New improvements of Jensen’s type inequalities via 4-convex functions with applications

Abstract

In this article, we present some new improvements of Jensen’s type inequalities via 4-convex and Green functions. These improvements are demonstrated in discrete as well as in integral versions. The aforesaid results enable us to give some improvements of Jensen’s and the Jensen–Steffensen inequalities. Also, we present some improvements of the reverse Jensen’s and the Jensen–Steffensen inequalities. Then as consequences of the improved Jensen’s inequality, we deduce some new bounds for the power, geometric and quasi-arithmetic means, also obtain bounds for the Hermite–Hadamard gap and improvements of the Hölder inequality. Finally as applications of the improved Jensen’s inequality, we present some new bounds for various divergences and Zipf–Mandelbrot entropy.

1 Introduction and preliminaries

Jensen’s inequality is one of the most significant inequality in the existing literature of mathematical inequalities for convex functions. Several well known mathematical inequalities for example Hölder’s, Minkowski’s, Ky Fan’s, Levinson’s, Hermite–Hadamard and Young’s inequalities etc can be deduced from this inequality. This inequality can be utilized for solving certain optimization problems in modern analysis. In more detail, this inequality can be used for estimation of Csiszár divergence and Zipf–Mandelbrot entropy [1, 4, 5, 7, 15, 16], it helps to investigate the stability of time-delayed systems [18], also dynamically consistent nonlinear evaluations in probability space, Rao-Blackwell estimates for certain parameters in their respective probability spaces and super linear expectations with its applications in economics can be investigated through this inequality [19, 20, 27]. Because of its significant role in modern applied analysis, several mathematicians have obtained some useful results related to Jensen’s inequality in the last couple of decades [2, 6, 8,9,10,11,12, 14, 17, 21, 22, 24]. In what follows, we present some improvements of Jensen’s type inequalities in discrete as well as in integral form via 4-convex and Green functions.

In the following theorem, the discrete form of Jensen’s inequality is given while its integral version in Riemann sense can be found in [15].

Theorem 1.1

Let $T:[\rho _1,\rho _2]\rightarrow \mathbb {R}$ be a convex function, $s_k\in [\rho _1,\rho _2]$, $u_k\ge 0$ for $k=1,2,\dots ,m$ with $U_m=\sum _{k=1}^{m}u_k>0$, then

$$\begin{aligned} T\Big (\frac{1}{U_m}\sum _{k=1}^{m}u_ks_k\Big )\le \frac{1}{U_m}\sum _{k=1}^{m}u_kT(s_k). \end{aligned}$$

(1.1)

About Jensen’s inequality, a question naturally comes into the mind that: is it possible to relax the condition of non-negativity of $u_k$ $(k=1,2,\dots ,m)$ at the expense of restricting $s_k$ $(k=1,2,\dots ,m)$ more severely?. The answer of this question was given by Steffensen in the following theorem [26].

Theorem 1.2

Let $T:[\rho _1,\rho _2]\rightarrow \mathbb {R}$ be a convex function, $s_k\in [\rho _1,\rho _2]$, $u_k\in \mathbb {R},$ $k=1,2,\dots ,m$. If $s_1\le s_2\le \cdots \le s_m$ or $s_1\ge s_2\ge \cdots \ge s_m$ and

$$\begin{aligned} 0\le \sum _{\gamma =1}^{k}u_\gamma \le \sum _{\gamma =1}^{m}u_\gamma , \,\,\, k=1,2,\dots , m,\,\,\,\,\,\sum _{\gamma =1}^{m}u_\gamma >0,\,\,\,\,\, \end{aligned}$$

then (1.1) holds.

The integral form of the above theorem can be seen in [13].

The following reverse of Jensen’s inequality has been given in [25, p. 83]:

Theorem 1.3

Let $T:[\rho _1,\rho _2]\rightarrow \mathbb {R}$ be a convex function, $s_k\in [\rho _1,\rho _2]$, $u_1>0$, $u_k\le 0$ for $k=2,3,\dots ,m$ with $U_m=\sum _{k=1}^{m}u_k>0$. Also, let $\frac{1}{U_m}\sum _{k=1}^{m}u_ks_k\in [\rho _1,\rho _2],$ then

$$\begin{aligned} T\Big (\frac{1}{U_m}\sum _{k=1}^{m}u_ks_k\Big )\ge \frac{1}{U_m}\sum _{k=1}^{m}u_kT(s_k). \end{aligned}$$

(1.2)

The following theorem presents a reverse of the Jensen–Steffensen inequality, also given in [25, p. 83]:

Theorem 1.4

Let $T:[\rho _1,\rho _2]\rightarrow \mathbb {R}$ be a convex function, $s_k\in [\rho _1,\rho _2],u_k\in \mathbb {R}$ for $k=1,2,\dots ,m$. Let $U_k=\sum _{j=1}^{k}u_j$ for $k=1,2,\dots ,m$ with $U_m>0$ and $\frac{1}{U_m}\sum _{k=1}^{m}u_ks_k\in [\rho _1,\rho _2]$. If the m-tuple $(s_1,s_2,\dots ,s_m)$ is monotonic, and there exists a number $p\in \{1,2,...m\}$ such that

$$\begin{aligned} U_k\le 0\quad \text{ for } \quad k<p\qquad and \qquad U_m-U_{k-1}\le 0\quad \text{ for } \quad k>p, \end{aligned}$$

then (1.2) holds.

To derive the main results, we need the following Green functions $G_{i}$ for $i=1,2,3,4,5,$ defined on $[\rho _{1},\rho _{2}]\times [\rho _{1},\rho _{2}]$ [23]:

$$\begin{aligned}&G_{1}(z,x)=\left\{ \begin{array}{lcl}\rho _{1}-x,&{}&{}\rho _{1}\le x\le z,\\ \rho _{1}-z,&{}&{}z\le x\le \rho _{2}.\\ \end{array}\right. \end{aligned}$$

(1.3)

$$\begin{aligned}&G_{2}(z,x)=\left\{ \begin{array}{lcl}z-\rho _{2},&{}&{}\rho _{1}\le x\le z,\\ x-\rho _{2},&{}&{}z\le x\le \rho _{2} .\\ \end{array}\right. \end{aligned}$$

(1.4)

$$\begin{aligned}&G_{3}(z,x)=\left\{ \begin{array}{lcl}z-\rho _{1},&{}&{}\rho _{1}\le x\le z,\\ x-\rho _{1},&{}&{}z\le x\le \rho _{2}.\\ \end{array}\right. \end{aligned}$$

(1.5)

$$\begin{aligned}&G_{4}(z,x)=\left\{ \begin{array}{lcl}\rho _{2}-x,&{}&{}\rho _{1}\le x\le z,\\ \rho _{2}-z,&{}&{}z\le x\le \rho _{2}.\\ \end{array}\right. \end{aligned}$$

(1.6)

$$\begin{aligned}&G_{5}(z,x)=\left\{ \begin{array}{lcl}\frac{(z-\rho _{2})(x-\rho _{1})}{\rho _{2}-\rho _{1}},&{}&{}\rho _{1}\le x\le z,\\ \frac{(x-\rho _{2})(z-\rho _{1})}{\rho _{2}-\rho _{1}},&{}&{}z\le x\le \rho _{2}.\\ \end{array}\right. \end{aligned}$$

(1.7)

These functions are continuous and convex with respect to both the variables z and x. Also, the following identities hold, for a function $T\in C^{2}[\rho _{1},\rho _{2}]$ [23]:

Lemma 1.5

Let $T\in C^{2}[\rho _{1},\rho _{2}],$ then the following identities hold.

$$\begin{aligned}&T(z)=T(\rho _{1})+(z-\rho _{1})T'(\rho _{2})+\int _{\rho _{1}}^{\rho _{2}}G_{1}(z,x)T''(x)dx,\end{aligned}$$

(1.8)

$$\begin{aligned}&T(z)=T(\rho _{2})+(z-\rho _{2})T'(\rho _{1})+\int _{\rho _{1}}^{\rho _{2}}G_{2}(z,x)T''(x)dx,\end{aligned}$$

(1.9)

$$\begin{aligned}&T(z)=T(\rho _{2})+(z-\rho _{1})T'(\rho _{1})-(\rho _{2}-\rho _{1})T'(\rho _{2})+\int _{\rho _{1}}^{\rho _{2}}G_{3}(z,x)T''(x)dx,\nonumber \\ \end{aligned}$$

(1.10)

$$\begin{aligned}&T(z)=T(\rho _{1})+(\rho _{2}-\rho _{1})T'(\rho _{1})-(\rho _{2}-z)T'(\rho _{2})+\int _{\rho _{1}}^{\rho _{2}}G_{4}(z,x)T''(x)dx,\nonumber \\ \end{aligned}$$

(1.11)

$$\begin{aligned}&T(z)=\frac{\rho _{2}-z}{\rho _{2}-\rho _{1}}T(\rho _{1})+\frac{z-\rho _{1}}{\rho _{2}-\rho _{1}}T(\rho _{2})+\int _{\rho _{1}}^{\rho _{2}}G_{5}(z,x)T''(x)dx, \end{aligned}$$

(1.12)

where $G_{i}$, for $i=1,2,3,4,5$ are given in (1.3)–(1.7) respectively.

In order to present the main results, the following inequality (1.13) will be useful, which is a simple consequence of Jensen’s inequality.

Lemma 1.6

Let $T:[\rho _{1},\rho _{2}]\rightarrow \mathbb {R}$ be a convex function and p(x) be a nonnegative weight function with $\int _{\rho _{1}}^{\rho _{2}}p(x)dx>0,$ then

$$\begin{aligned} T\left( \frac{1}{\int _{\rho _{1}}^{\rho _{2}}p(x)dx}\int _{\rho _{1}}^{\rho _{2}}xp(x)dx\right) \le \frac{1}{\int _{\rho _{1}}^{\rho _{2}}p(x)dx}\int _{\rho _{1}}^{\rho _{2}}T(x)p(x)dx. \end{aligned}$$

(1.13)

2 Main results

We begin to present our first main result.

Theorem 2.1

Let $T\in C^{2}[\rho _{1},\rho _{2}]$ be a 4-convex function and $s_{k}\in [\rho _{1},\rho _{2}],~u_{k}\in \mathbb {R}$ for $k=1,2,\dots ,m$ with $U_{m}:=\sum _{k=1}^{m}u_{k}\ne 0$ and $\frac{1}{U_{m}}\sum _{k=1}^{m}u_{k}s_{k}\in [\rho _{1},\rho _{2}]$. Also, let $G_{i} (i=1,2,3,4,5)$ be as defined in (1.3)–(1.7). If

$$\begin{aligned} \frac{1}{U_{m}}\sum _{k=1}^{m}u_{k}G_{i}(s_{k},x)-G_{i}\left( \frac{1}{U_{m}}\sum _{k=1}^{m}u_{k}s_{k},x\right) \ge 0,\,\,\,\, \text{ for } \,\,\,i\in \{1,2,3,4,5\}, \end{aligned}$$

(2.14)

then

$$\begin{aligned}&\frac{1}{U_{m}}\sum _{k=1}^{m}u_{k}T(s_{k})-T\left( \frac{1}{U_{m}}\sum _{k=1}^{m}u_{k}s_{k}\right) \nonumber \\&\quad \le \frac{T''(\rho _{2})-T''(\rho _{1})}{6(\rho _{2}-\rho _{1})}\left( \frac{1}{U_{m}}\sum _{k=1}^{m}u_{k}s^{3}_{k}-\Big (\frac{1}{U_{m}}\sum _{k=1}^{m}u_{k}s_{k}\Big )^{3} \right) \nonumber \\&\qquad +\frac{\rho _{2}T''(\rho _{1})-\rho _{1}T''(\rho _{2})}{2(\rho _{2}-\rho _{1})}\left( \frac{1}{U_{m}}\sum _{k=1}^{m}u_{k}s^{2}_{k}-\left( \frac{1}{U_{m}} \sum _{k=1}^{m}u_{k}s_{k}\right) ^{2}\right) . \end{aligned}$$

(2.15)

If the reverse inequality holds in (2.14), then the reverse inequality holds in (2.15).

If T is 4-concave function then the reverse inequality holds in (2.15).

Proof

Using (1.8)-(1.12) in $\frac{1}{U_{m}}\sum _{k=1}^{m}u_{k}T(s_{k})-T\left( \frac{1}{U_{m}}\sum _{k=1}^{m}u_{k}s_{k}\right)$, we obtain

$$\begin{aligned}&\frac{1}{U_{m}}\sum _{k=1}^{m}u_{k}T(s_{k})-T\left( \frac{1}{U_{m}} \sum _{k=1}^{m}u_{k}s_{k}\right) \nonumber \\&\quad =\int _{\rho _{1}}^{\rho _{2}}\left( \frac{1}{U_{m}}\sum _{k=1}^{m}u_{k}G_{i}(s_{k},x)-G_{i}\left( \frac{1}{U_{m}} \sum _{k=1}^{m}u_{k}s_{k},x\right) \right) T''(x)dx. \end{aligned}$$

(2.16)

Since (2.14) holds and T is 4-convex that is $T''$ is convex. Therefore by applying definition of convexity in the right hand side of (2.16) we obtain

$$\begin{aligned}&\frac{1}{U_{m}}\sum _{k=1}^{m}u_{k}T(s_{k})-T\left( \frac{1}{U_{m}} \sum _{k=1}^{m}u_{k}s_{k}\right) \nonumber \\&\quad \le \frac{T''(\rho _{1})}{\rho _{2}-\rho _{1}}\int _{\rho _{1}}^{\rho _{2}} \left( \frac{1}{U_{m}}\sum _{k=1}^{m}u_{k}G_{i}(s_{k},x)-G_{i}\left( \frac{1}{U_{m}}\sum _{k=1}^{m}u_{k}s_{k},x\right) \right) (\rho _{2}-x)dx \nonumber \\&\qquad +\frac{T''(\rho _{2})}{\rho _{2}-\rho _{1}}\int _{\rho _{1}}^{\rho _{2}} \left( \frac{1}{U_{m}}\sum _{k=1}^{m}u_{k}G_{i}(s_{k},x)-G_{i}\left( \frac{1}{U_{m}}\sum _{k=1}^{m}u_{k}s_{k},x\right) \right) (x-\rho _{1})dx.\qquad \qquad \end{aligned}$$

(2.17)

Now, if $T(x)=\frac{\rho _{2}x^{2}}{2}-\frac{x^{3}}{6},$ then $T''(x)=\rho _{2}-x$ and using (2.16) for these functions we get

$$\begin{aligned}&\int _{\rho _{1}}^{\rho _{2}} \left( \frac{1}{U_{m}}\sum _{k=1}^{m}u_{k}G_{i}(s_{k},x)-G_{i}\left( \frac{1}{U_{m}}\sum _{k=1}^{m}u_{k}s_{k},x\right) \right) (\rho _{2}-x)dx \nonumber \\&\quad =\frac{\rho _{2}}{2}\left( \frac{1}{U_{m}}\sum _{k=1}^{m}u_{k}s^{2}_{k}-\left( \frac{1}{U_{m}}\sum _{k=1}^{m}u_{k}s_{k}\right) ^{2}\right) -\frac{1}{6}\left( \frac{1}{U_{m}}\sum _{k=1}^{m}u_{k}s^{3}_{k}-\left( \frac{1}{U_{m}}\sum _{k=1}^{m}u_{k}s_{k}\right) ^{3}\right) .\nonumber \\ \end{aligned}$$

(2.18)

Similarly, using (2.16) for $T(x)=\frac{x^{3}}{6}-\frac{\rho _{1}x^{2}}{2},$ we get

$$\begin{aligned}&\int _{\rho _{1}}^{\rho _{2}} \left( \frac{1}{U_{m}}\sum _{k=1}^{m}u_{k}G_{i}(s_{k},x)-G_{i}\left( \frac{1}{U_{m}}\sum _{k=1}^{m}u_{k}s_{k},x\right) \right) (x-\rho _{1})dx \nonumber \\&\quad =\frac{1}{6}\left( \frac{1}{U_{m}}\sum _{k=1}^{m}u_{k}s^{3}_{k}-\left( \frac{1}{U_{m}}\sum _{k=1}^{m}u_{k}s_{k}\right) ^{3}\right) -\frac{\rho _{1}}{2}\left( \frac{1}{U_{m}}\sum _{k=1}^{m}u_{k}s^{2}_{k}-\left( \frac{1}{U_{m}}\sum _{k=1}^{m}u_{k}s_{k}\right) ^{2}\right) .\nonumber \\ \end{aligned}$$

(2.19)

Using (2.18) and (2.19) in (2.17), we get (2.15). $\square$

As an application of Theorem 2.1, we give an improvement of Jensen’s inequality.

Theorem 2.2

Let $T\in C^{2}[\rho _{1},\rho _{2}]$ be a 4-convex function and $s_{k}\in [\rho _{1},\rho _{2}],~u_{k}\ge 0$ for $k=1,2,\dots ,m$ with $\sum _{k=1}^{m}u_{k}=U_{m}>0,$ then (2.15) holds. If T is 4-concave function then the reverse inequality holds in (2.15).

Proof

Since $u_k\ge 0$ for all k with $U_m>0$ and the functions $G_{i}$ are convex for all i, therefore by Jensen’s inequality, the inequality (2.14) holds. So applying Theorem 2.1 for these facts, we have (2.15). $\square$

As applications of Theorem 2.2, we give two new upper bounds for the Hölder difference.

Corollary 2.3

Let $q>1,~p\not \in (2,3)$ such that $\frac{1}{q}+\frac{1}{p}=1.$ Also, let $[\rho _{1},\rho _{2}]$ be a positive interval and $(a_{1},a_{2},\dots ,a_{m}),(b_{1},b_{2},\dots ,b_{m})$ be two positive m-tuples with $\frac{\sum _{k=1}^{m}a_{k}b_{k}}{\sum _{k=1}^{m}b^{q}_{k}},~a_{k}b_{k}^{-\frac{q}{p}}\in [\rho _{1},\rho _{2}]$ for $k=1,2,\dots ,m,$ then

$$\begin{aligned}&\left( \sum _{k=1}^{m}a_{k}^{p}\right) ^{\frac{1}{p}} \left( \sum _{k=1}^{m}b_{k}^{q}\right) ^{\frac{1}{q}}-\sum _{k=1}^{m}a_{k}b_{k}\nonumber \\&\quad \le \Bigg (\frac{p(p-1)(\rho _{2}^{p-2}-\rho _{1}^{p-2})}{6(\rho _{2}-\rho _{1})}\left( \frac{1}{\sum _{k=1}^{m}b_{k}^{q}}\sum _{k=1}^{m}a_{k}^{3}b_{k}^{1-2\frac{q}{p}} -\left( \frac{1}{\sum _{k=1}^{m}b_{k}^{q}}\sum _{k=1}^{m}a_{k}b_{k}\right) ^{3}\right) \nonumber \\&\qquad +\frac{p(p-1)(\rho _{2}\rho _{1}^{p-2}-\rho _{1}\rho _{2}^{p-2})}{2(\rho _{2}-\rho _{1})} \Bigg (\frac{1}{\sum _{k=1}^{m}b_{k}^{q}}\sum _{k=1}^{m}a_{k}^{2}b_{k}^{1-\frac{q}{p}} \nonumber \\&\qquad -\left( \frac{1}{\sum _{k=1}^{m}b_{k}^{q}}\sum _{k=1}^{m}a_{k}b_{k}\right) ^{2}\Bigg )\Bigg )^{\frac{1}{p}}\sum _{k=1}^{m}b_{k}^{q}. \end{aligned}$$

(2.20)

Proof

Using (2.15) for $T(x)=x^{p},~u_{k}=b_{k}^{q}$ and $s_{k}=a_{k}b_{k}^{-\frac{q}{p}},$ we derive

$$\begin{aligned}&\left( \left( \sum _{k=1}^{m}a_{k}^{p}\right) \left( \sum _{k=1}^{m}b_{k}^{q}\right) ^{p-1}-\left( \sum _{k=1}^{m}a_{k}b_{k}\right) ^{p}\right) ^{\frac{1}{p}} \nonumber \\&\quad \le \Bigg (\frac{p(p-1)(\rho _{2}^{p-2}-\rho _{1}^{p-2})}{6(\rho _{2}-\rho _{1})}\left( \frac{1}{\sum _{k=1}^{m}b_{k}^{q}}\sum _{k=1}^{m}a_{k}^{3}b_{k}^{1-2\frac{q}{p}} -\left( \frac{1}{\sum _{k=1}^{m}b_{k}^{q}}\sum _{k=1}^{m}a_{k}b_{k}\right) ^{3}\right) \nonumber \\&\qquad +\frac{p(p-1)(\rho _{2}\rho _{1}^{p-2}-\rho _{1}\rho _{2}^{p-2})}{2(\rho _{2}-\rho _{1})} \Bigg (\frac{1}{\sum _{k=1}^{m}b_{k}^{q}}\sum _{k=1}^{m}a_{k}^{2}b_{k}^{1-\frac{q}{p}} \nonumber \\&\qquad -\left( \frac{1}{\sum _{k=1}^{m}b_{k}^{q}}\sum _{k=1}^{m}a_{k}b_{k}\right) ^{2}\Bigg )\Bigg )^{\frac{1}{p}}\sum _{k=1}^{m}b_{k}^{q}. \end{aligned}$$

(2.21)

By utilizing the inequality $\xi ^{e}-\zeta ^{e}\le (\xi -\zeta )^{e},~0\le \zeta \le \xi ,~e\in [0,1]$ for $\xi =\Big (\sum _{k=1}^{m}a_{k}^{p}\Big ) \Big (\sum _{k=1}^{m}b_{k}^{q}\Big )^{p-1},~\zeta =\Big (\sum _{k=1}^{m}a_{k}b_{k}\Big )^{p}$ and $e=\frac{1}{p},$ we obtain

$$\begin{aligned}&\left( \sum _{k=1}^{m}a_{k}^{p}\right) ^{\frac{1}{p}} \left( \sum _{k=1}^{m}b_{k}^{q}\right) ^{\frac{1}{q}}-\sum _{k=1}^{m}a_{k}b_{k} \nonumber \\&\quad \le \left( \left( \sum _{k=1}^{m}a_{k}^{p}\right) \left( \sum _{k=1}^{m}b_{k}^{q}\right) ^{p-1}-\left( \sum _{k=1}^{m}a_{k}b_{k}\right) ^{p}\right) ^{\frac{1}{p}}. \end{aligned}$$

(2.22)

Now using (2.22) in (2.21), we get (2.20). $\square$

Corollary 2.4

Let $0<p<1,~q=\frac{p}{p-1}$ such that $\frac{1}{p}\not \in (2,3).$ Also, $[\rho _{1},\rho _{2}]$ be a positive interval and $(a_{1},a_{2},\dots ,a_{m}),(b_{1},b_{2},\dots ,b_{m})$ be two positive m-tuples with $\frac{\sum _{k=1}^{m}a_{k}^{p}}{\sum _{k=1}^{m}b^{q}_{k}},~a_{k}^{p}b_{k}^{-q}\in [\rho _{1},\rho _{2}]$ for $k=1,2,\dots ,m,$ then

$$\begin{aligned}&\sum _{k=1}^{m}a_{k}b_{k}-\left( \sum _{k=1}^{m}a_{k}^{p}\right) ^{\frac{1}{p}} \left( \sum _{k=1}^{m}b_{k}^{q}\right) ^{\frac{1}{q}} \nonumber \\&\quad \le \frac{\left( 1-p\right) \left( \rho _{2}^{\frac{1}{p}-2}-\rho _{1}^{\frac{1}{p}-2}\right) }{6p^{2}\left( \rho _{2}-\rho _{1}\right) } \left( \sum _{k=1}^{m}a_{k}^{3p}b_{k}^{-2q} -\frac{\left( \sum _{k=1}^{m}a_{k}^{p}\right) ^{3}}{\left( \sum _{k=1}^{m}b_{k}^{q}\right) ^{2}}\right) \nonumber \\&\qquad +\frac{\left( 1-p\right) \left( \rho _{2}\rho _{1}^{\frac{1}{p}-2}-\rho _{1}\rho _{2}^{\frac{1}{p}-2}\right) }{2p^{2}\left( \rho _{2}-\rho _{1}\right) } \Bigg (\sum _{k=1}^{m}a_{k}^{2p}b_{k}^{-q} -\frac{\left( \sum _{k=1}^{m}a_{k}^{p}\right) ^{2}}{\left( \sum _{k=1}^{m}b_{k}^{q}\right) }\Bigg ). \end{aligned}$$

(2.23)

Proof

For the given values of p,  the function $T(x)=x^{\frac{1}{p}}$ for $x\in [\rho _{1},\rho _{2}],$ is convex as well as 4-convex. Therefore by using (2.15) for $T(x)=x^{\frac{1}{p}},~u_{k}=b_{k}^{q}$ and $s_{k}=a_{k}^{p}b_{k}^{-q},$ we get (2.23). $\square$

Definition 2.5

Let $\mathbf {u}=(u_{1},u_{2},\dots ,u_{m})$ and $\mathbf {s}=(s_{1},s_{2},\dots ,s_{m})$ be two positive m-tuples with $U_{m}=\sum _{k=1}^{m}u_{k}.$ Then the power mean of order $\alpha \in \mathbb {R}$ is defined as

$$\begin{aligned} \mathcal {M}_{\alpha }(\mathbf {u},\mathbf {s})=\left\{ \begin{array}{lcl}\left( \frac{1}{U_{m}}\sum _{k=1}^{m}u_{k}s^{\alpha }_{k}\right) ^{\frac{1}{\alpha }},&{}&{}\alpha \ne 0,\\ \left( \prod _{k=1}^{m}s^{u_{k}}_{k}\right) ^{\frac{1}{U_{m}}},&{}&{}\alpha =0.\\ \end{array}\right. \end{aligned}$$

As an application of Theorem 2.2, in the following corollary we present a bound for the power mean.

Corollary 2.6

Let $0<\rho _{1}<\rho _{2}$ and $\mathbf {u}=(u_{1},u_{2},\dots ,u_{m}),$ $\mathbf {s}=(s_{1},s_{2},\dots ,s_{m})$ be two positive m-tuples with $U_{m}=\sum _{k=1}^{m}u_{k}.$ Also, let r, t be two nonzero real numbers such that

• (i) if $r>0$ with $3r\le t$ or $r\le t\le 2r$ or $t<0,$ then we have

  $$\begin{aligned}&\mathcal {M}^{t}_{t}(\mathbf {u},\mathbf {s})-\mathcal {M}^{t}_{r}(\mathbf {u},\mathbf {s}) \nonumber \\&\quad \le \frac{t(t-r)}{6r^{2}(\rho _{2}-\rho _{1})}\left( \rho ^{\frac{t}{r}-2}_{2}-\rho ^{\frac{t}{r}-2}_{1}\right) \left( \mathcal {M}^{3r}_{3r}(\mathbf {u},\mathbf {s})-\mathcal {M}^{3r}_{r}(\mathbf {u},\mathbf {s})\right) \nonumber \\&\qquad +\frac{t(t-r)}{2r^{2}(\rho _{2}-\rho _{1})}\left( \rho _{2}\rho ^{\frac{t}{r}-2}_{1}-\rho _{1}\rho ^{\frac{t}{r}-2}_{2}\right) \left( \mathcal {M}^{2r}_{2r}(\mathbf {u},\mathbf {s})-\mathcal {M}^{2r}_{r}(\mathbf {u},\mathbf {s})\right) . \end{aligned}$$

  (2.24)

• (ii) If $r<0$ with $3r\ge t$ or $r\ge t\ge 2r$ or $t>0,$ then (2.24) holds.

• (iii) If $r>0$ with $2r<t<3r$ or $r<0$ with $3r<t<2r,$ then the reverse inequality holds in (2.24).

Proof

• (i) Let $T(x)=x^{\frac{t}{r}}$ for $x\in [\rho _{1},\rho _{2}],$ then the function T is 4-convex. Therefore using (2.15) for $T(x)=x^{\frac{t}{r}}$ and $s_{k}\rightarrow s^{r}_{k},$ we get (2.24).

• (ii) Also, in this case the function $T(x)=x^{\frac{t}{r}}$ for $x\in [\rho _{1},\rho _{2}]$ is 4-convex, therefore adopting the procedure of part (i), we obtain (2.24).

• (iii) For such values of r, t the function $T(x)=x^{\frac{t}{r}}$ for $x\in [\rho _{1},\rho _{2}]$ is 4-concave. Thus following the procedure of part (i) but for T as a 4-concave function, we obtain the reverse inequality in (2.24).

$\square$

The following corollary provides an interesting relation between different means as an application of Theorem 2.2.

Corollary 2.7

Let $0<\rho _{1}<\rho _{2}$ and $\mathbf {u}=(u_{1},u_{2},\dots ,u_{m}),$ $\mathbf {s}=(s_{1},s_{2},\dots ,s_{m})$ be two positive m-tuples with $U_{m}=\sum _{k=1}^{m}u_{k},$ then

$$\begin{aligned}&(i)\frac{\mathcal {M}_{1}(\mathbf {u},\mathbf {s})}{\mathcal {M}_{0}(\mathbf {u},\mathbf {s})}\nonumber \\&\quad \le \exp \left( \frac{\rho ^{2}_{1}+\rho _{1}\rho _{2}+\rho ^{2}_{2}}{2\rho ^{2}_{1}\rho ^{2}_{2}} \left( \mathcal {M}^{2}_{2}(\mathbf {u},\mathbf {s})-\mathcal {M}^{2}_{1}(\mathbf {u},\mathbf {s})\right) -\frac{\rho _{1}+\rho _{2}}{6\rho ^{2}_{1}\rho ^{2}_{2}} \left( \mathcal {M}^{3}_{3}(\mathbf {u},\mathbf {s})-\mathcal {M}^{3}_{1}(\mathbf {u},\mathbf {s})\right) \right) . \end{aligned}$$

(2.25)

$$\begin{aligned}&(ii)\mathcal {M}_{1}(\mathbf {u},\mathbf {s})-\mathcal {M}_{0}(\mathbf {u},\mathbf {s}) \nonumber \\&\quad \le \frac{e^{\rho _{2}}-e^{\rho _{1}}}{6(\rho _{2}-\rho _{1})} \left( \frac{1}{U_{m}}\sum _{k=1}^{m}u_{k}\ln ^{3}s_{k}-\ln ^{3}\mathcal {M}_{0}(\mathbf {u},\mathbf {s})\right) \nonumber \\&\qquad -\frac{\rho _{2}e^{\rho _{1}}-\rho _{1}e^{\rho _{2}}}{2(\rho _{2}-\rho _{1})} \left( \frac{1}{U_{m}}\sum _{k=1}^{m}u_{k}\ln ^{2}s_{k}-\ln ^{2}\mathcal {M}_{0}(\mathbf {u},\mathbf {s})\right) . \end{aligned}$$

(2.26)

Proof

• (i) Let $T(x)=-\ln x$ for $x\in [\rho _{1},\rho _{2}],$ then T is 4-convex. Therefore using (2.15) for this function, we get (2.25).

• (ii) Using (2.15) for the 4-convex function $T(x)=e^{x},~x\in [\rho _{1},\rho _{2}]$ and $s_{k}=\ln s_{k},$ we get (2.26).

$\square$

Definition 2.8

Let $\mathbf {u}=(u_{1},u_{2},\dots ,u_{m})$ and $\mathbf {s}=(s_{1},s_{2},\dots ,s_{m})$ be two positive m-tuples with $U_{m}=\sum _{k=1}^{m}u_{k}.$ Then for $\varphi$ as a strictly monotone, continuous function, the quasi arithmetic mean is defined as

$$\begin{aligned} \mathcal {M}_{\varphi }(\mathbf {u},\mathbf {s})=\varphi ^{-1}\left( \frac{1}{U_{m}}\sum _{k=1}^{m}u_{k}\varphi (s_{k})\right) . \end{aligned}$$

As an application of Theorem 2.2, in the following corollary we present a bound for the quasi arithmetic mean.

Corollary 2.9

Let $0<\rho _{1}<\rho _{2},$ and $\mathbf {u}=(u_{1},u_{2},\dots ,u_{m}),~\mathbf {s}=(s_{1},s_{2},\dots ,s_{m})$ be two positive m-tuples with $U_{m}=\sum _{k=1}^{m}u_{k}.$ Also, let $\varphi$ be a strictly monotone, continuous function and assume that $\beta \circ \varphi ^{-1}$ is a 4-convex function on $[\rho _{1},\rho _{2}],$ then the following inequality holds

$$\begin{aligned}&\frac{1}{U_{m}}\sum _{k=1}^{m}u_{k}\beta (s_{k})-\beta \left( \mathcal {M}_{\varphi }(\mathbf {u},\mathbf {s})\right) \nonumber \\&\quad \le \frac{(\beta \circ \varphi ^{-1})''(\rho _{2})-(\beta \circ \varphi ^{-1})''(\rho _{1})}{6(\rho _{2}-\rho _{1})} \left( \frac{1}{U_{m}}\sum _{k=1}^{m}u_{k}\varphi ^{3}(s_{k})-\varphi ^{3}\left( \mathcal {M}_{\varphi }(\mathbf {u},\mathbf {s})\right) \right) \nonumber \\&\qquad +\frac{\rho _{2}(\beta \circ \varphi ^{-1})''(\rho _{1})-\rho _{1}(\beta \circ \varphi ^{-1})''(\rho _{2})}{2(\rho _{2}-\rho _{1})} \left( \frac{1}{U_{m}}\sum _{k=1}^{m}u_{k}\varphi ^{2}(s_{k})-\varphi ^{2}\left( \mathcal {M}_{\varphi }(\mathbf {u},\mathbf {s})\right) \right) .\nonumber \\ \end{aligned}$$

(2.27)

Proof

(2.27) follows from (2.15) by assuming $s_{k}\rightarrow \varphi (s_{k})$ and $T\rightarrow \beta \circ \varphi ^{-1}.$ $\square$

As an application of Theorem 2.1, we obtain an improvement of the Jensen–Steffensen inequality.

Corollary 2.10

Let $T\in C^{2}[\rho _{1},\rho _{2}]$ be a 4-convex function and $s_{k}\in [\rho _{1},\rho _{2}],$ $u_{k}\in \mathbb {R}$ for $k=1,2,\dots ,m$. If $s_1\le s_2\le \cdots \le s_m$ or $s_1\ge s_2\ge \cdots \ge s_m$ and

$$\begin{aligned} 0\le \sum _{\gamma =1}^{k}u_\gamma \le \sum _{\gamma =1}^{m}u_\gamma , \,\,\, k=1,2,\dots , m,\,\,\,\,\,\sum _{\gamma =1}^{m}u_\gamma >0,\,\,\,\,\, \end{aligned}$$

then (2.15) holds. If T is 4-concave function then the reverse inequality holds in (2.15).

Proof

Since the Jensen–Steffensen conditions hold and the functions $G_{i}$ for all i are convex, therefore by the Jensen–Steffensen inequality, the inequality (2.14) holds. So, by applying Theorem 2.1, we get (2.15). $\square$

In the following corollary, we present a refinement of reverse of Jensen’s inequality under the conditions stated in Theorem 1.3.

Corollary 2.11

Let $T\in C^{2}[\rho _{1},\rho _{2}]$ be a 4-convex function, $s_k\in [\rho _1,\rho _2]$, $u_1>0$, $u_k\le 0$ for $k=2,3,\dots ,m$ with $U_m=\sum _{k=1}^{m}u_k>0$. Also, let $\frac{1}{U_m}\sum _{k=1}^{m}u_ks_k\in [\rho _1,\rho _2]$, then the reverse inequality in (2.15) holds.

Proof

Since for each $i=1,2,3,4,5$, $G_i$ is convex function, so by Theorem 1.3 we have $\bar{G}(x):=\frac{1}{U_{m}}\sum _{k=1}^{m}u_{k}G_{i}(s_{k},x)-G_{i}\Big (\frac{1}{U_{m}}\sum _{k=1}^{m}u_{k}s_{k},x\Big )\le 0$. Hence, using Theorem 2.1 we obtain reverse inequality in (2.15). $\square$

In the following corollary, we present a refinement of the reverse of the Jensen–Steffensen inequality under the conditions stated in Theorem 1.4.

Corollary 2.12

Let $T\in C^{2}[\rho _{1},\rho _{2}]$ be a 4-convex function, $s_k\in [\rho _1,\rho _2],u_k\in \mathbb {R}$ for $k=1,2,\dots ,m$. Let $U_k=\sum _{j=1}^{k}u_j$ for $k=1,2,\dots ,m$ with $U_m>0$ and $\frac{1}{U_m}\sum _{k=1}^{m}u_ks_k\in [\rho _1,\rho _2]$. If the m-tuple $(s_1,s_2,\dots ,s_m)$ is monotonic, and there exists a number $p\in \{1,2,...,m\}$ such that

$$\begin{aligned} U_k\le 0\quad \text{ for } \quad k<p\qquad and \qquad U_m-U_{k-1}\le 0\quad \text{ for } \quad k>p, \end{aligned}$$

then the reverse inequality in (2.15) holds.

Proof

The proof is similar to the proof of Corollary 2.11, but using Theorem 1.4 instead of Theorem 1.3. $\square$

The following theorem is the integral version of Theorem 2.1.

Theorem 2.13

Let $T\in C^{2}[\rho _{1},\rho _{2}]$ be a 4-convex function. Also, let $f_{1},f_{2}:[a_{1},a_{2}]\rightarrow \mathbb {R}$ be two integrable functions such that $f_{1}(y)\in [\rho _{1},\rho _{2}]$ for all $y\in [a_{1},a_{2}]$ with $D:=\int _{a_{1}}^{a_{2}}f_{2}(y)dy\ne 0$ and $\frac{1}{D}\int _{a_{1}}^{a_{2}}f_{1}(y)f_{2}(y)dy\in [\rho _{1},\rho _{2}]$. Suppose that $G_{i} (i=1,2,3,4,5)$ are defined as in (1.3)–(1.7), and

$$\begin{aligned} \frac{1}{D}\int _{a_{1}}^{a_{2}}f_{2}(y)G_{i}(f_{1}(y),x)dy-G_{i}\left( \frac{1}{D}\int _{a_{1}}^{a_{2}}f_{1}(y)f_{2}(y)dy,x\right) \ge 0,\, \text{ for } \,i\in \{1,2,3,4,5\}, \end{aligned}$$

(2.28)

then

$$\begin{aligned}&\frac{1}{D}\int _{a_{1}}^{a_{2}}(T\circ f_{1})(y)f_{2}(y)dy-T\left( \frac{1}{D}\int _{a_{1}}^{a_{2}}f_{1}(y)f_{2}(y)dy\right) \nonumber \\&\quad \le \frac{T''(\rho _{2})-T''(\rho _{1})}{6(\rho _{2}-\rho _{1})} \left( \frac{1}{D}\int _{a_{1}}^{a_{2}}f_{1}^{3}(y)f_{2}(y)dy-\left( \frac{1}{D}\int _{a_{1}}^{a_{2}}f_{1}(y)f_{2}(y)dy\right) ^{3}\right) \nonumber \\&\quad +\frac{\rho _{2}T''(\rho _{1})-\rho _{1}T''(\rho _{2})}{2(\rho _{2}-\rho _{1})}\nonumber \\&\quad \times \,\left( \frac{1}{D}\int _{a_{1}}^{a_{2}}f_{1}^{2}(y)f_{2}(y)dy-\left( \frac{1}{D}\int _{a_{1}}^{a_{2}}f_{1}(y)f_{2}(y)dy\right) ^{2}\right) . \end{aligned}$$

(2.29)

If the reverse inequality holds in (2.28), then the reverse inequality holds in (2.29).

If T is 4-concave function then the reverse inequality holds in (2.29).

The following corollary is the integral version of Theorem 2.2.

Corollary 2.14

Let $T\in C^{2}[\rho _{1},\rho _{2}]$ be a 4-convex function. Also, let $f_{1},f_{2}:[a_{1},a_{2}]\rightarrow \mathbb {R}$ be two integrable functions such that $f_{1}(y)\in [\rho _{1},\rho _{2}]$ and $f_{2}(y)$ is non negative for all $y\in [a_{1},a_{2}]$ with $D:=\int _{a_{1}}^{a_{2}}f_{2}(y)dy>0,$ then (2.29) holds. If T is 4-concave function then the reverse inequality holds in (2.29).

Remark 2.15

As applications of Corollary 2.14, two new bounds for the Hölder difference in integral form can be obtained. The procedure will be similar to those in Corollary 2.3 and Corollary 2.4.

Remark 2.16

As applications of Corollary 2.14, the integral versions of Corollary 2.6, Corollary 2.7 and Corollary 2.9 can be presented.

As an application of Corollary 2.14, we present a bound for the Hermite–Hadamard gap.

Corollary 2.17

Let $\psi \in C^{2}[a_{1},a_{2}]$ be a 4-convex function, then

$$\begin{aligned} \frac{1}{a_{2}-a_{1}}\int _{a_{1}}^{a_{2}}\psi (y)dy-\psi \left( \frac{a_{1}+a_{2}}{2}\right) \le \frac{\left( \psi ''(a_{1})+\psi ''(a_{2})\right) \left( a_{2}-a_{1}\right) ^{2}}{48}. \end{aligned}$$

(2.30)

Proof

Using (2.29) for $\psi =T,~[\rho _{1},\rho _{2}]=[a_{1},a_{2}]$ and $f_{2}(y)=1,~f_{1}(y)=y$ for all $y\in [a_{1},a_{2}],$ we get (2.30). $\square$

In the following corollary, we obtain a refinement of the integral Jensen–Steffensen inequality.

Corollary 2.18

Let $T\in C^{2}[\rho _{1},\rho _{2}]$ be a 4-convex function. Also, let $f_{1},f_{2}:[a_{1},a_{2}]\rightarrow \mathbb {R}$ be two integrable functions such that $f_{1}(y)\in [\rho _{1},\rho _{2}]$ for all $y\in [a_{1},a_{2}].$ If $f_1$ is monotonic function on $[a_{1},a_{2}]$ and $f_{2}$ satisfies

$$\begin{aligned} 0\le \int _{a_{1}}^{\lambda }f_{2}(y)dy\le \int _{a_{1}}^{a_{2}}f_{2}(y)dy,\,\,\, \lambda \in [a_{1},a_{2}],\,\,\,\,\,\int _{a_{1}}^{a_{2}}f_{2}(y)dy>0,\,\,\,\,\, \end{aligned}$$

then (2.29) holds.

If T is 4-concave function then the reverse inequality holds in (2.29).

In the following theorem, we present another improvement of Jensen’s type inequality in discrete form.

Theorem 2.19

Let $T\in C^{2}[\rho _{1},\rho _{2}]$ be a 4-convex function and $s_{k}\in [\rho _{1},\rho _{2}],~u_{k}\in \mathbb {R}$ for $k=1,2,\dots ,m$. If (2.14) holds, then

$$\begin{aligned}&\frac{1}{U_{m}}\sum _{k=1}^{m}u_{k}T(s_{k})-T\left( \frac{1}{U_{m}}\sum _{k=1}^{m}u_{k}s_{k}\right) \nonumber \\&\quad \ge \frac{1}{2}\left( \frac{1}{U_{m}}\sum _{k=1}^{m}u_{k}s^{2}_{k}-\left( \frac{1}{U_{m}}\sum _{k=1}^{m}u_{k}s_{k}\right) ^{2}\right) \nonumber \\&\quad \times T''\left( \frac{\frac{1}{U_{m}}\sum _{k=1}^{m}u_{k}s^{3}_{k} -\left( \frac{1}{U_{m}}\sum _{k=1}^{m}u_{k}s_{k}\right) ^{3}}{3\left( \frac{1}{U_{m}}\sum _{k=1}^{m}u_{k}s^{2}_{k}-\left( \frac{1}{U_{m}}\sum _{k=1}^{m}u_{k}s_{k}\right) ^{2}\right) }\right) . \end{aligned}$$

(2.31)

If the reverse inequality holds in (2.14), then the reverse inequality holds in (2.31).

If T is 4-concave function then the reverse inequality holds in (2.31).

Proof

Using (1.13) for $p(x)=\frac{1}{U_{m}}\sum _{k=1}^{m}u_{k}G_{i}(s_{k},x)-G_{i}\Big (\frac{1}{U_{m}}\sum _{k=1}^{m}u_{k}s_{k},x\Big )$ and $T=T'',$ then with the help of (2.16) we get the following inequality

$$\begin{aligned} \int _{\rho _{1}}^{\rho _{2}}\bar{G_{i}}(x)dxT''\left( \frac{\int _{\rho _{1}}^{\rho _{2}}\bar{G_{i}}(x)xdx}{\int _{\rho _{1}}^{\rho _{2}}\bar{G_{i}}(x)dx}\right) dx\le \frac{1}{U_{m}}\sum _{k=1}^{m}u_{k}T(s_{k})-T\left( \frac{1}{U_{m}} \sum _{k=1}^{m}u_{k}s_{k}\right) , \end{aligned}$$

(2.32)

where $\bar{G_{i}}(x)=\left( \frac{1}{U_{m}}\sum _{k=1}^{m}u_{k}G_{i}(s_{k},x)-G_{i}\left( \frac{1}{U_{m}}\sum _{k=1}^{m}u_{k}s_{k},x\right) \right) .$

Now, let $T(x)=\frac{x^{2}}{2},$ then $T''(x)=1$ and so using (2.16) for these functions we obtain

$$\begin{aligned} \int _{\rho _{1}}^{\rho _{2}}\bar{G_{i}}(x)dx=\frac{1}{2}\left( \frac{1}{U_{m}}\sum _{k=1}^{m}u_{k}s^{2}_{k}-\left( \frac{1}{U_{m}}\sum _{k=1}^{m}u_{k}s_{k}\right) ^{2}\right) . \end{aligned}$$

(2.33)

Also, let $T(x)=\frac{x^{3}}{6},$ then $T''(x)=x$ and so using (2.16) for these functions we obtain

$$\begin{aligned} \int _{\rho _{1}}^{\rho _{2}}\bar{G_{i}}(x)xdx=\frac{1}{6}\left( \frac{1}{U_{m}}\sum _{k=1}^{m}u_{k}s^{3}_{k}-\left( \frac{1}{U_{m}}\sum _{k=1}^{m}u_{k}s_{k}\right) ^{3}\right) . \end{aligned}$$

(2.34)

Using (2.33) and (2.34) in (2.32), we get (2.31). $\square$

As an application of the above theorem, we give a refinement of Jensen’s inequality.

Corollary 2.20

Let $T\in C^{2}[\rho _{1},\rho _{2}]$ be a 4-convex function and $s_{k}\in [\rho _{1},\rho _{2}],~u_{k}\ge 0$ for $k=1,2,\dots ,m$ with $\sum _{k=1}^{m}u_{k}=U_{m}>0,$ then (2.31) holds.

If T is 4-concave function then the reverse inequality holds in (2.31).

Proof

The proof is analogous to the proof of Theorem 2.2. $\square$

The following corollary provides a refinement of Hölder type inequality as an application of Corollary 2.20.

Corollary 2.21

Let $0<p<1,~q=\frac{p}{p-1}$ such that $\frac{1}{p}\not \in (2,3).$ Also, $[\rho _{1},\rho _{2}]$ be a positive interval and $(a_{1},a_{2},\dots ,a_{m}),(b_{1},b_{2},\dots ,b_{m})$ be two positive m-tuples with $\frac{\sum _{k=1}^{m}a_{k}^{p}}{\sum _{k=1}^{m}b^{q}_{k}},~a_{k}^{p}b_{k}^{-q}\in [\rho _{1},\rho _{2}]$ for $k=1,2,\dots ,m.$ Then

$$\begin{aligned}&\sum _{k=1}^{m}a_{k}b_{k}-\left( \sum _{k=1}^{m}a_{k}^{p}\right) ^{\frac{1}{p}} \left( \sum _{k=1}^{m}b_{k}^{q}\right) ^{\frac{1}{q}} \nonumber \\&\quad \ge \frac{1-p}{2p^{2}}\left( \sum _{k=1}^{m}a_{k}^{2p}b_{k}^{-q} -\left( \frac{\sum _{k=1}^{m}a_{k}^{p}}{\sum _{k=1}^{m}b_{k}^{q}}\right) ^{2}\right) \nonumber \\&\quad \times \left( \frac{\frac{\sum _{k=1}^{m}a_{k}^{3p}b_{k}^{-2q}}{\sum _{k=1}^{m}b_{k}^{q}}-\left( \frac{\sum _{k=1}^{m}a_{k}^{p}}{\sum _{k=1}^{m}b_{k}^{q}}\right) ^{3}}{3\left( \frac{\sum _{k=1}^{m}a_{k}^{2p}b_{k}^{-q}}{\sum _{k=1}^{m}b_{k}^{q}} -\left( \frac{\sum _{k=1}^{m}a_{k}^{p}}{\sum _{k=1}^{m}b_{k}^{q}}\right) ^{2}\right) }\right) ^{\frac{1}{p}-2}. \end{aligned}$$

(2.35)

Proof

For the given values of p,  the function $T(x)=x^{\frac{1}{p}}$ for $x\in [\rho _{1},\rho _{2}]$ is convex as well as 4-convex. Using (2.31) for $T(x)=x^{\frac{1}{p}},~u_{k}=b_{k}^{q}$ and $s_{k}=a_{k}^{p}b_{k}^{-q},$ we get (2.35). $\square$

As an application of Corollary 2.20, in the following corollary we present another bound for the power mean.

Corollary 2.22

Let $0<\rho _{1}<\rho _{2}$ and $\mathbf {u}=(u_{1},u_{2},\dots ,u_{m}),$ $\mathbf {s}=(s_{1},s_{2},\dots ,s_{m})$ be two positive m-tuples with $U_{m}=\sum _{k=1}^{m}u_{k}.$ Also, let r, t be two nonzero real numbers such that

• (i) if $r>0$ with $3r\le t$ or $r\le t\le 2r$ or $t<0,$ then

  $$\begin{aligned}&\mathcal {M}^{t}_{t}(\mathbf {u},\mathbf {s})-\mathcal {M}^{t}_{r}(\mathbf {u},\mathbf {s})\nonumber \\&\quad \ge \frac{t(t-r)}{2r^{2}}\left( \mathcal {M}^{2r}_{2r}(\mathbf {u},\mathbf {s})-\mathcal {M}^{2r}_{r}(\mathbf {u},\mathbf {s})\right) \left( \frac{\mathcal {M}^{3r}_{3r}(\mathbf {u},\mathbf {s})-\mathcal {M}^{3r}_{r}(\mathbf {u},\mathbf {s})}{3\left( \mathcal {M}^{2r}_{2r}(\mathbf {u},\mathbf {s})-\mathcal {M}^{2r}_{r}(\mathbf {u},\mathbf {s})\right) }\right) ^{\frac{t}{r}-2}. \end{aligned}$$

  (2.36)

• (ii) If $r<0$ with $3r\ge t$ or $r\ge t\ge 2r$ or $t>0,$ then we get again (2.36).

• (iii) If $r>0$ with $2r<t<3r$ or $r<0$ with $3r<t<2r,$ then the reverse inequality holds in (2.36).

Proof

• (i) Let $T(x)=x^{\frac{t}{r}}$ for $x\in [\rho _{1},\rho _{2}],$ then the function T is 4-convex. Therefore using (2.31) for $T(x)=x^{\frac{t}{r}}$ and $s_{k}\rightarrow s^{r}_{k},$ we get (2.36)

• (ii) Also, in this case the function $T(x)=x^{\frac{t}{r}}$ for $x\in [\rho _{1},\rho _{2}]$ is 4-convex, therefore adopting the procedure of part (i), we obtain (2.36).

• (iii) For such values of r, t the function $T(x)=x^{\frac{t}{r}}$ for $x\in [\rho _{1},\rho _{2}]$ is 4-concave. Thus following the procedure of part (i) but for T as a 4-concave function, we obtain the reverse inequality in (2.36).$\square$

The following corollary provides an interesting relationship between different means as an application of Corollary 2.20.

Corollary 2.23

Let $0<\rho _{1}<\rho _{2},$ and $\mathbf {u}=(u_{1},u_{2},\dots ,u_{m}),$ $\mathbf {s}=(s_{1},s_{2},\dots ,s_{m})$ be two positive m-tuples with $U_{m}=\sum _{k=1}^{m}u_{k},$ then

$$\begin{aligned}&(i)\frac{\mathcal {M}_{1}(\mathbf {u},\mathbf {s})}{\mathcal {M}_{0}(\mathbf {u},\mathbf {s})} \ge \exp \left( \frac{9}{2}\frac{\left( \mathcal {M}^{2}_{2}(\mathbf {u},\mathbf {s})-\mathcal {M}^{2}_{1}(\mathbf {u},\mathbf {s})\right) ^{3}}{\left( \mathcal {M}^{3}_{3}(\mathbf {u},\mathbf {s})-\mathcal {M}^{3}_{1}(\mathbf {u},\mathbf {s})\right) ^{2}}\right) . \end{aligned}$$

(2.37)

$$\begin{aligned}&(ii) \mathcal {M}_{1}(\mathbf {u},\mathbf {s})-\mathcal {M}_{0}(\mathbf {u},\mathbf {s}) \nonumber \\&\quad \ge \frac{1}{2}\left( \frac{1}{U_{m}}\sum _{k=1}^{m}u_{k}\ln ^{2}s_{k}-\ln ^{2}\mathcal {M}_{0}(\mathbf {u},\mathbf {s})\right) \nonumber \\&\quad \times \exp \left( \frac{\frac{1}{U_{m}}\sum _{k=1}^{m}u_{k}\ln ^{3}s_{k}-\ln ^{3}\mathcal {M}_{0}(\mathbf {u},\mathbf {s})}{3\left( \frac{1}{U_{m}}\sum _{k=1}^{m}u_{k}\ln ^{2}s_{k}-\ln ^{2}\mathcal {M}_{0}(\mathbf {u},\mathbf {s})\right) }\right) . \end{aligned}$$

(2.38)

Proof

• (i) Using (2.31) for the 4-convex function $T(x)=-\ln x,~x\in [\rho _{1},\rho _{2}],$ we get (2.37).

• (ii) Let $T(x)=e^{x},$ for $x\in [\rho _{1},\rho _{2}]$ then T is a 4-convex function. Thus using (2.31) for $T(x)=e^{x}$ and $s_{k}=\ln s_{k},$ we get (2.38).$\square$

As an application of Corollary 2.20, in the following corollary we present a bound for the quasi arithmetic mean.

Corollary 2.24

Let $\mathbf {u}=(u_{1},u_{2},\dots ,u_{m})$ and $\mathbf {s}=(s_{1},s_{2},\dots ,s_{m})$ be two positive m-tuples with $U_{m}=\sum _{k=1}^{m}u_{k}.$ Also, let $\varphi$ be a strictly monotone, continuous function and assume that $\beta \circ \varphi ^{-1}$ is a 4-convex function, then the following inequality holds

$$\begin{aligned}&\frac{1}{U_{m}}\sum _{k=1}^{m}u_{k}\beta (s_{k})-\beta \left( \mathcal {M}_{\varphi }(\mathbf {u},\mathbf {s})\right) \nonumber \\&\quad \ge \frac{1}{2}\Bigg (\frac{1}{U_{m}}\sum _{k=1}^{m}u_{k}\varphi ^{2}(s_{k})-\varphi ^{2}\left( \mathcal {M}_{\varphi }(\mathbf {u},\mathbf {s})\right) \Bigg ) \nonumber \\&\qquad \times (\beta \circ \varphi ^{-1})''\left( \frac{\frac{1}{U_{m}}\sum _{k=1}^{m}u_{k}\varphi ^{3}(s_{k}) -\varphi ^{3}\left( \mathcal {M}_{\varphi }(\mathbf {u},\mathbf {s})\right) }{3\left( \frac{1}{U_{m}}\sum _{k=1}^{m}u_{k}\varphi ^{2}(s_{k})-\varphi ^{2}\left( \mathcal {M}_{\varphi }(\mathbf {u},\mathbf {s})\right) \right) }\right) . \end{aligned}$$

(2.39)

Proof

(2.39) follows from (2.31) by assuming $s_{k}\rightarrow \varphi (s_{k})$ and $T\rightarrow \beta \circ \varphi ^{-1}.$ $\square$

As an application of Theorem 2.19, we give a refinement of the Jensen–Steffensen inequality.

Corollary 2.25

Let $T\in C^{2}[\rho _{1},\rho _{2}]$ be a 4-convex function and $s_{k}\in [\rho _{1},\rho _{2}],~u_{k}\in \mathbb {R}$ for $k=1,2,\dots ,m$. If $s_1\le s_2\le \cdots \le s_m$ or $s_1\ge s_2\ge \cdots \ge s_m$ and

$$\begin{aligned} 0\le \sum _{\gamma =1}^{k}u_\gamma \le \sum _{\gamma =1}^{m}u_\gamma , \,\,\, k=1,2,\dots , m,\,\,\,\,\,\sum _{\gamma =1}^{m}u_\gamma >0,\,\,\,\,\, \end{aligned}$$

then (2.31) holds.

If T is 4-concave function then the reverse inequality holds in (2.31).

Proof

The proof is analogous to the proof of Corollary 2.10. $\square$

Corollary 2.26

Under the assumptions of Corollary 2.11, the reverse inequality in (2.31) holds.

Proof

The idea of the proof is similar to the proof of Corollary 2.11. $\square$

Corollary 2.27

Under the assumptions of Corollary 2.12, the reverse inequality in (2.31) holds.

Proof

The idea of the proof is similar to the proof of Corollary 2.12. $\square$

In the following theorem, we state integral version of Theorem 2.19.

Theorem 2.28

Let $T\in C^{2}[\rho _{1},\rho _{2}]$ be a 4-convex function. Also, let $f_{1},f_{2}:[a_{1},a_{2}]\rightarrow \mathbb {R}$ be two integrable functions such that $f_{1}(y)\in [\rho _{1},\rho _{2}]$ for all $y\in [a_{1},a_{2}]$ with $D:=\int _{a_{1}}^{a_{2}}f_{2}(y)dy\ne 0$ and $\frac{1}{D}\int _{a_{1}}^{a_{2}}f_{1}(y)f_{2}(y)dy\in [\rho _{1},\rho _{2}]$. Suppose that the inequality (2.28) holds, then

$$\begin{aligned}&\frac{1}{D}\int _{a_{1}}^{a_{2}}(T\circ f_{1})(y)f_{2}(y)dy-T\left( \frac{1}{D}\int _{a_{1}}^{a_{2}}f_{1}(y)f_{2}(y)dy\right) \nonumber \\&\quad \ge \frac{1}{2}\left( \frac{1}{D}\int _{a_{1}}^{a_{2}}f_{1}^{2}(y)f_{2}(y)dy-\left( \frac{1}{D}\int _{a_{1}}^{a_{2}}f_{1}(y)f_{2}(y)dy\right) ^{2}\right) \nonumber \\&\qquad \times T''\left( \frac{\frac{1}{D}\int _{a_{1}}^{a_{2}}f_{1}^{3}(y)f_{2}(y)dy-\left( \frac{1}{D}\int _{a_{1}}^{a_{2}}f_{1}(y)f_{2}(y)dy\right) ^{3}}{3\left( \frac{1}{D}\int _{a_{1}}^{a_{2}}f_{1}^{2}(y)f_{2}(y)dy-\left( \frac{1}{D}\int _{a_{1}}^{a_{2}}f_{1}(y)f_{2}(y)dy\right) ^{2}\right) }\right) . \end{aligned}$$

(2.40)

If the reverse inequality holds in (2.28), then the reverse inequality holds in (2.40).

If T is 4-concave function then the reverse inequality holds in (2.40).

As an application of Theorem 2.28, we give a refinement of Jensen’s inequality.

Corollary 2.29

Let $T\in C^{2}[\rho _{1},\rho _{2}]$ be a 4-convex function. Also, let $f_{1}:[a_{1},a_{2}]\rightarrow \mathbb {R}$ be an integrable function such that $f_{1}(y)\in [\rho _{1},\rho _{2}]$ for all $y\in [a_{1},a_{2}]$ and $f_{2}:[a_{1},a_{2}]\rightarrow \mathbb {R}$ be a nonnegative function with $\int _{a_{1}}^{a_{2}}f_{2}(y)dy=D>0,$ then (2.40) holds. If T is 4-concave function then the reverse inequality holds in (2.40).

Remark 2.30

Similarly we can present integral version of Corollary 2.25.

Remark 2.31

Adopting the procedure of Corollary 2.21, one can present a refinement of the Hölder type inequality in integral form as an application of Corollary 2.29.

Remark 2.32

Integral versions of Corollary 2.22, Corollary 2.23 and of Corollary 2.24 can be presented as applications of Corollary 2.29.

As an application of Corollary 2.29, we present another bound for the Hermite–Hadamard gap.

Corollary 2.33

Let $\psi \in C^{2}[a_{1},a_{2}]$ be a 4-convex function, then

$$\begin{aligned} \frac{1}{a_{2}-a_{1}}\int _{a_{1}}^{a_{2}}\psi (y)dy-\psi \left( \frac{a_{1}+a_{2}}{2}\right) \ge \frac{\left( a_{2}-a_{1}\right) ^{2}}{24}\psi ''\left( \frac{a_{1}+a_{2}}{2}\right) . \end{aligned}$$

(2.41)

Proof

Using (2.40) for $\psi =T,~[\rho _{1},\rho _{2}]=[a_{1},a_{2}]$ and $f_{2}(y)=1,~f_{1}(y)=y$ for all $y\in [a_{1},a_{2}],$ we get (2.41). $\square$

3 Applications in information theory

Definition 3.1

(Csiszár divergence) Let $[\rho _{1},\rho _{2}]\subset \mathbb {R}$ and $f:[\rho _{1},\rho _{2}]\rightarrow \mathbb {R}$ be a function, then for $\mathbf {r}=(r_{1},r_{2},\dots ,r_{m})\in \mathbb {R}^{m}$ and $\mathbf {w}=(w_{1},w_{2},\dots ,w_{m})\in \mathbb {R}^{m}_{+}$ with $\frac{r_{k}}{w_{k}}\in [\rho _{1},\rho _{2}]~(k=1,2,\dots ,m),$ the Csiszár divergence is defined by

$$\begin{aligned} \bar{D}_{c}(\mathbf {r},\mathbf {w})=\sum _{k=1}^{m}w_{k}f\left( \frac{r_{k}}{w_{k}}\right) . \end{aligned}$$

Theorem 3.2

Let $f\in C^{2}[\rho _{1},\rho _{2}]$ be a 4-convex function and $\mathbf {r}=(r_{1},r_{2},\dots ,r_{m})\in \mathbb {R}^{m},$ $\mathbf {w}=(w_{1},w_{2},\dots ,w_{m})\in \mathbb {R}^{m}_{+}$ such that $\frac{\sum _{k=1}^{m}r_{k}}{\sum _{k=1}^{m}w_{k}},\frac{r_{k}}{w_{k}}\in [\rho _{1},\rho _{2}]$ for $k=1,2,\dots ,m,$ then

$$\begin{aligned}&\frac{1}{\sum _{k=1}^{m}w_{k}}\bar{D}_{c}(\mathbf {r},\mathbf {w})-f\left( \frac{\sum _{k=1}^{m}r_{k}}{\sum _{k=1}^{m}w_{k}}\right) \nonumber \\&\quad \le \frac{f''(\rho _{2})-f''(\rho _{1})}{6(\rho _{2}-\rho _{1})} \left( \frac{1}{\sum _{k=1}^{m}w_{k}}\sum _{k=1}^{m}\frac{r_{k}^{3}}{w^{2}_{k}}-\left( \frac{\sum _{k=1}^{m}r_{k}}{\sum _{k=1}^{m}w_{k}}\right) ^{3}\right) \nonumber \\&\qquad +\frac{\rho _{2}f''(\rho _{1})-\rho _{1}f''(\rho _{2})}{2(\rho _{2}-\rho _{1})} \left( \frac{1}{\sum _{k=1}^{m}w_{k}}\sum _{k=1}^{m}\frac{r_{k}^{2}}{w_{k}}-\left( \frac{\sum _{k=1}^{m}r_{k}}{\sum _{k=1}^{m}w_{k}}\right) ^{2}\right) . \end{aligned}$$

(3.42)

Proof

The result (3.42) can easily be deduced from (2.15) by choosing $T=f,~s_{k}=\frac{r_{k}}{w_{k}},~u_{k}=\frac{w_{k}}{\sum _{k=1}^{m}w_{k}}.$ $\square$

Theorem 3.3

Let $f\in C^{2}[\rho _{1},\rho _{2}]$ be a 4-convex function and $\mathbf {r}=(r_{1},r_{2},\dots ,r_{m})\in \mathbb {R}^{m},$ $\mathbf {w}=(w_{1},w_{2},\dots ,w_{m})\in \mathbb {R}^{m}_{+}$ such that $\frac{\sum _{k=1}^{m}r_{k}}{\sum _{k=1}^{m}w_{k}},\frac{r_{k}}{w_{k}}\in [\rho _{1},\rho _{2}]$ for $k=1,2,\dots ,m,$ then

$$\begin{aligned}&\frac{1}{\sum _{k=1}^{m}w_{k}}\bar{D}_{c}(\mathbf {r},\mathbf {w})-f\left( \frac{\sum _{k=1}^{m}r_{k}}{\sum _{k=1}^{m}w_{k}}\right) \nonumber \\&\quad \ge \frac{1}{2}\left( \frac{1}{\sum _{k=1}^{m}w_{k}}\sum _{k=1}^{m}\frac{r_{k}^{2}}{w_{k}} -\left( \frac{\sum _{k=1}^{m}r_{k}}{\sum _{k=1}^{m}w_{k}}\right) ^{2}\right) \nonumber \\&\qquad \times T''\left( \frac{\frac{1}{\sum _{k=1}^{m}w_{k}}\sum _{k=1}^{m}\frac{r_{k}^{3}}{w^{2}_{k}}-\left( \frac{\sum _{k=1}^{m}r_{k}}{\sum _{k=1}^{m}w_{k}}\right) ^{3}}{3\left( \frac{1}{\sum _{k=1}^{m}w_{k}}\sum _{k=1}^{m}\frac{r_{k}^{2}}{w_{k}} -\left( \frac{\sum _{k=1}^{m}r_{k}}{\sum _{k=1}^{m}w_{k}}\right) ^{2}\right) }\right) . \end{aligned}$$

(3.43)

Proof

The result (3.43) can easily be deduced from (2.31) by choosing $T=f,~s_{k}=\frac{r_{k}}{w_{k}},~u_{k}=\frac{w_{k}}{\sum _{k=1}^{m}w_{k}}.$ $\square$

Definition 3.4

(Rényi-divergence) For two positive probability distributions $\mathbf {r}=(r_{1},r_{2},\dots ,r_{m}),~\mathbf {w}=(w_{1},w_{2},\dots ,w_{m})$ and a nonnegative real number $\mu$ such that $\mu \ne 1,$ the Rényi-divergence is defined by

$$\begin{aligned} D_{re}(\mathbf {r},\mathbf {w})=\frac{1}{\mu -1}\log \left( \sum _{k=1}^{m}r_{k}^{\mu }w_{k}^{1-\mu }\right) . \end{aligned}$$

Corollary 3.5

Let $[\rho _{1},\rho _{2}]\subseteq \mathbb {R}^{+}$. Also, let $\mathbf {r}=(r_{1},r_{2},\dots ,r_{m}),~\mathbf {w}=(w_{1},w_{2},\dots ,w_{m})$ be positive probability distributions and $\mu >1$ such that $\sum _{k=1}^{m}w_{k}\left( \frac{r_{k}}{w_{k}}\right) ^{\mu },\left( \frac{r_{k}}{w_{k}}\right) ^{\mu -1}\in [\rho _{1},\rho _{2}]$ for $k=1,2,\dots ,m.$ Then

$$\begin{aligned}&D_{re}(\mathbf {r},\mathbf {w})-\frac{1}{\mu -1}\sum _{k=1}^{m}r_{k}\log \left( \frac{r_{k}}{w_{k}}\right) ^{\mu -1} \nonumber \\&\quad \le \frac{\rho _{1}+\rho _{2}}{6(1-\mu )\rho ^{2}_{1}\rho _{2}^{2}} \left( \sum _{k=1}^{m}r_{k}\left( \frac{r_{k}}{w_{k}}\right) ^{3(\mu -1)}-\left( \sum _{k=1}^{m}r_{k}^{\mu }w_{k}^{1-\mu }\right) ^{3}\right) \nonumber \\&\qquad +\frac{\rho ^{2}_{1}+\rho _{1}\rho _{2}+\rho ^{2}_{2}}{2(\mu -1)\rho ^{2}_{1}\rho _{2}^{2}} \left( \sum _{k=1}^{m}r_{k}\left( \frac{r_{k}}{w_{k}}\right) ^{2(\mu -1)}-\left( \sum _{k=1}^{m}r_{k}^{\mu }w_{k}^{1-\mu }\right) ^{2}\right) . \end{aligned}$$

(3.44)

Proof

Let $T(x)=-\frac{1}{\mu -1}\log x,~x\in [\rho _{1},\rho _{2}],$ then $T''(x)=\frac{1}{(\mu -1)x^{2}}>0$ and $T''''(x)=\frac{6}{(\mu -1)x^{4}}>0.$ This verifies that T is a 4-convex function, therefore using (2.15) for $T(x)=-\frac{1}{\mu -1}\log x,~u_{k}=r_{k}$ and $s_{k}=\left( \frac{r_{k}}{w_{k}}\right) ^{\mu -1}$, we obtain (3.44). $\square$

Corollary 3.6

Let $[\rho _{1},\rho _{2}]\subseteq \mathbb {R}^{+}$. Also, let $\mathbf {r}=(r_{1},r_{2},\dots ,r_{m}),~\mathbf {w}=(w_{1},w_{2},\dots ,w_{m})$ be positive probability distributions and $\mu >1$ with $\sum _{k=1}^{m}w_{k}\left( \frac{r_{k}}{w_{k}}\right) ^{\mu },\left( \frac{r_{k}}{w_{k}}\right) ^{\mu -1}\in [\rho _{1},\rho _{2}]$ for $k=1,2,\dots ,m.$ Then

$$\begin{aligned}&D_{re}(\mathbf {r},\mathbf {w})-\frac{1}{\mu -1}\sum _{k=1}^{m}r_{k}\log \left( \frac{r_{k}}{w_{k}}\right) ^{\mu -1} \nonumber \\&\quad \ge \frac{9}{2(\mu -1)}\frac{\left( \sum _{k=1}^{m}r_{k}\left( \frac{r_{k}}{w_{k}}\right) ^{2(\mu -1)}-\left( \sum _{k=1}^{m}r_{k}^{\mu }w_{k}^{1-\mu }\right) ^{2}\right) ^{3}}{\left( \sum _{k=1}^{m}r_{k}\left( \frac{r_{k}}{w_{k}}\right) ^{3(\mu -1)}-\left( \sum _{k=1}^{m}r_{k}^{\mu }w_{k}^{1-\mu }\right) ^{3}\right) ^{2}}. \end{aligned}$$

(3.45)

Proof

Using (2.31) for the 4-convex function $T(x)=-\frac{1}{\mu -1}\log x,~u_{k}=r_{k}$ and $s_{k}=\left( \frac{r_{k}}{w_{k}}\right) ^{\mu -1}$, we get (3.45). $\square$

Definition 3.7

(Shannon-entropy) If $\mathbf {w}=(w_{1},w_{2},\dots ,w_{m}),$ is a positive probability distribution, then the Shannon-entropy (information divergence) is defined by

$$\begin{aligned} E_{s}(\mathbf {w})=-\sum _{k=1}^{m}w_{k}\log w_{k}. \end{aligned}$$

Corollary 3.8

Let $[\rho _{1},\rho _{2}]\subseteq \mathbb {R}^{+}$ and $\mathbf {w}=(w_{1},w_{2},\dots ,w_{m})$ be a positive probability distribution such that $\frac{1}{w_{k}}\in [\rho _{1},\rho _{2}]$ for $k=1,2,\dots ,m,$ then

$$\begin{aligned} \log m-E_{s}(\mathbf {w})\le \frac{\rho ^{2}_{1}+\rho _{1}\rho _{2}+\rho ^{2}_{2}}{2\rho ^{2}_{1}\rho _{2}^{2}} \left( \sum _{k=1}^{m}\frac{1}{w_{k}}-m^{2}\right) -\frac{\rho _{1}+\rho _{2}}{6\rho ^{2}_{1}\rho _{2}^{2}}\left( \sum _{k=1}^{m}\frac{1}{w^{2}_{k}}-m^{3}\right) . \end{aligned}$$

(3.46)

Proof

Let $f(x)=-\log x,~x\in [\rho _{1},\rho _{2}],$ then $f''''(x)=\frac{6}{x^{4}}>0.$ This shows that f is a 4-convex function, therefore using (3.42) for $f(x)=-\log x$ and $(r_{1},r_{2},\dots ,r_{m})=(1,1,\dots ,1),$ we get (3.46). $\square$

Corollary 3.9

Let $[\rho _{1},\rho _{2}]\subseteq \mathbb {R}^{+}$ and $\mathbf {w}=(w_{1},w_{2},\dots ,w_{m})$ be a positive probability distribution such that $\frac{1}{w_{k}}\in [\rho _{1},\rho _{2}]$ for $k=1,2,\dots ,m,$ then

$$\begin{aligned} \log m-E_{s}(\mathbf {w})\ge \frac{9\left( \sum _{k=1}^{m}\frac{1}{w_{k}}-m^{2}\right) ^{3}}{2\left( \sum _{k=1}^{m}\frac{1}{w_{k}^{2}}-m^{3}\right) ^{2}}. \end{aligned}$$

(3.47)

Proof

Using (3.43) for the 4-convex function $f(x)=-\log x$ and $(r_{1},r_{2},\dots ,r_{m})=(1,1,\dots ,1),$ we get (3.47). $\square$

Definition 3.10

(Kullback-Leibler divergence) If $\mathbf {r}=(r_{1},r_{2},\dots ,r_{m})$ and $\mathbf {w}=(w_{1},w_{2},\dots ,w_{m}),$ are two positive probability distributions, then the Kullback-Leibler divergence is defined by

$$\begin{aligned} D_{kl}(\mathbf {r},\mathbf {w})=\sum _{k=1}^{m}r_{k}\log \frac{r_{k}}{w_{k}}. \end{aligned}$$

Corollary 3.11

Let $[\rho _{1},\rho _{2}]\subseteq \mathbb {R}^{+}$ and $\mathbf {r}=(r_{1},r_{2},\dots ,r_{m}),~\mathbf {w}=(w_{1},w_{2},\dots ,w_{m})$ be positive probability distributions such that $\frac{r_{k}}{w_{k}}\in [\rho _{1},\rho _{2}]$ for $k=1,2,\dots ,m,$ then

$$\begin{aligned} D_{kl}(\mathbf {r},\mathbf {w})\le \frac{\rho _{1}+\rho _{2}}{2\rho _{1}\rho _{2}} \left( \sum _{k=1}^{m}\frac{r_{k}^{2}}{w_{k}}-1\right) -\frac{1}{6\rho _{1}\rho _{2}}\left( \sum _{k=1}^{m}\frac{r_{k}^{3}}{w^{2}_{k}}-1\right) . \end{aligned}$$

(3.48)

Proof

Let $f(x)=x\log x,~x\in [\rho _{1},\rho _{2}],$ then $f''''(x)=\frac{2}{x^{3}}>0,$ which shows that f is a 4-convex function. So we get (3.48) by using (3.42) for $f(x)=x\log x.$ $\square$

Corollary 3.12

Let $[\rho _{1},\rho _{2}]\subseteq \mathbb {R}^{+}$ and $\mathbf {r}=(r_{1},r_{2},\dots ,r_{m}),~\mathbf {w}=(w_{1},w_{2},\dots ,w_{m})$ be positive probability distributions such that $\frac{r_{k}}{w_{k}}\in [\rho _{1},\rho _{2}]$ for $k=1,2,\dots ,m,$ then

$$\begin{aligned} D_{kl}(\mathbf {r},\mathbf {w})\ge \frac{3\left( \sum _{k=1}^{m}\frac{r_{k}^{2}}{w_{k}}-1\right) ^{2}}{2\left( \sum _{k=1}^{m}\frac{r_{k}^{3}}{w_{k}^{2}}-1\right) }. \end{aligned}$$

(3.49)

Proof

We get (3.49) by using (3.43) for the 4-convex function $f(x)=x\log x.$ $\square$

Definition 3.13

($\chi ^{2}$-divergence) For two positive probability distributions $\mathbf {r}=(r_{1},r_{2},\dots ,r_{m}), \mathbf {w}=(w_{1},w_{2},\dots ,w_{m})$, the $\chi ^{2}$-divergence is defined by

$$\begin{aligned} D_{\chi ^{2}}(\mathbf {r},\mathbf {w})=\sum _{k=1}^{m}\frac{(r_{k}-w_{k})^{2}}{w_{k}}. \end{aligned}$$

Corollary 3.14

If $[\rho _{1},\rho _{2}]\subseteq \mathbb {R}^{+}$ and $\mathbf {r}=(r_{1},r_{2},\dots ,r_{m}),~\mathbf {w}=(w_{1},w_{2},\dots ,w_{m})$ are two positive probability distributions such that $\frac{r_{k}}{w_{k}}\in [\rho _{1},\rho _{2}]$ for $k=1,2,\dots ,m,$ then

$$\begin{aligned} D_{\chi ^{2}}(\mathbf {r},\mathbf {w})\le \sum _{k=1}^{m}\frac{r_{k}^{2}}{w_{k}}-1. \end{aligned}$$

(3.50)

Proof

Let $f(x)=(x-1)^{2}$ for $x\in [\rho _{1},\rho _{2}],$ then $f''''(x)=0.$ This shows that f is a 4-convex function, therefore inequality (3.50) follows by using (3.42) for $f(x)=(x-1)^{2}.$ $\square$

Corollary 3.15

If $[\rho _{1},\rho _{2}]\subseteq \mathbb {R}^{+}$ and $\mathbf {r}=(r_{1},r_{2},\dots ,r_{m}),~\mathbf {w}=(w_{1},w_{2},\dots ,w_{m})$ are two positive probability distributions with $\frac{r_{k}}{w_{k}}\in [\rho _{1},\rho _{2}]$ for $k=1,2,\dots ,m,$ then

$$\begin{aligned} D_{\chi ^{2}}(\mathbf {r},\mathbf {w})\ge \sum _{k=1}^{m}\frac{r_{k}^{2}}{w_{k}}-1. \end{aligned}$$

(3.51)

Proof

The inequality (3.51) follows by using (3.43) for $f(x)=(x-1)^{2}.$ $\square$

Definition 3.16

(Bhattacharyya-coefficient) Bhattacharyya-coefficient for two positive probability distributions $\mathbf {r}=(r_{1},r_{2},\dots ,r_{m})$ and $\mathbf {w}=(w_{1},w_{2},\dots ,w_{m})$ is defined by

$$\begin{aligned} C_{b}(\mathbf {r},\mathbf {w})=\sum _{k=1}^{m}\sqrt{r_{k}w_{k}}. \end{aligned}$$

Corollary 3.17

Let $[\rho _{1},\rho _{2}]\subseteq \mathbb {R}^{+}$ and $\mathbf {r}=(r_{1},r_{2},\dots ,r_{m}),~\mathbf {w}=(w_{1},w_{2},\dots ,w_{m})$ be two positive probability distributions such that $\frac{r_{k}}{w_{k}}\in [\rho _{1},\rho _{2}]$ for $k=1,2,\dots ,m.$ Then

$$\begin{aligned}&1-C_{b}(\mathbf {r},\mathbf {w}) \nonumber \\&\quad \le \frac{\rho ^{\frac{3}{2}}_{1}-\rho ^{\frac{3}{2}}_{2}}{24\rho ^{\frac{3}{2}}_{1}\rho _{2}^{\frac{3}{2}}(\rho _{2}-\rho _{1})} \left( \sum _{k=1}^{m}\frac{r_{k}^{3}}{w^{2}_{k}}-1\right) \nonumber \\&\qquad +\frac{\rho ^{\frac{5}{2}}_{2}-\rho ^{\frac{5}{2}}_{1}}{8\rho ^{\frac{3}{2}}_{1}\rho _{2}^{\frac{3}{2}}(\rho _{2}-\rho _{1})} \left( \sum _{k=1}^{m}\frac{r_{k}^{2}}{w_{k}}-1\right) . \end{aligned}$$

(3.52)

Proof

Let $f(x)=-\sqrt{x},~x\in [\rho _{1},\rho _{2}].$ Then $f''''(x)=\frac{15}{16x^{\frac{7}{2}}}>0,$ which shows that f is a 4-convex function. Thus we get (3.52) by following (3.42) for $f(x)=-\sqrt{x}.$ $\square$

Corollary 3.18

Let $[\rho _{1},\rho _{2}]\subseteq \mathbb {R}^{+}$ and $\mathbf {r}=(r_{1},r_{2},\dots ,r_{m}),~\mathbf {w}=(w_{1},w_{2},\dots ,w_{m})$ be two positive probability distributions such that $\frac{r_{k}}{w_{k}}\in [\rho _{1},\rho _{2}]$ for $k=1,2,\dots ,m,$ then

$$\begin{aligned} 1-C_{b}(\mathbf {r},\mathbf {w})\ge \frac{3^{\frac{3}{2}} \left( \sum _{k=1}^{m}\frac{r_{k}^{2}}{w_{k}}-1\right) ^{\frac{5}{2}}}{8\left( \sum _{k=1}^{m}\frac{r_{k}^{3}}{w^{2}_{k}}-1\right) ^{\frac{3}{2}}}. \end{aligned}$$

(3.53)

Proof

Inequality (3.53) can be obtained by using (3.43) for the 4-convex function $f(x)=-\sqrt{x}$. $\square$

3.1 Applications for the Zipf–Mandelbrot entropy

The probability mass function for the Zipf–Mandelbrot law can be written as:

$$\begin{aligned} f_{(k,m,\theta ,s)}=\frac{1/(k+\theta )^{s}}{M_{m,\theta ,s}}, \end{aligned}$$

for $k=1,2,\dots ,m,$ $m\in \{1,2,\dots \}$, $\theta \ge 0,$ $s>0$ and $M_{m,\theta ,s}=\sum _{k=1}^{m}\frac{1}{(k+\theta )^{s}}$ is a generalized harmonic number. In connection to the attitude of information theory, we utilize entropies to compute the amount of information in a written text. The Zipf–Mandelbrot entropy mentioned in [3] is given by:

$$\begin{aligned} Z(M,\theta ,s)=\frac{s}{M_{m,\theta ,s}}\sum _{k=1}^{m}\frac{\log (k+\theta )}{(k+\theta )^{s}}+\log M_{m,\theta ,s}. \end{aligned}$$

Corollary 3.19

Let $0<\rho _{1}<\rho _{2},$ $\theta \ge 0,~s>0$ and $w_{k}\ge 0$ for $k=1,2,\dots ,m$ with $\sum _{k=1}^{m}w_{k}=1$. Then

$$\begin{aligned}&-Z(M,\theta ,s)-\frac{1}{M_{m,\theta ,s}}\sum _{k=1}^{m}\frac{\log w_{k}}{(k+\theta )^{s}}\nonumber \\&\quad \le \frac{\rho _{1}+\rho _{2}}{2\rho _{1}\rho _{2}}\left( \sum _{k=1}^{m}\frac{1}{w_{k}(k+\theta )^{2s}M^{2}_{m,\theta ,s}}-1\right) -\frac{1}{6\rho _{1}\rho _{2}}\left( \sum _{k=1}^{m}\frac{1}{w_{k}^{2}(k+\theta )^{3s}M^{3}_{m,\theta ,s}}-1\right) .\nonumber \\ \end{aligned}$$

(3.54)

Proof

For $r_{k}=\frac{1}{(k+\theta )^{s}M_{m,\theta ,s}},~k=1,2,\dots ,m,$ we have

$$\begin{aligned}&\sum _{k=1}^{m}r_{k}\log \frac{r_{k}}{w_{k}}=\sum _{k=1}^{m}\frac{1}{(k+\theta )^{s}M_{m,\theta ,s}}\left( -s\log (k+\theta )-\log M_{m,\theta ,s}-\log w_{k}\right) \nonumber \\&\quad =-Z(M,\theta ,s)-\frac{1}{M_{m,\theta ,s}}\sum _{k=1}^{m}\frac{\log w_{k}}{(k+\theta )^{s}}. \end{aligned}$$

(3.55)

Also,

$$\begin{aligned}&\frac{\rho _{1}+\rho _{2}}{2\rho _{1}\rho _{2}}\left( \sum _{k=1}^{m}\frac{r_{k}^{2}}{w_{k}}-1\right) -\frac{1}{6\rho _{1}\rho _{2}}\left( \sum _{k=1}^{m}\frac{r_{k}^{3}}{w^{2}_{k}}-1\right) \nonumber \\&\quad =\frac{\rho _{1}+\rho _{2}}{2\rho _{1}\rho _{2}}\left( \sum _{k=1}^{m}\frac{1}{w_{k}(k+\theta )^{2s}M^{2}_{m,\theta ,s}}-1\right) -\frac{1}{6\rho _{1}\rho _{2}}\left( \sum _{k=1}^{m}\frac{1}{w_{k}^{2}(k+\theta )^{3s}M^{3}_{m,\theta ,s}}-1\right) .\nonumber \\ \end{aligned}$$

(3.56)

Now using (3.55) and (3.56) in (3.48), we get (3.54). $\square$

Corollary 3.20

Let $0<\rho _{1}<\rho _{2},$ $\theta _{1},\theta _{2}\ge 0,~s_{1},s_{2}>0,$ then

$$\begin{aligned}&-Z(M,\theta _{1},s_{1})+\sum _{k=1}^{m}\frac{\log (k+\theta _{2})^{s_{2}}M_{m,\theta _{2},s_{2}}}{(k+\theta _{1})^{s_{1}}M_{m,\theta _{1},s_{1}}} \nonumber \\&\quad \le \frac{\rho _{1}+\rho _{2}}{2\rho _{1}\rho _{2}}\left( \sum _{k=1}^{m}\frac{(k+\theta _{2})^{s_{2}}M_{m,\theta _{2},s_{2}}}{(k+\theta _{1})^{2s_{1}}M^{2}_{m,\theta _{1},s_{1}}}-1\right) -\frac{1}{6\rho _{1}\rho _{2}}\left( \sum _{k=1}^{m}\frac{(k+\theta _{2})^{2s_{2}}M^{2}_{m,\theta _{2},s_{2}}}{(k+\theta _{1})^{3s_{1}}M^{3}_{m,\theta _{1},s_{1}}}-1\right) .\nonumber \\ \end{aligned}$$

(3.57)

Proof

For $r_{k}=\frac{1}{(k+\theta _{1})^{s_{1}}M_{m,\theta _{1},s_{1}}},~w_{k}=\frac{1}{(k+\theta _{2})^{s_{2}}M_{m,\theta _{2},s_{2}}},~k=1,2,\dots ,m,$ we have

$$\begin{aligned} \sum _{k=1}^{m}r_{k}\log \frac{r_{k}}{w_{k}}= & {} \sum _{k=1}^{m}\frac{1}{(k+\theta _{1})^{s_{1}}M_{m,\theta _{1},s_{1}}} \left( \log (k+\theta _{2})^{s_{2}}M_{m,\theta _{2},s_{2}}-\log (k+\theta _{1})^{s_{1}}M_{m,\theta _{1},s_{1}}\right) \nonumber \\= & {} -Z(M,\theta _{1},s_{1})+\sum _{k=1}^{m}\frac{\log (k+\theta _{2})^{s_{2}}M_{m,\theta _{2},s_{2}}}{(k+\theta _{1})^{s_{1}}M_{m,\theta _{1},s_{1}}}. \end{aligned}$$

(3.58)

Also,

$$\begin{aligned}&\frac{\rho _{1}+\rho _{2}}{2\rho _{1}\rho _{2}}\left( \sum _{k=1}^{m}\frac{r_{k}^{2}}{w_{k}}-1\right) -\frac{1}{6\rho _{1}\rho _{2}}\left( \sum _{k=1}^{m}\frac{r_{k}^{3}}{w^{2}_{k}}-1\right) \nonumber \\&\quad =\frac{\rho _{1}+\rho _{2}}{2\rho _{1}\rho _{2}}\left( \sum _{k=1}^{m}\frac{(k+\theta _{2})^{s_{2}}M_{m,\theta _{2},s_{2}}}{(k+\theta _{1})^{2s_{1}}M^{2}_{m,\theta _{1},s_{1}}}-1\right) -\frac{1}{6\rho _{1}\rho _{2}}\left( \sum _{k=1}^{m}\frac{(k+\theta _{2})^{2s_{2}}M^{2}_{m,\theta _{2},s_{2}}}{(k+\theta _{1})^{3s_{1}}M^{3}_{m,\theta _{1},s_{1}}}-1\right) .\nonumber \\ \end{aligned}$$

(3.59)

Now utilizing (3.58) and (3.59) in (3.48), we get (3.57). $\square$

Corollary 3.21

Let $0<\rho _{1}<\rho _{2},$ $\theta \ge 0,~s>0$ and $w_{k}\ge 0$ for $k=1,2,\dots ,m$ with $\sum _{k=1}^{m}w_{k}=1$. Then

$$\begin{aligned}&-Z(M,\theta ,s)-\frac{1}{M_{m,\theta ,s}}\sum _{k=1}^{m}\frac{\log w_{k}}{(k+\theta )^{s}} \nonumber \\&\quad \ge \frac{3\left( \sum _{k=1}^{m}\frac{1}{w_{k}(k+\theta )^{2s}M^{2}_{m,\theta ,s}}-1\right) ^{2}}{2\left( \sum _{k=1}^{m}\frac{1}{w_{k}^{2}(k+\theta )^{3s}M^{3}_{m,\theta ,s}}-1\right) }. \end{aligned}$$

(3.60)

Proof

Using (3.49) for $r_{k}=\frac{1}{(k+\theta )^{s}M_{m,\theta ,s}},~k=1,2,\dots ,m,$ we get (3.60). $\square$

Corollary 3.22

Let $0<\rho _{1}<\rho _{2},$ $\theta _{1},\theta _{2}\ge 0,~s_{1},s_{2}>0,$ then

$$\begin{aligned}&-Z(M,\theta _{1},s_{1})+\sum _{k=1}^{m}\frac{\log (k+\theta _{2})^{s_{2}}M_{m,\theta _{2},s_{2}}}{(k+\theta _{1})^{s_{1}}M_{m,\theta _{1},s_{1}}} \nonumber \\&\quad \ge \frac{3\left( \sum _{k=1}^{m}\frac{(k+\theta _{2})^{s_{2}}M_{m,\theta _{2},s_{2}}}{(k+\theta _{1})^{2s_{1}}M^{2}_{m,\theta _{1},s_{1}}}-1\right) ^{2}}{2\left( \sum _{k=1}^{m}\frac{(k+\theta _{2})^{2s_{2}}M^{2}_{m,\theta _{2},s_{2}}}{(k+\theta _{1})^{3s_{1}}M^{3}_{m,\theta _{1},s_{1}}}-1\right) }. \end{aligned}$$

(3.61)

Proof

Using (3.49) for $r_{k}=\frac{1}{(k+\theta _{1})^{s_{1}}M_{m,\theta _{1},s_{1}}},~w_{k}=\frac{1}{(k+\theta _{2})^{s_{2}}M_{m,\theta _{2},s_{2}}},~k=1,2,\dots ,m,$ we get (3.61). $\square$