Schur-geometric convexity of the generalized Gini-Heronian means involving three parameters

Abstract

In this paper, we give a unified generalization of the Gini means and Heronian means. The Schur-geometric convexity of the generalized Gini-Heronian means are investigated. Our result generalizes an earlier result given by Shi et al. (J. Inequal. Appl. 2008:879273, 2008). At the end of the paper, two new inequalities related to the generalized Gini-Heronian means are established to illustrate the applicability of the given result.

MSC:26D15, 26E60, 26A51.

1 Introduction

The Schur convexity of functions relating to special means have been investigated by many mathematicians, a number of results can be found in the monograph of Marshall and Olkin [1]. As a supplement to the Schur convexity of functions, the Schur-geometric convexity of functions was recently studied by Shi and Zhang [2–4], Zhang and Yang [5] and Chu et al. [6], some related results have been found to have an important application in discovering and proving the inequalities for special means. The purpose of this paper is to investigate the Schur-geometric convexity of functions related to Gini means and Heronian means. Besides, as application, we establish two new inequalities for generalized Gini-Heronian means. Our result generalizes an earlier result given by Shi et al. in [7].

In what follows, we denote the set of real numbers by ℝ, the set of nonnegative real numbers by ${\mathbb{R}}_{+}$, the set of positive real numbers by ${\mathbb{R}}_{++}$, and the set of nonpositive real numbers by ${\mathbb{R}}_{-}$.

Let $(r,s)\in {\mathbb{R}}^2$, $(x,y)\in {\mathbb{R}}_{++}^2$; the classical Gini means are defined by (see [8])

$$G(r,s;x,y)=\{\begin{array}{ll}{(\frac{x^r+y^r}{x^s+y^s})}^{1/(r-s)},& r\ne s,\\ exp(\frac{x^{r}lnx+y^{r}lny}{x^r+y^r}),& r=s.\end{array}$$

In 2007, Sándor [9] investigated the Schur convexity of $G(r,s;x,y)$ with respect to $(r,s)$, and obtained the following result.

Theorem A For fixed $(x,y)\in {\mathbb{R}}_{++}^2$ and $x\ne y$, the Gini means $G(r,s;x,y)$ is Schur concave with respect to $(r,s)$ on ${\mathbb{R}}_{+}^2$, and $G(r,s;x,y)$ is Schur convex with respect to $(r,s)$ on ${\mathbb{R}}_{-}^2$.

In the same year, Wang [10] proved the Schur convexity and the Schur-geometric convexity of $G(r,s;x,y)$ with respect to $(x,y)$ on ${\mathbb{R}}_{++}^2$, as follows.

Theorem B The Gini means $G(r,s;x,y)$ is Schur convex with respect to $(x,y)$ on ${\mathbb{R}}_{++}^2$ if and only if $(r,s)\in \{(r,s)|r\ge 0,s\ge 0,r+s\ge 1\}$, and $G(r,s;x,y)$ is Schur concave with respect to $(x,y)$ on ${\mathbb{R}}_{++}^2$ if and only if $(r,s)\in \{(r,s)|r\le 0,r+s\le 1\}\cup \{(r,s)|s\le 0,r+s\le 1\}$.

Theorem C The Gini means $G(r,s;x,y)$ is Schur-geometric convex with respect to $(x,y)$ on ${\mathbb{R}}_{++}^2$ if and only if $(r,s)\in \{(r,s)|r+s\ge 0\}$, and $G(r,s;x,y)$ is Schur-geometric concave with respect to $(x,y)$ on ${\mathbb{R}}_{++}^2$ if and only if $(r,s)\in \{(r,s)|r+s\le 0\}$.

Some different proofs concerning the Schur convexity of $G(r,s;x,y)$ were given by Shi et al. [11], Chu and Xia [12], respectively.

Xia and Chu [13] presented the necessary and sufficient condition for the Schur-harmonic-convexity of $G(r,s;x,y)$ with respect to $(x,y)$ on ${\mathbb{R}}_{++}^2$.

A further discussion on the Schur-power-convexity of $G(r,s;x,y)$ with respect to $(x,y)$ on ${\mathbb{R}}_{++}^2$ was given by Yang [14]. Meanwhile, the necessary and sufficient condition for the Schur-power-convexity of $G(r,s;x,y)$ was obtained.

Let $(x,y)\in {\mathbb{R}}_{++}^2$; the classical Heronian means is defined by (see [15])

$$H_e(x,y)=\frac{x+\sqrt{xy}+y}{3}.$$

In 1999, Mao [16] gave the definition of the dual Heronian means as follows:

$${\tilde{H}}_e(x,y)=\frac{x+4\sqrt{xy}+y}{6}.$$

In 2001, Janous [17] considered a unified generalization of the Heronian means $H_e(x,y)$ and dual Heronian means ${\tilde{H}}_e(x,y)$, and presented the following Heronian-type means with a parameter w:

$$H_w(x,y)=\{\begin{array}{ll}\frac{x+w\sqrt{xy}+y}{w+2},& 0\le w<\mathrm{\infty },\\ \sqrt{xy},& w=\mathrm{\infty }.\end{array}$$

Jia and Cao [18] investigated the exponential generalization of the Heronian means,

$$H_p(x,y)=\{\begin{array}{ll}{(\frac{x^p+{(xy)}^{p/2}+y^p}{3})}^{1/p},& p\ne 0,\\ \sqrt{xy},& p=0,\end{array}$$

and established some related inequalities. Moreover, the monotonicity and Schur convexity of the Heronian means $H_p(x,y)$ were discussed by Li et al. in [19].

Shi et al. [7] discussed the Schur convexity and Schur-geometric convexity of a further generalization of the Heronian means given by

$$H_{p,w}(x,y)=\{\begin{array}{ll}{(\frac{x^p+w{(xy)}^{p/2}+y^p}{w+2})}^{1/p},& p\ne 0,\\ \sqrt{xy},& p=0,\end{array}$$

and obtained some significant results, asserted by Theorems D and E below.

Theorem D For fixed $(p,w)\in {\mathbb{R}}^2$,

• (1) if $(p,w)\in \{(p,w)|p\ge 2,0\le w\le 2\}$, then $H_{p,w}(x,y)$ is Schur convex for $(x,y)\in {\mathbb{R}}_{+}^2$;

• (2) if $(p,w)\in \{(p,w)|p\le 1,w\ge 0\}\cup \{(p,w)|1<p\le 3/2,w\ge 1\}\cup \{(p,w)|3/2<p\le 2,w\ge 2\}$, then $H_{p,w}(x,y)$ is Schur concave for $(x,y)\in {\mathbb{R}}_{+}^2$.

Theorem E For fixed $(p,w)\in {\mathbb{R}}^2$,

• (1) if $(p,w)\in \{(p,w)|p>0,w\ge 0\}$, then $H_{p,w}(x,y)$ is Schur-geometric convex for $(x,y)\in {\mathbb{R}}_{++}^2$;

• (2) if $(p,w)\in \{(p,w)|p<0,w\ge 0\}$, then $H_{p,w}(x,y)$ is Schur-geometric concave for $(x,y)\in {\mathbb{R}}_{++}^2$.

As a further investigation of Theorem D, Fu et al. [20] gave the necessary and sufficient condition for the Schur convexity of the generalized Heronian means $H_{p,w}(x,y)$. Yang [21] investigated the Schur-power-convexity of $H_{p,w}(x,y)$ with respect to $(x,y)\in {\mathbb{R}}_{++}^2$. Mortici [22] studied certain special means relating to convex functions.

In this paper, we shall generalize the Gini means $G(r,s;x,y)$ and the Heronian means $H_{p,w}(x,y)$ in a unified form. For this purpose we define a generalized Gini-Heronian means containing three parameters p, q, and w, as follows:

$$H_{p,q,w}(x,y)=\{\begin{array}{ll}{(\frac{x^p+w{(xy)}^{p/2}+y^p}{x^q+w{(xy)}^{q/2}+y^q})}^{1/(p-q)},& p\ne q,\\ exp(\frac{x^{p}lnx+(w/2){(xy)}^{p/2}ln(xy)+y^{p}lny}{x^p+w{(xy)}^{p/2}+y^p}),& p=q,\end{array}$$

where $(p,q)\in {\mathbb{R}}^2$, $(x,y)\in {\mathbb{R}}_{++}^2$.

The Schur-geometric convexity of the generalized Gini-Heronian means will be discussed in Section 3. As applications, several inequalities related to generalized Gini-Heronian means are established in Section 4.

2 Definitions and lemmas

We introduce and establish several definitions and lemmas, which will be used in the proofs of the main results in Sections 3 and 4.

Definition 1 (see [1])

For any $x=(x_1,x_2,\dots ,x_n),y=(y_1,y_2,\dots ,y_n)\in {\mathbb{R}}^n$, let $x_{[1]}\ge x_{[2]}\ge \cdots \ge x_{[n]}$ and $y_{[1]}\ge y_{[2]}\ge \cdots \ge y_{[n]}$ denote the components of x and y in decreasing order, respectively.

The n-tuple y is said to majorize x (or x is to be majorized by y), in symbols $x\prec y$, if

$$\sum _{i=1}^k{x}_{[i]}\le \sum _{i=1}^k{y}_{[i]}\text{ holds}\text{for }k=1,2,\dots ,n-1\text{and}\sum _{i=1}^n{x}_i=\sum _{i=1}^n{y}_i.$$

Definition 2 (see [23])

For any $x=(x_1,x_2,\dots ,x_n),y=(y_1,y_2,\dots ,y_n)\in \mathrm{\Omega }$ ($\mathrm{\Omega }\subset {\mathbb{R}}_{++}^n$), $\varphi :\mathrm{\Omega }\to \mathbb{R}$ is said to be a Schur-geometric convex function on Ω if $(ln{x}_1,ln{x}_2,\dots ,ln{x}_n)\prec (ln{y}_1,ln{y}_2,\dots ,ln{y}_n)$ on Ω implies $\varphi (x)\le \varphi (y)$, ϕ is said to be a Schur-geometric concave function on Ω if and only if −ϕ is a Schur-geometric convex function.

Definition 3 (see [23])

For any $x=(x_1,x_2,\dots ,x_n),y=(y_1,y_2,\dots ,y_n)\in \mathrm{\Omega }$ ($\mathrm{\Omega }\subset {\mathbb{R}}_{++}^n$), Ω is said to be a geometrically convex set if $(x_1^{\alpha }{y}_1^{\beta },x_2^{\alpha }{y}_2^{\beta },\dots ,x_n^{\alpha }{y}_n^{\beta })\in \mathrm{\Omega }$ for all $x,y\in \mathrm{\Omega }$, $\alpha ,\beta \in [0,1]$ with $\alpha +\beta =1$.

Lemma 1 (see [23])

Let Ω ($\mathrm{\Omega }\subset {\mathbb{R}}_{++}^n$) be symmetric and have a nonempty interior set ${\mathrm{\Omega }}^o$, and let $\varphi :\mathrm{\Omega }\to \mathbb{R}$ be continuous on Ω and differentiable in ${\mathrm{\Omega }}^o$. If ϕ is symmetric on Ω and

$$(x_1-x_2)(x_1\frac{\partial \varphi }{\partial x_1}-x_2\frac{\partial \varphi }{\partial x_2})\ge 0(\le 0)$$

holds for any $(x_1,x_2,\dots ,x_n)\in {\mathrm{\Omega }}^o$, then ϕ is a Schur-geometric convex (Schur-geometric concave) function.

Lemma 2 (see [7])

Let $a\le b$, $u(t)=tb+(1-t)a$, $v(t)=ta+(1-t)b$, $1/2\le t_2\le t_1\le 1$ or $0\le t_1\le t_2\le 1/2$, then

$$(\frac{a+b}{2},\frac{a+b}{2})\prec (u(t_2),v(t_2))\prec (u(t_1),v(t_1))\prec (a,b).$$

Lemma 3 Let $p,q\in \mathbb{R}$, $p>q$, $\lambda \ge 1$, and let

$$g_1(\lambda )=p{\lambda }^p-q{\lambda }^{p/2+q/2}+q{\lambda }^{p/2-q/2}-p.$$

Then $g_1(\lambda )\ge 0$ for $p+q\ge 0$, and $g_1(\lambda )\le 0$ for $p+q\le 0$.

Proof Let $f_1(\lambda )=2{\lambda }^{1-(p/2-q/2)}{g}_1^{\prime }(\lambda )$, $f_2(\lambda )={\lambda }^{1-q}{f}_1^{\prime }(\lambda )$. Straightforward computation yields

$$\begin{array}{c}{f}_1(\lambda )=2p^2{\lambda }^{p/2+q/2}-q(p+q){\lambda }^q+q(p-q),\hfill \\ f_1(1)=2(p-q)(p+q),\hfill \\ f_2(\lambda )={\lambda }^{1-q}{f}_1^{\prime }(\lambda )=p^2(p+q){\lambda }^{p/2-q/2}-q^2(p+q),\hfill \\ f_2(1)=(p-q){(p+q)}^2,\hfill \\ f_2^{\prime }(\lambda )=\frac{1}{2}{p}^2(p+q)(p-q){\lambda }^{p/2-q/2-1}.\hfill \end{array}$$

Case 1. $pq\ne 0$.

• (1) If $p+q=0$, then

  $$g_1(\lambda )=p{\lambda }^p+p{\lambda }^{p/2-p/2}-p{\lambda }^{p/2+p/2}-p=0.$$

• (2) If $p+q>0$, then from the expressions $f_2^{\prime }(\lambda )$, $f_2(1)$, $f_1(1)$ above we find that, for $\lambda \ge 1$,

  $$f_2^{\prime }(\lambda )>0,f_2(1)>0,f_1(1)>0.$$

  We thus conclude that the functions $f_2(\lambda )$, $f_1(\lambda )$, and $g_1(\lambda )$ are increasing for $\lambda \in [1,+\mathrm{\infty })$. In fact, for $\lambda \ge 1$, one has

  $$\begin{array}{rl}{f}_2^{\prime }(\lambda )>0& ⟹f_2(\lambda )>0⟹f_1^{\prime }(\lambda )>0⟹f_1(\lambda )>0\\ ⟹g_1^{\prime }(\lambda )>0⟹g_1(\lambda )\ge g_1(1)=0.\end{array}$$

• (3) If $p+q<0$, then

  $$f_2^{\prime }(\lambda )<0,f_2(1)>0,\underset{\lambda \to +\mathrm{\infty }}{lim}\frac{f_2(\lambda )}{{\lambda }^{p/2-q/2}}=p^2(p+q)<0.$$

According to the above relations and the continuity of $f_2(\lambda )$, we deduce that there exists ${\lambda }_1\in (1,+\mathrm{\infty })$ such that $f_2({\lambda }_1)=0$, which leads us to $f_2(\lambda )>0$ for $\lambda \in [1,{\lambda }_1)$, and $f_2(\lambda )<0$ for $\lambda \in ({\lambda }_1,+\mathrm{\infty })$.

Thus, we deduce that $f_1(\lambda )$ is increasing on $[1,{\lambda }_1)$ and decreasing on $({\lambda }_1,+\mathrm{\infty })$, and thereby we get

$$f_1(\lambda )\le f_1({\lambda }_1)\text{for }{\lambda }_1\in (1,+\mathrm{\infty }).$$

On the other hand, we deduce from $f_2({\lambda }_1)=0$ that ${\lambda }_1^{p/2-q/2}=q^2/p^2$, thus, we have

$$\begin{array}{rcl}{f}_1({\lambda }_1)& =& 2p^2{\lambda }_1^{p/2-q/2}{\lambda }_1^q-q(p+q){\lambda }_1^q+q(p-q)\\ =& 2p^2\cdot \frac{q^2}{p^2}\cdot {\lambda }_1^q-q(p+q){\lambda }_1^q+q(p-q)\\ =& q(p-q)(1-{\lambda }_1^q)<0,\end{array}$$

which implies $f_1(\lambda )<0$, i.e., $g_1^{\prime }(\lambda )<0$ for $\lambda \in [1,+\mathrm{\infty })$.

We conclude that $g_1(\lambda )$ is decreasing on $[1,+\mathrm{\infty })$. It, therefore, follows that $g_1(\lambda )\le g_1(1)\le 0$.

Case 2. $pq=0$.

It is easy to verify that:

• If $p>q=0$ (it implies that $p+q>0$), then $g_1(?)=p({?}^p-1)=0$.

• If $0=p>q$ (it implies that $p+q<0$), then $g_1(?)=q{?}^{-q/2}(1-{?}^q)=0$.

• If $0=p=q$ (it implies that $p+q=0$), then $g_1(?)=0$.

The proof of Lemma 3 is complete. □

Lemma 4 Let $p,q\in \mathbb{R}$, $p>q$, $\lambda \ge 1$, and let

$$g_2(\lambda )=(p-q){\lambda }^{p+q}+(p+q){\lambda }^p-(p+q){\lambda }^q-(p-q).$$

Then $g_2(\lambda )\ge 0$ for $p+q\ge 0$, and $g_2(\lambda )\le 0$ for $p+q\le 0$.

Proof Let $h(\lambda )={\lambda }^{1-q}{g}_2^{\prime }(\lambda )$. It follows from a simple computation that

$$\begin{array}{c}h(\lambda )=(p-q)(p+q){\lambda }^p+p(p+q){\lambda }^{p-q}-q(p+q),\hfill \\ h(1)=2(p-q)(p+q),\hfill \\ h^{\prime }(\lambda )=p(p-q)(p+q){\lambda }^{p-1}(1+{\lambda }^{-q}).\hfill \end{array}$$

Case 1. $pq\ne 0$.

• (1) If $p+q=0$, then

  $$g_2(\lambda )=(p+p){\lambda }^{p-p}+(p-p){\lambda }^p-(p-p){\lambda }^{-p}-(p+p)=0.$$

• (2) If $p+q>0$, then from the hypothesis $p>q$ we find that $p>(p+q)/2>0$, and then we derive from the expressions $h^{\prime }(\lambda )$, $h(1)$ that, for $\lambda \ge 1$,

  $$h^{\prime }(\lambda )>0,h(1)>0,$$

  which shows that $h(\lambda )$ is increasing on $[1,+\mathrm{\infty })$, so, we have $h(\lambda )\ge h(1)>0$ for $\lambda \ge 1$.

  Thus, we obtain $g_2^{\prime }(\lambda )>0$ for $\lambda \ge 1$. Now, from the fact that $g_2(\lambda )$ is increasing on $[1,+\mathrm{\infty })$, we obtain $g_2(\lambda )\ge g_2(1)=0$ ($\lambda \ge 1$).

• (3) If $p+q<0$ and $p>0$, then

  $$h^{\prime }(\lambda )\le 0,h(1)<0.$$

  Note that $h(\lambda )$ is decreasing on $[1,+\mathrm{\infty })$, we get $h(\lambda )\le h(1)<0$ for $\lambda \ge 1$, and then we get $g_2^{\prime }(\lambda )<0$ for $\lambda \ge 1$.

  Finally, in view of the fact that $g_2(\lambda )$ is decreasing on $[1,+\mathrm{\infty })$, we deduce $g_2(\lambda )\le g_2(1)=0$ ($\lambda \ge 1$).

• (4) If $p+q<0$ and $p<0$, then

  $$h^{\prime }(\lambda )>0,h(1)<0,\underset{\lambda \to +\mathrm{\infty }}{lim}\frac{h(\lambda )}{{\lambda }^{p-q}}=p(p+q)>0.$$

By the monotonicity and the continuity of $h(\lambda )$, we find that there exists ${\lambda }_2\in (1,+\mathrm{\infty })$ such that $h({\lambda }_2)=0$, which implies that $h(\lambda )<0$ for $\lambda \in [1,{\lambda }_2)$, and $h(\lambda )>0$ for $\lambda \in ({\lambda }_2,+\mathrm{\infty })$.

We hereby deduce that $g_2^{\prime }(\lambda )<0$ for $\lambda \in [1,{\lambda }_2)$, and $g_2^{\prime }(\lambda )>0$ for $\lambda \in ({\lambda }_2,+\mathrm{\infty })$. Further, we conclude that $g_2(\lambda )$ is decreasing on $[1,{\lambda }_2)$ and increasing on $({\lambda }_2,+\mathrm{\infty })$.

Therefore, we obtain

$$g_2(\lambda )\le max\{g_2(1),\underset{\lambda \to +\mathrm{\infty }}{lim}{g}_2(\lambda )\}=max\{0,-(p-q)\}=0.$$

Case 2. $pq=0$.

It is easy to verify that:

• If $p>q=0$ (it implies that $p+q>0$), then $g_2(?)=2p({?}^p-1)=0$.

• If $0=p>q$ (it implies that $p+q<0$), then $g_2(?)=2q(1-{?}^q)=0$.

• If $0=p=q$ (it implies that $p+q=0$), then $g_2(?)=0$.

This completes the proof of Lemma 4. □

3 Main result

The main result of this paper is given by Theorem 1 below.

Theorem 1 For fixed $(p,q,w)\in {\mathbb{R}}^3$,

• (1) if $p+q\ge 0$ and $w\ge 0$, then the generalized Gini-Heronian means $H_{p,q,w}(x,y)$ are Schur-geometric convex for $(x,y)\in {\mathbb{R}}_{++}^2$;

• (2) if $p+q\le 0$ and $w\ge 0$, then the generalized Gini-Heronian means $H_{p,q,w}(x,y)$ are Schur-geometric concave for $(x,y)\in {\mathbb{R}}_{++}^2$.

Proof We consider the following two cases.

Case 1. If $p=q$, then

$$H_{p,q,w}(x,y)=exp\left(\frac{x^{p}lnx+(w/2){(xy)}^{p/2}ln(xy)+y^{p}lny}{x^p+w{(xy)}^{p/2}+y^p}\right).$$

Differentiating $H_{p,q,w}(x,y)$ with respect to x and y, respectively, we obtain

$$\frac{\partial H}{\partial x}=\frac{H_{p,q,w}(x,y)G_1(x,y,p,w)}{x{[x^p+w{(xy)}^{p/2}+y^p]}^2},\frac{\partial H}{\partial y}=\frac{H_{p,q,w}(x,y)G_2(x,y,p,w)}{y{[x^p+w{(xy)}^{p/2}+y^p]}^2},$$

where

$$\begin{array}{c}{G}_1(x,y,p,w)=[p{x}^{p}lnx+x^p+(pw/4){(xy)}^{p/2}ln(xy)+(w/2){(xy)}^{p/2}][x^p+w{(xy)}^{p/2}+y^p]\hfill \\ \phantom{G_1(x,y,p,w)=}-p[x^{p}lnx+(w/2){(xy)}^{p/2}ln(xy)+y^{p}lny][x^p+(w/2){(xy)}^{p/2}],\hfill \\ G_2(x,y,p,w)=[p{y}^{p}lny+y^p+(pw/4){(xy)}^{p/2}ln(xy)+(w/2){(xy)}^{p/2}][x^p+w{(xy)}^{p/2}+y^p]\hfill \\ \phantom{G_2(x,y,p,w)=}-p[x^{p}lnx+(w/2){(xy)}^{p/2}ln(xy)+y^{p}lny][y^p+(w/2){(xy)}^{p/2}].\hfill \end{array}$$

By calculation, it follows that

$$\mathrm{\Lambda }:=(x-y)(x\frac{\partial H}{\partial x}-y\frac{\partial H}{\partial y})=\frac{(x-y)H_{p,q,w}(x,y)F(x,y,p,w)}{{[x^p+w{(xy)}^{p/2}+y^p]}^2},$$

where

$$\begin{array}{rcl}F(x,y,p,w)& =& G_1(x,y,p,w)-G_2(x,y,p,w)\\ =& (p{x}^{p}lnx+x^p-p{y}^{p}lny-y^p)[x^p+w{(xy)}^{p/2}+y^p]\\ -p[x^{p}lnx+(w/2){(xy)}^{p/2}ln(xy)+y^{p}lny](x^p-y^p)\\ =& {(xy)}^{p/2}[x^p-y^p+(p/2)(x^p+y^p)ln(x/y)]w+2p{x}^p{y}^{p}ln(x/y)+x^{2p}-y^{2p}.\end{array}$$

Note that the expression Λ is symmetric in x and y, without loss of generality we assume that $x\ge y$.

Setting $\lambda =x/y$, $\lambda \ge 1$, we have

$$F(x,y,p,w)={(xy)}^{p/2}{y}^p[{\lambda }^p-1+(p/2)({\lambda }^p+1)ln\lambda ]w+y^{2p}(2p{\lambda }^{p}ln\lambda +{\lambda }^{2p}-1).$$

In addition, it is easy to show that

$$\begin{array}{c}[{\lambda }^p-1+(p/2)({\lambda }^p+1)ln\lambda ]w+y^{2p}(2p{\lambda }^{p}ln\lambda +{\lambda }^{2p}-1)\ge 0\text{for }p\ge 0\text{ and }w\ge 0,\hfill \\ [{\lambda }^p-1+(p/2)({\lambda }^p+1)ln\lambda ]w+y^{2p}(2p{\lambda }^{p}ln\lambda +{\lambda }^{2p}-1)\le 0\text{for }p\le 0\text{ and }w\ge 0.\hfill \end{array}$$

This yields

$$\mathrm{\Lambda }\ge 0\text{for }p\ge 0,w\ge 0\text{and}\mathrm{\Lambda }\le 0\text{for }p\le 0,w\ge 0.$$

Case 2. If $p\ne q$, then

$$H_{p,q,w}(x,y)={\left(\frac{x^p+w{(xy)}^{p/2}+y^p}{x^q+w{(xy)}^{q/2}+y^q}\right)}^{1/(p-q)}.$$

Differentiating $H_{p,q,w}(x,y)$ with respect to x and y, respectively, we get

$$\begin{array}{c}\frac{\partial H}{\partial x}=\frac{H_{p,q,w}(x,y)}{p-q}[\frac{p{x}^{p-1}+(pw/2)y{(xy)}^{(p/2)-1}}{x^p+w{(xy)}^{p/2}+y^p}-\frac{q{x}^{q-1}+(qw/2)y{(xy)}^{(q/2)-1}}{x^q+w{(xy)}^{q/2}+y^q}],\hfill \\ \frac{\partial H}{\partial y}=\frac{H_{p,q,w}(x,y)}{p-q}[\frac{p{y}^{p-1}+(pw/2)x{(xy)}^{(p/2)-1}}{x^p+w{(xy)}^{p/2}+y^p}-\frac{q{y}^{q-1}+(qw/2)x{(xy)}^{(q/2)-1}}{x^q+w{(xy)}^{q/2}+y^q}].\hfill \end{array}$$

Direct calculation gives

$$\begin{array}{rcl}\mathrm{\Lambda }& :=& (x-y)(x\frac{\partial H}{\partial x}-y\frac{\partial H}{\partial y})\\ =& \frac{H_{p,q,w}(x,y)(x-y)}{p-q}[\frac{p(x^p-y^p)}{x^p+w{(xy)}^{p/2}+y^p}-\frac{q(x^q-y^q)}{x^q+w{(xy)}^{q/2}+y^q}].\end{array}$$

It is obvious that the expression Λ is symmetric with respect to p and q (it is also symmetric with respect to x and y), and without loss of generality we assume that $x\ge y$ and $p>q$ in the following discussion.

Simplifying the expression Λ, we obtain

$$\mathrm{\Lambda }=\frac{H_{p,q,w}(x,y)(x-y)F(x,y,p,q,w)}{(p-q)(x^p+w{(xy)}^{p/2}+y^p)(x^q+w{(xy)}^{q/2}+y^q)},$$

where

$$\begin{array}{rcl}F(x,y,p,q,w)& =& p(x^p-y^p)(x^q+w{(xy)}^{q/2}+y^q)-q(x^q-y^q)(x^p+w{(xy)}^{p/2}+y^p)\\ =& (p{x}^{p+(q/2)}{y}^{q/2}-q{x}^{q+(p/2)}{y}^{p/2}+q{x}^{p/2}{y}^{q+(p/2)}-p{x}^{q/2}{y}^{p+(q/2)})w\\ +(p-q)x^{p+q}+(p+q)x^p{y}^q-(p+q)x^q{y}^p-(p-q)y^{p+q}\\ =& y^{p+q}[p{(x/y)}^{p+(q/2)}-q{(x/y)}^{q+(p/2)}+q{(x/y)}^{p/2}-p{(x/y)}^{q/2}]w\\ +y^{p+q}[(p-q){(x/y)}^{p+q}+(p+q){(x/y)}^p-(p+q){(x/y)}^q-(p-q)].\end{array}$$

Setting $\lambda =x/y$, $\lambda \ge 1$, then

$$F(x,y,p,q,w)=y^{p+q}[{\lambda }^{q/2}{g}_1(\lambda )w+g_2(\lambda )],$$

where

$$\begin{array}{c}{g}_1(\lambda )=p{\lambda }^p-q{\lambda }^{p/2+q/2}+q{\lambda }^{p/2-q/2}-p,\hfill \\ g_2(\lambda )=(p-q){\lambda }^{p+q}+(p+q){\lambda }^p-(p+q){\lambda }^q-(p-q).\hfill \end{array}$$

It follows from Lemmas 3 and 4 that

$$\begin{array}{c}F(x,y,p,q,w)\ge 0\text{for }p+q\ge 0\text{ and }w\ge 0,\hfill \\ F(x,y,p,q,w)\le 0\text{for }p+q\le 0\text{ and }w\ge 0.\hfill \end{array}$$

This evidently implies that

$$\mathrm{\Lambda }\ge 0\text{for }p+q\ge 0,w\ge 0\text{and}\mathrm{\Lambda }\le 0\text{for }p+q\le 0,w\ge 0.$$

By using the conclusions obtained in Cases 1 and 2 together with an application of Lemma 1, we are led to the desired results:

$H_{p,q,w}(x,y)$ is Schur-geometric convex on ${\mathbb{R}}_{++}^2$ when $p+q\ge 0$ and $w\ge 0$. Furthermore, $H_{p,q,w}(x,y)$ is Schur-geometric concave on ${\mathbb{R}}_{++}^2$ when $p+q\le 0$ and $w\ge 0$.

The proof of Theorem 1 is thus completed. □

Remark 1 The main result of Theorem E would follow as a special case of Theorem 1 ($q=0$). Namely, the result stated in Theorem 1 is an extension of the result given in [7].

4 An application

As an application of Theorem 1, we establish the following interesting inequalities for generalized Gini-Heronian means.

Theorem 2 Let $0<x\le y$, and let $1/2\le t_2\le t_1\le 1$ or $0\le t_1\le t_2\le 1/2$.

If $p+q\ge 0$ and $w\ge 0$, then

$$\begin{array}{rcl}{H}_{p,q,w}(\sqrt{xy},\sqrt{xy})& \le & H_{p,q,w}(y^{t_2}{x}^{1-t_2},x^{t_2}{y}^{1-t_2})\\ \le & H_{p,q,w}(y^{t_1}{x}^{1-t_1},x^{t_1}{y}^{1-t_1})\le H_{p,q,w}(x,y).\end{array}$$

(4.1)

If $p+q\le 0$ and $w\ge 0$, then

$$\begin{array}{rcl}{H}_{p,q,w}(\sqrt{xy},\sqrt{xy})& \ge & H_{p,q,w}(y^{t_2}{x}^{1-t_2},x^{t_2}{y}^{1-t_2})\\ \ge & H_{p,q,w}(y^{t_1}{x}^{1-t_1},x^{t_1}{y}^{1-t_1})\ge H_{p,q,w}(x,y),\end{array}$$

(4.2)

where $H_{p,q,w}(x,y)$ is given by

$$H_{p,q,w}(x,y)=\{\begin{array}{ll}{(\frac{x^p+w{(xy)}^{p/2}+y^p}{x^q+w{(xy)}^{q/2}+y^q})}^{1/(p-q)},& p\ne q,\\ exp(\frac{x^{p}lnx+(w/2){(xy)}^{p/2}ln(xy)+y^{p}lny}{x^p+w{(xy)}^{p/2}+y^p}),& p=q.\end{array}$$

Proof Using Lemma 2 with a substitution $a=lnx$, $b=lny$ gives

$$\begin{array}{rcl}(\frac{lnx+lny}{2},\frac{lnx+lny}{2})& \prec & (t_{2}lny+(1-t_2)lnx,t_{2}lnx+(1-t_2)lny)\\ \prec & (t_{1}lny+(1-t_1)lnx,t_{1}lnx+(1-t_1)lny)\prec (lnx,lny),\end{array}$$

which is equivalent to

$$\begin{array}{rcl}(ln\sqrt{xy},ln\sqrt{xy})& \prec & (ln\left(y^{t_2}{x}^{1-t_2}\right),ln\left(x^{t_2}{y}^{1-t_2}\right))\\ \prec & (ln\left(y^{t_1}{x}^{1-t_1}\right),ln\left(x^{t_1}{y}^{1-t_1}\right))\prec (lnx,lny).\end{array}$$

On the other hand, we derive from Theorem 1 that

$H_{p,q,w}(x,y)$ is Schur-geometric convex for $p+q\ge 0$ and $w\ge 0$, $H_{p,q,w}(x,y)$ is Schur-geometric concave for $p+q\le 0$ and $w\ge 0$.

Thus, from Definition 2, it follows that

$$\begin{array}{rcl}{H}_{p,q,w}(\sqrt{xy},\sqrt{xy})& \le & H_{p,q,w}(y^{t_2}{x}^{1-t_2},x^{t_2}{y}^{1-t_2})\\ \le & H_{p,q,w}(y^{t_1}{x}^{1-t_1},x^{t_1}{y}^{1-t_1})\le H_{p,q,w}(x,y)\end{array}$$

for $p+q\ge 0$ and $w\ge 0$; and that

$$\begin{array}{rcl}{H}_{p,q,w}(\sqrt{xy},\sqrt{xy})& \ge & H_{p,q,w}(y^{t_2}{x}^{1-t_2},x^{t_2}{y}^{1-t_2})\\ \ge & H_{p,q,w}(y^{t_1}{x}^{1-t_1},x^{t_1}{y}^{1-t_1})\ge H_{p,q,w}(x,y)\end{array}$$

for $p+q\le 0$ and $w\ge 0$.

The above-mentioned inequalities are the required inequalities in Theorem 2. This completes the proof of Theorem 2. □

Putting $q=0$ in Theorem 2 gives the following inequalities.

Theorem 3 Let $0<x\le y$, $w\ge 0$, and let $1/2\le t_2\le t_1\le 1$ or $0\le t_1\le t_2\le 1/2$. Then, for $p\ge 0$, we have the inequality

$$\begin{array}{rcl}G(x,y)& \le & H_{p,w}(y^{t_2}{x}^{1-t_2},x^{t_2}{y}^{1-t_2})\\ \le & H_{p,w}(y^{t_1}{x}^{1-t_1},x^{t_1}{y}^{1-t_1})\le H_{p,w}(x,y).\end{array}$$

(4.3)

Furthermore, for $p\le 0$ we have the inequality

$$\begin{array}{rcl}G(x,y)& \ge & H_{p,w}(y^{t_2}{x}^{1-t_2},x^{t_2}{y}^{1-t_2})\\ \ge & H_{p,w}(y^{t_1}{x}^{1-t_1},x^{t_1}{y}^{1-t_1})\ge H_{p,w}(x,y),\end{array}$$

(4.4)

where $G(x,y)=\sqrt{xy}$, $H_{p,w}(x,y)$ is given by

$$H_{p,w}(x,y)=\{\begin{array}{ll}{(\frac{x^p+w{(xy)}^{p/2}+y^p}{w+2})}^{1/p},& p\ne 0,\\ \sqrt{xy},& p=0.\end{array}$$

Remark 2 Inequalities (4.3) and (4.4) were first presented by Shi et al. in [7]. It is obvious that the inequalities given in Theorem 2 provide the generalized versions of these inequalities.