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.