Improvements of bounds for the Sándor–Yang means

Abstract

In the article, we provide new bounds for two Sándor–Yang means in terms of the arithmetic and contraharmonic means. Our results are the improvements of the previously known results.

1 Introduction

Let $p\in\mathbb{R}$ and $x,y > 0$ with $x \ne y$. Then the arithmetic mean $A(x, y)$, quadratic mean $Q(x, y)$, contraharmonic mean $C(x, y)$, Neuman–Sándor mean $NS(x, y)$ [1], Seiffert mean $T(x, y)$ [2,3,4,5], pth power mean $M_{p}(x, y)$ [6,7,8,9,10,11,12,13], and Schwab–Borchardt mean $SB(x, y)$ [14, 15] are defined by

$$\begin{aligned} &A(x,y)=\frac{x+y}{2},\qquad Q(x,y)=\sqrt{\frac{x^{2}+y^{2}}{2}},\qquad C(x, y)= \frac {x^{2}+y^{2}}{x+y}, \end{aligned}$$

(1.1)

$$\begin{aligned} &NS(x,y)=\frac{x-y}{2\sinh^{-1} (\frac{x-y}{x+y} )},\qquad T(x, y)=\frac{x-y}{2\arctan (\frac{x-y}{x+y} )}, \\ &M_{p}(x, y)= \textstyle\begin{cases} (\frac{x^{p}+y^{p}}{2} )^{1/p}, & p\neq0,\\ \sqrt{xy}, & p=0, \end{cases}\displaystyle \end{aligned}$$

(1.2)

and

$$ SB(x,y)= \textstyle\begin{cases} \frac{\sqrt{y^{2}-x^{2}}}{\arccos{(x/y)}}, & x< y,\\ \frac{\sqrt{x^{2}-y^{2}}}{\cosh^{-1}{(x/y)}}, & x>y, \end{cases} $$

respectively, where $\sinh^{-1}(t)=\log(t+\sqrt{t^{2}+1})$ and $\cosh ^{-1}(t)=\log(t+\sqrt{t^{2}-1})$ are the inverse hyperbolic sine and cosine functions.

Let $U(x,y)$ and $V(x,y)$ be the symmetric bivariate means. Then Yang [16] introduced the Sándor–Yang mean

$$ R_{UV}(x, y)=:V(x,y)e^{\frac{U(x, y)}{SB[U(x,y), V(x,y)]}-1}, $$

and provided the explicit formulas for $R_{AQ}(x,y)$ and $R_{QA}(x, y)$ as follows:

$$\begin{aligned} &R_{AQ}(x,y)=Q(x,y)e^{A(x,y)/T(x,y)-1}, \end{aligned}$$

(1.3)

$$\begin{aligned} &R_{QA}(x,y)=A(x,y)e^{Q(x,y)/NS(x,y)-1}. \end{aligned}$$

(1.4)

Recently, the bounds and properties for certain bivariate means and related special functions have attracted the attention of many researchers [17,18,19,20,21,22,23,24,25,26,27,28].

Zhao, Qian, and Song [29] proved that the double inequalities

$$\begin{aligned} &M_{\alpha}(a,b)< R_{QA}(a,b)< M_{\beta}(a,b), \\ &M_{\lambda}(a,b)< R_{AQ}(a,b)< M_{\mu}(a,b) \end{aligned}$$

hold for all $a, b>0$ with $a\neq b$ if and only if $\alpha\leq\log 2/[1+\log2-\log(1+\sqrt{2})]=1.5517\ldots$ , $\beta\geq5/3$, $\lambda\leq4\log2/(4+2\log2-\pi)=1.2351\ldots$ , and $\mu\geq4/3$.

Xu [30], and Xu, Chu, and Qian [31] proved that the two-sided inequalities

$$\begin{aligned} &C^{1/6}(x, y)A^{5/6}(x,y)< R_{AQ}(x,y)< \frac{1}{6}C(x,y)+\frac{5}{6}A(x,y), \end{aligned}$$

(1.5)

$$\begin{aligned} &C^{1/3}(x, y)A^{2/3}(x,y)< R_{QA}(x,y)< \frac{1}{3}C(x,y)+\frac{2}{3}A(x,y) \end{aligned}$$

(1.6)

are valid for all $x, y>0$ with $x\neq y$.

The main purpose of this paper is to improve the bounds for $R_{AQ}(x,y)$ and $R_{QA}(x,y)$ given by (1.5) and (1.6).

2 Lemmas

In order to prove our main results, we need four lemmas which we present in this section.

Lemma 2.1

(see [32, Theorem 1.25])

Let $a, b\in\mathbb{R}$ with $a< b$, $f, g: [a, b]\rightarrow\mathbb {R}$ be continuous on $[a, b]$ and differentiable on $(a, b)$, and $g^{\prime}(x)\neq0$ on $(a, b)$. If $f^{\prime}(x)/g^{\prime}(x)$ is increasing (decreasing) on $(a, b)$, then so are the functions

$$ \frac{f(x)-f(a)}{g(x)-g(a)},\qquad \frac{f(x)-f(b)}{g(x)-g(b)}. $$

If $f^{\prime}(x)/g^{\prime}(x)$ is strictly monotone, then the monotonicity in the conclusion is also strict.

Lemma 2.2

(see [33, Lemma 1.1])

Suppose that the power series $f(x)=\sum_{n=0}^{\infty}a_{n}x^{n}$ and $g(x)=\sum_{n=0}^{\infty}b_{n}x^{n}$ have the radius of convergence $r>0$, and $b_{n}>0$ for all $n=0, 1, 2, \ldots$ . If there exists $n_{0}\geq1$ such that the non-constant sequence $\{a_{n}/b_{n}\} _{n=0}^{\infty}$ is increasing (decreasing) for $0\leq n\leq n_{0}$ and decreasing (increasing) for $n\geq n_{0}$, then there exists $x_{0}\in(0,r)$ such that the function $f(x)/g(x)$ is strictly increasing (decreasing) on $(0,x_{0})$ and decreasing (increasing) on $(x_{0},r)$.

Lemma 2.3

The function

$$ f(t)=\frac{\frac{2}{3}\log[\sec(t)]+\frac{t}{\tan(t)}-1}{\log [\frac{\sec^{2}(t)+5}{6} ]-\frac{\log[\sec(t)]}{3}} $$

(2.1)

is strictly decreasing from $(0, \pi/4)$ onto $(3\pi+4\log 2-12)/[2(6\log7-7\log2-6\log3)], 12/25)$.

Proof

Let

$$\begin{aligned} &f_{1}(t)=\frac{2}{3}\log\bigl[\sec(t)\bigr]+ \frac{t}{\tan(t)}-1, \\ &f_{2}(t)=\log \biggl[\frac{\sec^{2}(t)+5}{6} \biggr]-\frac{\log[\sec(t)]}{3}, \\ &f_{3}(t)=2\tan(t)+\sin(t)\cos(t)-3t,\qquad f_{4}(t)= \frac{5\sin ^{5}(t)}{\cos(t)[1+5\cos^{2}(t)]}. \end{aligned}$$

Then it is not difficult to verify that

$$\begin{aligned} &f_{1}\bigl(0^{+}\bigr)=f_{2} \bigl(0^{+}\bigr)=f_{3}\bigl(0^{+} \bigr)=f_{4}\bigl(0^{+}\bigr)=0, \end{aligned}$$

(2.2)

$$\begin{aligned} &f(t)=\frac{f_{1}(t)}{f_{2}(t)},\qquad \frac{f^{\prime}_{1}(t)}{f^{\prime }_{2}(t)}=\frac{f_{3}(t)}{f_{4}(t)}, \end{aligned}$$

(2.3)

$$\begin{aligned} &\frac{f^{\prime}_{3}(t)}{f^{\prime}_{4}(t)}=\frac{2 [1+5\cos ^{2}(t) ]^{2}}{5 [10\cos^{4}(t)+19\cos^{2}(t)+1 ]}, \end{aligned}$$

(2.4)

$$\begin{aligned} &\biggl[\frac{f^{\prime}_{3}(t)}{f^{\prime}_{4}(t)} \biggr]^{\prime} =-\frac{2\sin(2t)[1+5\cos^{2}(t)][75\cos^{2}(t)-9]}{5 [10\cos ^{4}(t)+19\cos^{2}(t)+1 ]^{2}}< 0 \end{aligned}$$

(2.5)

for $t\in(0, \pi/4)$.

Therefore, the function $f(t)$ is strictly decreasing on $(0, \pi/4)$ follows easily from Lemma 2.1, (2.2), (2.3), and (2.5).

It follows from (2.1)–(2.4) that

$$ f\bigl(0^{+}\bigr)=\lim_{t\rightarrow0^{+}}\frac{f^{\prime}_{3}(t)}{f^{\prime }_{4}(t)}= \frac{12}{25} $$

and

$$ f \biggl(\frac{\pi}{4} \biggr)=\lim_{t\rightarrow\pi/4} \frac{\frac {2}{3}\log[\sec(t)]+\frac{t}{\tan(t)}-1}{\log [\frac{\sec ^{2}(t)+5}{6} ]-\frac{\log[\sec(t)]}{3}} =\frac{3\pi+4\log2-12}{2(6\log7-7\log2-6\log3)}=0.4258\ldots. $$

 □

Lemma 2.4

The function

$$ g(t)=\frac{t\coth(t)-\frac{2\log[\cosh(t)]}{3}-1}{\log [\frac {\cosh^{2}(t)+2}{3} ]-\frac{2\log[\cosh(t)]}{3}} $$

(2.6)

is strictly decreasing from $(0, \log(1+\sqrt{2}))$ onto $([3\sqrt {2}\log(1+\sqrt{2})-\log2-3]/(5\log2-3\log3), 3/10)$.

Proof

Let

$$\begin{aligned} &g_{1}(t)=t\coth(t)-\frac{2\log[\cosh(t)]}{3}-1, \\ &g_{2}(t)=\log \biggl[\frac{\cosh^{2}(t)+2}{3} \biggr]-\frac{2\log [\cosh(t)]}{3}, \\ &g_{3}(t)= \bigl[3\sinh(t)+\sinh^{3}(t)-3t\cosh(t) \bigr] \bigl[\cosh^{2}(t)+2 \bigr], \\ &g_{4}(t)=4\sinh^{5}(t). \end{aligned}$$

Then we clearly see that

$$\begin{aligned} &g_{1}\bigl(0^{+}\bigr)=g_{2} \bigl(0^{+}\bigr)=g_{3}\bigl(0^{+} \bigr)=g_{4}\bigl(0^{+}\bigr)=0, \end{aligned}$$

(2.7)

$$\begin{aligned} &g(t)=\frac{g_{1}(t)}{g_{2}(t)}, \qquad\frac{g^{\prime}_{1}(t)}{g^{\prime }_{2}(t)}=\frac{g_{3}(t)}{g_{4}(t)}. \end{aligned}$$

(2.8)

Elaborate computations lead to

$$\begin{aligned} \frac{g^{\prime}_{3}(t)}{g^{\prime}_{4}(t)}&=\frac{5\sinh(4t)+50\sinh (2t)-84t-36t\cosh(2t)}{20[\sinh(4t)-2\sinh(2t)]} \\ &=\frac{5\sum_{n=0}^{\infty}\frac{(4t)^{2n+1}}{(2n+1)!}+50\sum_{n=0}^{\infty}\frac{(2t)^{2n+1}}{(2n+1)!}-84t-36t\sum_{n=0}^{\infty }\frac{(2t)^{2n}}{(2n)!}}{20\sum_{n=0}^{\infty}\frac {(4t)^{2n+1}}{(2n+1)!}-40\sum_{n=0}^{\infty}\frac {(2t)^{2n+1}}{(2n+1)!}}=:\frac{\sum_{n=0}^{\infty}a_{n}}{\sum_{n=0}^{\infty}b_{n}}, \end{aligned}$$

(2.9)

where

$$ a_{n}=\frac{ (5\times2^{2n+1}-9n-1 )2^{2n+5}}{(2n+3)!},\qquad b_{n}=\frac{10 (2^{2n+2}-1 )2^{2n+5}}{(2n+3)!}. $$

(2.10)

From (2.10) we clearly see that

$$\begin{aligned} &\frac{a_{1}}{b_{1}}-\frac{a_{0}}{b_{0}}=-\frac{1}{10}< 0, \end{aligned}$$

(2.11)

$$\begin{aligned} &\frac{a_{n+1}}{b_{n+1}}-\frac{a_{n}}{b_{n}}=\frac{9 [3(2n-1)2^{2n+1}+1 ]}{10 (2^{2n+2}-1 ) (2^{2n+4}-1 )}>0 \end{aligned}$$

(2.12)

for all $n\geq1$, and

$$ b_{n}>0 $$

(2.13)

for all $n\geq0$.

It follows from Lemma 2.2 and (2.9)–(2.13) that there exists $t_{0}\in (0, \infty)$ such that the function $g^{\prime}_{3}(t)/g^{\prime}_{4}(t)$ is strictly decreasing on $(0, t_{0})$ and strictly increasing on $(t_{0}, \infty)$.

Note that

$$\begin{aligned} &\biggl[\frac{g^{\prime}_{3}(t)}{g^{\prime}_{4}(t)} \biggr]^{\prime} =\frac{4\cosh(4t)+16\cosh(2t)-18t\sinh(2t)-21}{5[\sinh(4t)-2\sinh(2t)]} \\ &\phantom{\biggl[\frac{g^{\prime}_{3}(t)}{g^{\prime}_{4}(t)} \biggr]^{\prime} =}{}-\frac{[5\sinh(4t)+50\sinh(2t)-84t-36t\cosh(2t)][\cosh(4t)-\cosh (2t)]}{5[\sinh(4t)-2\sinh(2t)]^{2}}, \\ &\biggl[\frac{g^{\prime}_{3}(t)}{g^{\prime}_{4}(t)} \biggr]^{\prime }_{t=\log(1+\sqrt{2})} = \frac{(13{,}464\sqrt{2}+19{,}041)\log(1+\sqrt{2})-12{,}117\sqrt {2}-17{,}136}{5770+4080\sqrt{2}} \\ &\phantom{\biggl[\frac{g^{\prime}_{3}(t)}{g^{\prime}_{4}(t)} \biggr]^{\prime }_{t=\log(1+\sqrt{2})} }=-0.0613\ldots< 0. \end{aligned}$$

(2.14)

From (2.14) and piecewise monotonicity of the function $g^{\prime }_{3}(t)/g^{\prime}_{4}(t)$, we clearly see that $t_{0}>\log(1+\sqrt {2})$ and the function $g^{\prime}_{3}(t)/g^{\prime}_{4}(t)$ is strictly decreasing on $(0, \log(1+\sqrt{2}))$. Then Lemma 2.1 together with (2.7) and (2.8) leads to the conclusion that $g(t)$ is strictly decreasing on $(0, \log(1+\sqrt{2}))$.

It follows from (2.6)–(2.10) that

$$ g\bigl(0^{+}\bigr)=\frac{a_{0}}{b_{0}}=\frac{3}{10},\qquad g\bigl( \log(1+\sqrt{2})\bigr)=\frac {3\sqrt{2}\log(1+\sqrt{2})-\log2-3}{5\log2-3\log3}=0.2719\ldots. $$

 □

3 Main results

Theorem 3.1

The double inequality

$$\begin{aligned} &\biggl[\frac{1}{6}C(x,y)+\frac{5}{6}A(x,y) \biggr]^{\alpha_{1}} \bigl[C^{1/6}(x,y)A^{5/6}(x,y) \bigr]^{1-\alpha_{1}}\\ &\quad < R_{AQ}(x,y) \\ &\quad< \biggl[\frac{1}{6}C(x,y)+\frac{5}{6}A(x,y) \biggr]^{\beta_{1}} \bigl[C^{1/6}(x,y)A^{5/6}(x,y) \bigr]^{1-\beta_{1}} \end{aligned}$$

holds for all $x, y>0$ with $x\neq y$ if and only if $\alpha_{1}\leq (3\pi+4\log2-12)/[2(6\log7-7\log2-6\log3)]=0.4258\ldots$ and $\beta_{1}\geq12/25$.

Proof

Since $A(x,y)$, $R_{AQ}(x, y)$, and $C(x,y)$ are symmetric and homogenous of degree one, without loss of generality, we assume that $x> y> 0$. Let $\nu=(x-y)/(x+y)\in(0, 1)$ and $t=\arctan(\nu)\in(0, \pi/4)$. Then (1.1)–(1.3) lead to

$$\begin{aligned} &\frac{\log [R_{AQ}(x,y) ]-\log [C^{1/6}(x,y)A^{5/6}(x,y) ]}{\log [C(x,y)/6+5A(x, y)/6 ]-\log [C^{1/6}(x,y)A^{5/6}(x,y) ]} \\ &\quad=\frac{\log(\sqrt{1+\nu^{2}})+\arctan(\nu)/\nu-1-\log (\sqrt [6]{1+\nu^{2}} )}{\log [(1+\nu^{2})/6+5/6 ]-\log (\sqrt[6]{1+\nu^{2}} )} \\ &\quad=\frac{\frac{2}{3}\log[\sec(t)]+\frac{t}{\tan(t)}-1}{\log [\frac{\sec^{2}(t)+5}{6} ]-\frac{\log[\sec(t)]}{3}}. \end{aligned}$$

(3.1)

Therefore, Theorem 3.1 follows easily from Lemma 2.3 and (3.1). □

Theorem 3.2

The two-sided inequalities

$$\begin{aligned} &\biggl[\frac{1}{3}C(x,y)+\frac{2}{3}A(x,y) \biggr]^{\alpha_{2}} \bigl[C^{1/3}(x,y)A^{2/3}(x,y) \bigr]^{1-\alpha_{2}}\\ &\quad< R_{QA}(x,y) \\ &\quad< \biggl[\frac{1}{3}C(x,y)+\frac{2}{3}A(x,y) \biggr]^{\beta_{2}} \bigl[C^{1/3}(x,y)A^{2/3}(x,y) \bigr]^{1-\beta_{2}} \end{aligned}$$

are valid for all $x, y>0$ with $x\neq y$ if and only if $\alpha _{2}\leq[3\sqrt{2}\log(1+\sqrt{2})-\log2-3]/(5\log2-3\log 3)=0.2719\ldots$ and $\beta_{2}\geq3/10$.

Proof

Since $A(x,y)$, $R_{QA}(x,y)$, and $C(x,y)$ are symmetric and homogenous of degree one, without loss generality, we assume that $x> y>0$. Let $v=(x-y)/(x+y)\in(0, 1)$ and $t=\sinh^{-1}(v)\in(0, \log(1+\sqrt {2})$. Then from (1.1), (1.3), and (1.4) we clearly see that

$$\begin{aligned} &\frac{\log [R_{QA}(x,y) ]-\log [C^{1/3}(x,y)A^{2/3}(x,y) ]}{\log [C(x,y)/3+2A(x,y)/3 ]-\log [C^{1/3}(x,y)A^{2/3}(x,y) ]} \\ &\quad=\frac{ [\sqrt{1+v^{2}}\sinh^{-1}(v) ]/v-1-\log (\sqrt [3]{1+v^{2}} )}{\log [(1+v^{2})/3+2/3 ]-\log (\sqrt[3]{1+v^{2}} )} \\ &\quad=\frac{t\coth(t)-\frac{2\log[\cosh(t)]}{3}-1}{\log [\frac {\cosh^{2}(t)+2}{3} ]-\frac{2\log[\cosh(t)]}{3}}. \end{aligned}$$

(3.2)

Therefore, Theorem 3.2 follows easily from Lemma 2.4 and (3.2). □

From (1.3), (1.4), and Theorems 3.1 and 3.2 we get Corollary 3.3 immediately.

Corollary 3.3

Let

$$\begin{aligned} \lambda(\alpha; a, b)={}&6\alpha\log\bigl[C(a,b)+5A(a,b)\bigr] \\ &{}+(1-\alpha)\bigl[\log C(a,b)+5\log A(a,b)\bigr]-6\log Q(a,b)+6(1-\alpha\log6), \\ \mu(\alpha; a, b)={}&3\alpha\log\bigl[C(a,b)+2A(a,b)\bigr] \\ &{}+(1-\alpha)\log C(a,b)-(1+2\alpha)\log A(a,b)+3(1-\alpha\log3). \end{aligned}$$

Then the double inequalities

$$\begin{aligned} &\frac{6A(a,b)}{\lambda(\beta_{1}; a, b)}< T(a,b)< \frac {6A(a,b)}{\lambda(\alpha_{1}; a, b)}, \\ &\frac{3Q(a,b)}{\mu(\beta_{2}; a, b)}< NS(a,b)< \frac{3Q(a,b)}{\mu (\alpha_{2}; a, b)} \end{aligned}$$

hold for all $a, b>0$ with $a\neq b$ if and only if $\alpha_{1}\leq (3\pi+4\log2-12)/[2(6\log7-7\log2-6\log3)]=0.4258\ldots$ , $\beta_{1}\geq12/25$, $\alpha_{2}\leq[3\sqrt{2}\log(1+\sqrt {2})-\log2-3]/(5\log2-3\log3)=0.2719\ldots$ , and $\beta_{2}\geq3/10$.

4 Results and discussion

In the article, we present the best possible parameters $\alpha_{1}$, $\beta_{1}$, $\alpha_{2}$, and $\beta_{2}$ such that the double inequalities

$$\begin{aligned} &\biggl[\frac{1}{6}C(x,y)+\frac{5}{6}A(x,y) \biggr]^{\alpha_{1}} \bigl[C^{1/6}(x,y)A^{5/6}(x,y) \bigr]^{1-\alpha_{1}}\\ &\quad< R_{AQ}(x,y) \\ &\quad< \biggl[\frac{1}{6}C(x,y)+\frac{5}{6}A(x,y) \biggr]^{\beta_{1}} \bigl[C^{1/6}(x,y)A^{5/6}(x,y) \bigr]^{1-\beta_{1}}, \\ &\biggl[\frac{1}{3}C(x,y)+\frac{2}{3}A(x,y) \biggr]^{\alpha_{2}} \bigl[C^{1/3}(x,y)A^{2/3}(x,y) \bigr]^{1-\alpha_{2}}\\ &\quad< R_{QA}(x,y) \\ &\quad< \biggl[\frac{1}{3}C(x,y)+\frac{2}{3}A(x,y) \biggr]^{\beta_{2}} \bigl[C^{1/3}(x,y)A^{2/3}(x,y) \bigr]^{1-\beta_{2}} \end{aligned}$$

hold for all $x, y>0$ with $x\neq y$. Our results are the improvements of the inequalities given by (1.5) and (1.6).

5 Conclusion

We present sharp upper and lower bounds for the Sándor–Yang means $R_{AQ}$ and $R_{QA}$ in terms of the arithmetic and contraharmonic means and provide new bounds for the Seiffert mean T and Neuman–Sándor mean NS. Our approach may have further applications in the theory of bivariate means and special functions.