New Inequalities of Hermite-Hadamard Type for $HA$-Convex Functions

Silvestru Sever Dragomir$^\ast $

April 11, 2017. Accepted: September 11, 2017. Published online: August 6, 2018.

$^\ast $Mathematics, College of Engineering & 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: sever.dragomir@vu.edu.au.

Some new inequalities of Hermite-Hadamard type for $HA$-convex functions defined on positive intervals are given.

MSC. 26D15; 25D10.

Keywords. Convex functions, Integral inequalities, $HA$-Convex functions.

1 Introduction

Following [ 1 ] (see also [ 41 ] ) we say that the function $f:I\subset R\setminus \left\{ 0\right\} \rightarrow \mathbb {R}$ is HA-convex or harmonically convex if

\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}

for all $x,y\in I$ and $t\in \left[ 0,1\right] $. If the inequality in (1) is reversed, then $f$ is said to be HA-concave or harmonically concave.

In order to avoid any confusion with the class of AH-convex functions, namely the functions satisfying the condition

\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}

we call the class of functions satisfying (1) as HA-convex functions.

If $I\subset \left( 0,\infty \right) $ and $f$ is convex and nondecreasing function then $f$ is HA-convex and if $f$ is HA-convex and nonincreasing function then $f$ is convex.

The following simple but important fact is as follows:

Criterion 1

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 HA-convex on $\left[ a,b\right] $ if and only if $g$ is convex in the usual sense on $\big[ \frac{1}{b},\frac{1}{a}\big] .$

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

\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}

for any convex function $h:\left[ c,d\right] \rightarrow \mathbb {R}$.

For related results, see [ 1 ] – [ 18 ] , [ 21 ] – [ 26 ] , [ 27 ] – [ 37 ] and [ 38 ] – [ 49 ] .

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

\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}

Using the change of variable $s=\frac{1}{t},$ we have

\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*}

and by (4) we get

\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}

The inequality (5) has been obtained in a different manner in [ 41 ] by I. Işcan.

The identric mean $I\left( a,b\right) $ is defined by

\begin{equation*} I\left( a,b\right) :=\tfrac {1}{e}\big( \tfrac {b^{b}}{a^{a}}\big) ^{\frac{1}{b-a}} \end{equation*}

while the logarithmic mean is defined by

\begin{equation*} L\left( a,b\right) :=\tfrac {b-a}{\ln b-\ln a}. \end{equation*}

In the recent paper [ 25 ] we established the following inequalities for HA-convex functions:

Theorem 2

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

\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}

and

Theorem 3

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

\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}

Motivated by the above results, we establish in this paper some new inequalities of Hermite-Hadamard type for HA-convex functions. Some applications for special means are also given.

2 Further Results

We start with the following characterization of HA-convex functions.

Theorem 4

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 $\left[ a,b\right] .$

Proof â–¼

Assume that $f$ is HA-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 $f\left( t\right) =g\big( \frac{1}{t}\big) .$

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

\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*}

which shows that $h$ is convex on $\left[ a,b\right] .$

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

\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*}

which shows that $f$ is HA-convex on the interval $\left[ a,b\right] .$

Proof â–¼

Remark 5

If $f$ is HA-convex on the interval $\left[ a,b\right] ,$ then by Theorem 4 the function $h\left( t\right) =tf\left( t\right) $ is convex on $\left[ a,b\right] $ and by Hermite-Hadamard inequality (3) we get the inequality (7). This gives a direct proof of (7) and it is simpler than in [ 25 ] . â–¡

In 1994, [ 11 ] (see also [ 32 , p. 22 ] ) we proved the following refinement of Hermite-Hadamard inequality. For a direct proof that is different from the one in [ 11 ] , see the recent paper [ 24 ] .

Lemma 6

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

\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}

We can state the following result:

Theorem 7

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

\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}

Follows by Lemma 6 for the convex function $p\left( x\right) =xf\left( x\right) ,$ $x\in \left[ a,b\right] .$

If we take $n=2$ and $x\in \left[ a,b\right] ,$ then by (9) we have

\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}

If in this inequality we choose $x=\frac{a+b}{2},$ then we get the inequality

\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}

If we take in (10) $x=\frac{2ab}{a+b},$ then we get

\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}

We also have:

Theorem 8

\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}

Proof â–¼

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 $\big[ \frac{1}{b},\frac{1}{a}\big] .$

Using the inequality (8) we get

\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}

that is equivalent to

\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}

\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}

By re-indexing the sums and taking into account that

\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*}

we obtain the desired result (13).

Proof â–¼

Remark 9

If we take $n=2$ and $x\in \left[ a,b\right] ,$ then by (13) we have, after appropriate calculations, that

\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}

If we take in (16) $x=\frac{2ab}{a+b}\in \left[ a,b\right] ,$ then we get

\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}

If we take in (16) $x=\frac{a+b}{2}\in \left[ a,b\right] ,$ then we get

\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}

â–¡

3 Related Results

We recall some facts on the lateral derivatives of a convex function.

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.

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

\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}

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

\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*}

In particular, $\varphi $ is a nondecreasing function. If $f$ is differentiable and convex on $\mathring {I}$, then $\partial f=\left\{ f^{\prime }\right\} .$

Lemma 10

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

\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}

for any $s\in \left( a,b\right) $ and $t\in \left[ a,b\right] .$

Also, we have

\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}

and

\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}

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.

Proof â–¼

If $f$ is HA-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

\begin{equation*} h_{\pm }^{\prime }\left( t\right) =f\left( t\right) +tf_{\pm }^{\prime }\left( t\right) \end{equation*}

for any $t\in \left( a,b\right) .$ Also, if $f_{+}^{\prime }\left( a\right) $ and $f_{-}^{\prime }\left( b\right) $ are finite then

\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*}

Writing the gradient inequality for the convex function $h,$ namely

\begin{equation*} h\left( t\right) -h\left( s\right) \geq h_{\pm }^{\prime }\left( s\right) \left( t-s\right) \end{equation*}

for any $s\in \left( a,b\right) $ and $t\in \left[ a,b\right] ,$ we have

\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*}

that is equivalent to

\begin{equation*} tf\left( t\right) -tf\left( s\right) \geq sf_{\pm }^{\prime }\left( s\right) \left( t-s\right) \end{equation*}

for any $s\in \left( a,b\right) $ and $t\in \left[ a,b\right] .$

Now, by dividing with $t{\gt}0$ we get the desired result (20).

The rest follows by the corresponding properties of convex function $h.$

Proof â–¼

We use the following results obtained by the author in [ 19 ] and [ 20 ]

Lemma 11

Let $h:\left[ \alpha ,\beta \right] \rightarrow \mathbb {R}$ be a convex function on $\left[ \alpha ,\beta \right] .$ Then we have the inequalities

\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}

and

\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}

The constant $\frac{1}{8}$ is best possible in 23 and 24.

The following result holds:

Theorem 12

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

\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}

and

\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}

Proof â–¼

Making use of inequality (23) in Lemma 11 for the convex function $h\left( t\right) =tf\left( t\right) $ we have

\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*}

which proves the inequality (25).

The inequality (26) follows by (24).

Proof â–¼

Corollary 13

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

\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}

and

\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}

We remark that from (27) we have

\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}

and from (28) we have

\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}

The identric mean $I\left( a,b\right) $ is defined by

\begin{equation*} I\left( a,b\right) :=\tfrac {1}{e}\Big( \tfrac {b^{b}}{a^{a}}\Big) ^{\frac{1}{b-a}} \end{equation*}

while the logarithmic mean is defined by

\begin{equation*} L\left( a,b\right) :=\tfrac {b-a}{\ln b-\ln a}. \end{equation*}

The following result also holds:

Theorem 14

Let $f:\left[ a,b\right] \subset \left( 0,\infty \right) \rightarrow \mathbb {R}$ be an HA-convex function on the interval $\left[ a,b\right] .$

(i) If $bf\left( b\right) -af\left( a\right) \neq \int _{a}^{b}f\left( s\right) ds$ and

\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}

then

\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}

(ii) If $f\left( b\right) \neq f\left( a\right) $ and

\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}

then

\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}

(iii) If $af\left( b\right) \neq bf\left( a\right) $ and

\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}

then

\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}

Proof â–¼

We know that if $f:\left[ a,b\right] \subset \left( 0,\infty \right) \rightarrow \mathbb {R}$ is an HA-convex function on the interval $\left[ a,b\right] $ then the functions is differentiable except for at most countably many points. Then, from (20) we have

\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}

for any $t\in \left[ a,b\right] $ and almost every $s\in \left( a,b\right) .$

(i) If we take the Lebesgue integral mean in (37), then we get

\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}

for any $t\in \left[ a,b\right] .$

If we take $t=\alpha _{f}$ in (38) then we get the desired inequality (32).

(ii) If we divide the inequality (37) by $s$ then we get

\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}

for any $t\in \left[ a,b\right] $ and almost every $s\in \left( a,b\right) .$

If we take the Lebesgue integral mean in (39), then we get

\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*}

that is equivalent to

\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}

for any $t\in \left[ a,b\right] $

If we take $t=\beta _{f}$ in (40) then we get the desired result (34).

(iii) If we divide the inequality (37) by $s^{2}$ then we get

\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}

for any $t\in \left[ a,b\right] $ and almost every $s\in \left( a,b\right) .$

If we take the Lebesgue integral mean in (41), then we get

\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*}

which is equivalent to

\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*}

or, to

\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*}

Proof â–¼

Remark 15

We observe that a sufficient condition for (31) and (33) 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 (35) also holds.

We also have the following result:

Theorem 16

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

\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}

Proof â–¼

Since the function $h\left( t\right) =tf\left( t\right) $ is convex, then we have

\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*}

for any $x,y\in \left[ a,b\right] .$

If we divide this inequality by $xy{\gt}0$ we get

\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}

for any $x,y\in \left[ a,b\right] .$

If we replace $x$ by $\left( 1-t\right) a+tb$ and $y$ by $ta+\left( 1-t\right) b$ in (43), then we get

\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}

for any $t\in \left[ 0,1\right] .$

Integrating (44) on $\left[ 0,1\right] $ over $t$ we get

\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}

Observe that, by the appropriate change of variable,

\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*}

and

\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*}

and by (45) we get the first inequality in (42).

From the convexity of $h$ we also have

\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*}

and

\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*}

for any $t\in \left[ 0,1\right] .$

Add these inequalities to get

\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*}

for any $t\in \left[ 0,1\right] .$

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

\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}

for any $t\in \left[ 0,1\right] .$

If we integrate the inequality (46) over $t$ on $\left[ 0,1\right] ,$ then we obtain

\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}

Since

\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*}

and

\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*}

then

\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*}

By (47) we then have

\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*}

which proves the second inequality in (42).

Proof â–¼

4 Applications

We consider the arithmetic mean $A\left( a,b\right) =\frac{a+b}{2},$ the geometric mean $G\left( a,b\right)$ = $\sqrt{ab}$ and harmonic mean $H\left( a,b\right) =\frac{2ab}{a+b}$ for the positive numbers $a,b{\gt}0.$

If we use the inequalities (13) for the HA-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

\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}

In particular, we have

\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}

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 HA-concave on $\left( 0,\infty \right) .$

If we write the inequality (11) for the HA-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

\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}

In particular, we have

\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}

The interested reader may apply the above inequalities for other HA-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.

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New Inequalities of Hermite-Hadamard Type for \(HA\)-Convex Functions

1 Introduction

2 Further Results

3 Related Results

4 Applications

Bibliography