f_{+}^{\prime }\in \partial f\) and if \(\varphi \in \partial f\), then </p>
<div class="displaymath" id="a0000000016">
  \begin{equation*}  f_{-}^{\prime }\left( x\right) \leq \varphi \left( x\right) \leq f_{+}^{\prime }\left( x\right) \text{ for any }x\in \mathring {I}\text{.} \end{equation*}
</div>
<p>In particular, \(\varphi \) is a nondecreasing function. If \(f\) is differentiable and convex on \(\mathring {I}\), then \(\partial f=\left\{  f^{\prime }\right\}  .\) </p>
<p><div class="lemma_thmwrapper " id="l.3.1">
  <div class="lemma_thmheading">
    <span class="lemma_thmcaption">
    Lemma
    </span>
    <span class="lemma_thmlabel">10</span>
  </div>
  <div class="lemma_thmcontent">
  <p>Let \(f:\left[ a,b\right] \subset \left( 0,\infty \right) \rightarrow \mathbb {R}\) be an HA-convex function on the interval \(\left[ a,b\right] .\) Then \(f\) has lateral derivatives in every point of \(\left( a,b\right) \) and </p>
<div class="equation" id="e.3.1">
<p>
  <div class="equation_content">
    \begin{equation}  f\left( t\right) -f\left( s\right) \geq sf_{\pm }^{\prime }\left( s\right) \left( 1-\tfrac {s}{t}\right) \label{e.3.1} \end{equation}
  </div>
  <span class="equation_label">20</span>
</p>
</div>
<p>for any \(s\in \left( a,b\right) \) and \(t\in \left[ a,b\right] .\) </p>
<p>Also, we have </p>
<div class="equation" id="e.3.2">
<p>
  <div class="equation_content">
    \begin{equation}  f\left( t\right) -f\left( a\right) \geq af_{+}^{\prime }\left( a\right) \left( 1-\tfrac {a}{t}\right) \label{e.3.2} \end{equation}
  </div>
  <span class="equation_label">21</span>
</p>
</div>
<p>and</p>
<div class="equation" id="e.3.3">
<p>
  <div class="equation_content">
    \begin{equation}  f\left( t\right) -f\left( b\right) \geq bf_{-}^{\prime }\left( b\right) \left( 1-\tfrac {b}{t}\right) \label{e.3.3} \end{equation}
  </div>
  <span class="equation_label">22</span>
</p>
</div>
<p>for any \(t\in \left[ a,b\right] \) provided the lateral derivatives \(f_{+}^{\prime }\left( a\right) \) and \(f_{-}^{\prime }\left( b\right) \) are finite. </p>

  </div>
</div> </p>
<p><div class="proof_wrapper" id="a0000000017">
  <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 \(f\) is <i class="itshape">HA</i>-convex function on the interval \(\left[ a,b\right] ,\) then the function \(h\left( t\right) =tf\left( t\right) \) is convex on \(\left[ a,b\right] \), therefore the function \(f\) has lateral derivatives in each point of \(\left( a,b\right) \) and</p>
<div class="displaymath" id="a0000000018">
  \begin{equation*}  h_{\pm }^{\prime }\left( t\right) =f\left( t\right) +tf_{\pm }^{\prime }\left( t\right) \end{equation*}
</div>
<p>for any \(t\in \left( a,b\right) .\) Also, if \(f_{+}^{\prime }\left( a\right) \) and \(f_{-}^{\prime }\left( b\right) \) are finite then </p>
<div class="displaymath" id="a0000000019">
  \begin{equation*}  h_{+}^{\prime }\left( a\right) =f\left( a\right) +af_{+}^{\prime }\left( a\right) \text{ and }h_{-}^{\prime }\left( b\right) =f\left( b\right) +bf_{-}^{\prime }\left( b\right) . \end{equation*}
</div>
<p>Writing the gradient inequality for the convex function \(h,\) namely</p>
<div class="displaymath" id="a0000000020">
  \begin{equation*}  h\left( t\right) -h\left( s\right) \geq h_{\pm }^{\prime }\left( s\right) \left( t-s\right) \end{equation*}
</div>
<p>for any \(s\in \left( a,b\right) \) and \(t\in \left[ a,b\right] ,\) we have</p>
<div class="displaymath" id="a0000000021">
  \begin{equation*}  tf\left( t\right) -sf\left( s\right) \geq \left[ f\left( s\right) +sf_{\pm }^{\prime }\left( s\right) \right] \left( t-s\right) =f\left( s\right) \left( t-s\right) +sf_{\pm }^{\prime }\left( s\right) \left( t-s\right) \end{equation*}
</div>
<p>that is equivalent to</p>
<div class="displaymath" id="a0000000022">
  \begin{equation*}  tf\left( t\right) -tf\left( s\right) \geq sf_{\pm }^{\prime }\left( s\right) \left( t-s\right) \end{equation*}
</div>
<p>for any \(s\in \left( a,b\right) \) and \(t\in \left[ a,b\right] .\) </p>
<p>Now, by dividing with \(t{\gt}0\) we get the desired result (<a href="#e.3.1">20</a>). </p>
<p>The rest follows by the corresponding properties of convex function \(h.\) <div class="proof_wrapper" id="a0000000023">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div> </p>
<p>We use the following results obtained by the author in <span class="cite">
	[
	<a href="#SSD1.a" >19</a>
	]
</span> and <span class="cite">
	[
	<a href="#SSD2.a" >20</a>
	]
</span> </p>
<p><div class="lemma_thmwrapper " id="l.2.2">
  <div class="lemma_thmheading">
    <span class="lemma_thmcaption">
    Lemma
    </span>
    <span class="lemma_thmlabel">11</span>
  </div>
  <div class="lemma_thmcontent">
  <p>Let \(h:\left[ \alpha ,\beta \right] \rightarrow \mathbb {R}\) be a convex function on \(\left[ \alpha ,\beta \right] .\) Then we have the inequalities</p>
<div class="displaymath" id="e.3.4">
  \begin{align}  \tfrac {1}{8}\left[ h_{+}^{\prime }\left( \tfrac {\alpha +\beta }{2}\right) -h_{-}^{\prime }\big( \tfrac {\alpha +\beta }{2}\big) \right] \left( \beta -\alpha \right)& \leq \tfrac {h\left( \alpha \right) +h\left( \beta \right) }{2}-\tfrac {1}{\beta -\alpha }\int _{\alpha }^{\beta }h\left( t\right) dt \label{e.3.4}\\ & \leq \tfrac {1}{8}\left[ h_{-}^{\prime }\left( \beta \right) -h_{+}^{\prime }\left( \alpha \right) \right] \left( \beta -\alpha \right) \notag \end{align}
</div>
<p>and</p>
<div class="displaymath" id="e.3.5">
  \begin{align}  \tfrac {1}{8}\left[ h_{+}^{\prime }\big( \tfrac {\alpha +\beta }{2}\big) -h_{-}^{\prime }\big( \tfrac {\alpha +\beta }{2}\big) \right] \left( \beta -\alpha \right)& \leq \label{e.3.5} \tfrac {1}{\beta -\alpha }\int _{\alpha }^{\beta }h\left( t\right) dt-h\big( \tfrac {\alpha +\beta }{2}\big)\\ & \leq \tfrac {1}{8}\left[ h_{-}^{\prime }\left( \beta \right) -h_{+}^{\prime }\left( \alpha \right) \right] \left( \beta -\alpha \right) . \notag \end{align}
</div>
<p>The constant \(\frac{1}{8}\) is best possible in <a href="#e.3.4" class="eqref">23</a> and <a href="#e.3.5" class="eqref">24</a>. </p>

  </div>
</div> </p>
<p>The following result holds: </p>
<p><div class="theorem_thmwrapper " id="t.3.1">
  <div class="theorem_thmheading">
    <span class="theorem_thmcaption">
    Theorem
    </span>
    <span class="theorem_thmlabel">12</span>
  </div>
  <div class="theorem_thmcontent">
  <p>Let \(f:\left[ a,b\right] \subset \left( 0,\infty \right) \rightarrow \mathbb {R}\) be an HA-convex function on the interval \(\left[ a,b\right] .\) Then we have</p>
<div class="displaymath" id="e.3.6">
  \begin{align} &  \tfrac {1}{16}\left[ f_{+}^{\prime }\left( \tfrac {a+b}{2}\right) -f_{-}^{\prime }\left( \tfrac {a+b}{2}\right) \right] \big( b^{2}-a^{2}\big)\leq \label{e.3.6} \\ &  \leq \tfrac {af\left( a\right) +bf\left( b\right) }{2}-\tfrac {1}{b-a}\int _{a}^{b}tf\left( t\right) dt \notag \\ &  \leq \tfrac {1}{8}\left[ f\left( b\right) -f\left( a\right) \right] \left( b-a\right) +\tfrac {1}{8}\left[ bf_{-}^{\prime }\left( b\right) -af_{+}^{\prime }\left( a\right) \right] \left( b-a\right) \notag \end{align}
</div>
<p>and</p>
<div class="displaymath" id="e.3.7">
  \begin{align} &  \tfrac {1}{16}\left[ f_{+}^{\prime }\left( \tfrac {a+b}{2}\right) -f_{-}^{\prime }\left( \tfrac {a+b}{2}\right) \right] \left( b^{2}-a^{2}\right) \label{e.3.7}\leq \\ &  \leq \tfrac {1}{b-a}\int _{a}^{b}tf\left( t\right) dt-\tfrac {a+b}{2}f\left( \tfrac {a+b}{2}\right) \notag \\ &  \leq \tfrac {1}{8}\left[ f\left( b\right) -f\left( a\right) \right] \left( b-a\right) +\tfrac {1}{8}\left[ bf_{-}^{\prime }\left( b\right) -af_{+}^{\prime }\left( a\right) \right] \left( b-a\right) . \notag \end{align}
</div>

  </div>
</div> </p>
<p><div class="proof_wrapper" id="a0000000024">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div> Making use of inequality (<a href="#e.3.4">23</a>) in Lemma <a href="#l.2.2">11</a> for the convex function \(h\left( t\right) =tf\left( t\right) \) we have</p>
<div class="displaymath" id="a0000000025">
  \begin{align*} &  \tfrac {1}{8}\left[ \tfrac {a+b}{2}f_{+}^{\prime }\left( \tfrac {a+b}{2}\right) -\tfrac {a+b}{2}f_{-}^{\prime }\left( \tfrac {a+b}{2}\right) \right] \left( b-a\right)\leq \\ &  \leq \tfrac {af\left( a\right) +bf\left( b\right) }{2}-\tfrac {1}{b-a}\int _{a}^{b}tf\left( t\right) dt \\ &  \leq \tfrac {1}{8}\left[ f\left( b\right) +bf_{-}^{\prime }\left( b\right) -f\left( a\right) -af_{+}^{\prime }\left( a\right) \right] \left( b-a\right) , \end{align*}
</div>
<p>which proves the inequality (<a href="#e.3.6">25</a>). </p>
<p>The inequality (<a href="#e.3.7">26</a>) follows by (<a href="#e.3.5">24</a>). <div class="proof_wrapper" id="a0000000026">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div> </p>
<p><div class="corollary_thmwrapper " id="c.3.1">
  <div class="corollary_thmheading">
    <span class="corollary_thmcaption">
    Corollary
    </span>
    <span class="corollary_thmlabel">13</span>
  </div>
  <div class="corollary_thmcontent">
  <p>Let \(f:\left[ a,b\right] \subset \left( 0,\infty \right) \rightarrow \mathbb {R}\) be a differentiable HA-convex function on the interval \(\left[ a,b\right] .\) Then we have</p>
<div class="displaymath" id="e.3.8">
  \begin{align}  0&  \leq \tfrac {af\left( a\right) +bf\left( b\right) }{2}-\tfrac {1}{b-a}\int _{a}^{b}tf\left( t\right) dt \label{e.3.8} \\ &  \leq \tfrac {1}{8}\left[ f\left( b\right) -f\left( a\right) \right] \left( b-a\right) +\tfrac {1}{8}\left[ bf_{-}^{\prime }\left( b\right) -af_{+}^{\prime }\left( a\right) \right] \left( b-a\right) \notag \end{align}
</div>
<p>and</p>
<div class="displaymath" id="e.3.9">
  \begin{align}  0&  \leq \tfrac {1}{b-a}\int _{a}^{b}tf\left( t\right) dt-\tfrac {a+b}{2}f\left( \tfrac {a+b}{2}\right) \label{e.3.9} \\ &  \leq \tfrac {1}{8}\left[ f\left( b\right) -f\left( a\right) \right] \left( b-a\right) +\tfrac {1}{8}\left[ bf_{-}^{\prime }\left( b\right) -af_{+}^{\prime }\left( a\right) \right] \left( b-a\right) . \notag \end{align}
</div>

  </div>
</div> </p>
<p>We remark that from (<a href="#e.3.8">27</a>) we have</p>
<div class="displaymath" id="e.3.10">
  \begin{align} &  \tfrac {\left( 3a+b\right) f\left( a\right) +\left( a+3b\right) f\left( b\right) }{8}-\tfrac {1}{8}\left[ bf_{-}^{\prime }\left( b\right) -af_{+}^{\prime }\left( a\right) \right] \left( b-a\right)\leq \label{e.3.10} \\ &  \leq \tfrac {1}{b-a}\int _{a}^{b}tf\left( t\right) dt\leq \tfrac {af\left( a\right) +bf\left( b\right) }{2} \notag \end{align}
</div>
<p>and from (<a href="#e.3.9">28</a>) we have</p>
<div class="displaymath" id="e.3.11">
  \begin{align}  \tfrac {a+b}{2}f\left( \tfrac {a+b}{2}\right) \leq &  \tfrac {1}{b-a}\int _{a}^{b}tf\left( t\right) dt \label{e.3.11} \\ \leq &  \tfrac {a+b}{2}f\left( \tfrac {a+b}{2}\right) +\tfrac {1}{8}\left[ f\left( b\right) -f\left( a\right) \right] \left( b-a\right) \notag \\ &  +\tfrac {1}{8}\left[ bf_{-}^{\prime }\left( b\right) -af_{+}^{\prime }\left( a\right) \right] \left( b-a\right) . \notag \end{align}
</div>
<p>The <i class="itshape">identric mean</i> \(I\left( a,b\right) \) is defined by</p>
<div class="displaymath" id="a0000000027">
  \begin{equation*}  I\left( a,b\right) :=\tfrac {1}{e}\Big( \tfrac {b^{b}}{a^{a}}\Big) ^{\frac{1}{b-a}} \end{equation*}
</div>
<p>while the <i class="itshape">logarithmic mean</i> is defined by</p>
<div class="displaymath" id="a0000000028">
  \begin{equation*}  L\left( a,b\right) :=\tfrac {b-a}{\ln b-\ln a}. \end{equation*}
</div>
<p>The following result also holds: </p>
<p><div class="theorem_thmwrapper " id="t.3.2">
  <div class="theorem_thmheading">
    <span class="theorem_thmcaption">
    Theorem
    </span>
    <span class="theorem_thmlabel">14</span>
  </div>
  <div class="theorem_thmcontent">
  <p>Let \(f:\left[ a,b\right] \subset \left( 0,\infty \right) \rightarrow \mathbb {R}\) be an HA-convex function on the interval \(\left[ a,b\right] .\) </p>
<p>(i) If \(bf\left( b\right) -af\left( a\right) \neq \int _{a}^{b}f\left( s\right) ds\) and</p>
<div class="equation" id="e.3.12">
<p>
  <div class="equation_content">
    \begin{equation}  \alpha _{f}:=\tfrac {\int _{a}^{b}s^{2}f^{\prime }\left( s\right) ds}{\int _{a}^{b}sf^{\prime }\left( s\right) ds}=\tfrac {b^{2}f\left( b\right) -a^{2}f\left( a\right) -2\int _{a}^{b}sf\left( s\right) ds}{bf\left( b\right) -af\left( a\right) -\int _{a}^{b}f\left( s\right) ds}\in \left[ a,b\right] \label{e.3.12} \end{equation}
  </div>
  <span class="equation_label">31</span>
</p>
</div>
<p>then</p>
<div class="equation" id="e.3.13">
<p>
  <div class="equation_content">
    \begin{equation}  f\left( \alpha _{f}\right) \geq \tfrac {1}{b-a}\int _{a}^{b}f\left( s\right) ds. \label{e.3.13} \end{equation}
  </div>
  <span class="equation_label">32</span>
</p>
</div>
<p>(ii) If \(f\left( b\right) \neq f\left( a\right) \) and</p>
<div class="equation" id="e.3.14">
<p>
  <div class="equation_content">
    \begin{equation}  \beta _{f}=\tfrac {\int _{a}^{b}sf^{\prime }\left( s\right) ds}{\int _{a}^{b}f^{\prime }\left( s\right) ds}=\tfrac {bf\left( b\right) -af\left( a\right) -\int _{a}^{b}f\left( s\right) ds}{f\left( b\right) -f\left( a\right) }\in \left[ a,b\right] \label{e.3.14} \end{equation}
  </div>
  <span class="equation_label">33</span>
</p>
</div>
<p>then</p>
<div class="equation" id="e.3.15">
<p>
  <div class="equation_content">
    \begin{equation}  f\left( \beta _{f}\right) \geq \tfrac {1}{\ln b-\ln a}\int _{a}^{b}f\left( s\right) ds. \label{e.3.15} \end{equation}
  </div>
  <span class="equation_label">34</span>
</p>
</div>
<p>(iii) If \(af\left( b\right) \neq bf\left( a\right) \) and </p>
<div class="equation" id="e.3.16">
<p>
  <div class="equation_content">
    \begin{equation}  \gamma _{f}:=\tfrac {\left( f\left( b\right) -f\left( a\right) \right) ab}{af\left( b\right) -bf\left( a\right) }\in \left[ a,b\right] \label{e.3.16} \end{equation}
  </div>
  <span class="equation_label">35</span>
</p>
</div>
<p>then</p>
<div class="equation" id="e.3.17">
<p>
  <div class="equation_content">
    \begin{equation}  f\left( \gamma _{f}\right) \geq \tfrac {2ab}{b-a}\int _{a}^{b}\tfrac {f\left( s\right) }{s^{2}}ds. \label{e.3.17} \end{equation}
  </div>
  <span class="equation_label">36</span>
</p>
</div>

  </div>
</div> </p>
<p><div class="proof_wrapper" id="a0000000029">
  <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 know that if \(f:\left[ a,b\right] \subset \left( 0,\infty \right) \rightarrow \mathbb {R}\) is an <i class="itshape">HA</i>-convex function on the interval \(\left[ a,b\right] \) then the functions is differentiable except for at most countably many points. Then, from (<a href="#e.3.1">20</a>) we have </p>
<div class="equation" id="e.3.18">
<p>
  <div class="equation_content">
    \begin{equation}  f\left( t\right) -f\left( s\right) \geq sf^{\prime }\left( s\right) \left( 1-\tfrac {s}{t}\right) \label{e.3.18} \end{equation}
  </div>
  <span class="equation_label">37</span>
</p>
</div>
<p>for any \(t\in \left[ a,b\right] \) and almost every \(s\in \left( a,b\right) .\) </p>
<p>(i) If we take the Lebesgue integral mean in (<a href="#e.3.18">37</a>), then we get</p>
<div class="equation" id="e.3.19">
<p>
  <div class="equation_content">
    \begin{equation}  f\left( t\right) -\tfrac {1}{b-a}\int _{a}^{b}f\left( s\right) ds\geq \tfrac {1}{b-a}\int _{a}^{b}sf^{\prime }\left( s\right) ds-\tfrac {1}{t}\tfrac {1}{b-a}\int _{a}^{b}s^{2}f^{\prime }\left( s\right) ds \label{e.3.19} \end{equation}
  </div>
  <span class="equation_label">38</span>
</p>
</div>
<p>for any \(t\in \left[ a,b\right] .\) </p>
<p>If we take \(t=\alpha _{f}\) in (<a href="#e.3.19">38</a>) then we get the desired inequality (<a href="#e.3.13">32</a>). </p>
<p>(ii) If we divide the inequality (<a href="#e.3.18">37</a>) by \(s\) then we get</p>
<div class="equation" id="e.3.20">
<p>
  <div class="equation_content">
    \begin{equation}  \tfrac {1}{s}f\left( t\right) -\tfrac {f\left( s\right) }{s}\geq f^{\prime }\left( s\right) -\tfrac {1}{t}sf^{\prime }\left( s\right) \label{e.3.20} \end{equation}
  </div>
  <span class="equation_label">39</span>
</p>
</div>
<p>for any \(t\in \left[ a,b\right] \) and almost every \(s\in \left( a,b\right) .\) </p>
<p>If we take the Lebesgue integral mean in (<a href="#e.3.20">39</a>), then we get</p>
<div class="displaymath" id="a0000000030">
  \begin{align*}  f\left( t\right) \tfrac {1}{b-a}\int _{a}^{b}\tfrac {1}{s}ds-\tfrac {1}{b-a}\int _{a}^{b}\tfrac {f\left( s\right) }{s}ds\tfrac {1}{b-a}\int _{a}^{b}f^{\prime }\left( s\right) ds-\tfrac {1}{t}\tfrac {1}{b-a}\int _{a}^{b}sf^{\prime }\left( s\right) ds \end{align*}
</div>
<p>that is equivalent to</p>
<div class="displaymath" id="e.3.21">
  \begin{align} &  \tfrac {f\left( t\right) }{L\left( a,b\right) }-\tfrac {1}{b-a}\int _{a}^{b}\tfrac {f\left( s\right) }{s}ds \geq \tfrac {f\left( b\right) -f\left( a\right) }{b-a}-\tfrac {1}{t}\tfrac {bf\left( b\right) -af\left( a\right) -\int _{a}^{b}f\left( s\right) ds}{b-a} \label{e.3.21} \end{align}
</div>
<p>for any \(t\in \left[ a,b\right] \) </p>
<p>If we take \(t=\beta _{f}\) in (<a href="#e.3.21">40</a>) then we get the desired result (<a href="#e.3.15">34</a>). </p>
<p>(iii) If we divide the inequality (<a href="#e.3.18">37</a>) by \(s^{2}\) then we get</p>
<div class="equation" id="e.3.22">
<p>
  <div class="equation_content">
    \begin{equation}  \tfrac {1}{s^{2}}f\left( t\right) -\tfrac {f\left( s\right) }{s^{2}}\geq \tfrac {f^{\prime }\left( s\right) }{s}-\tfrac {1}{t}f^{\prime }\left( s\right) \label{e.3.22} \end{equation}
  </div>
  <span class="equation_label">41</span>
</p>
</div>
<p>for any \(t\in \left[ a,b\right] \) and almost every \(s\in \left( a,b\right) .\) </p>
<p>If we take the Lebesgue integral mean in (<a href="#e.3.22">41</a>), then we get</p>
<div class="displaymath" id="a0000000031">
  \begin{align*} &  f\left( t\right) \tfrac {1}{b-a}\int _{a}^{b}\tfrac {1}{s^{2}}ds-\tfrac {1}{b-a}\int _{a}^{b}\tfrac {f\left( s\right) }{s^{2}}ds\geq \tfrac {1}{b-a}\int _{a}^{b}\tfrac {f^{\prime }\left( s\right) }{s}ds-\tfrac {1}{t}\tfrac {1}{b-a}\int _{a}^{b}f^{\prime }\left( s\right) ds, \end{align*}
</div>
<p>which is equivalent to</p>
<div class="displaymath" id="a0000000032">
  \begin{align*} &  f\left( t\right) \tfrac {1}{ab}-\tfrac {1}{b-a}\int _{a}^{b}\tfrac {f\left( s\right) }{s^{2}}ds\geq \tfrac {1}{b-a}\Big[ \tfrac {f\left( b\right) }{b}-\tfrac {f\left( a\right) }{a}+\int _{a}^{b}\tfrac {f\left( s\right) }{s^{2}}ds\Big] -\tfrac {1}{t}\tfrac {f\left( b\right) -f\left( a\right) }{b-a} \end{align*}
</div>
<p>or, to</p>
<div class="displaymath" id="a0000000033">
  \begin{align*} &  f\left( t\right) \tfrac {1}{ab}-\tfrac {2}{b-a}\int _{a}^{b}\tfrac {f\left( s\right) }{s^{2}}ds\geq \tfrac {1}{b-a}\tfrac {af\left( b\right) -bf\left( a\right) }{ba}-\tfrac {1}{t}\tfrac {f\left( b\right) -f\left( a\right) }{b-a}. \end{align*}
</div>
<p> <div class="proof_wrapper" id="a0000000034">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div> </p>
<p><div class="remark_thmwrapper " id="r.3.1">
  <div class="remark_thmheading">
    <span class="remark_thmcaption">
    Remark
    </span>
    <span class="remark_thmlabel">15</span>
  </div>
  <div class="remark_thmcontent">
  <p>We observe that a sufficient condition for (<a href="#e.3.12">31</a>) and (<a href="#e.3.14">33</a>) to hold is that \(f\) is increasing on \(\left[ a,b\right] .\) If \(f\left( a\right) {\lt}0{\lt}f\left( b\right) ,\) then the inequality (<a href="#e.3.16">35</a>) also holds. </p>

  </div>
</div> </p>
<p>We also have the following result: </p>
<p><div class="theorem_thmwrapper " id="t.3.3">
  <div class="theorem_thmheading">
    <span class="theorem_thmcaption">
    Theorem
    </span>
    <span class="theorem_thmlabel">16</span>
  </div>
  <div class="theorem_thmcontent">
  <p>Let \(f:\left[ a,b\right] \subset \left( 0,\infty \right) \rightarrow \mathbb {R}\) be an HA-convex function on the interval \(\left[ a,b\right] .\) Then we have</p>
<div class="equation" id="e.3.23">
<p>
  <div class="equation_content">
    \begin{equation}  f\big( \tfrac {a+b}{2}\big) \leq \tfrac {1}{\ln b-\ln a}\int _{a}^{b}\tfrac {f\left( t\right) }{a+b-t}dt\leq \tfrac {af\left( a\right) +bf\left( b\right) }{a+b}. \label{e.3.23} \end{equation}
  </div>
  <span class="equation_label">42</span>
</p>
</div>

  </div>
</div> </p>
<p><div class="proof_wrapper" id="a0000000035">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div> Since the function \(h\left( t\right) =tf\left( t\right) \) is convex, then we have</p>
<div class="displaymath" id="a0000000036">
  \begin{equation*}  \tfrac {x+y}{2}f\big( \tfrac {x+y}{2}\big) \leq \tfrac {xf\left( x\right) +yf\left( y\right) }{2} \end{equation*}
</div>
<p>for any \(x,y\in \left[ a,b\right] .\) </p>
<p>If we divide this inequality by \(xy{\gt}0\) we get</p>
<div class="equation" id="e.3.24">
<p>
  <div class="equation_content">
    \begin{equation}  \tfrac {1}{2}\big( \tfrac {1}{x}+\tfrac {1}{y}\big) f\big( \tfrac {x+y}{2}\big) \leq \tfrac {1}{2}\left( \tfrac {f\left( x\right) }{y}+\tfrac {f\left( y\right) }{x}\right) , \label{e.3.24} \end{equation}
  </div>
  <span class="equation_label">43</span>
</p>
</div>
<p>for any \(x,y\in \left[ a,b\right] .\) </p>
<p>If we replace \(x\) by \(\left( 1-t\right) a+tb\) and \(y\) by \(ta+\left( 1-t\right) b\) in (<a href="#e.3.24">43</a>), then we get</p>
<div class="displaymath" id="e.3.25">
  \begin{align} &  \tfrac {1}{2}\left( \tfrac {1}{\left( 1-t\right) a+tb}+\tfrac {1}{ta+\left( 1-t\right) b}\right) f\big( \tfrac {a+b}{2}\big)\leq \tfrac {1}{2}\left( \tfrac {f\left( \left( 1-t\right) a+tb\right) }{ta+\left( 1-t\right) b}+\tfrac {f\left( ta+\left( 1-t\right) b\right) }{\left( 1-t\right) a+tb}\right),\label{e.3.25} \end{align}
</div>
<p>for any \(t\in \left[ 0,1\right] .\) </p>
<p>Integrating (<a href="#e.3.25">44</a>) on \(\left[ 0,1\right] \) over \(t\) we get</p>
<div class="displaymath" id="e.3.26">
  \begin{align} &  \tfrac {1}{2}\left( \int _{0}^{1}\tfrac {1}{\left( 1-t\right) a+tb}dt+\int _{0}^{1}\tfrac {1}{ta+\left( 1-t\right) b}dt\right) f\big( \tfrac {a+b}{2}\big)\leq \label{e.3.26} \\ &  \leq \tfrac {1}{2}\left( \int _{0}^{1}\tfrac {f\left( \left( 1-t\right) a+tb\right) }{ta+\left( 1-t\right) b}dt+\int _{0}^{1}\tfrac {f\left( ta+\left( 1-t\right) b\right) }{\left( 1-t\right) a+tb}dt\right) . \notag \end{align}
</div>
<p>Observe that, by the appropriate change of variable,</p>
<div class="displaymath" id="a0000000037">
  \begin{equation*}  \int _{0}^{1}\tfrac {1}{\left( 1-t\right) a+tb}dt=\int _{0}^{1}\tfrac {1}{ta+\left( 1-t\right) b}dt=\tfrac {1}{b-a}\int _{a}^{b}\tfrac {du}{u}=\tfrac {\ln b-\ln a}{b-a} \end{equation*}
</div>
<p>and</p>
<div class="displaymath" id="a0000000038">
  \begin{equation*}  \int _{0}^{1}\tfrac {f\left( \left( 1-t\right) a+tb\right) }{ta+\left( 1-t\right) b}dt=\int _{0}^{1}\tfrac {f\left( ta+\left( 1-t\right) b\right) }{\left( 1-t\right) a+tb}=\tfrac {1}{b-a}\int _{a}^{b}\tfrac {f\left( u\right) }{a+b-u}du \end{equation*}
</div>
<p>and by (<a href="#e.3.26">45</a>) we get the first inequality in (<a href="#e.3.23">42</a>). </p>
<p>From the convexity of \(h\) we also have</p>
<div class="displaymath" id="a0000000039">
  \begin{equation*}  \left( \left( 1-t\right) a+tb\right) f\left( \left( 1-t\right) a+tb\right) \leq \left( 1-t\right) af\left( a\right) +tbf\left( b\right) \end{equation*}
</div>
<p>and</p>
<div class="displaymath" id="a0000000040">
  \begin{equation*}  \left( ta+\left( 1-t\right) b\right) f\left( ta+\left( 1-t\right) b\right) \leq taf\left( a\right) +\left( 1-t\right) bf\left( b\right) \end{equation*}
</div>
<p>for any \(t\in \left[ 0,1\right] .\) </p>
<p>Add these inequalities to get</p>
<div class="displaymath" id="a0000000041">
  \begin{align*} &  \left( \left( 1-t\right) a+tb\right) f\left( \left( 1-t\right) a+tb\right) +\left( ta+\left( 1-t\right) b\right) f\left( ta+\left( 1-t\right) b\right)\leq \\ &  \leq af\left( a\right) +bf\left( b\right) \end{align*}
</div>
<p>for any \(t\in \left[ 0,1\right] .\) </p>
<p>If we divide this inequality by \(\left( \left( 1-t\right) a+tb\right) \left( ta+\left( 1-t\right) b\right) ,\) then we get</p>
<div class="equation" id="e.3.27">
<p>
  <div class="equation_content">
    \begin{equation}  \tfrac {f\left( \left( 1-t\right) a+tb\right) }{ta+\left( 1-t\right) b}+\tfrac {f\left( ta+\left( 1-t\right) b\right) }{\left( 1-t\right) a+tb}\leq \tfrac {af\left( a\right) +bf\left( b\right) }{\left( \left( 1-t\right) a+tb\right) \left( ta+\left( 1-t\right) b\right) } \label{e.3.27} \end{equation}
  </div>
  <span class="equation_label">46</span>
</p>
</div>
<p>for any \(t\in \left[ 0,1\right] .\) </p>
<p>If we integrate the inequality (<a href="#e.3.27">46</a>) over \(t\) on \(\left[ 0,1\right] ,\) then we obtain</p>
<div class="displaymath" id="e.3.28">
  \begin{align} &  \int _{0}^{1}\tfrac {f\left( \left( 1-t\right) a+tb\right) }{ta+\left( 1-t\right) b}dt+\int _{0}^{1}\tfrac {f\left( ta+\left( 1-t\right) b\right) }{\left( 1-t\right) a+tb}dt\leq \label{e.3.28} \\ &  \leq \left[ af\left( a\right) +bf\left( b\right) \right] \int _{0}^{1}\tfrac {dt}{\left( \left( 1-t\right) a+tb\right) \left( ta+\left( 1-t\right) b\right) }. \notag \end{align}
</div>
<p>Since</p>
<div class="displaymath" id="a0000000042">
  \begin{equation*}  \int _{0}^{1}\tfrac {dt}{\left( \left( 1-t\right) a+tb\right) \left( ta+\left( 1-t\right) b\right) }=\tfrac {1}{b-a}\int _{a}^{b}\tfrac {du}{u\left( a+b-u\right) } \end{equation*}
</div>
<p>and</p>
<div class="displaymath" id="a0000000043">
  \begin{equation*}  \tfrac {1}{u\left( a+b-u\right) }=\tfrac {1}{a+b}\big( \tfrac {1}{u}+\tfrac {1}{a+b-u}\big) , \end{equation*}
</div>
<p>then</p>
<div class="displaymath" id="a0000000044">
  \begin{align*}  \int _{a}^{b}\tfrac {du}{u\left( a+b-u\right) }&  =\tfrac {1}{a+b}\int _{a}^{b}\big( \tfrac {1}{u}+\tfrac {1}{a+b-u}\big) du =\tfrac {2}{a+b}\left( \ln b-\ln a\right) . \end{align*}
</div>
<p>By (<a href="#e.3.28">47</a>) we then have</p>
<div class="displaymath" id="a0000000045">
  \begin{equation*}  \tfrac {2}{b-a}\int _{a}^{b}\tfrac {f\left( u\right) }{a+b-u}du\leq 2\left[ \tfrac {af\left( a\right) +bf\left( b\right) }{a+b}\right] \tfrac {\ln b-\ln a}{b-a}, \end{equation*}
</div>
<p>which proves the second inequality in (<a href="#e.3.23">42</a>). <div class="proof_wrapper" id="a0000000046">
  <div class="proof_heading">
    <span class="proof_caption">
    Proof
    </span>
    <span class="expand-proof">â–¼</span>
  </div>
  <div class="proof_content">
  
  </div>
</div> </p>
<h1 id="a0000000047">4 Applications</h1>
<p>We consider the <i class="itshape">arithmetic mean</i> \(A\left( a,b\right) =\frac{a+b}{2},\) the <i class="itshape">geometric mean</i> \(G\left( a,b\right)\) = \(\sqrt{ab}\) and <i class="itshape">harmonic mean </i>\(H\left( a,b\right) =\frac{2ab}{a+b}\) for the positive numbers \(a,b{\gt}0.\) </p>
<p>If we use the inequalities (<a href="#e.2.5">13</a>) for the <i class="itshape">HA</i>-convex function \(f\left( t\right) =t\) on the interval \(\left[ a,b\right] \subset \left( 0,\infty \right) \) then for any division \(a=x_{0}{\lt}x_{1}{\lt}...{\lt}x_{n-1}{\lt}x_{n}=b\) with \(n\geq 1\) we have the inequalities</p>
<div class="displaymath" id="e.4.1">
  \begin{align}  \tfrac {2ab}{a+b}&  \leq \tfrac {2ab}{b-a}\sum _{j=0}^{n-1}\tfrac {x_{j+1}-x_{j}}{x_{j+1}+x_{j}}\leq \tfrac {G^{2}\left( a,b\right) }{L\left( a,b\right) } \leq \tfrac {ab}{2\left( b-a\right) }\sum _{i=0}^{n-1}\tfrac {x_{j+1}^{2}-x_{j}^{2}}{x_{j+1}x_{j}}\leq A\left( a,b\right) . \label{e.4.1}\end{align}
</div>
<p>In particular, we have </p>
<div class="displaymath" id="e.4.2">
  \begin{align}  H\left( a,b\right) &  \leq 2ab\big( \tfrac {1}{a+3b}+\tfrac {1}{3a+b}\big) \leq \tfrac {G^{2}\left( a,b\right) }{L\left( a,b\right) } \leq \tfrac {H\left( a,b\right) +A\left( a,b\right) }{2}\left( \leq A\left( a,b\right) \right). \label{e.4.2}\end{align}
</div>
<p>Consider the function \(f:\left( 0,\infty \right) \rightarrow \mathbb {R}\), \(f\left( t\right) =\frac{\ln t}{t}.\) Observe that \(g\left( t\right) =f\big( \frac{1}{t}\big) =-t\ln t,\) which shows that \(f\) is <i class="itshape">HA</i>-concave on \(\left( 0,\infty \right) .\) </p>
<p>If we write the inequality (<a href="#e.2.4">11</a>) for the <i class="itshape">HA</i>-concave function \(f\left( t\right) =\tfrac {\ln t}{t}\) on \(\left( 0,\infty \right) ,\) then we have for any division \(a=x_{0}{\lt}x_{1}{\lt}...{\lt}x_{n-1}{\lt}x_{n}=b\) with \(n\geq 1\) that</p>
<div class="displaymath" id="e.4.3">
  \begin{align}  A\left( a,b\right) &  \geq \prod \limits _{i=0}^{n-1}\big( \tfrac {x_{i+1}+x_{i}}{2}\big) ^{\frac{x_{i+1}-x_{i}}{b-a}}\geq I\left( a,b\right) \geq \prod \limits _{i=0}^{n-1}\left( x_{i}x_{i+1}\right) ^{\frac{x_{i+1}-x_{i}}{2\left( b-a\right) }}\geq G\left( a,b\right).\label{e.4.3} \end{align}
</div>
<p>In particular, we have</p>
<div class="displaymath" id="e.4.4">
  \begin{align}  A\left( a,b\right) &  \geq \big( \tfrac {b+3a}{4}\big) ^{\frac{1}{2\left( b-a\right) }}\big( \tfrac {a+3b}{4}\big) ^{\frac{1}{2\left( b-a\right) }} \label{e.4.4} \\ &  \geq I\left( a,b\right) \geq \sqrt{A\left( a,b\right) G\left( a,b\right) }\left( \geq G\left( a,b\right) \right) . \notag \end{align}
</div>
<p>The interested reader may apply the above inequalities for other <i class="itshape">HA</i>-convex functions such as \(f\left( t\right) =\frac{h\left( t\right) }{t},\) \(t{\gt}0\) with \(h\) any convex function on an interval \(I\subset \left( 0,\infty \right) \) etc. The details are omitted. </p>
<p><small class="footnotesize">  </small></p>
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  <dd><p><a href ="https://doi.org/10.1186/1029-242X-2013-326"> <i class="sc">M. Tunç</i>, <i class="itshape">Ostrowski-type inequalities via h-convex functions with applications to special means</i>, J. Inequal. Appl., <b class="bfseries">2013</b>, 2013:326. <img src="img-0001.png" alt="\includegraphics[scale=0.1]{ext-link.png}" style="width:12.0px; height:10.700000000000001px" />
</a> </p>
</dd>
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</dd>
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  <dd><p><a href ="https://doi.org/10.1155/2010/507560"> <i class="sc">X.-M. Zhang, Y.-M. Chu</i>, <i class="sc">X.-H. Zhang</i>, <i class="itshape">The Hermite-Hadamard type inequality of GA-convex functions and its application</i>, J. Ineq. Appl., <b class="bfseries">2010</b>, Article ID 507560, 11 pp. <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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