Some Inequalities Related to Jensen-Type Results with Applications

Abstract

The class of harmonic convex functions has acquired a very useful and significant placement among the non-convex functions, since this class not only reinforces some major results of the class of convex functions, but also has supported the development of some remarkable results in analysis where the class of convex functions is silent. Therefore, many researchers have deployed themselves to explore valuable results for this class of non-convex functions. This paper obtains new discrete inequalities for univariate harmonic convex functions on linear spaces related to a Jensen-type and a variant of the Jensen-type results. Our results are refinements of very important recent inequalities presented by Dragomir and Baloch et al. Furthermore, we provide the natural applications of our results.

1. Introduction

Within the last several decades, the study of mathematical inequalities and their applications has advanced rapidly and exponentially, having a major impact on numerous research fields [1,2]. Notably, convexity may be used to obtain various novel notions concerning mathematical inequalities and their applications [3,4,5]. Among these mathematical inequalities, Jensen’s inequality is one of the most important inequalities enabled by convexity [6,7,8].

Many inequalities are direct results of Jensen’s inequality, such as Holder, Hermite–Hadamard, Ky Fan’s, Young’s inequalities, and so on [4,9]. Jensen’s inequality also played an important part in statistics, with several applications including estimations for different divergences [10,11], as well as several estimations for Zipf–Mandelbrot law [12,13] and Shannon entropy [14]. In 2003, A. McD. Mercer [15] gave a variant of Jensen’s inequality, which hugely influences the theory of inequalities. S.S. Dragomir [16] introduced the discrete Jensen-type inequality for harmonic convex functions (HCF), and I.A. Baloch et al. in [17] presented a variant of the discrete Jensen-type inequality for harmonic convex functions. Further, I.A. Baloch et al. furnished many results as a direct application of discrete Jensen-type and a variant of the discrete Jensen-type inequality for harmonic-convex functions, see for example [18,19,20]. Many researchers are working on harmonic-convex functions and trying to produce remarkable results and applications in various fields, see [21,22,23,24,25,26,27,28,29].

Definition 1.

Give a function $f:\mathcal{D}\subseteq \mathbb{R}\setminus \left\{0\right\}\to \mathbb{R}$, it is said to be harmonic convex function on $\mathcal{D}$, if

$$f\left(\frac{uv}{\alpha u+(1-\alpha )v}\right)\le \alpha f\left(v\right)+\left(1-\alpha \right)f\left(u\right)$$

(1)

holds for all $u,v\in \mathcal{D}$ and $\alpha \in \left[0,1\right]$. Furthermore, if the inequality is reversed, then f is called harmonic concave.

In [16], S.S. Dragomir proved the following result, which is known as Jensen-type inequality for harmonic convex functions.

Theorem 1.

Let $\mathcal{D}\subseteq (0,\infty )$ be an interval. If $f:\mathcal{D}\to \mathbb{R}$ is a harmonic convex function, then

$$f\left(\frac{1}{{\sum }_{r=1}^n\frac{{\omega }_r}{u_r}}\right)\le {\displaystyle \sum _{r=1}^n}{\omega }_{r}f\left(u_r\right)$$

(2)

holds for all $u_1,\dots ,u_n\in \mathcal{D}$ and ${\omega }_r\in [0,1]$ with ${\displaystyle \sum _{r=1}^n}{\omega }_r=1$.

In [17], I.A. Baloch et al. proved the subsequent result:

Theorem 2.

Let $[m,M]\subseteq (0,\infty )$ be an interval. If $f:[m,M]\to \mathbb{R}$ is harmonic convex function, then for any finite sequence ${\left(u_r\right)}_{r=1}^n\in [m,M]$ and ${\omega }_r\in [0,1]$ with ${\sum }_{r=1}^n{\omega }_r=1$, we have

$$f\left(\frac{1}{\frac{1}{m}+\frac{1}{M}-{\sum }_{r=1}^n\frac{{\omega }_r}{u_r}}\right)\le f\left(m\right)+f\left(M\right)-{\displaystyle \sum _{r=1}^n}{\omega }_{r}f\left(u_r\right).$$

(3)

In [30], I.A. Baloch et al. proved the subsequent result:

Theorem 3.

If $\mathsf{\Phi }:\mathcal{D}\to \mathbb{R}$ is harmonic convex function on harmonic convex subset $\mathcal{D}\subseteq \mathbb{R}\setminus \left\{0\right\}$, then for any finite positive sequence ${\left\{u_r\right\}}_{r=1}^n\in \mathcal{D}$ and ${\omega }_r$ with $W_n:={\sum }_{r=1}^n{\omega }_r>0$, we have

$$\begin{array}{cc}\hfill n& \underset{1\le r\le n}{min}\left\{{\omega }_r\right\}\left[\frac{1}{n}{\displaystyle \sum _{r=1}^n}\mathsf{\Phi }\left(u_r\right)-\mathsf{\Phi }\left(\frac{1}{\frac{1}{n}{\sum }_{r=1}^n\frac{1}{u_r}}\right)\right]\hfill \\ \hfill & \le \frac{1}{W_n}{\displaystyle \sum _{r=1}^n}{\omega }_r\mathsf{\Phi }\left(u_r\right)-\mathsf{\Phi }\left(\frac{1}{\frac{1}{W_n}{\sum }_{r=1}^n\frac{{\omega }_r}{u_r}}\right)\hfill \\ \hfill & \le n\underset{1\le r\le n}{max}\left\{{\omega }_r\right\}\left[\frac{1}{n}{\displaystyle \sum _{r=1}^n}\mathsf{\Phi }\left(u_r\right)-\mathsf{\Phi }\left(\frac{1}{\frac{1}{n}{\sum }_{r=1}^n\frac{1}{u_r}}\right)\right].\hfill \end{array}$$

(4)

This paper is organized as follows: we firstly derive the refinements of Jensen-type and a variant of Jensen-type. Next, we further refine our presented refinements and point out more or less direct consequences of our results. Secondly, we derive refinements of the well-known weighted HGA and an infinite version of weighted HGA as applications. Finally, we discuss the importance, source of motivation and significance of this research for the class of harmonic convex functions.

2. Main Results

In this section, we firstly establish a refinement of the Jensen-type inequality presented in Theorem 1.

Theorem 4.

Let $\mathcal{D}\subseteq \mathbb{R}\setminus \left\{0\right\}$ be an interval. If $f:\mathcal{D}\to \mathbb{R}$ is harmonic convex function, then for any finite positive sequence ${\left\{u_r\right\}}_{r=1}^n\in \mathcal{D}$ and ${\omega }_r\ge 0$ with ${\sum }_{r=1}^n{\omega }_r=1$, we have

$$\begin{array}{cc}\hfill f\left(\frac{1}{{\displaystyle \sum _{s=1}^n}\frac{{\omega }_s}{u_s}}\right)& \le \underset{1\le r\le n}{min}\left[(1-{\omega }_r)f\left(\frac{1-{\omega }_r}{{\sum }_{s=1}^n\frac{{\omega }_s}{u_s}-\frac{{\omega }_r}{u_r}}\right)+{\omega }_{r}f\left(u_r\right)\right]\hfill \\ \hfill & \le \frac{1}{n}\left[{\displaystyle \sum _{r=1}^n}(1-{\omega }_r)f\left(\frac{1-{\omega }_r}{{\sum }_{s=1}^n\frac{{\omega }_s}{u_s}-\frac{{\omega }_r}{u_r}}\right)+{\displaystyle \sum _{r=1}^n}{\omega }_{r}f\left(u_r\right)\right]\hfill \\ \hfill & \le \underset{1\le r\le n}{max}\left[(1-{\omega }_r)f\left(\frac{1-{\omega }_r}{{\sum }_{s=1}^n\frac{{\omega }_s}{u_s}-\frac{{\omega }_r}{u_r}}\right)+{\omega }_{r}f\left(u_r\right)\right]\hfill \\ \hfill & \le {\displaystyle \sum _{s=1}^n}{\omega }_{s}f\left(u_s\right).\hfill \end{array}$$

(5)

In particular

$$\begin{array}{cc}\hfill f\left(\frac{n}{{\sum }_{s=1}^n\frac{1}{u_s}}\right)& \le \frac{1}{n}\underset{1\le r\le n}{min}\left[(n-1)f\left(\frac{n-1}{{\displaystyle \sum _{s=1}^n}\frac{1}{u_s}-\frac{1}{u_r}}\right)+f\left(u_r\right)\right]\hfill \\ \hfill & \le \frac{1}{n^2}\left[{\displaystyle \sum _{r=1}^n}(n-1)f\left(\frac{n-1}{{\sum }_{s=1}^n\frac{1}{u_s}-\frac{1}{u_r}}\right)+{\displaystyle \sum _{r=1}^n}f\left(u_r\right)\right]\hfill \\ \hfill & \le \frac{1}{n}\underset{1\le r\le n}{max}\left[(n-1)f\left(\frac{n-1}{{\sum }_{s=1}^n\frac{1}{u_s}-\frac{1}{u_r}}\right)+f\left(u_r\right)\right]\hfill \\ \hfill & \le \frac{1}{n}{\displaystyle \sum _{s=1}^n}f\left(u_s\right).\hfill \end{array}$$

(6)

Proof. 

For any $r\in \{1,\dots ,n\}$, we have

$${\displaystyle \sum _{s=1}^n}\frac{{\omega }_s}{u_s}-\frac{{\omega }_r}{u_r}={\displaystyle \sum _{\begin{array}{c}s=1\\ s\ne r\end{array}}^n}\frac{{\omega }_s}{u_s}={\displaystyle \sum _{\begin{array}{c}s=1\\ s\ne r\end{array}}^n}{\omega }_s\frac{1}{{\displaystyle \sum _{\begin{array}{c}s=1\\ s\ne r\end{array}}^n}{\omega }_s}{\displaystyle \sum _{\begin{array}{c}s=1\\ s\ne r\end{array}}^n}\frac{{\omega }_s}{u_s}=(1-{\omega }_r)\frac{1}{{\displaystyle \sum _{\begin{array}{c}s=1\\ s\ne r\end{array}}^n}{\omega }_s}{\displaystyle \sum _{\begin{array}{c}s=1\\ s\ne r\end{array}}^n}\frac{{\omega }_s}{u_s},$$

which implies that

$$\frac{{\displaystyle \sum _{s=1}^n}\frac{{\omega }_s}{u_s}-\frac{{\omega }_r}{u_r}}{1-{\omega }_r}=\frac{1}{{\displaystyle \sum _{\begin{array}{c}s=1\\ s\ne r\end{array}}^n}{\omega }_s}{\displaystyle \sum _{\begin{array}{c}s=1\\ s\ne r\end{array}}^n}\frac{{\omega }_s}{u_s}\in D$$

(7)

for each $r\in \{1,\dots ,n\}$, since right side of (7) is a reciprocal of harmonic convex combinations of elements $u_s\in D,s\in \{1,\dots ,n\}\setminus \left\{r\right\}.$ Taking the function f on reciprocal of (7) and applying Jensen-type inequality (2), we get successively

$$\begin{array}{cc}\hfill f\left(\frac{1-{\omega }_r}{{\displaystyle \sum _{s=1}^n}\frac{{\omega }_s}{u_s}-\frac{{\omega }_r}{u_r}}\right)& =f\left(\frac{1}{\frac{1}{{\displaystyle \sum _{\begin{array}{c}s=1\\ s\ne r\end{array}}^n}{\omega }_s}{\displaystyle \sum _{\begin{array}{c}s=1\\ s\ne r\end{array}}^n}\frac{{\omega }_s}{u_s}}\right)\le \frac{1}{{\displaystyle \sum _{\begin{array}{c}s=1\\ s\ne r\end{array}}^n}{\omega }_s}{\displaystyle \sum _{\begin{array}{c}s=1\\ s\ne r\end{array}}^n}{\omega }_{s}f\left(u_s\right)\hfill \\ \hfill & =\frac{1}{1-{\omega }_r}\left[{\displaystyle \sum _{s=1}^n}{\omega }_{s}f\left(u_s\right)-{\omega }_{r}f\left(u_r\right)\right]\hfill \end{array}$$

for any $r\in \{1,\dots ,n\},$ which implies

$$(1-{\omega }_r)f\left(\frac{1-{\omega }_r}{{\displaystyle \sum _{s=1}^n}\frac{{\omega }_s}{u_s}-\frac{{\omega }_r}{u_r}}\right)+{\omega }_{r}f\left(u_r\right)\le {\displaystyle \sum _{s=1}^n}{\omega }_{s}f\left(u_s\right)$$

(8)

for each $r\in \{1,\dots ,n\}.$

Utilizing the harmonic convexity of f, we also have

$$(1-{\omega }_r)f\left(\frac{1-{\omega }_r}{{\displaystyle \sum _{s=1}^n}\frac{{\omega }_s}{u_s}-\frac{{\omega }_r}{u_r}}\right)+{\omega }_{r}f\left(u_r\right)\ge f\left[\frac{1}{(1-{\omega }_r)\frac{{\displaystyle \sum _{s=1}^n}\frac{{\omega }_s}{u_s}-\frac{{\omega }_r}{u_r}}{1-{\omega }_r}+\frac{{\omega }_r}{u_r}}\right]=f\left(\frac{1}{{\displaystyle \sum _{s=1}^n}\frac{{\omega }_s}{u_s}}\right)$$

(9)

for each $r\in \{1,\dots ,n\}.$

Taking the minimum over r in (9) and utilizing the fact that

$$\underset{r\in \{1,\dots ,n\}}{min}{a}_r\le \frac{1}{n}{\displaystyle \sum _{r=1}^n}{a}_r\le \underset{r\in \{1,\dots ,n\}}{max}{a}_r$$

and then taking maximum over r in (8), we deduce our result (5). □

Theorem 5.

Let $f:(0,\infty )\to \mathbb{R}$ be a harmonic convex function and $u,v\in (0,\infty ).$ If $u_r:=\frac{uv}{(1-{\alpha }_r)v+{\alpha }_{r}u}$ with ${\alpha }_r\in [0,1]$ and ${\omega }_r\ge 0,r\in \{1,\dots ,n\}$ with ${\displaystyle \sum _{r=1}^n}{\omega }_r=1,$ then

$$\begin{array}{cc}\hfill f\left(\frac{1}{\frac{1}{u}+\frac{1}{v}-{\sum }_{r=1}^n\frac{{\omega }_r}{u_r}}\right)& \le {\displaystyle \sum _{r=1}^n}{\omega }_{r}f\left(\frac{1}{\frac{1}{u}+\frac{1}{v}-\frac{1}{u_r}}\right)\hfill \\ \hfill & \le \left({\displaystyle \sum _{r=1}^n}{\omega }_r{\alpha }_r\right)f\left(u\right)+\left(1-{\displaystyle \sum _{r=1}^n}{\omega }_r{\alpha }_r\right)f\left(v\right)\hfill \\ \hfill & \le f\left(u\right)+f\left(v\right)-{\displaystyle \sum _{r=1}^n}{\omega }_{r}f\left(u_r\right).\hfill \end{array}$$

(10)

Proof. 

By Jensen-type discrete inequality (2), we have

$$\begin{array}{cc}\hfill f\left(\frac{1}{\frac{1}{u}+\frac{1}{v}-{\sum }_{r=1}^n\frac{{\omega }_r}{u_r}}\right)& =f\left(\frac{1}{{\sum }_{r=1}^n{\omega }_r\left(\frac{1}{u}+\frac{1}{v}-\frac{1}{u_r}\right)}\right)\le {\displaystyle \sum _{r=1}^n}{\omega }_{r}f\left(\frac{1}{\frac{1}{u}+\frac{1}{v}-\frac{1}{u_r}}\right)\hfill \\ \hfill & ={\displaystyle \sum _{r=1}^n}{\omega }_{r}f\left(\frac{1}{\frac{1}{u}+\frac{1}{v}-\frac{(1-{\alpha }_r)v+{\alpha }_{r}u}{uv}}\right)\hfill \\ \hfill & ={\displaystyle \sum _{r=1}^n}{\omega }_{r}f\left(\frac{uv}{(1-{\alpha }_r)u+{\alpha }_{r}v}\right)\hfill \end{array}$$

which proves the first inequality in (10).

By utilizing harmonic convexity of f, we derive

$$\begin{array}{cc}\hfill {\displaystyle \sum _{r=1}^n}{\omega }_{r}f\left(\frac{uv}{(1-{\alpha }_r)u+{\alpha }_{r}v}\right)& \le {\displaystyle \sum _{r=1}^n}{\omega }_r[{\alpha }_{r}f\left(u\right)+(1-{\alpha }_r)f\left(v\right)]\hfill \\ \hfill & =\left({\displaystyle \sum _{r=1}^n}{\omega }_r{\alpha }_r\right)f\left(u\right)+\left(1-{\displaystyle \sum _{r=1}^n}{\omega }_r{\alpha }_r\right)f\left(v\right)\hfill \end{array}$$

which proves the second inequality in (10).

Now, again by harmonic convexity of f, we also have

$$f\left(\frac{uv}{(1-{\alpha }_r)v+{\alpha }_{r}u}\right)\le (1-{\alpha }_r)f\left(u\right)+{\alpha }_{r}f\left(v\right)$$

for $r\in \{1,\dots ,n\}.$ Multiply this inequality by ${\omega }_r\ge 0,r\in \{1,\dots ,n\}$ and sum over r from 1 to n, then we deduce

$$\begin{array}{cc}\hfill {\displaystyle \sum _{r=1}^n}{\omega }_{r}f\left(u_r\right)& \le {\displaystyle \sum _{r=1}^n}{\omega }_r[(1-{\alpha }_r)f\left(u\right)+{\alpha }_{r}f\left(v\right)]\hfill \\ \hfill & =\left(1-{\displaystyle \sum _{r=1}^n}{\omega }_r{\alpha }_r\right)f\left(u\right)+\left({\displaystyle \sum _{r=1}^n}{\omega }_r{\alpha }_r\right)f\left(v\right).\hfill \end{array}$$

(11)

Therefore, by (11)

$$\begin{array}{cc}\hfill & \left({\displaystyle \sum _{r=1}^n}{\omega }_r{\alpha }_r\right)f\left(u\right)+\left(1-{\displaystyle \sum _{r=1}^n}{\omega }_r{\alpha }_r\right)f\left(v\right)\hfill \\ \hfill & =f\left(u\right)+f\left(v\right)-\left[\left(1-{\displaystyle \sum _{r=1}^n}{\omega }_r{\alpha }_r\right)f\left(u\right)+\left({\displaystyle \sum _{r=1}^n}{\omega }_r{\alpha }_r\right)f\left(v\right)\right]\hfill \\ \hfill & \le f\left(u\right)+f\left(v\right)-{\displaystyle \sum _{r=1}^n}{\omega }_{r}f\left(u_r\right),\hfill \end{array}$$

which proves the last part of inequality in (10). □

Remark 1.

We observe that the inequality (10) is equivalent to the inequality

$$\begin{array}{cc}\hfill {\displaystyle \sum _{r=1}^n}{\omega }_{r}f\left(u_r\right)& \le \left[\left(1-{\displaystyle \sum _{r=1}^n}{\omega }_r{\alpha }_r\right)f\left(u\right)\right]+\left({\displaystyle \sum _{r=1}^n}{\omega }_r{\alpha }_r\right)f\left(v\right)\hfill \\ \hfill & \le f\left(u\right)+f\left(v\right)-{\displaystyle \sum _{r=1}^n}{\omega }_{r}f\left(\frac{1}{\frac{1}{u}+\frac{1}{v}-\frac{1}{u_r}}\right)\hfill \\ \hfill & \le f\left(u\right)+f\left(v\right)-f\left(\frac{1}{\frac{1}{u}+\frac{1}{v}-{\sum }_{r=1}^n\frac{{\omega }_r}{u_r}}\right),\hfill \end{array}$$

(12)

where $f:(0,\infty )\to \mathbb{R}$ is a harmonic convex function, $u,v\in (0,\infty ),u_r=\frac{uv}{(1-{\alpha }_r)v+{\alpha }_{r}u}$ with ${\alpha }_r\in 0,1$ and ${\omega }_r\ge 0,$ $r\in \{1,\dots ,n\}$ with ${\sum }_{r=1}^n{\omega }_r=1$.

Since the distance between the extreme terms is greater than the distance between internal ones, we can state the following corollary as well:

Corollary 1.

Under the assumptions of Theorem 5, we have

$$\begin{array}{cc}\hfill 0& \le \left({\displaystyle \sum _{r=1}^n}{\omega }_r{\alpha }_r\right)f\left(u\right)+\left(1-{\displaystyle \sum _{r=1}^n}{\omega }_r{\alpha }_r\right)f\left(v\right)-{\displaystyle \sum _{r=1}^n}{\omega }_{r}f\left(\frac{1}{\frac{1}{u}+\frac{1}{v}-\frac{1}{u_r}}\right)\hfill \\ \hfill & \le f\left(u\right)+f\left(v\right)-f\left(\frac{1}{\frac{1}{u}+\frac{1}{v}-{\sum }_{r=1}^n\frac{{\omega }_r}{u_r}}\right)-{\displaystyle \sum _{r=1}^n}{\omega }_{r}f\left(u_r\right).\hfill \end{array}$$

(13)

Theorem 6.

Under the assumptions of Theorem 5, we also have

$$\begin{array}{cc}\hfill & f\left(\frac{1}{\frac{1}{u}+\frac{1}{v}-{\sum }_{r=1}^n\frac{{\omega }_r}{u_r}}\right)\hfill \\ \hfill & \le \underset{1\le r\le n}{min}\left\{{\omega }_r\right\}\left[{\displaystyle \sum _{r=1}^n}f\left(\frac{1}{\frac{1}{u}+\frac{1}{v}-\frac{1}{u_r}}\right)-nf\left(\frac{1}{\frac{1}{u}+\frac{1}{v}-\frac{1}{n}{\sum }_{r=1}^n\frac{1}{u_r}}\right)\right]\hfill \\ \hfill & +f\left(\frac{1}{\frac{1}{u}+\frac{1}{v}-{\sum }_{r=1}^n\frac{{\omega }_r}{u_r}}\right)\hfill \\ \hfill & \le {\displaystyle \sum _{r=1}^n}{\omega }_{r}f\left(\frac{1}{\frac{1}{u}+\frac{1}{v}-\frac{1}{u_r}}\right)\le f\left(u\right)+f\left(v\right)-{\displaystyle \sum _{r=1}^n}{\omega }_{r}f\left(u_r\right).\hfill \end{array}$$

(14)

Proof. 

By the harmonic convexity of f, we have (see [17])

$$f\left(\frac{1}{\frac{1}{u}+\frac{1}{v}-\frac{1}{u_r}}\right)+f\left(u_r\right)\le f\left(u\right)+f\left(v\right)$$

for $r\in \{1,\dots ,n\}$.

If we multiply this inequality by ${\omega }_r\ge 0$ and sum over 1 to n, then we get

$${\displaystyle \sum _{r=1}^n}{\omega }_{r}f\left(\frac{1}{\frac{1}{u}+\frac{1}{v}-\frac{1}{u_r}}\right)\le f\left(u\right)+f\left(v\right)-{\displaystyle \sum _{r=1}^n}{\omega }_{r}f\left(u_r\right).$$

(15)

If we apply the first inequality in (4) for harmonic convex function $\mathsf{\Phi }\left(t\right)=f\left(\frac{1}{\frac{1}{u}+\frac{1}{v}-\frac{1}{t}}\right)$ for $t\in [0,1]$, we have

$$\begin{array}{cc}\hfill 0& \le \underset{1\le r\le n}{min}\left\{{\omega }_r\right\}\left[{\displaystyle \sum _{r=1}^n}f\left(\frac{1}{\frac{1}{u}+\frac{1}{v}-\frac{1}{u_r}}\right)-nf\left(\frac{1}{\frac{1}{u}+\frac{1}{v}-\frac{1}{n}{\sum }_{r=1}^n\frac{1}{u_r}}\right)\right]\hfill \\ \hfill & \le {\displaystyle \sum _{r=1}^n}{\omega }_{r}f\left(\frac{1}{\frac{1}{u}+\frac{1}{v}-\frac{1}{u_r}}\right)-f\left(\frac{1}{\frac{1}{u}+\frac{1}{v}-{\sum }_{r=1}^n\frac{{\omega }_r}{u_r}}\right).\hfill \end{array}$$

(16)

By making the use of (15) and (16), we get the desired result (14). □

We also have: The proof follows by Theorem 6 observing that the difference between the extreme terms is more significant than the difference between the internal ones.

Corollary 2.

With the assumptions of Theorem 5, we have

$$\begin{array}{cc}\hfill 0& \le {\displaystyle \sum _{r=1}^n}{\omega }_{r}f\left(\frac{1}{\frac{1}{u}+\frac{1}{v}-\frac{1}{u_r}}\right)-f\left(\frac{1}{\frac{1}{u}+\frac{1}{v}-{\sum }_{r=1}^n\frac{{\omega }_r}{u_r}}\right)\hfill \\ \hfill & -\underset{1\le r\le n}{min}\left\{{\omega }_r\right\}\left[{\displaystyle \sum _{r=1}^n}f\left(\frac{1}{\frac{1}{u}+\frac{1}{v}-\frac{1}{u_r}}\right)-nf\left(\frac{1}{\frac{1}{u}+\frac{1}{v}-\frac{1}{n}{\sum }_{r=1}^n\frac{1}{u_r}}\right)\right]\hfill \\ \hfill & \le f\left(u\right)+f\left(v\right)-{\displaystyle \sum _{r=1}^n}{\omega }_{r}f\left(u_r\right)-f\left(\frac{1}{\frac{1}{u}+\frac{1}{v}-{\sum }_{r=1}^n\frac{{\omega }_r}{u_r}}\right).\hfill \end{array}$$

(17)

Theorem 7.

Let $f:(0,\infty )\to \mathbb{R}$ be a harmonic convex function and $u_r\in (0,\infty ),r\in \{1,\dots ,n\}$ and $a,b\in \mathbb{R}\setminus \left\{0\right\}$ with $\frac{a{u}_r}{u_r-a},\frac{b{u}_r}{u_r-b}\in (0,\infty )$ for $r\in \{1,\dots ,n\}$, then for all $\lambda \in [0,1]$ and ${\omega }_r\ge 0$ for $r\in \{1,\dots ,n\}$ with ${\displaystyle \sum _{r=1}^n}{\omega }_r=1,$ we have

$$\begin{array}{cc}\hfill f\left(\frac{1}{\frac{1-\lambda }{a}+\frac{\lambda }{b}-{\sum }_{r=1}^n\frac{{\omega }_r}{u_r}}\right)& \le {\displaystyle \sum _{r=1}^n}{\omega }_{r}f\left(\frac{1}{\frac{1-\lambda }{a}+\frac{\lambda }{b}-\frac{1}{u_r}}\right)\hfill \\ \hfill & \le (1-\lambda ){\displaystyle \sum _{r=1}^n}{\omega }_{r}f\left(\frac{a{u}_r}{u_r-a}\right)+\lambda {\displaystyle \sum _{r=1}^n}{\omega }_{r}f\left(\frac{b{u}_r}{u_r-b}\right)\hfill \\ \hfill & \le {\displaystyle \sum _{r=1}^n}{\omega }_{r}f\left(\frac{a{u}_r}{u_r-a}\right)+{\displaystyle \sum _{r=1}^n}{\omega }_{r}f\left(\frac{b{u}_r}{u_r-b}\right)\hfill \\ \hfill & -{\displaystyle \sum _{r=1}^n}{\omega }_{r}f\left(\frac{1}{\frac{1-\lambda }{b}+\frac{\lambda }{a}-\frac{1}{u_r}}\right)\hfill \end{array}$$

(18)

and

$$\begin{array}{cc}\hfill 0& \le (1-\lambda ){\displaystyle \sum _{r=1}^n}{\omega }_{r}f\left(\frac{a{u}_r}{u_r-a}\right)+\lambda {\displaystyle \sum _{r=1}^n}{\omega }_{r}f\left(\frac{b{u}_r}{u_r-b}\right)-{\displaystyle \sum _{r=1}^n}{\omega }_{r}f\left(\frac{1}{\frac{1-\lambda }{a}+\frac{\lambda }{b}-\frac{1}{u_r}}\right)\hfill \\ \hfill & \le {\displaystyle \sum _{r=1}^n}{\omega }_{r}f\left(\frac{a{u}_r}{u_r-a}\right)+{\displaystyle \sum _{r=1}^n}{\omega }_{r}f\left(\frac{b{u}_r}{u_r-b}\right)-{\displaystyle \sum _{r=1}^n}{\omega }_{r}f\left(\frac{1}{\frac{1-\lambda }{b}+\frac{\lambda }{a}-\frac{1}{u_r}}\right)\hfill \\ \hfill & -f\left(\frac{1}{\frac{1-\lambda }{a}+\frac{\lambda }{b}-{\sum }_{r=1}^n\frac{{\omega }_r}{u_r}}\right).\hfill \end{array}$$

(19)

Proof. 

By the Jensen’s type inequality (2), we have

$$\begin{array}{cc}\hfill f\left(\frac{1}{\frac{1-\lambda }{a}+\frac{\lambda }{b}-{\sum }_{r=1}^n\frac{{\omega }_r}{u_r}}\right)& =f\left(\frac{1}{{\sum }_{r=1}^n{\omega }_r[\frac{1-\lambda }{a}+\frac{\lambda }{b}-\frac{1}{u_r}]}\right)\hfill \\ \hfill & =f\left(\frac{1}{{\sum }_{r=1}^n{\omega }_r[(1-\lambda )(\frac{1}{a}-\frac{1}{u_r})+\lambda (\frac{1}{a}-\frac{1}{u_r})]}\right)\hfill \\ \hfill & \le {\displaystyle \sum _{r=1}^n}{\omega }_{r}f\left(\frac{1}{(1-\lambda )(\frac{1}{a}-\frac{1}{u_r})+\lambda (\frac{1}{a}-\frac{1}{u_r})}\right)\hfill \\ \hfill & ={\displaystyle \sum _{r=1}^n}{\omega }_{r}f\left(\frac{1}{\frac{1-\lambda }{a}+\frac{\lambda }{b}-\frac{1}{u_r}}\right),\hfill \end{array}$$

which proves the first inequality in (18).

By the harmonic convexity of f

$$\begin{array}{cc}\hfill f\left(\frac{1}{\frac{1-\lambda }{a}+\frac{\lambda }{b}-\frac{1}{u_r}}\right)& =f\left(\frac{1}{(1-\lambda )(\frac{1}{a}-\frac{1}{u_r})+\lambda (\frac{1}{b}-\frac{1}{u_r})}\right)\hfill \\ \hfill & \le (1-\lambda )f\left(\frac{1}{\frac{1}{a}-\frac{1}{u_r}}\right)+\lambda f\left(\frac{1}{\frac{1}{b}-\frac{1}{u_r}}\right)\hfill \\ \hfill & =(1-\lambda )f\left(\frac{a{u}_r}{u_r-a}\right)+\lambda f\left(\frac{b{u}_r}{u_r-b}\right)\hfill \\ \hfill & =f\left(\frac{a{u}_r}{u_r-a}\right)+f\left(\frac{b{u}_r}{u_r-b}\right)-\left[\lambda f\left(\frac{a{u}_r}{u_r-a}\right)+(1-\lambda )f\left(\frac{b{u}_r}{u_r-b}\right)\right]\hfill \end{array}$$

for $r\in \{1,\dots ,n\}$.

If we multiply this inequality by ${\omega }_r\ge 0$ and sum over r from 1 to n, then

$$\begin{array}{cc}\hfill & {\displaystyle \sum _{r=1}^n}{\omega }_{r}f\left(\frac{1}{\frac{1-\lambda }{a}+\frac{\lambda }{b}-\frac{1}{u_r}}\right)\hfill \\ \hfill & \le (1-\lambda ){\displaystyle \sum _{r=1}^n}{\omega }_{r}f\left(\frac{a{u}_r}{u_r-a}\right)+\lambda {\displaystyle \sum _{r=1}^n}{\omega }_{r}f\left(\frac{b{u}_r}{u_r-b}\right)\hfill \\ \hfill & \le {\displaystyle \sum _{r=1}^n}{\omega }_{r}f\left(\frac{a{u}_r}{u_r-a}\right)+{\displaystyle \sum _{r=1}^n}{\omega }_{r}f\left(\frac{b{u}_r}{u_r-b}\right)-{\displaystyle \sum _{r=1}^n}{\omega }_r\left[\lambda f\left(\frac{a{u}_r}{u_r-a}\right)+(1-\lambda )f\left(\frac{b{u}_r}{u_r-b}\right)\right],\hfill \end{array}$$

which proves the second inequality in (18).

By the harmonic convexity of f, we also have

$$\begin{array}{cc}\hfill \lambda f\left(\frac{a{u}_r}{u_r-a}\right)+(1-\lambda )f\left(\frac{b{u}_r}{u_r-b}\right)& \ge f\left(\frac{1}{(1-\lambda )(\frac{1}{b}-\frac{1}{u_r})+\lambda (\frac{1}{a}-\frac{1}{u_r})}\right)\hfill \\ \hfill & =f\left(\frac{1}{\frac{1-\lambda }{b}+\frac{\lambda }{a}-\frac{1}{u_r}}\right)\hfill \end{array}$$

for $r\in \{1,\dots ,n\}$.

If we multiply this inequality by ${\omega }_r\ge 0$ and sum over r from 1 to n, then

$${\displaystyle \sum _{r=1}^n}{\omega }_r[\lambda f\left(\frac{a{u}_r}{u_r-a}\right)+(1-\lambda )f\left(\frac{b{u}_r}{u_r-b}\right)]\ge {\displaystyle \sum _{r=1}^n}{\omega }_{r}f\left(\frac{1}{\frac{1-\lambda }{b}+\frac{\lambda }{a}-\frac{1}{u_r}}\right)$$

and

$$\begin{array}{cc}\hfill & {\displaystyle \sum _{r=1}^n}{\omega }_{r}f\left(\frac{a{u}_r}{u_r-a}\right)+{\displaystyle \sum _{r=1}^n}{\omega }_{r}f\left(\frac{b{u}_r}{u_r-b}\right)-{\displaystyle \sum _{r=1}^n}{\omega }_r\left[\lambda f\left(\frac{a{u}_r}{u_r-a}\right)+(1-\lambda )f\left(\frac{b{u}_r}{u_r-b}\right)\right]\hfill \\ \hfill & \le {\displaystyle \sum _{r=1}^n}{\omega }_{r}f\left(\frac{a{u}_r}{u_r-a}\right)+{\displaystyle \sum _{r=1}^n}{\omega }_{r}f\left(\frac{b{u}_r}{u_r-b}\right)-{\displaystyle \sum _{r=1}^n}{\omega }_{r}f\left(\frac{1}{\frac{1-\lambda }{b}+\frac{\lambda }{a}-\frac{1}{u_r}}\right),\hfill \end{array}$$

which proves the last inequality in (18).

The inequality (19) follows from the fact that distance between extreme terms is greater then the distance between the internal ones. □

Corollary 3.

Let $f:(0,\infty )\to \mathbb{R}$ be a harmonic convex function and $u_r\in (0,\infty ),r\in \{1,\dots ,n\}$ and $a,b\in \mathbb{R}\setminus \left\{0\right\}$ with $\frac{a{u}_r}{u_r-a},\frac{b{u}_r}{u_r-b}\in (0,\infty )$ for $r\in \{1,\dots ,n\}$, then for all $\lambda \in [0,1]$ and ${\omega }_r\ge 0$ for $r\in \{1,\dots ,n\}$ with ${\displaystyle \sum _{r=1}^n}{\omega }_r=1,$ we have

$$\begin{array}{cc}\hfill f\left(\frac{1}{\frac{a+b}{2ab}-{\sum }_{r=1}^n\frac{{\omega }_r}{u_r}}\right)& \le {\displaystyle \sum _{r=1}^n}{\omega }_{r}f\left(\frac{1}{\frac{a+b}{2ab}-\frac{1}{u_r}}\right)\hfill \\ \hfill & \le \frac{1}{2}{\displaystyle \sum _{r=1}^n}{\omega }_{r}f\left(\frac{a{u}_r}{u_r-a}\right)+\frac{1}{2}{\displaystyle \sum _{r=1}^n}{\omega }_{r}f\left(\frac{b{u}_r}{u_r-b}\right)\hfill \\ \hfill & \le {\displaystyle \sum _{r=1}^n}{\omega }_{r}f\left(\frac{a{u}_r}{u_r-a}\right)+{\displaystyle \sum _{r=1}^n}{\omega }_{r}f\left(\frac{b{u}_r}{u_r-b}\right)\hfill \\ \hfill & -{\displaystyle \sum _{r=1}^n}{\omega }_{r}f\left(\frac{1}{\frac{a+b}{2ab}-\frac{1}{u_r}}\right)\hfill \end{array}$$

(20)

and

$$\begin{array}{cc}\hfill 0& \le \frac{1}{2}{\displaystyle \sum _{r=1}^n}{\omega }_{r}f\left(\frac{a{u}_r}{u_r-a}\right)+\frac{1}{2}{\displaystyle \sum _{r=1}^n}{\omega }_{r}f\left(\frac{b{u}_r}{u_r-b}\right)-{\displaystyle \sum _{r=1}^n}{\omega }_{r}f\left(\frac{1}{\frac{a+b}{2ab}-\frac{1}{u_r}}\right)\hfill \\ \hfill & \le {\displaystyle \sum _{r=1}^n}{\omega }_{r}f\left(\frac{a{u}_r}{u_r-a}\right)+{\displaystyle \sum _{r=1}^n}{\omega }_{r}f\left(\frac{b{u}_r}{u_r-b}\right)-{\displaystyle \sum _{r=1}^n}{\omega }_{r}f\left(\frac{1}{\frac{a+b}{2ab}-\frac{1}{u_r}}\right)\hfill \\ \hfill & -f\left(\frac{1}{\frac{a+b}{2ab}-{\sum }_{r=1}^n\frac{{\omega }_r}{u_r}}\right).\hfill \end{array}$$

(21)

Now, we give the improvement of (18) as follows:

Corollary 4.

Under assumptions of Theorem 7, we have

$$\begin{array}{cc}\hfill & f\left(\frac{1}{\frac{a+b}{2ab}-{\sum }_{r=1}^n\frac{{\omega }_r}{u_r}}\right)\hfill \\ \hfill & \le \frac{1}{2}\left[f\left(\frac{1}{\frac{\lambda }{a}+\frac{1-\lambda }{b}-{\sum }_{r=1}^n\frac{{\omega }_r}{u_r}}\right)+f\left(\frac{1}{\frac{1-\lambda }{a}+\frac{\lambda }{b}-{\sum }_{r=1}^n\frac{{\omega }_r}{u_r}}\right)\right]\hfill \\ \hfill & \le \frac{1}{2}\left[{\displaystyle \sum _{r=1}^n}{\omega }_{r}f\left(\frac{1}{\frac{1-\lambda }{a}+\frac{\lambda }{b}-\frac{1}{u_r}}\right)+{\displaystyle \sum _{r=1}^n}{\omega }_{r}f\left(\frac{1}{\frac{\lambda }{a}+\frac{1-\lambda }{b}-\frac{1}{u_r}}\right)\right]\hfill \\ \hfill & \le \frac{1}{2}\left[{\displaystyle \sum _{r=1}^n}{\omega }_{r}f\left(\frac{a{u}_r}{u_r-a}\right)+{\displaystyle \sum _{r=1}^n}{\omega }_{r}f\left(\frac{b{u}_r}{u_r-b}\right)\right]\hfill \\ \hfill & \le {\displaystyle \sum _{r=1}^n}{\omega }_{r}f\left(\frac{a{u}_r}{u_r-a}\right)+{\displaystyle \sum _{r=1}^n}{\omega }_{r}f\left(\frac{b{u}_r}{u_r-b}\right)\hfill \\ \hfill & -\frac{1}{2}\left[{\displaystyle \sum _{r=1}^n}{\omega }_{r}f\left(\frac{1}{\frac{\lambda }{b}+\frac{1-\lambda }{a}-\frac{1}{u_r}}\right)+{\displaystyle \sum _{r=1}^n}{\omega }_{r}f\left(\frac{1}{\frac{1-\lambda }{b}+\frac{\lambda }{a}-\frac{1}{u_r}}\right)\right]\hfill \\ \hfill & \le {\displaystyle \sum _{r=1}^n}{\omega }_{r}f\left(\frac{a{u}_r}{u_r-a}\right)+{\displaystyle \sum _{r=1}^n}{\omega }_{r}f\left(\frac{b{u}_r}{u_r-b}\right)-{\displaystyle \sum _{r=1}^n}{\omega }_{r}f\left(\frac{1}{\frac{a+b}{2ab}-\frac{1}{u_r}}\right).\hfill \end{array}$$

(22)

Proof. 

If we write (18) for $1-\lambda$ instead of $\lambda$, we obtain

$$\begin{array}{cc}\hfill f\left(\frac{1}{\frac{\lambda }{a}+\frac{1-\lambda }{b}-{\sum }_{r=1}^n\frac{{\omega }_r}{u_r}}\right)& \le {\displaystyle \sum _{r=1}^n}{\omega }_{r}f\left(\frac{1}{\frac{\lambda }{a}+\frac{1-\lambda }{b}-\frac{1}{u_r}}\right)\hfill \\ \hfill & \le \lambda {\displaystyle \sum _{r=1}^n}{\omega }_{r}f\left(\frac{a{u}_r}{u_r-a}\right)+(1-\lambda ){\displaystyle \sum _{r=1}^n}{\omega }_{r}f\left(\frac{b{u}_r}{u_r-b}\right)\hfill \\ \hfill & \le {\displaystyle \sum _{r=1}^n}{\omega }_{r}f\left(\frac{a{u}_r}{u_r-a}\right)+{\displaystyle \sum _{r=1}^n}{\omega }_{r}f\left(\frac{b{u}_r}{u_r-b}\right)\hfill \\ \hfill & -{\displaystyle \sum _{r=1}^n}{\omega }_{r}f\left(\frac{1}{\frac{\lambda }{b}+\frac{1-\lambda }{a}-\frac{1}{u_r}}\right).\hfill \end{array}$$

(23)

Now, by adding (18) and (23), we obtain

$$\begin{array}{cc}\hfill & f\left(\frac{1}{\frac{\lambda }{a}+\frac{1-\lambda }{b}-{\sum }_{r=1}^n\frac{{\omega }_r}{u_r}}\right)+f\left(\frac{1}{\frac{1-\lambda }{a}+\frac{\lambda }{b}-{\sum }_{r=1}^n\frac{{\omega }_r}{u_r}}\right)\hfill \\ \hfill & \le {\displaystyle \sum _{r=1}^n}{\omega }_{r}f\left(\frac{1}{\frac{1-\lambda }{a}+\frac{\lambda }{b}-\frac{1}{u_r}}\right)+{\displaystyle \sum _{r=1}^n}{\omega }_{r}f\left(\frac{1}{\frac{\lambda }{a}+\frac{1-\lambda }{b}-\frac{1}{u_r}}\right)\hfill \\ \hfill & \le {\displaystyle \sum _{r=1}^n}{\omega }_{r}f\left(\frac{a{u}_r}{u_r-a}\right)+{\displaystyle \sum _{r=1}^n}{\omega }_{r}f\left(\frac{b{u}_r}{u_r-b}\right)\hfill \\ \hfill & \le 2\left[{\displaystyle \sum _{r=1}^n}{\omega }_{r}f\left(\frac{a{u}_r}{u_r-a}\right)+{\displaystyle \sum _{r=1}^n}{\omega }_{r}f\left(\frac{b{u}_r}{u_r-b}\right)\right]-{\displaystyle \sum _{r=1}^n}{\omega }_{r}f\left(\frac{1}{\frac{\lambda }{b}+\frac{1-\lambda }{a}-\frac{1}{u_r}}\right)\hfill \\ \hfill & -{\displaystyle \sum _{r=1}^n}{\omega }_{r}f\left(\frac{1}{\frac{1-\lambda }{b}+\frac{\lambda }{a}-\frac{1}{u_r}}\right),\hfill \end{array}$$

namely

$$\begin{array}{cc}\hfill & \frac{1}{2}\left[f\left(\frac{1}{\frac{\lambda }{a}+\frac{1-\lambda }{b}-{\sum }_{r=1}^n\frac{{\omega }_r}{u_r}}\right)+f\left(\frac{1}{\frac{1-\lambda }{a}+\frac{\lambda }{b}-{\sum }_{r=1}^n\frac{{\omega }_r}{u_r}}\right)\right]\hfill \\ \hfill & \le \frac{1}{2}\left[{\displaystyle \sum _{r=1}^n}{\omega }_{r}f\left(\frac{1}{\frac{1-\lambda }{a}+\frac{\lambda }{b}-\frac{1}{u_r}}\right)+{\displaystyle \sum _{r=1}^n}{\omega }_{r}f\left(\frac{1}{\frac{\lambda }{a}+\frac{1-\lambda }{b}-\frac{1}{u_r}}\right)\right]\hfill \\ \hfill & \le \frac{1}{2}\left[{\displaystyle \sum _{r=1}^n}{\omega }_{r}f\left(\frac{a{u}_r}{u_r-a}\right)+{\displaystyle \sum _{r=1}^n}{\omega }_{r}f\left(\frac{b{u}_r}{u_r-b}\right)\right]\hfill \\ \hfill & \le {\displaystyle \sum _{r=1}^n}{\omega }_{r}f\left(\frac{a{u}_r}{u_r-a}\right)+{\displaystyle \sum _{r=1}^n}{\omega }_{r}f\left(\frac{b{u}_r}{u_r-b}\right)\hfill \\ \hfill & -\frac{1}{2}\left[{\displaystyle \sum _{r=1}^n}{\omega }_{r}f\left(\frac{1}{\frac{\lambda }{b}+\frac{1-\lambda }{a}-\frac{1}{u_r}}\right)+{\displaystyle \sum _{r=1}^n}{\omega }_{r}f\left(\frac{1}{\frac{1-\lambda }{b}+\frac{\lambda }{a}-\frac{1}{u_r}}\right)\right],\hfill \end{array}$$

which proves the second, the third and the fourth inequalities in (22).

By the harmonic convexity of f

$$\begin{array}{cc}\hfill & \frac{1}{2}\left[f\left(\frac{1}{\frac{\lambda }{a}+\frac{1-\lambda }{b}-{\sum }_{r=1}^n\frac{{\omega }_r}{u_r}}\right)+f\left(\frac{1}{\frac{1-\lambda }{a}+\frac{\lambda }{b}-{\sum }_{r=1}^n\frac{{\omega }_r}{u_r}}\right)\right]\hfill \\ \hfill & \ge f\left(\frac{1}{\frac{a+b}{2ab}-{\sum }_{r=1}^n\frac{{\omega }_r}{u_r}}\right),\hfill \end{array}$$

which proves the first inequality in (22).

Moreover,

$$\begin{array}{cc}\hfill & \frac{1}{2}\left[{\displaystyle \sum _{r=1}^n}{\omega }_{r}f\left(\frac{1}{\frac{\lambda }{a}+\frac{1-\lambda }{b}-\frac{1}{u_r}}\right)+{\displaystyle \sum _{r=1}^n}{\omega }_{r}f\left(\frac{1}{\frac{1-\lambda }{a}+\frac{\lambda }{b}-\frac{1}{u_r}}\right)\right]\hfill \\ \hfill & ={\displaystyle \sum _{r=1}^n}{\omega }_r\left[\frac{1}{2}f\left(\frac{1}{\frac{\lambda }{a}+\frac{1-\lambda }{b}-\frac{1}{u_r}}\right)+\frac{1}{2}f\left(\frac{1}{\frac{1-\lambda }{a}+\frac{\lambda }{b}-\frac{1}{u_r}}\right)\right]\hfill \\ \hfill & \ge {\displaystyle \sum _{r=1}^n}{\omega }_{r}f\left(\frac{1}{\frac{a+b}{2ab}-\frac{1}{u_r}}\right),\hfill \end{array}$$

which implies that

$$\begin{array}{cc}\hfill & {\displaystyle \sum _{r=1}^n}{\omega }_{r}f\left(\frac{a{u}_r}{u_r-a}\right)+{\displaystyle \sum _{r=1}^n}{\omega }_{r}f\left(\frac{b{u}_r}{u_r-b}\right)\hfill \\ \hfill & -\frac{1}{2}\left[{\displaystyle \sum _{r=1}^n}{\omega }_{r}f\left(\frac{1}{\frac{\lambda }{b}+\frac{1-\lambda }{a}-\frac{1}{u_r}}\right)+{\displaystyle \sum _{r=1}^n}{\omega }_{r}f\left(\frac{1}{\frac{1-\lambda }{b}+\frac{\lambda }{a}-\frac{1}{u_r}}\right)\right]\hfill \\ \hfill & \le {\displaystyle \sum _{r=1}^n}{\omega }_{r}f\left(\frac{a{u}_r}{u_r-a}\right)+{\displaystyle \sum _{r=1}^n}{\omega }_{r}f\left(\frac{b{u}_r}{u_r-b}\right)-{\displaystyle \sum _{r=1}^n}{\omega }_{r}f\left(\frac{1}{\frac{a+b}{2ab}-\frac{1}{u_r}}\right),\hfill \end{array}$$

and the last part of (22) is proved. □

Remark 2.

Let $f:(0,\infty )\subset \mathbb{R}\setminus \left\{0\right\}\to \mathbb{R}$ be harmonic convex. Now, if we take

$$y_r=\frac{1}{\frac{1-\lambda }{a}+\frac{\lambda }{b}-\frac{1}{u_r}},$$

namely

$$u_r=\frac{1}{\frac{1-\lambda }{a}+\frac{\lambda }{b}-\frac{1}{y_r}},$$

then, by (20) we get for ${\omega }_r\ge 0$ for $r\in \{1,\dots ,n\}$ with ${\sum }_{r=1}^n{\omega }_r=1$ that

$$\begin{array}{cc}\hfill f\left(\frac{1}{(2\lambda -1)\frac{b-a}{2ab}+{\sum }_{r=1}^n\frac{{\omega }_r}{y_r}}\right)& \le {\displaystyle \sum _{r=1}^n}{\omega }_{r}f\left(\frac{1}{(2\lambda -1)\frac{b-a}{2ab}+\frac{1}{y_r}}\right)\hfill \\ \hfill & \le (1-\lambda ){\displaystyle \sum _{r=1}^n}{\omega }_{r}f\left(\frac{1}{\lambda \left(\frac{b-a}{ab}\right)+\frac{1}{y_r}}\right)\hfill \\ \hfill & +\lambda {\displaystyle \sum _{r=1}^n}{\omega }_{r}f\left(\frac{1}{(1-\lambda )\left(\frac{a-b}{ab}\right)+\frac{1}{y_r}}\right)\hfill \\ \hfill & \le {\displaystyle \sum _{r=1}^n}{\omega }_{r}f\left(\frac{1}{\lambda \left(\frac{b-a}{ab}\right)+\frac{1}{y_r}}\right)+{\displaystyle \sum _{r=1}^n}{\omega }_{r}f\left(\frac{1}{(1-\lambda )\left(\frac{a-b}{ab}\right)+\frac{1}{y_r}}\right)\hfill \\ \hfill & -{\displaystyle \sum _{r=1}^n}{\omega }_{r}f\left(\frac{1}{(2\lambda -1)\frac{b-a}{2ab}+\frac{1}{y_r}}\right)\hfill \end{array}$$

(24)

provided that $y_r\in (0,\infty )$ and $\frac{1}{\lambda \left(\frac{b-a}{ab}\right)+\frac{1}{y_r}},\frac{1}{(1-\lambda )\left(\frac{a-b}{ab}\right)+\frac{1}{y_r}}\in (0,\infty )$ for $r\in \{1,\dots ,n\}$ and $\lambda \in [0,1]$.

For $\lambda =\frac{1}{2}$, we derive the inequality

$$\begin{array}{cc}\hfill f\left(\frac{1}{{\sum }_{r=1}^n\frac{{\omega }_r}{y_r}}\right)& \le {\displaystyle \sum _{r=1}^n}{\omega }_{r}f\left(y_r\right)\hfill \\ \hfill & \le \frac{1}{2}{\displaystyle \sum _{r=1}^n}{\omega }_{r}f\left(\frac{1}{\frac{b-a}{2ab}+\frac{1}{y_r}}\right)+\frac{1}{2}{\displaystyle \sum _{r=1}^n}{\omega }_{r}f\left(\frac{1}{\frac{a-b}{2ab}+\frac{1}{y_r}}\right)\hfill \\ \hfill & \le {\displaystyle \sum _{r=1}^n}{\omega }_{r}f\left(\frac{1}{\frac{b-a}{2ab}+\frac{1}{y_r}}\right)+{\displaystyle \sum _{r=1}^n}{\omega }_{r}f\left(\frac{1}{\frac{a-b}{2ab}+\frac{1}{y_r}}\right)\hfill \\ \hfill & -{\displaystyle \sum _{r=1}^n}{\omega }_{r}f\left(y_r\right)\hfill \end{array}$$

(25)

provided that $y_r\in (0,\infty )$ and $\frac{1}{\lambda \left(\frac{b-a}{ab}\right)+\frac{1}{y_r}},\frac{1}{(1-\lambda )\left(\frac{a-b}{ab}\right)+\frac{1}{y_r}}\in (0,\infty )$ for $r\in \{1,\dots ,n\}$.

From (25), we also obtain

$$\begin{array}{cc}\hfill 0& \le \frac{1}{2}{\displaystyle \sum _{r=1}^n}{\omega }_{r}f\left(\frac{1}{\frac{b-a}{2ab}+\frac{1}{y_r}}\right)+\frac{1}{2}{\displaystyle \sum _{r=1}^n}{\omega }_{r}f\left(\frac{1}{\frac{a-b}{2ab}+\frac{1}{y_r}}\right)-{\displaystyle \sum _{r=1}^n}{\omega }_{r}f\left(y_r\right)\hfill \\ \hfill & \le {\displaystyle \sum _{r=1}^n}{\omega }_{r}f\left(\frac{1}{\frac{b-a}{2ab}+\frac{1}{y_r}}\right)+{\displaystyle \sum _{r=1}^n}{\omega }_{r}f\left(\frac{1}{\frac{a-b}{2ab}+\frac{1}{y_r}}\right)\hfill \\ \hfill & -{\displaystyle \sum _{r=1}^n}{\omega }_{r}f\left(y_r\right)-f\left(\frac{1}{{\sum }_{r=1}^n\frac{{\omega }_r}{y_r}}\right)\hfill \end{array}$$

(26)

provided that $y_r\in (0,\infty )$ and $\frac{1}{\lambda \left(\frac{b-a}{ab}\right)+\frac{1}{y_r}},\frac{1}{(1-\lambda )\left(\frac{a-b}{ab}\right)+\frac{1}{y_r}}\in (0,\infty )$ for $r\in \{1,\dots ,n\}$.

Theorem 8.

Let $f:(0,\infty )\to \mathbb{R}$ be a harmonic convex function and $u_r\in (0,\infty ),r\in \{1,\dots ,n\}$ and $a,b\in \mathbb{R}\setminus \left\{0\right\}$ with $\frac{a{u}_r}{u_r-a},\frac{b{u}_r}{u_r-b}\in (0,\infty )$ for $r\in \{1,\dots ,n\}$, then for all $\lambda \in [0,1]$ and ${\omega }_r\ge 0$ for $r\in \{1,\dots ,n\}$ with ${\displaystyle \sum _{r=1}^n}{\omega }_r=1,$ we have

$$\begin{array}{cc}\hfill 0& \le (1-\lambda ){\displaystyle \sum _{r=1}^n}{\omega }_{r}f\left(\frac{a{u}_r}{u_r-a}\right)+\lambda {\displaystyle \sum _{r=1}^n}{\omega }_{r}f\left(\frac{b{u}_r}{u_r-b}\right)-{\displaystyle \sum _{r=1}^n}{\omega }_{r}f\left(\frac{1}{\frac{1-\lambda }{b}+\frac{\lambda }{a}-\frac{1}{u_r}}\right)\hfill \\ \hfill & \le (1-\lambda ){\displaystyle \sum _{r=1}^n}{\omega }_{r}f\left(\frac{a{u}_r}{u_r-a}\right)+\lambda {\displaystyle \sum _{r=1}^n}{\omega }_{r}f\left(\frac{b{u}_r}{u_r-b}\right)-f\left(\frac{1}{\frac{1-\lambda }{a}+\frac{\lambda }{b}-{\sum }_{r=1}^n\frac{{\omega }_r}{u_r}}\right)\hfill \\ & -\underset{1\le r\le n}{min}\left\{{\omega }_r\right\}\left[{\displaystyle \sum _{r=1}^n}f\left(\frac{1}{\frac{1-\lambda }{a}+\frac{\lambda }{b}-\frac{1}{u_r}}\right)-nf\left(\frac{1}{\frac{1-\lambda }{a}+\frac{\lambda }{b}-\frac{1}{n}{\sum }_{r=1}^n\frac{1}{u_r}}\right)\right]\hfill \\ \hfill & \le (1-\lambda ){\displaystyle \sum _{r=1}^n}{\omega }_{r}f\left(\frac{a{u}_r}{u_r-a}\right)+\lambda {\displaystyle \sum _{r=1}^n}{\omega }_{r}f\left(\frac{b{u}_r}{u_r-b}\right)-f\left(\frac{1}{\frac{1-\lambda }{a}+\frac{\lambda }{b}-{\sum }_{r=1}^n\frac{{\omega }_r}{u_r}}\right).\hfill \end{array}$$

(27)

Proof. 

Using (4), we have

$$\begin{array}{cc}\hfill & {\displaystyle \sum _{r=1}^n}{\omega }_{r}f\left(\frac{1}{\frac{1-\lambda }{a}+\frac{\lambda }{b}-\frac{1}{u_r}}\right)-f\left(\frac{1}{\frac{1-\lambda }{a}+\frac{\lambda }{b}-{\sum }_{r=1}^n\frac{{\omega }_r}{u_r}}\right)\hfill \\ \hfill & \ge \underset{1\le r\le n}{min}\left\{{\omega }_r\right\}\left[{\displaystyle \sum _{r=1}^n}f\left(\frac{1}{\frac{1-\lambda }{a}+\frac{\lambda }{b}-\frac{1}{u_r}}\right)-nf\left(\frac{1}{\frac{1-\lambda }{a}+\frac{\lambda }{b}-\frac{1}{n}{\sum }_{r=1}^n\frac{1}{u_r}}\right)\right].\hfill \end{array}$$

(28)

Therefore,

$$\begin{array}{cc}\hfill & (1-\lambda ){\displaystyle \sum _{r=1}^n}{\omega }_{r}f\left(\frac{a{u}_r}{u_r-a}\right)+\lambda {\displaystyle \sum _{r=1}^n}{\omega }_{r}f\left(\frac{b{u}_r}{u_r-b}\right)-{\displaystyle \sum _{r=1}^n}{\omega }_{r}f\left(\frac{1}{\frac{1-\lambda }{b}+\frac{\lambda }{a}-\frac{1}{u_r}}\right)\hfill \\ \hfill & =(1-\lambda ){\displaystyle \sum _{r=1}^n}{\omega }_{r}f\left(\frac{a{u}_r}{u_r-a}\right)+\lambda {\displaystyle \sum _{r=1}^n}{\omega }_{r}f\left(\frac{b{u}_r}{u_r-b}\right)-{\displaystyle \sum _{r=1}^n}{\omega }_{r}f\left(\frac{1}{\frac{1-\lambda }{b}+\frac{\lambda }{a}-\frac{1}{u_r}}\right)\hfill \\ \hfill & +f\left(\frac{1}{\frac{1-\lambda }{a}+\frac{\lambda }{b}-{\sum }_{r=1}^n\frac{{\omega }_r}{u_r}}\right)-f\left(\frac{1}{\frac{1-\lambda }{a}+\frac{\lambda }{b}-{\sum }_{r=1}^n\frac{{\omega }_r}{u_r}}\right)\hfill \\ \hfill & =(1-\lambda ){\displaystyle \sum _{r=1}^n}{\omega }_{r}f\left(\frac{a{u}_r}{u_r-a}\right)+\lambda {\displaystyle \sum _{r=1}^n}{\omega }_{r}f\left(\frac{b{u}_r}{u_r-b}\right)-f\left(\frac{1}{\frac{1-\lambda }{a}+\frac{\lambda }{b}-{\sum }_{r=1}^n\frac{{\omega }_r}{u_r}}\right)\hfill \\ \hfill & -\left({\displaystyle \sum _{r=1}^n}{\omega }_{r}f\left(\frac{1}{\frac{1-\lambda }{b}+\frac{\lambda }{a}-\frac{1}{u_r}}\right)-f\left(\frac{1}{\frac{1-\lambda }{a}+\frac{\lambda }{b}-{\sum }_{r=1}^n\frac{{\omega }_r}{u_r}}\right)\right)\hfill \\ \hfill & \le (1-\lambda ){\displaystyle \sum _{r=1}^n}{\omega }_{r}f\left(\frac{a{u}_r}{u_r-a}\right)+\lambda {\displaystyle \sum _{r=1}^n}{\omega }_{r}f\left(\frac{b{u}_r}{u_r-b}\right)-f\left(\frac{1}{\frac{1-\lambda }{a}+\frac{\lambda }{b}-{\sum }_{r=1}^n\frac{{\omega }_r}{u_r}}\right)\hfill \\ \hfill & -\underset{1\le r\le n}{min}\left\{{\omega }_r\right\}\left[{\displaystyle \sum _{r=1}^n}f\left(\frac{1}{\frac{1-\lambda }{a}+\frac{\lambda }{b}-\frac{1}{u_r}}\right)-nf\left(\frac{1}{\frac{1-\lambda }{a}+\frac{\lambda }{b}-\frac{1}{n}{\sum }_{r=1}^n\frac{1}{u_r}}\right)\right]\hfill \\ \hfill & \le (1-\lambda ){\displaystyle \sum _{r=1}^n}{\omega }_{r}f\left(\frac{a{u}_r}{u_r-a}\right)+\lambda {\displaystyle \sum _{r=1}^n}{\omega }_{r}f\left(\frac{b{u}_r}{u_r-b}\right)-f\left(\frac{1}{\frac{1-\lambda }{a}+\frac{\lambda }{b}-{\sum }_{r=1}^n\frac{{\omega }_r}{u_r}}\right),\hfill \end{array}$$

which proves the desired inequality (27). □

Corollary 5.

With the assumption of Theorem 8, we have

$$\begin{array}{cc}\hfill 0& \le \frac{1}{2}\left[{\displaystyle \sum _{r=1}^n}{\omega }_{r}f\left(\frac{a{u}_r}{u_r-a}\right)+{\displaystyle \sum _{r=1}^n}{\omega }_{r}f\left(\frac{b{u}_r}{u_r-b}\right)\right]-{\displaystyle \sum _{r=1}^n}{\omega }_{r}f\left(\frac{1}{\frac{a+b}{2ab}-\frac{1}{u_r}}\right)\hfill \\ \hfill & \le \frac{1}{2}\left[{\displaystyle \sum _{r=1}^n}{\omega }_{r}f\left(\frac{a{u}_r}{u_r-a}\right)+{\displaystyle \sum _{r=1}^n}{\omega }_{r}f\left(\frac{b{u}_r}{u_r-b}\right)\right]-f\left(\frac{1}{\frac{a+b}{2ab}-{\sum }_{r=1}^n\frac{{\omega }_r}{u_r}}\right)\hfill \\ \hfill & -\underset{1\le r\le n}{min}\left\{{\omega }_r\right\}\left[{\displaystyle \sum _{r=1}^n}f\left(\frac{1}{\frac{a+b}{2ab}-\frac{1}{u_r}}\right)-nf\left(\frac{1}{\frac{a+b}{2ab}-\frac{1}{n}{\sum }_{r=1}^n\frac{1}{u_r}}\right)\right]\hfill \\ \hfill & \le \frac{1}{2}\left[{\displaystyle \sum _{r=1}^n}{\omega }_{r}f\left(\frac{a{u}_r}{u_r-a}\right)+{\displaystyle \sum _{r=1}^n}{\omega }_{r}f\left(\frac{b{u}_r}{u_r-b}\right)\right]-f\left(\frac{1}{\frac{a+b}{2ab}-{\sum }_{r=1}^n\frac{{\omega }_r}{u_r}}\right).\hfill \end{array}$$

(29)

3. Applications

Let $u_r,{\omega }_r>0,r\in \{1,\dots ,n\}$ with ${\sum }_{r=1}^n{\omega }_r=1$. Then, the following inequality is well-known in the literature as the weighted Harmonic–Geometric–Arithmetic (HGA) mean inequality:

$$\frac{1}{{\sum }_{r=1}^n\frac{{\omega }_r}{u_r}}\le {\displaystyle \prod _{r=1}^n}{u}_r^{{\omega }_r}\le {\displaystyle \sum _{r=1}^n}{\omega }_r{u}_r.$$

(30)

• The equality holds in (30) if and only if $u_1=\cdots =u_n$. The inequality (30) was nicely proved in [31].
  By applying the inequality (5) for harmonic convex function $f\left(u\right)=lnu,u\in (0,\infty )$ and performing the necessary calculations, we obtain the following refinement of the first inequality of (30):

  $$\begin{array}{cc}\hfill \frac{1}{{\sum }_{r=1}^n\frac{{\omega }_r}{u_r}}& \le \underset{1\le s\le n}{min}\left\{{\left(\frac{1-{\omega }_s}{{\sum }_{r=1}^n{\omega }_r{u}_r-{\omega }_s{u}_s}\right)}^{1-{\omega }_s}.u_s^{{\omega }_s}\right\}\hfill \\ \hfill & \le {\displaystyle \prod _{s=1}^n}{\left[{\left(\frac{1-{\omega }_s}{{\sum }_{r=1}^n{\omega }_r{u}_r-{\omega }_s{u}_s}\right)}^{1-{\omega }_s}.u_s^{{\omega }_s}\right]}^{\frac{1}{n}}\hfill \\ \hfill & \le \underset{1\le s\le n}{max}\left\{{\left(\frac{1-{\omega }_s}{{\sum }_{r=1}^n{\omega }_r{u}_r-{\omega }_s{u}_s}\right)}^{1-{\omega }_s}.u_s^{{\omega }_s}\right\}\hfill \\ \hfill & \le {\displaystyle \prod _{r=1}^n}{u}_r^{{\omega }_r}.\hfill \end{array}$$

  (31)
• For $I=[a,b]$ and $u_r\in I$, $(r\ge 1)$ with ${\omega }_r>0$ such that ${\sum }_{r=1}^{\infty }{\omega }_r{u}_r$, affine combination $a+b-{\sum }_{r=1}^{\infty }{\omega }_r{u}_r$ and ${(a^{-1}+b^{-1}-{\sum }_{r=1}^{\infty }{\omega }_r{u}_r^{-1})}^{-1}$ converges in I, the following result was proved in [31] as

  $${\left(a^{-1}+b^{-1}-{\displaystyle \sum _{r=1}^{\infty }}{\omega }_r{u}_r^{-1}\right)}^{-1}\le ab{\displaystyle \prod _{r=1}^{\infty }}{u}_r^{-{\omega }_r}\le a+b-{\displaystyle \sum _{r=1}^{\infty }}{\omega }_r{u}_r.$$

  (32)
  This inequality was firstly proved by Pavić in [32] without considering the class of harmonic convex functions. By applying the inequality (10) for harmonic convex function $f\left(u_r\right)=ln{u}_r,u_r\in I=[a,b]\subset (0,\infty )$ and performing the necessary calculations with assumption that all infinite sum converge, we have the following refinement of first inequality (32).

  $$\begin{array}{cc}\hfill {\left(a^{-1}+b^{-1}-{\displaystyle \sum _{r=1}^{\infty }}{\omega }_r{u}_r^{-1}\right)}^{-1}& \le {\displaystyle \prod _{r=1}^{\infty }}{(a^{-1}+b^{-1}-u_r^{-1})}^{-{\omega }_r}\hfill \\ \hfill & \le a^{\left({\sum }_{r=}^{\infty }{\omega }_r{\alpha }_r\right)}{b}^{\left(1-{\sum }_{r=1}^{\infty }{\omega }_r{\alpha }_r\right)}\hfill \\ \hfill & \le ab{\displaystyle \prod _{r=1}^{\infty }}{u}_r^{-{\omega }_r}.\hfill \end{array}$$

  (33)

4. Conclusions

Jensen-type and a variant of Jensen-type inequalities are dominant among numerous results for the class of harmonic convex functions, which gave fascinating consequences in the field of mathematical inequalities, for example smart, shorter and easy proofs of Hölder, weighted HGA and infinite version weighted HGA inequalities. In addition, this class also provided sharp bounds, refinements in inequalities and solved definite integrals, which were having no or tedious analytical solutions. This advancement in analysis with the class of harmonic convex functions is strong source of motivation for further development in this area. Keeping in view the significance of the class of harmonic convex functions, we presented several refinements of Jensen-type and variants of Jensen-type inequalities. Further, we gave a refinement of weighted HG inequality and a refinement of the infinite version weighted HG inequality as applications of our main results. We hope techniques and consequences of this article will inspire researchers to explore more interesting follow-ups in this area.