1. Introduction
During the investigation of convexity, many researchers founded new classes of functions which are not convex in general. Some of them are the so called harmonic convex functions [
1], harmonic \((\alpha, m)\)-convex functions [
2], harmonic \((s,m)\)-convex functions [
4,
5] and harmonic \((p,(s,m))\)-convex functions [
3]. For a quick glance on importance of these classes and applications, see [
1,
2,
3,
4,
5,
6,
7,
8,
9,
10,
11,
12,
13,
14,
15,
16,
17,
18,
19] and references therein.
Definition1.
A function \(f:I \subseteq \mathbb{R}\backslash \{0\} \rightarrow \mathbb{R}\) is said to be harmonic convex function on \(I\) if
\begin{equation}\label{IE1}
f\left( \frac{xy}{tx+(1-t)y}\right) \leq tf\left( y\right) +\left(
1-t\right) f\left( x\right)
\end{equation}
(1)
holds for all \(x,y\in I\) and \(t\in \left[ 0,1\right] \). If the inequality is reversed, then \(f\) is said to be harmonic concave.
In [
5,
20], Baloch
et al. and Noor
et al. also gave the definition of harmonic log-convex functions as follow:
Definition 2.
A function \(f:I \subseteq \mathbb{R}\backslash \{0\} \rightarrow (0,\infty)\) is said to be harmonic log-convex function on \(I\) if
\begin{equation}\label{IE2}
f\left( \frac{xy}{tx+(1-t)y}\right) \leq [f\left( x\right)]^{1 – t} [f\left( y\right)]^{t}
\end{equation}
(2)
%
holds for all \(x,y\in I\) and \(t\in \left[ 0,1\right] \). If the inequality is reversed, then \(f\) is said to be harmonic log-concave.
In [
20], Noor
et al. proved the following result for harmonic log-convex functions:
Theorem 3.
Let \(I \subseteq \mathbb{R}\backslash \{0\}\) be an interval. If \(f: I \rightarrow (0,\infty)\) is harmonic convex function, then
\begin{equation}\label{IE3}
f \left( \frac{2ab}{a + b} \right) \leq \exp \left[\frac{ab}{b – a} \int_{a}^{b}\log \left(\frac{f(x)}{x^{2}}\right)dx \right] \leq \sqrt{f(a) f(b)}
\end{equation}
(3)
for all \(a, b \in I\) and \( a < b\).
Here, motivated by the above result we study the class of harmonic log-convex functions and present some new inequalities for this class of functions.
2. Main Results
The following result holds.
Theorem 4.
Let \(f:I \subseteq \mathbb{R} \backslash \{0\} \rightarrow (0,\infty)\) be harmonic log-convex function. Then, for every \(t \in [0,1]\), we have
\begin{eqnarray}\label{MI1}
\int_{a}^{b}f(x)dx &\geq& \int_{a}^{b}[f(x)]^{1 – t} \left[\frac{a^{2}b^{2}}{[(a + b)x – ab]^{2}}f\left(\frac{abx}{(a + b)x – ab}\right)\right]^{t}dx \nonumber \\
&\geq& \left\{
\begin{array}{ll}
(1 – 2t) a^{2}b^{2}\int_{\frac{ab}{ta + (1 – t)b}}^{\frac{ab}{(1 – t)a + tb}} \frac{[(a + b)tu – ab]^{2(t – 1)}}{[ab – (1 – t)(a + b)u]^{2t}}f(u)du & \mbox{if } t \neq \frac{1}{2}; \\
\frac{2ab}{a + b}\ln(\frac{b}{a}) f\left( \frac{2ab}{a + b} \right) & \mbox{if } t = \frac{1}{2}.
\end{array}
\right.
\end{eqnarray}
(4)
Proof.
The cases \(t = 0, \frac{1}{2}, 1\) are obvious. Assume that \(t \in (0,1) \backslash \left\{\frac{1}{2}\right\}.\) By the harmonic log-convexity of \(f\) we have
\begin{eqnarray}\label{MI2}
[f(x)]^{1 – t} \left[f\left(\frac{abx}{(a + b)x – ab}\right)\right]^{t} &\geq& f\left( \frac{\frac{abx^{2}}{(a + b)x – ab}}{tx + (1 – t)\frac{abx}{(a + b)x – ab}} \right) f\left( \frac{abx}{(a + b)tx – (2t – 1)ab} \right)
\end{eqnarray}
(5)
for any \(x \in [a,b]\). This allows that
\begin{equation}\label{MI3}
[f(x)]^{1 – t} \left[\frac{a^{2}b^{2}}{[(a + b)x – ab]^{2}}f\left(\frac{abx}{(a + b)x – ab}\right)\right]^{t} \geq \frac{a^{2t}b^{2t}}{[(a + b)x – ab]^{2t}}f\left( \frac{abx}{(a + b)tx – (2t – 1)ab} \right).
\end{equation}
(6)
Integrating the inequality (6) over \(x\) on \([a,b]\), we have
\begin{equation*}
\int_{a}^{b}[f(x)]^{1 – t} \left[\frac{a^{2}b^{2}}{[(a + b)x – ab]^{2}}f\left(\frac{abx}{(a + b)x – ab}\right)\right]^{t}dx \geq \int_{a}^{b}\frac{a^{2t}b^{2t}}{[(a + b)x – ab]^{2t}}f\left( \frac{abx}{(a + b)tx – (2t – 1)ab} \right)dx.
\end{equation*}
Since \(t \neq \frac{1}{2},\) then \(u = \frac{abx}{(a + b)tx – (2t – 1)ab}\) is the change of variable with \( dx = \frac{(1 – 2t) a^{2}b^{2}}{[(a + b)tu – ab]^{2}}du\).
For \(x = a\), we get \(u = \frac{ab}{ta + (1 – t)b}\) and for \(x = b\), we get \(u = \frac{ab}{(1 – t)a + tb}.\) Therefore,
$$
\int_{a}^{b}\frac{a^{2t}b^{2t}}{[(a + b)x – ab]^{2t}}f\left( \frac{abx}{(a + b)tx – (2t – 1)ab} \right)dx= (1 – 2t) a^{2}b^{2}\int_{\frac{ab}{ta + (1 – t)b}}^{\frac{ab}{(1 – t)a + tb}} \frac{[(a + b)tu – ab]^{2(t – 1)}}{[ab – (1 – t)(a + b)u]^{2t}}f(u)du,
$$
and hence the second inequality (4) is proved. By the Holder integral inequality for \(p = \frac{1}{1 – t}\), \(q = \frac{1}{t}\), we have
\begin{eqnarray*}
&&\int_{a}^{b}[f(x)]^{1 – t} \left[\frac{a^{2}b^{2}}{[(a + b)x – ab]^{2}}f\left(\frac{abx}{(a + b)x – ab}\right)\right]^{t}dx\\
&&\leq \left(\int_{a}^{b}\left([f(x)]^{1 – t}\right)^{\frac{1}{1 – t}}dx\right)^{1 – t} \left(\int_{a}^{b}\left(\left[\frac{a^{2}b^{2}}{[(a + b)x – ab]^{2}}f\left(\frac{abx}{(a + b)x – ab}\right)\right]^{t}\right)^{\frac{1}{t}}dx\right)^{t}\\
&&=\left(\int_{a}^{b}f(x)dx\right)^{1 – t} \left(\int_{a}^{b}\frac{a^{2}b^{2}}{[(a + b)x – ab]^{2}}f\left(\frac{abx}{(a + b)x – ab}\right)dx\right)^{t}\\
&&=\left(\int_{a}^{b}f(x)dx\right)^{1 – t} \left(\int_{a}^{b}f(x)dx\right)^{t}=\int_{a}^{b}f(x)dx.\end{eqnarray*}
This proves the first part of inequality (4).
Author Contributions
All authors contributed equally to the writing of this paper. All authors read and approved the
final manuscript.
Competing Interests
The author(s) do not have any competing interests in the manuscript.
