1 Introduction

Let \(\nu\in(-\infty, \infty)\) and \(\sigma, \tau>0\) with \(\sigma\neq\tau \). Then we denote by

$$ \begin{gathered} \mathbf{G}(\sigma, \tau)=\sigma^{1/2}\tau^{1/2}, \qquad \mathbf{U}(\sigma, \tau)=\frac{\sqrt{2}(\sigma-\tau)}{2\arctan(\frac {\sqrt{2}(\sigma-\tau)}{2\sqrt{\sigma\tau}} )}, \\\mathbf{Q}(\sigma , \tau)= \biggl(\frac{\sigma^{2}+\tau^{2}}{2} \biggr)^{1/2},\end{gathered} $$
(1.1)

and

$$ \mathbf{H}_{\nu}(\sigma, \tau)= \biggl(\frac{\sigma^{\nu}+\tau^{\nu }}{2} \biggr)^{1/\nu}\quad(\nu\neq0), \qquad\mathbf{ H}_{0}(\sigma, \tau)=\sigma^{1/2}\tau^{1/2} $$

the geometric mean, Yang mean [1], quadratic mean [2], and νth Hölder mean [3] of σ and τ, respectively.

It is not difficult to verify that the νth Hölder mean \(H_{\nu }(\sigma, \tau)\) is strictly increasing with respect to \(\nu\in(-\infty , \infty)\) for all distinct positive real numbers σ and τ, and

$$\begin{gathered} \mathbf{H}_{-1}(\sigma, \tau)=\frac{2\sigma\tau}{\sigma+\tau}=\mathbf {H}(\sigma, \tau), \qquad\mathbf{H}_{0}(\sigma, \tau)=\sigma^{1/2} \tau^{1/2}=\mathbf{G}(\sigma, \tau), \\ \mathbf{H}_{1}(\sigma, \tau)=\frac{\sigma+\tau}{2}=\mathbf{A}(\sigma, \tau), \qquad\mathbf{H}_{2}(\sigma, \tau)= \biggl(\frac{\sigma^{2}+\tau ^{2}}{2} \biggr)^{1/2}=\mathbf{Q}(\sigma, \tau)\end{gathered} $$

are the classical harmonic, geometric, arithmetic, and quadratic means of σ and τ, respectively.

The bivariate means have in the past decades been the subject of intense research activity [4,5,6,7,8,9,10,11,12,13] because many important special functions can be expressed by the bivariate means [14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31] and they have wide applications in mathematics, statistics, physics, economics [32,33,34,35,36,37,38,39,40,41,42,43,44,45,46,47,48,49,50,51,52,53,54,55], and many other natural and human social sciences [56,57,58,59,60,61,62,63,64,65,66,67,68,69,70,71,72,73,74,75,76].

Yang, Wu, and Chu [77] proved that \(\kappa_{1}= 2\log2/(2\log\pi-\log 2)\simeq0.8684\) is the largest possible value and \(\kappa_{2}=4/3\) is the least possible value such that the two-sided inequality

$$ \mathbf{H}_{\kappa_{1}}(\sigma, \tau)< \mathbf{U}(\sigma, \tau)< \mathbf{H}_{\kappa_{2}}(\sigma, \tau) $$

takes place for all distinct positive real numbers σ and τ, which leads to the conclusion that

$$ \mathbf{G}(\sigma, \tau)< \mathbf{U}(\sigma, \tau)< \mathbf{Q}(\sigma, \tau) $$

for \(\sigma, \tau>0\) with \(\sigma\neq\tau\).

In [78], Qian and Chu found that \(\lambda=\lambda_{0}\simeq0.5451\) and \(\mu=2\) are the best possible parameters such that the double inequality

$$ \mathcal{L}_{\lambda}(\sigma, \tau)< \mathbf{U}(\sigma, \tau)< \mathcal{L}_{\mu}(\sigma, \tau) $$

holds for all unequal positive real numbers σ and τ, where

$$\begin{gathered} \mathcal{L}_{\nu}(\sigma, \tau)= \biggl[\frac{\sigma^{\nu+1}-\tau^{\nu +1}}{(\nu+1)(\sigma-\tau)} \biggr]^{1/\nu} \quad(\nu\neq-1, 0) \\ \mathcal{L}_{-1}(\sigma, \tau)=\frac{\sigma-\tau}{\log\sigma-\log\tau}, \qquad \mathcal{L}_{0}(\sigma, \tau)=\frac{1}{e} \biggl( \frac{\sigma^{\sigma}}{\tau^{\tau}} \biggr)^{1/(\sigma-\tau)}\end{gathered} $$

is the νth generalized logarithmic mean of σ and τ.

The Sándor–Yang mean \(\mathbf{SY}(\sigma, \tau)\) [1] and two-parameter geometric and arithmetic mean \(\mathbf{GA}_{\eta, \nu}(\sigma, \tau)\) [79] are defined by

$$ \mathbf{SY}(\sigma, \tau)=\mathbf{Q}(\sigma, \tau)e^{\mathbf{G}(\sigma, \tau)/\mathbf{U}(\sigma, \tau)-1} $$
(1.2)

and

$$ \mathbf{GA}_{\eta, \nu}(\sigma, \tau)=\mathbf{G}^{\nu} \bigl[\eta \sigma+(1-\eta)\tau, \eta\tau+(1-\eta)\sigma \bigr]\mathbf{A}^{1-\nu}( \sigma, \tau), $$
(1.3)

respectively.

Identity (1.3) leads to the conclusion that

$$\begin{aligned}& \mathbf{GA}_{p, 1}(\sigma, \tau)=\mathbf{G} \bigl[p\sigma+(1-p)\tau, p\tau+(1-p) \sigma \bigr], \end{aligned}$$
(1.4)
$$\begin{aligned}& \mathbf{GA}_{p, 2}(\sigma, \tau)=\mathbf{H} \bigl[p\sigma+(1-p)\tau, p \tau+(1-p)\sigma \bigr], \end{aligned}$$
(1.5)

and

$$ \mathbf{GA}_{p, 0}(\sigma, \tau)=\mathbf{GA}_{1/2, 1/2}(\sigma, \tau)=\mathbf{A}(\sigma, \tau). $$
(1.6)

Chu et al. [79] proved that the inequalities

$$ \mathbf{GA}_{\eta_{1}, \nu}(\sigma, \tau)>\mathbf{AGM}(\sigma, \tau) $$

and

$$ \mathbf{GA}_{\eta_{2}, \nu}(\sigma, \tau)>\mathbf{L}(\sigma, \tau) $$

are valid for all distinct positive real numbers σ and τ if and only if

$$ \eta_{1}\geq\frac{1}{2}-\frac{\sqrt{2\nu}}{4\nu}, \qquad \eta_{2}\geq\frac{1}{2}-\frac{\sqrt{6\nu}}{6\nu} $$

if \(\nu\in[1, \infty)\) and \(0<\eta_{1}, \eta_{2}<1/2\), where

$$ \mathbf{L}(\sigma, \tau)=\mathcal{L}_{-1}(\sigma, \tau)= \frac{\sigma-\tau}{\log\sigma-\log\tau} $$

and

$$ \mathbf{AGM}(\sigma, \tau)=\frac{\pi}{2\int_{0}^{\pi}\frac{dt}{\sqrt {\sigma^{2}\cos^{2}t+\tau^{2}\sin^{2}t}}} $$

are the logarithmic and Gaussian arithmetic-geometric means of σ and τ, respectively.

Zhang, Yang, and Qian [80], and He et al. [81] proved that

$$ \lambda_{1}=\lambda_{2}=\frac{\sqrt{2}}{e}\simeq0.5203, \qquad\lambda_{3}=\frac{2\log2}{2+\log2}\simeq0.5147, \qquad \nu_{1}=\frac{5}{6}, \qquad\nu_{2}=\nu_{3}= \frac{2}{3} $$

are the best possible parameters such that the double inequalities

$$\begin{gathered} \lambda_{1}\mathbf{A}(\sigma, \tau)+(1-\lambda_{1}) \mathbf{H}(\sigma, \tau)< \mathbf{SY}(\sigma, \tau)< \nu_{1}\mathbf{A}( \sigma, \tau)+(1-\nu_{1})\mathbf{H}(\sigma, \tau), \\ \lambda_{2}\mathbf{A}(\sigma, \tau)+(1-\lambda_{1}) \mathbf{G}(\sigma, \tau)< \mathbf{SY}(\sigma, \tau)< \nu_{2}\mathbf{A}( \sigma, \tau)+(1-\nu_{2})\mathbf{G}(\sigma, \tau),\end{gathered} $$

and

$$ \mathbf{H}_{\lambda_{3}}(\sigma, \tau)< \mathbf{SY}(\sigma, \tau)< \mathbf{H}_{\nu_{3}}(\sigma, \tau) $$
(1.7)

hold for all \(\sigma, \tau>0\) with \(\sigma\neq\tau\).

From (1.4)–(1.7) and the monotonicity of the function \(\nu\rightarrow \mathbf{H}_{\nu}(\sigma, \tau)\), we clearly see that

$$ \begin{aligned}[b] \mathbf{GA}_{1, 2}(\sigma, \tau)&=\mathbf{H}(\sigma, \tau)= \mathbf{H}_{-1}(\sigma, \tau)< \mathbf{G}(\sigma, \tau)= \mathbf{H}_{0}(\sigma, \tau) \\ &< \mathbf{SY}(\sigma, \tau)< \mathbf{H}_{1}(\sigma, \tau)=\mathbf{A}( \sigma, \tau)=\mathbf{GA}_{p, 0}(\sigma, \tau)=\mathbf{GA}_{1/2, 1/2}( \sigma, \tau)\end{aligned} $$
(1.8)

for all \(\sigma, \tau>0\) with \(\sigma\neq\tau\).

Motivated by inequality (1.8), we naturally ask the question: For fixed \(p\in\mathbb{R}\), what are the best possible parameters λ and μ on the interval \((0, 1/2)\) or \((1/2, 1)\) depending only on the parameter p such that the double inequality

$$ \mathbf{GA}_{\lambda, p}(\sigma, \tau)< \mathbf{SY}(\sigma, \tau)< \mathbf{GA}_{\mu, p}(\sigma, \tau) $$

is valid for all unequal positive real numbers σ and τ?

It is the aim of the article to answer the question in the case of \(p\in[1, \infty)\) and \(\lambda, \mu\in(0, 1/2)\).

2 Lemmas

Lemma 2.1

(see [82, Theorem 1.25])

Let \(\kappa_{1}, \kappa_{2}\in\mathbb{R}\)with \(\kappa_{1}<\kappa _{2}\), \(\mathcal{F}, \mathcal{G}: [\kappa_{1}, \kappa_{2}]\rightarrow \mathbb{R}\)be continuous on \([\kappa_{1}, \kappa_{2}]\)and differentiable on \((\kappa_{1}, \kappa_{2})\)with \(\mathcal{G}^{\prime }(t)\neq0\)on \((\kappa_{1}, \kappa_{2})\). Then both the functions

$$ \frac{\mathcal{F}(t)-\mathcal{F}(\kappa_{1})}{\mathcal{G}(t)-\mathcal {G}(\kappa_{1})} $$

and

$$ \frac{\mathcal{F}(t)-\mathcal{F}(\kappa_{2})}{\mathcal{G}(t)-\mathcal {G}(\kappa_{2})} $$

are (strictly) increasing (decreasing) on \((\kappa_{1}, \kappa_{2})\)if \(\mathcal{F}^{\prime}(t)/\mathcal{G}^{\prime}(t)\)is (strictly) increasing (decreasing) on \((\kappa_{1}, \kappa_{2})\).

Lemma 2.2

The inequality

$$ \frac{1}{3p}+ \biggl(\frac{2}{e^{2}} \biggr)^{1/p}< 1 $$

holds for all \(p\geq1\).

Proof

Let \(p\in[1, \infty)\) and

$$ f_{1}(p)=\frac{1}{3p}+ \biggl(\frac{2}{e^{2}} \biggr)^{1/p}. $$
(2.1)

Then (2.1) leads to

$$\begin{aligned}& \lim_{p\rightarrow\infty}f_{1}(p)=1, \end{aligned}$$
(2.2)
$$\begin{aligned}& \begin{aligned}[b] f_{1}^{\prime}(p)&=\frac{2}{p^{2}}\log \biggl( \frac{\sqrt{2}e}{2} \biggr) \biggl[ \biggl(\frac{\sqrt{2}}{e} \biggr)^{2/p} -\frac{1}{6\log(\frac{\sqrt{2}e}{2} )} \biggr] \\ &\geq\frac{2}{p^{2}}\log \biggl(\frac{\sqrt{2}e}{2} \biggr) \biggl[ \biggl( \frac{\sqrt{2}}{e} \biggr)^{2} -\frac{1}{6\log(\frac{\sqrt{2}e}{2} )} \biggr] \\ &=\frac{12\log(\frac{\sqrt{2}e}{2} )-e^{2}}{3e^{2}p^{2}}.\end{aligned} \end{aligned}$$
(2.3)

Note that

$$ 12\log \biggl(\frac{\sqrt{2}e}{2} \biggr)-e^{2}\simeq0.4521>0. $$
(2.4)

Therefore, Lemma 2.2 follows easily from (2.1)–(2.4). □

Lemma 2.3

The function

$$ f_{2}(x)=\frac{4(x^{2}+1)\arctan(x)+x(x^{2}+2)}{x(3x^{2}+2)} $$
(2.5)

is strictly decreasing from \((0, \infty)\)on \((1/3, 3)\).

Proof

It follows from (2.5) that

$$ f_{2} \bigl(0^{+} \bigr)=3, \qquad\lim_{x\rightarrow\infty}f_{2}(x)= \frac{1}{3}, $$
(2.6)

where and in what follows \(f (\lambda^{+} )\) denotes the right limit of the function f at λ.

Let

$$ \varphi_{1}(x)=4\arctan(x)+\frac{x(x^{2}+2)}{x^{2}+1}, \qquad \varphi_{2}(x)=\frac{x(3x^{2}+2)}{x^{2}+1}. $$

Then we clearly see that

$$\begin{aligned}& \varphi_{1} \bigl(0^{+} \bigr)=\varphi_{2} \bigl(0^{+} \bigr)=0, \qquad f_{2}(x)=\frac{\varphi_{1}(x)}{\varphi_{2}(x)}, \\& \frac{\varphi_{1}^{\prime}(x)}{\varphi_{2}^{\prime}(x)}=\frac {x^{2}+3}{3x^{2}+1}. \end{aligned}$$
(2.7)

It is not difficult to verify that the function \(x\rightarrow\varphi _{1}^{\prime}(x)/\varphi_{2}^{\prime}(x)\) is strictly decreasing on \((0, \infty)\).

Therefore, Lemma 2.3 follows from (2.6), (2.7), and Lemma 2.1 together with the monotonicity of the function \(\varphi_{1}^{\prime}(x)/\varphi _{2}^{\prime}(x)\) on the interval \((0, \infty)\). □

Lemma 2.4

Let \(0< u<1\), \(p\geq1\), and

$$ g(u, p; x)=\frac{p}{2}\log \biggl(\frac{(1-u)x^{2}+2}{x^{2}+2} \biggr)+ \frac{1}{2}\log \biggl(\frac{x^{2}+2}{2(x^{2}+1)} \biggr)-\frac{\arctan (x)}{x}+1. $$
(2.8)

Then the following statements are true:

  1. (1)

    \(g(u, p; x)>0\)for all \(x\in(0, \infty)\)if and only if \(u\leq1/(3p)\);

  2. (2)

    \(g(u, p; x)<0\)for all \(x\in(0, \infty)\)if and only if \(u\geq 1-(2/e^{2})^{1/p}\).

Proof

From (2.8) we clearly see that

$$\begin{aligned}& g \bigl(u, p; 0^{+} \bigr)=0, \end{aligned}$$
(2.9)
$$\begin{aligned}& \lim_{x\rightarrow\infty}g(u, p; x)=\frac{p}{2}\log(1-u)+1- \frac{1}{2}\log2. \end{aligned}$$
(2.10)

Let

$$ g_{0}(p, x)=\frac{(x^{2}+2)[(x^{2}+2)\arctan (x)-2x]}{x^{2}[(x^{2}+2)\arctan(x)+2(p-1)x]}. $$
(2.11)

Then

$$ g_{0} \bigl(p, 0^{+} \bigr)=\frac{1}{3p}, \qquad\lim _{x\rightarrow\infty}g_{0}(p, x)=1. $$
(2.12)

Differentiating \(g(u, p; x)\) with respect to x gives

$$ \frac{\partial g(u, p; x)}{\partial x}=\frac{(x^{2}+2)\arctan (x)+2(p-1)x}{(x^{2}+2)[(1-u)x^{2}+2]} \bigl[g_{0}(p, x)-u \bigr]. $$
(2.13)

Let

$$ g_{1}(x)=\arctan(x)-\frac{2x}{x^{2}+2}, \qquad g_{2}(x)= \frac{x^{2}}{x^{2}+2} \biggl[\arctan(x)+\frac{2(p-1)x}{x^{2}+2} \biggr]. $$

Then we clearly see that

$$\begin{aligned}& g_{0}(p, x)=\frac{g_{1}(x)}{g_{2}(x)}, \qquad g_{1} \bigl(0^{+} \bigr)=g_{2} \bigl(0^{+} \bigr)=0, \end{aligned}$$
(2.14)
$$\begin{aligned}& \begin{aligned}[b] \frac{g_{1}^{\prime}(x)}{g_{2}^{\prime}(x)}&=\frac {x(x^{2}+2)(3x^{2}+2)}{4(x^{2}+1)(x^{2}+2)\arctan (x)+x[(3-2p)x^{4}+2(5p-3)x^{2}+4(3p-2)]} \\ &=\frac{1}{\frac{4(x^{2}+1)\arctan(x)+x(x^{2}+2)}{x(3x^{2}+2)}+\frac {2(p-1)}{3}\frac{23x^{2}+22}{(x^{2}+2)(3x^{2}+2)}-\frac{2(p-1)}{3}}.\end{aligned} \end{aligned}$$
(2.15)

It is not difficult to verify that the function \(x\rightarrow (23x^{2}+22)/[(x^{2}+2)(3x^{2}+2)]\) is strictly decreasing on \((0, \infty)\). Then from Lemma 2.3 and (2.15) we know that \(g_{1}^{\prime }(x)/g_{2}^{\prime}(x)\) is strictly increasing on \((0, \infty)\). Therefore, the fact that the function \(x\rightarrow g_{0}(x, p)\) is strictly increasing on \((0, \infty)\) follows from Lemma 2.1 and (2.14) together with the monotonicity of \(g_{1}^{\prime}(x)/g_{2}^{\prime}(x)\) on the interval \((0, \infty)\).

From Lemma 2.2 we know that the interval \((0, 1)\) can be expressed by

$$ (0, 1)= \biggl(0, \frac{1}{3p} \biggr]\cup \biggl(\frac{1}{3p}, 1- \biggl( \frac{2}{e^{2}} \biggr)^{1/p} \biggr)\cup\biggl[1- \biggl( \frac{2}{e^{2}} \biggr)^{1/p}, 1 \biggr). $$

We divide the proof into three cases.

Case 1: \(0< u\leq1/(3p)\). Then (2.12) and (2.13) together with the monotonicity of the function \(x\rightarrow g_{0}(x, p)\) lead to the conclusion that the function \(x\rightarrow g(u, p; x)\) is strictly increasing on \((0, \infty)\). Therefore \(g(u, p; x)>0\) for all \(x\in(0, \infty)\) follows from (2.9) and the monotonicity of the function \(x\rightarrow g(u, p; x)\) on the interval \((0, \infty)\).

Case 2: \(1-(2/e^{2})^{1/p}\leq u<1\). Then from (2.10), (2.12), (2.13), Lemma 2.2, and the monotonicity of the function \(x\rightarrow g_{0}(x, p)\), we clearly see that

$$ \lim_{x\rightarrow\infty}g(u, p; x)\leq0, $$
(2.16)

and there exists \(x_{0}\in(0, \infty)\) such that the function \(x\rightarrow g(u, p; x)\) is strictly decreasing on \((0, x_{0})\) and strictly increasing on \((x_{0}, \infty)\). Therefore \(g(u, p; x)<0\) for all \(x\in(0, \infty)\) follows from (2.9) and (2.16) together with the piecewise monotonicity of the function \(x\rightarrow g(u, p; x)\) on the interval \((0, \infty)\).

Case 3: \(1/(3p)< u<1-(2/e^{2})^{1/p}\). Then it follows from (2.10), (2.12), (2.13), and the monotonicity of the function \(x\rightarrow g_{0}(x, p)\) that

$$ \lim_{x\rightarrow\infty}g(u, p; x)>0, $$
(2.17)

and there exists \(x^{\ast}\in(0, \infty)\) such that the function \(x\rightarrow g(u, p; x)\) is strictly decreasing on \((0, x^{\ast})\) and strictly increasing on \((x^{\ast}, \infty)\). Therefore, there exists \(\tau\in(0, \infty)\) such that \(g(u, p; x)<0\) for \(x\in(0, \tau)\) and \(g(u, p; x)>0\) for \(x\in(\tau, \infty)\) follows from (2.9) and (2.17) together with the piecewise monotonicity of the function \(x\rightarrow g(u, p; x)\) on the interval \((0, \infty)\). □

3 Main result

Theorem 3.1

Let \(p\geq1\), \(0<\lambda, \mu<1/2\), andσandτbe any two different positive real numbers. Then the double inequality

$$ \mathbf{GA}_{\lambda, p}(\sigma, \tau)< \mathbf{SY}(\sigma, \tau)< \mathbf{GA}_{\mu, p}(\sigma, \tau) $$

holds if and only if

$$ \lambda\leq\frac{1}{2}-\frac{1}{2}\sqrt{1- \biggl( \frac{2}{e^{2}} \biggr)^{1/p}}, \qquad\mu\geq\frac{1}{2}- \frac{\sqrt{3p}}{6p}. $$

Proof

From (1.1)–(1.3) we clearly see that both \(\mathbf{GA}_{\theta, p}(\sigma, \tau)\) and \(\mathbf{SY}(\sigma, \tau)\) are symmetric and homogenous of degree one with respect to their variables σ and τ. Without loss of generality, we assume that \(\sigma>\tau>0\). Let \(0<\theta<1/2\) and \(x=(\sigma-\tau)/\sqrt{2\sigma\tau}>0\). Then (1.1)–(1.3) lead to

$$\begin{aligned}& \mathbf{SY}(\sigma, \tau)=\mathbf{G}(\sigma, \tau)\sqrt{1+x^{2}}e^{\frac {\arctan(x)}{x}-1}, \\& \mathbf{GA}_{\theta, p}(\sigma, \tau)=\mathbf{G}(\sigma, \tau)\sqrt{1+ \frac{x^{2}}{2}} \biggl[\frac{ (1-(1-2\theta)^{2} )x^{2}+2}{x^{2}+2} \biggr]^{p/2}, \\& \begin{aligned}[b] &\log \bigl[\mathbf{GA}_{\theta, p}(\sigma, \tau) \bigr]-\log \bigl[ \mathbf{SY}( \sigma, \tau) \bigr] \\ &\quad=\frac{p}{2}\log \biggl[\frac{ (1-(1-2\theta)^{2} )x^{2}+2}{x^{2}+2} \biggr] +\frac{1}{2} \log \biggl(\frac{x^{2}+2}{2(x^{2}+1)} \biggr)-\frac{\arctan(x)}{x}+1 \\ &\quad=g \bigl((1-2\theta)^{2}, p;x \bigr),\end{aligned} \end{aligned}$$
(3.1)

where \(g(\cdot, p; x)\) is defined by (2.8).

Therefore, Theorem 3.1 follows easily from Lemma 2.4 and (3.1). □

4 Applications

Let \(p=1, 2\). Then Theorem 3.1 leads to Theorem 4.1 immediately, which provides the sharp bounds for the Sándor–Yang mean in terms of the one-parameter geometric and harmonic means.

Theorem 4.1

Let \(0<\lambda_{1}, \lambda_{2}, \mu_{1}, \mu_{2}<1/2\), andσandτbe any two distinct positive real numbers. Then the double inequalities

$$ \mathbf{G} \bigl[\lambda_{1}\sigma+(1-\lambda_{1})\tau, \lambda_{1}\tau+(1-\lambda_{1})\sigma \bigr]< \mathbf{SY}( \sigma, \tau)< \mathbf{G} \bigl[\mu_{1}\sigma+(1-\mu_{1}) \tau, \mu_{1}\tau+(1-\mu_{1})\sigma \bigr] $$

and

$$ \mathbf{H} \bigl[\lambda_{2}\sigma+(1-\lambda_{2})\tau, \lambda_{2}\tau+(1-\lambda_{2})\sigma \bigr]< \mathbf{SY}( \sigma, \tau)< \mathbf{H} \bigl[\mu_{2}\sigma+(1-\mu_{2})\tau, \mu_{2}\tau+(1-\mu_{2})\sigma \bigr] $$

hold if and only if

$$\begin{gathered} \lambda_{1}\leq\frac{1}{2}-\frac{1}{2}\sqrt{1- \frac{2}{e^{2}}}\simeq0.0730, \qquad\mu_{1}\geq\frac{1}{2}- \frac{\sqrt{3}}{6}\simeq0.2113, \\ \lambda_{2}\leq\frac{1}{2}-\frac{1}{2}\sqrt{1- \frac{\sqrt{2}}{e}}\simeq0.1537, \qquad\mu_{2}\geq\frac{1}{2}- \frac{\sqrt{6}}{12}\simeq0.2959.\end{gathered} $$

Theorem 3.1 and (1.2) also lead to Theorem 4.2, which gives the sharp bounds for the Yang mean in terms of the two-parameter geometric and arithmetic mean and the quadratic and geometric means.

Theorem 4.2

Let \(p\geq1\), \(0<\alpha, \beta<1/2\), andσandτbe any two different positive real numbers. Then the two-sided inequality

$$ \frac{\mathbf{G}(\sigma, \tau)}{\log[\mathbf{GA}_{\alpha, p}(\sigma, \tau)]-\log[\mathbf{Q}(\sigma, \tau)]+1}< \mathbf{U}(\sigma, \tau)< \frac {\mathbf{G}(\sigma, \tau)}{\log[\mathbf{GA}_{\beta, p}(\sigma, \tau )]-\log[\mathbf{Q}(\sigma, \tau)]+1} $$

takes place if and only if

$$ \alpha\geq\frac{1}{2}-\frac{\sqrt{3p}}{6p}, \qquad\beta\leq \frac{1}{2}-\frac{1}{2}\sqrt{1- \biggl(\frac{2}{e^{2}} \biggr)^{1/p}}. $$

Let \(\sigma>\tau=1/2\), \(\alpha=1/2-\sqrt{3p}/(6p)\), and \(\beta=1/2-\sqrt {1-(2/e^{2})^{1/p}}/2\). Then it follows from (1.1), (1.3) that

$$\begin{aligned}& \mathbf{U} \biggl(\sigma, \frac{1}{2} \biggr)=\frac{2\sigma-1}{2\sqrt {2}\arctan(\frac{2\sigma-1}{2\sqrt{\sigma}} )}, \end{aligned}$$
(4.1)
$$\begin{aligned}& \begin{aligned}[b] &\mathbf{GA}_{1/2-\sqrt{3p}/(6p), p} \biggl(\sigma, \frac{1}{2} \biggr) \\ &\quad= \biggl[\frac{4(3p-1)\sigma^{2}+4(3p+1)\sigma+3p-1}{48p} \biggr]^{p} \biggl(\frac{2\sigma+1}{4} \biggr)^{1-p},\end{aligned} \end{aligned}$$
(4.2)
$$\begin{aligned}& \begin{aligned}[b] &\mathbf{GA}_{1/2-\sqrt{1-(2/e^{2})^{1/p}}/2, p} \biggl(\sigma, \frac {1}{2} \biggr) \\ &\quad= \biggl[\frac{4\times2^{1/p}\sigma^{2}+4 (2e^{2/p}-2^{1/p} )\sigma +2^{1/p}}{16e^{2/p}} \biggr]^{p} \biggl(\frac{2\sigma+1}{4} \biggr)^{1-p}.\end{aligned} \end{aligned}$$
(4.3)

From Theorem 4.2 and (4.1)–(4.3) we obtain Theorem 4.3, which presents new one-parameter bounds for the inverse tangent function.

Theorem 4.3

The double inequality

$$\begin{gathered} \frac{2\sigma-1}{2\sqrt{\sigma}} \biggl[p\log \bigl(4\times2^{1/p}\sigma^{2}+4 \bigl(2e^{2/p}-2^{1/p} \bigr)\sigma+2^{1/p} \bigr) \\ \quad\quad{}+(1-p)\log(2\sigma+1)-\frac{1}{2}\log \bigl(4\sigma^{2}+1 \bigr)-p\log4-1-\frac{1}{2}\log2 \biggr] \\ \quad< \arctan \biggl(\frac{2\sigma-1}{2\sqrt{\sigma}} \biggr)< \frac{2\sigma -1}{2\sqrt{\sigma}} \biggl[p\log \bigl(4(3p-1)\sigma^{2}+4(3p+1)\sigma+3p-1 \bigr) \\ \qquad{}+(1-p)\log(2\sigma+1)-\frac{1}{2}\log \bigl(4\sigma^{2}+1 \bigr)-p (\log p+\log3+2\log2 )+1-\frac{1}{2}\log2 \biggr]\end{gathered} $$

holds for all \(\sigma>1/2\)and \(p\geq1\).

5 Consequences and discussion

In the article, we have given the sharp bounds for the Sándor–Yang mean

$$ \mathbf{SY}(\sigma, \tau)=\mathbf{Q}(\sigma, \tau)e^{\mathbf{G}(\sigma, \tau)/\mathbf{U}(\sigma, \tau)-1} $$

in terms of the two-parameter geometric and arithmetic mean

$$ \mathbf{GA}_{\eta, \nu}(\sigma, \tau)=\mathbf{G}^{\nu} \bigl[\eta \sigma+(1-\eta)\tau, \eta\tau+(1-\eta)\sigma \bigr]\mathbf{A}^{1-\nu}( \sigma, \tau) $$

and the one-parameter geometric and harmonic means

$$ \mathbf{G} \bigl[\lambda\sigma+(1-\lambda)\tau, \lambda\tau+(1-\lambda )\sigma \bigr] $$

and

$$ \mathbf{H} \bigl[\mu\sigma+(1-\mu)\tau, \mu\tau+(1-\mu)\sigma \bigr], $$

and have found the new bounds for the Yang mean

$$ \mathbf{U}(\sigma, \tau)=\frac{\sqrt{2}(\sigma-\tau)}{2\arctan(\frac {\sqrt{2}(\sigma-\tau)}{2\sqrt{\sigma\tau}} )} $$

and the inverse tangent function \(\arctan[(2\sigma-1)/(2\sqrt{\sigma})]\).

6 Conclusion

In the article, we have proved that the double inequalities

$$ \mathbf{GA}_{\lambda, p}(\sigma, \tau)< \mathbf{SY}(\sigma, \tau)< \mathbf{GA}_{\mu, p}(\sigma, \tau) $$

and

$$ \frac{\mathbf{G}(\sigma, \tau)}{\log[\mathbf{GA}_{\mu, p}(\sigma, \tau )]-\log[\mathbf{Q}(\sigma, \tau)]+1}< \mathbf{U}(\sigma, \tau)< \frac {\mathbf{G}(\sigma, \tau)}{\log[\mathbf{GA}_{\lambda, p}(\sigma, \tau )]-\log[\mathbf{Q}(\sigma, \tau)]+1} $$

are valid for all distinct positive real numbers σ and τ if and only if

$$ \lambda\leq\frac{1}{2}-\frac{1}{2}\sqrt{1- \biggl( \frac{2}{e^{2}} \biggr)^{1/p}}, \qquad\mu\geq\frac{1}{2}- \frac{\sqrt{3p}}{6p}. $$

if \(p\geq1\) and \(\lambda, \mu\in(0, 1/2)\).