# Sharp bounds for the Sándor–Yang means in terms of arithmetic and contra-harmonic means

## Abstract

In the article, we provide several sharp upper and lower bounds for two Sándor–Yang means in terms of combinations of arithmetic and contra-harmonic means.

## Preliminaries

Let $$a, b>0$$ with $$a\neq b$$. Then the arithmetic mean $$A(a,b)$$ , the quadratic mean $$Q(a,b)$$ , the contra-harmonic mean $$C(a, b)$$ , the Neuman–Sándor mean $$NS(a, b)$$ , the second Seiffert mean $$T(a, b)$$ [13, 14], and the Schwab–Borchardt mean $$SB(a, b)$$ [15, 16] of a and b are defined by

\begin{aligned} &A(a,b)=\frac{a+b}{2}, \qquad Q(a,b)=\sqrt{\frac{a^{2}+b^{2}}{2}}, \qquad C(a,b)=\frac{a^{2}+b^{2}}{a+b}, \end{aligned}
(1.1)
\begin{aligned} \ &NS(a,b)=\frac{a-b}{2\sinh^{-1} (\frac{a-b}{a+b} )}, \end{aligned}
(1.2)
\begin{aligned} &T(a,b)=\frac{a-b}{2\arctan (\frac{a-b}{a+b} )}, \\ &SB(a,b)=\textstyle\begin{cases} \frac{\sqrt{b^{2}-a^{2}}}{\arccos(a/b)}, & a< b,\\ \frac{\sqrt{a^{2}-b^{2}}}{\cosh^{-1}(a/b)}, & a>b, \end{cases}\displaystyle \end{aligned}
(1.3)

respectively, where $$\sinh^{-1}(x)=\log(x+\sqrt{x^{2}+1})$$ and $$\cosh^{-1}(x)=\log(x+\sqrt{x^{2}-1})$$ are respectively the inverse hyperbolic sine and cosine functions. The Schwab–Borchardt mean $$SB(a,b)$$ is strictly increasing, non-symmetric and homogeneous of degree one with respect to its variables. It can be expressed by the degenerated completely symmetric elliptic integral of the first kind . Recently, the Schwab–Borchardt mean has attracted the attention of many researchers. In particular, many remarkable inequalities for the Schwab–Borchardt mean and its generated means can be found in the literature .

Let $$X(a,b)$$ and $$Y(a,b)$$ denote symmetric bivariate means of a and b. Then Yang  introduced the Sándor–Yang mean

$$R_{XY}(a,b)=Y(a,b)e^{\frac{X(a,b)}{SB[X(a,b), Y(a,b)]}-1}$$

and presented the explicit formulas for $$R_{QA}(a,b)$$ and $$R_{AQ}(a,b)$$ as follows:

\begin{aligned} &R_{QA}(a,b)=A(a,b)e^{\frac{Q(a,b)}{NS(a,b)}-1}, \end{aligned}
(1.4)
\begin{aligned} &R_{AQ}(a,b)=Q(a,b)e^{\frac{A(a,b)}{T(a,b)}-1}. \end{aligned}
(1.5)

Very recently, the bounds involving the Sándor–Yang means have been the subject of intensive research. Numerous interesting results and inequalities for $$R_{QA}(a,b)$$ and $$R_{AQ}(a,b)$$ can be found in the literature .

Neuman  established the inequality

$$R_{AQ}(a,b)< R_{QA}(a,b)$$
(1.6)

for $$a,b > 0$$ with $$a\neq b$$.

In , Xu proved that the double inequalities

\begin{aligned} &\alpha_{1}C(a,b)+(1- \alpha_{1})A(a,b)< R_{QA}(a,b)< \beta_{1}C(a,b)+(1- \beta_{1})A(a,b), \\ &\alpha_{2}C(a,b)+(1-\alpha_{2})A(a,b)< R_{AQ}(a,b)< \beta_{2}C(a,b)+(1-\beta_{2})A(a,b) \end{aligned}
(1.7)

hold for all $$a,b > 0$$ with $$a\neq b$$ if and only if $$\alpha_{1}\leq (1+\sqrt{2})^{\sqrt{2}}/e-1=0.2794\ldots$$ , $$\beta_{1}\geq 1/3$$, $$\alpha_{2}\leq \sqrt{2}e^{\pi/4-1}-1=0.1410\ldots$$ and $$\beta_{2}\geq 1/6$$.

From (1.6) and (1.7), together the well-known inequalities

$$C(a,b)> Q(a,b)> A(a,b), \qquad Q(a,b)>\frac{1}{3}C(a,b)+ \frac{2}{3}A(a,b),$$

we clearly see that

$$A(a,b)< R_{AQ}(a,b)< R_{QA}(a,b)< Q(a,b)< C(a,b)$$
(1.8)

for all $$a, b >0$$ with $$a\neq b$$.

The main purpose of this paper is to find the best possible parameters $$\alpha_{i}, \beta_{i}\in (0, 1)$$ $$(i=1, 2, 3, 4)$$ such that the double inequalities

\begin{aligned} &C^{\alpha_{1}}(a,b)A^{1-\alpha_{1}}(a,b)< R_{QA}(a,b)< C^{\beta_{1}}(a,b)A^{1-\beta_{1}}(a,b), \\ &C^{\alpha_{2}}(a,b)A^{1-\alpha_{2}}(a,b)< R_{AQ}(a,b)< C^{\beta_{2}}(a,b)A^{1-\beta_{2}}(a,b), \\ &\alpha_{3} \biggl[\frac{1}{3}C(a,b)+\frac{2}{3}A(a,b) \biggr]+(1-\alpha_{3})C^{1/3}(a,b)A^{2/3}(a,b) \\ &\quad < R_{QA}(a,b)< \beta_{3} \biggl[\frac{1}{3}C(a,b)+ \frac{2}{3}A(a,b) \biggr]+(1-\beta_{3})C^{1/3}(a,b)A^{2/3}(a,b), \\ &\alpha_{4} \biggl[\frac{1}{6}C(a,b)+\frac{5}{6}A(a,b) \biggr]+(1-\alpha_{4})C^{1/6}(a,b)A^{5/6}(a,b) \\ &\quad < R_{AQ}(a,b)< \beta_{4} \biggl[\frac{1}{6}C(a,b)+ \frac{5}{6}A(a,b) \biggr]+(1-\beta_{4})C^{1/6}(a,b)A^{5/6}(a,b) \end{aligned}

hold for all $$a, b>0$$ with $$a\neq b$$.

## Lemmas

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

### Lemma 2.1

(see )

Let $$a, b\in \mathbb{R}$$ with $$a< b$$, $$f, g: [a, b]\mapsto \mathbb{R}$$ be continuous on $$[a, b]$$ and differentiable on $$(a, b)$$, and $$g^{\prime}(x)\neq 0$$ 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 )

Let $$A(t)=\sum_{k=0}^{\infty}a_{k}t^{k}$$ and $$B(t)=\sum_{k=0}^{\infty }b_{k}t^{k}$$ be two real power series converging on $$(-r,r)$$ ($$r>0$$) with $$b_{k}>0$$ for all k. If the non-constant sequence $$\{a_{k}/b_{k}\}_{k=0}^{\infty}$$ is increasing (decreasing) for all k, then the function $$t\mapsto A(t)/B(t)$$ is strictly increasing (decreasing) on $$(0,r)$$.

### Lemma 2.3

The function

$$\phi(x)=\frac{x\coth(x)-1}{2\log[\cosh(x)]}$$

is strictly increasing from $$(0, \log(1+\sqrt{2})$$ onto $$(1/3, [\sqrt{2}\log(1+\sqrt{2})-1]/\log 2)$$.

### Proof

Let $${\phi _{1}} ( x ) = x\coth ( x ) - 1$$, $${\phi _{2}} ( x ) = 2\log [ {\cosh ( x )} ]$$. Then elaborate computations lead to

\begin{aligned} &\phi ( x ) = \frac{{{\phi _{1}} ( x )}}{{{\phi _{2}} ( x )}} = \frac{{{\phi _{1}} ( x ) - {\phi _{1}} ( {{0^{+} }} )}}{{{\phi _{2}} ( x ) - {\phi _{2}} ( 0 )}}, \end{aligned}
(2.1)
\begin{aligned} &\begin{aligned}[b] \frac{{{{\phi }'_{1}} ( x )}}{{{{\phi }'_{2}} ( x )}} &= \frac{{\sinh ( x ){{\cosh }^{2}} ( x ) - x\cosh ( x )}}{{2{{\sinh }^{3}} ( x )}} \\ &= \frac{{\sinh ( {3x} ) + \sinh ( x ) - 4x\cosh ( x )}}{{2\sinh ( {3x} ) - 6\sinh ( x )}} = \frac{{\sum_{n = 0}^{\infty}{\frac{{{3^{2n + 1}} - 8n - 3}}{{ ( {2n + 1} )!}}{x^{2n + 1}}} }}{{\sum_{n = 0}^{\infty}{\frac{{6 ( {{3^{2n}} - 1} )}}{{ ( {2n + 1} )!}}{x^{2n + 1}}} }} \\ &= \frac{{\sum_{n = 1}^{\infty}{\frac{{{3^{2n + 1}} - 8n - 3}}{{ ( {2n + 1} )!}}{x^{2n + 1}}} }}{{\sum_{n = 1}^{\infty}{\frac{{6 ( {{3^{2n}} - 1} )}}{{ ( {2n + 1} )!}}{x^{2n + 1}}} }} = \frac{{\sum_{n = 0}^{\infty}{\frac{{{3^{2n + 3}} - 8n - 11}}{{ ( {2n + 3} )!}}{x^{2n + 3}}} }}{{\sum_{n = 0}^{\infty}{\frac{{6 ( {{3^{2n{\mathrm{ + }}2}} - 1} )}}{{ ( {2n + 3} )!}}{x^{2n + 3}}} }}. \end{aligned} \end{aligned}
(2.2)

Let

$${a_{n}} = \frac{{{3^{2n + 3}} - 8n - 11}}{{ ( {2n + 3} )!}}, \qquad {b_{n}} = \frac{{6 ( {{3^{2n{\mathrm{ + }}2}} - 1} )}}{{ ( {2n + 3} )!}}.$$
(2.3)

Then

$${b_{n}} > 0$$
(2.4)

and

$$\frac{{{a_{n + 1}}}}{{{b_{n + 1}}}} - \frac{{{a_{n}}}}{{{b_{n}}}} = \frac{{4 [ { ( {72n + 63} ){3^{2n}} + 1} ]}}{ {3 ( {{3^{2n + 4}} - 1} ) ( {{3^{2n{\mathrm{ + }}2}} - 1} )}} > 0$$
(2.5)

for all $$n \ge 0$$.

It follows from Lemma 2.2 and (2.2)–(2.5) that $${\phi '_{1}} ( x )/{\phi '_{2}} ( x )$$ is strictly increasing on $$( {0,\log ( {1 + \sqrt {2} } )} )$$.

Note that

$$\phi \bigl( {{0^{+} }} \bigr) = \frac{{{a_{0}}}}{{{b_{0}}}} = \frac{1}{3}, \qquad \phi \bigl( {\log ( {1 + \sqrt {2} } )} \bigr) = \frac{{\sqrt {2} \log ( {1 + \sqrt {2} } ) - 1}}{{\log 2}} = 0.3555 \ldots.$$
(2.6)

Therefore, Lemma 2.3 follows from Lemma 2.1, (2.1), and (2.6) together with the monotonicity of $${\phi '_{1}} ( x )/{\phi '_{2}} ( x )$$. □

### Lemma 2.4

The function

$$\varphi(x)=\frac{\log\sec(x)+x\cot(x)-1}{2\log\sec(x)}$$

is strictly increasing from $$(0, \pi/4)$$ onto $$(1/6, 1/2-(4-\pi)(4\log 2))$$.

### Proof

Let $${\varphi _{1}} ( x ) = \log \sec ( x ) + x\cot ( x ) - 1$$, $${\varphi _{2}} ( x ) = 2\log [ {\sec ( x )} ]$$, $${\varphi _{3}} ( x ) = \sin ( x ) - x\cos ( x )$$, and $${\varphi _{4}} ( x ) = 2{\sin ^{3}} ( x )$$. Then elaborate computations lead to

\begin{aligned} &\varphi ( x ) = \frac{{{\varphi _{1}} ( x )}}{{{\varphi _{2}} ( x )}} = \frac{{{\varphi _{1}} ( x ) - {\varphi _{1}} ( {{0^{+} }} )}}{{{\varphi _{2}} ( x ) - {\varphi _{2}} ( 0 )}}, \end{aligned}
(2.7)
\begin{aligned} &\frac{{{{\varphi }'_{1}} ( x )}}{{{{\varphi }'_{2}} ( x )}} = \frac{{{\varphi _{3}} ( x )}}{{{\varphi _{4}} ( x )}} = \frac{{{\varphi _{3}} ( x ) - {\varphi _{3}} ( 0 )}}{{{\varphi _{4}} ( x ) - {\varphi _{4}} ( 0 )}} \end{aligned}
(2.8)

and

$$\frac{{{{\varphi }'_{3}} ( x )}}{{{{\varphi }'_{4}} ( x )}} = \frac{x}{{3\sin ( {2x} )}} = \frac{1}{6} \times \frac{1}{{\sin (2x)/ ( {2x} )}}.$$
(2.9)

It is well known that the function $$x \to \sin ( x )/x$$ is strictly decreasing on $$( {0,\pi /2} )$$, hence equation (2.9) leads to the conclusion that the function $${\varphi '_{3}} ( x )/{\varphi '_{4}} ( x )$$ is strictly increasing on $$( {0,\pi /4} )$$.

Note that

\begin{aligned} &\varphi \bigl( {{0^{+} }} \bigr) = \lim_{x \to {0^{+} }} \frac{{{{\varphi }'_{3}} ( x )}}{{{{\varphi }'_{4}} ( x )}} = \frac{1}{6}, \\ &\varphi \biggl( { \frac{\pi }{4}} \biggr) = \frac{1}{2} - \frac{{4 - \pi }}{{4\log 2}} = 0.1903 \dots. \end{aligned}
(2.10)

Therefore, Lemma 2.4 follows from Lemma 2.1 and (2.7)–(2.9) together with the monotonicity of $${\varphi '_{3}} ( x )/{\varphi '_{4}} ( x )$$. □

### Lemma 2.5

Let $$p \in ( {0,1} )$$ and

$$f ( x ) = 3{p^{2}} {x^{10}} + 14p ( {1 - p} ){x^{6}} + 18{p^{2}} {x^{4}} - 9{ ( {1 - p} )^{2}} {x^{2}} - 2p ( {1 - p} ).$$

Then the following statements are true:

1. (1)

If $$p = 3/10$$, then $$f ( x ) > 0$$ for all $$x \in ( {1,\sqrt{2}} )$$;

2. (2)

If $$p = 3 [ {{{ ( {1 + \sqrt {2} } )}^{\sqrt {2} }}/e - \sqrt{2}} ]/ ( {4 - 3\sqrt{2}} ) = 0.2663 \dots$$ , then there exists $${\lambda _{0}}( = 1.0808 \dots ) \in ( {1,\sqrt{2}} )$$ such that $$f ( x ) < 0$$ for $$x \in ( {1,{\lambda _{0}}} )$$ and $$f ( x ) > 0$$ for $$x \in ( {{\lambda _{0}},\sqrt{2}} )$$.

### Proof

Part $$(1)$$ follows easily from

$$f ( x ) = \frac{3}{{100}} \bigl( {{x^{2}} - 1} \bigr) \bigl( {9{x^{8}} + 9{x^{6}} + 107{x^{4}} + 161{x^{2}} + 14} \bigr) > 0$$

for all $$x \in ( {1,\sqrt{2}} )$$ if $$p = 3/10$$.

For part $$(2)$$, if $$p = 3 [ {{{ ( {1 + \sqrt {2} } )}^{\sqrt {2} }}/e - \sqrt{2}} ]/ ( {4 - 3\sqrt{2}} )$$, then numerical computations lead to

\begin{aligned} &20p - 3 = 2.3273 \dots > 0, \end{aligned}
(2.11)
\begin{aligned} &f ( 1 ) = 3 ( {10p - 3} ) = - 1.008 \dots < 0, \end{aligned}
(2.12)
\begin{aligned} &f \bigl( {\sqrt{2}} \bigr) = 1.6809 \dots > 0, \end{aligned}
(2.13)
\begin{aligned} &f' ( x ) = 30{p^{2}} {x^{9}} + 84p ( {1 - p} ){x^{5}} + 72{p^{2}} {x^{3}} - 18{ ( {1 - p} )^{2}}x. \end{aligned}
(2.14)

It follows from (2.11) and (2.14) that

$$f' ( x ) > \bigl[ {30{p^{2}} + 84p ( {1 - p} ) + 72{p^{2}} - 18{{ ( {1 - p} )}^{2}}} \bigr]x = 6 ( {20p - 3} )x > 0$$
(2.15)

for all $$x \in ( {1,\sqrt{2}} )$$.

Therefore, part $$(2)$$ follows easily from (2.12), (2.13), (2.15), and the numerical results $$f ( {1.0808} ) < 0$$ and $$f ( {1.0809} ) > 0$$. □

### Lemma 2.6

Let $$p \in ( {0,1} )$$ and

$$g ( x ) = 3{p^{2}} {x^{11}} + 56p ( {1 - p} ){x^{6}} + 75{p^{2}} {x^{5}} - 72{ ( {1 - p} )^{2}}x - 50p ( {1 - p} ).$$

Then the following statements are true:

1. (1)

If $$p = 12/25$$, then $$g ( x ) > 0$$ for all $$x \in ( {1,\sqrt{2}} )$$;

2. (2)

If $$p = 6 [ {\sqrt {2} {e^{\pi/4- 1}} - \sqrt{2}} ]/ ( {7 - 6\sqrt{2}} ) = 0.4210 \dots$$ , then there exists $${\mu _{0}}( = 1.0577 \dots ) \in ( {1,\sqrt{2}} )$$ such that $$g ( x ) < 0$$ for $$x \in ( {1,{\mu _{0}}} )$$ and $$g ( x ) > 0$$ for $$x \in ( {{\mu _{0}},\sqrt{2}} )$$.

### Proof

Part $$(1)$$ follows easily from

\begin{aligned} g ( x ) =& \frac{{24}}{{625}} ( {x - 1} ) \bigl( 18{x^{10}} + 18{x^{9}} + 18{x^{8}} + 18{x^{7}} + 18{x^{6}} + 382{x^{5}} + 832{x^{4}} \\ &{} + 832{x^{3}} + 832{x^{2}} + 832x + 325 \bigr) > 0 \end{aligned}

for all $$x \in ( {1,\sqrt{2}} )$$ if $$p = 12/25$$.

For part $$(2)$$, if $$p =6[\sqrt{2}e^{\pi/4-1}-\sqrt{2}]/(7-6\sqrt{2}) = 0.4210 \ldots$$ , then numerical computations lead to

\begin{aligned} &20p - 3 = 5.4217 \dots > 0, \end{aligned}
(2.16)
\begin{aligned} &g ( 1 ) = 6 ( {25p - 12} ) = - 8.8367 \dots < 0, \end{aligned}
(2.17)
\begin{aligned} &g \bigl( {\sqrt{2}} \bigr) = 13.6200 \dots > 0, \end{aligned}
(2.18)
\begin{aligned} &g' ( x ) = 3 \bigl[ {11{p^{2}} {x^{10}} + 112p ( {1 - p} ){x^{5}} + 125{p^{2}} {x^{4}} - 24{{ ( {1 - p} )}^{2}}} \bigr]. \end{aligned}
(2.19)

It follows from (2.16) and (2.19) that

\begin{aligned} g' ( x ) &> 11{p^{2}} + 112p ( {1 - p} ) + 125{p^{2}} - 24{ ( {1 - p} )^{2}} \\ &= 24 ( {20p - 3} ) > 0 \end{aligned}
(2.20)

for $$x \in ( {1,\sqrt{2}} )$$.

Therefore, part $$(2)$$ follows easily from (2.17), (2.18), and (2.20) together with the numerical results $$g ( {1.0577} ) < 0$$ and $$g ( {1.0578} ) > 0$$. □

## Main results

We are now in a position to state and prove our main results.

### Theorem 3.1

The double inequality

$${C^{{\alpha _{1}}}} ( {a,b} ){A^{1 - {\alpha _{1}}}} ( {a,b} ) < {R_{QA}} ( {a,b} ) < {C^{{\beta _{1}}}} ( {a,b} ){A^{1 - {\beta _{1}}}} ( {a,b} )$$
(3.1)

holds for all $$a,b>0$$ with $$a\ne b$$ if and only if $${\alpha _{1}} \le 1/3$$ and $${\beta _{1}} \ge [ {\sqrt {2} \log ( {1 + \sqrt {2} } ) - 1} ]/\log 2$$.

### Proof

Clearly, inequality (3.1) can be rewritten as

$${ \biggl[ {\frac{{C(a,b)}}{{A(a,b)}}} \biggr]^{{\alpha _{1}}}} < \frac{{{R_{QA}} ( {a,b} )}}{{A(a,b)}} < { \biggl[ {\frac{{C(a,b)}}{{A(a,b)}}} \biggr]^{{\beta _{1}}}}.$$
(3.2)

Since $$A(a,b)$$, $${R_{QA}} ( {a,b} )$$, and $$C(a,b)$$ are symmetric and homogenous of degree one, we assume that $$a > b > 0$$. Let $$v = ( {a - b} )/ ( {a + b} ) \in ( {0,1} )$$. Then from (1.1), (1.2), and (1.4) we know that inequality (3.2) is equivalent to

$${\alpha _{1}} < \frac{{ [ {\sqrt {1 + {v^{2}}}\sinh^{-1} ( v )} ]/v - 1}}{{\log ( {1 + {v^{2}}} )}} < {\beta _{1}}.$$
(3.3)

Let $$x =\sinh^{-1} ( v )$$. Then $$x \in ( {0,\log ( {1 + \sqrt {2} } )} )$$ and

$$\frac{{ [ {\sqrt {1 + {v^{2}}}\sinh^{-1} ( v )} ]/v - 1}}{{\log ( {1 + {v^{2}}} )}} = \frac{{x\coth ( x ) - 1}}{{2\log [ {\cosh ( x )} ]}}: = \phi ( x ).$$
(3.4)

Therefore, inequality (3.1) holds for all $$a,b > 0$$ with $$a \ne b$$ if and only if $${\alpha _{1}} \le 1/3$$ and $${\beta _{1}} \ge [ {\sqrt {2} \log ( {1 + \sqrt {2} } ) - 1} ]/\log 2$$ follows from (3.2)–(3.4) and Lemma 2.3. □

### Theorem 3.2

The double inequality

$${C^{{\alpha _{2}}}} ( {a,b} ){A^{1 - {\alpha _{2}}}} ( {a,b} ) < {R_{AQ}} ( {a,b} ) < {C^{{\beta _{2}}}} ( {a,b} ){A^{1 - {\beta _{2}}}} ( {a,b} )$$
(3.5)

holds for all $$a,b>0$$ with $$a\ne b$$ if and only if $${\alpha _{2}} \le 1/6$$ and $${\beta _{2}} \ge 1/2 - ( {4 - \pi } )/ ( {4\log 2} ) = 0.1903 \dots$$ .

### Proof

Clearly, inequality (3.5) can be rewritten as

$$\biggl[ {\frac{{C(a,b)}}{{A(a,b)}}} \biggr]^{{\alpha _{2}}} < \frac{{{R_{AQ}} ( {a,b} )}}{{A(a,b)}} < { \biggl[ {\frac{{C(a,b)}}{{A(a,b)}}} \biggr]^{{\beta _{2}}}}.$$
(3.6)

Since $$A(a,b)$$, $${R_{AQ}} ( {a,b} )$$, and $$C(a,b)$$ are symmetric and homogenous of degree one, we assume that $$a > b > 0$$. Let $$v = ( {a - b} )/ ( {a + b} ) \in ( {0,1} )$$. Then from (1.1), (1.3), and (1.5) we see that inequality (3.6) is equivalent to

$${\alpha _{2}} < \frac{{\log \sqrt {1 + {v^{2}}} + [ {\arctan ( v )} ]/v - 1}}{{\log ( {1 + {v^{2}}} )}} < {\beta _{2}}.$$
(3.7)

Let $$x = \arctan ( v )$$. Then $$x \in ( {0,\pi /4} )$$ and

\begin{aligned} &\frac{{\log \sqrt {1 + {v^{2}}} + [ {\arctan ( v )} ]/v - 1}}{{\log ( {1 + {v^{2}}} )}} \\ &\quad = \frac{{\log \sec ( x ) + x\cot ( x ) - 1}}{{2\log \sec ( x )}}: = \varphi ( x ). \end{aligned}
(3.8)

Therefore, inequality (3.5) holds for all $$a,b > 0$$ with $$a \ne b$$ if and only if $${\alpha _{2}} \le 1/6$$ and $${\beta _{2}} \ge 1/2 - ( {4 - \pi } )/ ( {4\log 2} ) = 0.1903 \dots$$ follows from (3.6)–(3.8) and Lemma 2.4. □

### Theorem 3.3

The double inequality

\begin{aligned} &{\alpha _{3}} \biggl[ {\frac{1}{3}C ( {a,b} ) + \frac{2}{3}A ( {a,b} )} \biggr] + ( {1 - {\alpha _{3}}} ){C^{1/3}} ( {a,b} ){A^{2/3}} ( {a,b} ) \\ &\quad < {R_{QA}} ( {a,b} ) < {\beta _{3}} \biggl[ { \frac{1}{3}C ( {a,b} ) + \frac{2}{3}A ( {a,b} )} \biggr] + ( {1 - { \beta _{3}}} ){C^{1/3}} ( {a,b} ){A^{2/3}} ( {a,b} ) \end{aligned}

holds for all $$a,b>0$$ with $$a\ne b$$ if and only if $${\alpha _{3}} \le 3 [ {{{ ( {1 + \sqrt {2} } )}^{\sqrt {2} }}/e - \sqrt{2}} ]/ ( {4 - 3\sqrt{2}} ) = 0.2663 \dots$$ and $${\beta _{3}} \ge 3/10$$.

### Proof

Since $${R_{QA}} ( {a,b} )$$, $$A(a,b)$$, and $$C(a,b)$$ are symmetric and homogenous of degree one, without loss generality, we assume that $$a > b > 0$$. Let $$v = ( {a - b} )/ ( {a + b} )$$, $$x = \sqrt{{1 + {v^{2}}}}$$, and $$p \in ( {0,1} )$$. Then $$v \in ( {0,1} )$$, $$x \in ( {1,\sqrt{2}} )$$, and (1.1), (1.2), and (1.4) lead to

\begin{aligned} &\log \frac{{{R_{QA}} ( {a,b} )}}{{p [ {\frac{1}{3}C ( {a,b} ) + \frac{2}{3}A ( {a,b} )} ] + ( {1 - p} ){C^{1/3}} ( {a,b} ){A^{2/3}} ( {a,b} )}} \\ &\quad = \frac{{\sqrt {1 + {v^{2}}} \sinh^{-1} ( v )}}{v} - \log \biggl[ {p \biggl( {\frac{1}{3}{v^{2}} + 1} \biggr) + ( {1 - p} )\sqrt{{1 + {v^{2}}}}} \biggr] - 1 \\ &\quad = \frac{{{x^{3}}\sinh^{-1} ( {\sqrt {{x^{6}} - 1} } )}}{{\sqrt {{x^{6}} - 1} }} - \log \biggl[ {p \biggl( {\frac{1}{3}{x^{6}} + \frac{2}{3}} \biggr) + ( {1 - p} ){x^{2}}} \biggr] - 1. \end{aligned}
(3.9)

Let

$$F ( x ) = \frac{{{x^{3}}\sinh^{-1} ( {\sqrt {{x^{6}} - 1} } )}}{{\sqrt {{x^{6}} - 1} }} - \log \biggl[ {p \biggl( { \frac{1}{3}{x^{6}} + \frac{2}{3}} \biggr) + ( {1 - p} ){x^{2}}} \biggr] - 1.$$
(3.10)

\begin{aligned} &F \bigl( {{1^{+} }} \bigr) = 0, \qquad F \bigl( {\sqrt{2}} \bigr) = \sqrt {2} \log ( {1 + \sqrt {2} } ) - \log \biggl[ {\frac{4}{3}p + \sqrt{2} ( {1 - p} )} \biggr] - 1, \end{aligned}
(3.11)
\begin{aligned} &F' ( x ) = \frac{{3{x^{2}}}}{{{{ ( {{x^{6}} - 1} )}^{3/2}}}}{F_{1}} ( x ), \end{aligned}
(3.12)

where

\begin{aligned} &{F_{1}} ( x ) = \frac{{\sqrt {{x^{6}} - 1} [ { - p{x^{10}} + ( {1 - p} ){x^{6}} + 4p{x^{4}} + 2 ( {1 - p} )} ]}}{{x [ {p ( {{x^{6}} + 2} ) + 3 ( {1 - p} ){x^{2}}} ]}} - \sinh^{-1} \bigl( { \sqrt {{x^{6}} - 1} } \bigr), \\ &{F_{1}} ( 1 ) = 0, \qquad {F_{1}} \bigl( { \sqrt{2}} \bigr) = \frac{{ ( {\sqrt{{2048}} - 2\sqrt {2} } )p - \sqrt{{2048}}}}{{3\sqrt{2}p - 3\sqrt{2} - 4p}} - \log ( {1 + \sqrt {2} } ), \end{aligned}
(3.13)
\begin{aligned} &{F'_{1}} ( x ) = - \frac{{2{{ ( {{x^{6}} - 1} )}^{3/2}}}}{{{x^{2}}{{ [ {p ( {{x^{6}} + 2} ) + 3 ( {1 - p} ){x^{2}}} ]}^{2}}}}f ( x ), \end{aligned}
(3.14)

where $$f ( x )$$ is defined as in Lemma 2.5.

We divide the proof into four cases.

Case 1 $$p = 3/10$$. Then it follows from (3.9)–(3.14) and Lemma 2.5(1) that

$${R_{QA}} ( {a,b} ) < \frac{3}{{10}} \biggl[ {\frac{1}{3}C ( {a,b} ) + \frac{2}{3}A ( {a,b} )} \biggr] + \frac{7}{{10}}{C^{1/3}} ( {a,b} ){A^{2/3}} ( {a,b} ).$$

Case 2 $$0 < p < 3/10$$. Let $$v > 0$$ and $$v \to {0^{+} }$$. Then power series expansion leads to

\begin{aligned} &\frac{{\sqrt {1 + {v^{2}}} \sinh^{-1} ( v )}}{v} - \log \biggl[ {p \biggl( { \frac{1}{3}{v^{2}} + 1} \biggr) + ( {1 - p} )\sqrt{{1 + {v^{2}}}}} \biggr] - 1 \\ &\quad = \biggl( {\frac{1}{{30}} - \frac{1}{9}p} \biggr){v^{4}} + O \bigl( {{v^{6}}} \bigr). \end{aligned}
(3.15)

Equations (3.9), (3.10), and (3.15) lead to the conclusion that there exists $$0 < {\delta _{1}} < 1$$ such that

$${R_{QA}} ( {a,b} ) > p \biggl[ {\frac{1}{3}C ( {a,b} ) + \frac{2}{3}A ( {a,b} )} \biggr] + ( {1 - p} ){C^{1/3}} ( {a,b} ){A^{2/3}} ( {a,b} )$$

for all $$a > b > 0$$ with $$( {a - b} )/ ( {a + b} ) \in ( {0,{\delta _{1}}} )$$.

Case 3 $$p = 3 [ {{{ ( {1 + \sqrt {2} } )}^{\sqrt {2} }}/e - \sqrt{2}} ]/ ( {4 - 3\sqrt{2}} )$$. Then (3.13) leads to

$${F_{1}} \bigl( {\sqrt{2}} \bigr) = - 0.0039 \dots < 0.$$
(3.16)

Let $${\lambda _{0}} = 1.0808 \dots$$ be the number given in Lemma 2.5(2). Then we divide the discussion into two subcases.

Subcase 1 $$x \in ( {1,{\lambda _{0}}} ]$$. Then $${F_{1}} ( x ) > 0$$ for $$x \in ( {1,{\lambda _{0}}} ]$$ follows easily from (3.13) and (3.14) together with Lemma 2.5(2).

Subcase 2 $$x \in ( {{\lambda _{0}},\sqrt{2}} )$$. Then Lemma 2.5(2) and (3.14) lead to the conclusion that $${F_{1}} ( x )$$ is strictly decreasing on the interval $$[ {{\lambda _{0}},\sqrt{2}} )$$. Then, from (3.16) and Subcase 1, we know that there exists $${\lambda _{1}} \in ( {{\lambda _{0}},\sqrt{2}} )$$ such that $${F_{1}} ( x ) > 0$$ for $$x \in [ {{\lambda _{0}},{\lambda _{1}}} )$$ and $${F_{1}} ( x ) < 0$$ for $$x \in ( {{\lambda _{1}},\sqrt{2}} )$$.

It follows from Subcases 1 and 2 together with (3.12) that $$F ( x )$$ is strictly increasing on $$( {1,{\lambda _{1}}} ]$$ and strictly decreasing on $$[ {{\lambda _{1}},\sqrt{2}} )$$. Therefore,

$${R_{QA}} ( {a,b} ) > p \biggl[ {\frac{1}{3}C ( {a,b} ) + \frac{2}{3}A ( {a,b} )} \biggr] + ( {1 - p} ){C^{1/3}} ( {a,b} ){A^{2/3}} ( {a,b} )$$

follows from (3.9)–(3.11) and (3.16) together with the piecewise monotonicity of $$F ( x )$$.

Case 4 $$3 [ {{{ ( {1 + \sqrt {2} } )}^{\sqrt {2} }}/e - \sqrt{2}} ]/ ( {4 - 3\sqrt{2}} ) < p < 1$$. Then (3.11) leads to

$$F \bigl( {\sqrt{2}} \bigr) = \sqrt {2} \log ( {1 + \sqrt {2} } ) - \log \biggl[ {\frac{4}{3}p + \sqrt{2} ( {1 - p} )} \biggr] - 1 < 0.$$
(3.17)

Equations (3.9) and (3.10) together with inequality (3.17) imply that there exists $$0 <\delta_{1}^{\ast} < 1$$ such that

$${R_{QA}} ( {a,b} ) < p \biggl[ {\frac{1}{3}C ( {a,b} ) + \frac{2}{3}A ( {a,b} )} \biggr] + ( {1 - p} ){C^{1/3}} ( {a,b} ){A^{2/3}} ( {a,b} )$$

for all $$a > b > 0$$ with $$( {a - b} )/ ( {a + b} ) \in ( {1 -\delta_{1}^{\ast} ,1} )$$. □

### Theorem 3.4

The double inequality

\begin{aligned} &{\alpha _{4}} \biggl[ {\frac{1}{6}C ( {a,b} ) + \frac{5}{6}A ( {a,b} )} \biggr] + ( {1 - {\alpha _{4}}} ){C^{1/6}} ( {a,b} ){A^{5/6}} ( {a,b} ) \\ &\quad< {R_{AQ}} ( {a,b} ) < {\beta _{4}} \biggl[ {\frac{1}{6}C ( {a,b} ) + \frac{5}{6}A ( {a,b} )} \biggr] + ( {1 - {\beta _{4}}} ){C^{1/6}} ( {a,b} ){A^{5/6}} ( {a,b} ) \end{aligned}

holds for all $$a,b>0$$ with $$a\ne b$$ if and only if $${\alpha _{4}} \le 6 [ {\sqrt {2} {{\mathrm{e}} ^{ ( {\pi /4 - 1} )}} - \sqrt{2}} ]/ ( {7 - 6\sqrt{2}} ) = 0.4210 \dots$$ and $${\beta _{4}} \ge 12/25$$.

### Proof

Since $${R_{AQ}} ( {a,b} )$$, $$A(a,b)$$, and $$C(a,b)$$ are symmetric and homogenous of degree one, without loss generality, we assume that $$a > b > 0$$. Let $$v = ( {a - b} )/ ( {a + b} )$$, $$x = \sqrt{{1 + {v^{2}}}}$$, and $$p \in ( {0,1} )$$. Then $$v \in ( {0,1} )$$, $$x \in ( {1,\sqrt{2}} )$$ and (1.1), (1.3), and (1.5) lead to

\begin{aligned} &\log \frac{{{R_{AQ}} ( {a,b} )}}{{p [ {\frac{1}{6}C ( {a,b} ) + \frac{5}{6}A ( {a,b} )} ] + ( {1 - p} ){C^{1/6}} ( {a,b} ){A^{5/6}} ( {a,b} )}} \\ &\quad = \log \sqrt {1 + {v^{2}}} + \frac{{\arctan ( v )}}{v} - \log \biggl[ {p \biggl( {\frac{1}{6}{v^{2}} + 1} \biggr) + ( {1 - p} ) \sqrt{{1 + {v^{2}}}}} \biggr] - 1 \\ &\quad = 3\log ( x ) + \frac{{\arctan ( {\sqrt {{x^{6}} - 1} } )}}{{\sqrt {{x^{6}} - 1} }} - \log \biggl[ {p \biggl( { \frac{1}{6}{x^{6}} + \frac{5}{6}} \biggr) + ( {1 - p} )x} \biggr] - 1. \end{aligned}
(3.18)

Let

$$G ( x ) = 3\log ( x ) + \frac{{\arctan ( {\sqrt {{x^{6}} - 1} } )}}{{\sqrt {{x^{6}} - 1} }} - \log \biggl[ {p \biggl( {\frac{1}{6}{x^{6}} + \frac{5}{6}} \biggr) + ( {1 - p} )x} \biggr] - 1.$$
(3.19)

\begin{aligned} &G \bigl( {{1^{+} }} \bigr) = 0, \qquad G \bigl( {\sqrt{2}} \bigr) = \log ( {\sqrt {2} } ) + \frac{\pi }{4} - \log \biggl[ { \frac{7}{6}p + \sqrt{2} ( {1 - p} )} \biggr] - 1, \end{aligned}
(3.20)
\begin{aligned} &G' ( x ) = \frac{{3{x^{5}}}}{{{{ ( {{x^{6}} - 1} )}^{3/2}}}}{G_{1}} ( x ), \end{aligned}
(3.21)

where

\begin{aligned} &{G_{1}} ( x ) = \frac{{\sqrt {{x^{6}} - 1} [ { - p{x^{11}} + 4 ( {1 - p} ){x^{6}} + 7p{x^{5}} + 2 ( {1 - p} )} ]}}{{{x^{5}} [ {p ( {{x^{6}} + 5} ) + 6 ( {1 - p} )x} ]}} - \arctan \bigl( {\sqrt {{x^{6}} - 1} } \bigr), \\ &{G_{1}} ( 1 ) = 0, \qquad {G_{1}} \bigl( { \sqrt{2}} \bigr) = \frac{{5 [ {\sqrt{2} ( {1 - p} ) + p} ]}}{{6\sqrt{2} ( {1 - p} ) + 7p}} - \frac{\pi }{4}, \end{aligned}
(3.22)
\begin{aligned} &{G'_{1}} ( x ) = - \frac{{{{ ( {{x^{6}} - 1} )}^{3/2}}}}{{{x^{6}}{{ [ {p ( {{x^{6}} + 5} ) + 6 ( {1 - p} )x} ]}^{2}}}}g ( x ), \end{aligned}
(3.23)

where $$g ( x )$$ is defined as in Lemma 2.6.

We divide the proof into four cases.

Case 1 $$p = 12/25$$. Then it follows from (3.18)–(3.23) and Lemma 2.6(1) that

$${R_{AQ}} ( {a,b} ) < \frac{{12}}{{25}} \biggl[ {\frac{1}{3}C ( {a,b} ) + \frac{2}{3}A ( {a,b} )} \biggr] + \frac{{13}}{{25}}{C^{1/3}} ( {a,b} ){A^{2/3}} ( {a,b} ).$$

Case 2 $$0 < p < 12/25$$. Let $$v > 0$$ and $$v \to {0^{+} }$$, then power series expansion leads to

\begin{aligned} &\log \sqrt {1 + {v^{2}}} + \frac{{\arctan ( v )}}{v} - \log \biggl[ {p \biggl( {\frac{1}{6}{v^{2}} + 1} \biggr) + ( {1 - p} ) \sqrt{{1 + {v^{2}}}}} \biggr] - 1 \\ &\quad = \biggl( {\frac{1}{{30}} - \frac{5}{{72}}p} \biggr){v^{4}} + O \bigl( {{v^{6}}} \bigr). \end{aligned}
(3.24)

Equations (3.18), (3.19), and (3.24) lead to the conclusion that there exists $$0 < {\delta _{2}} < 1$$ such that

$${R_{AQ}} ( {a,b} ) > p \biggl[ {\frac{1}{3}C ( {a,b} ) + \frac{2}{3}A ( {a,b} )} \biggr] + ( {1 - p} ){C^{1/3}} ( {a,b} ){A^{2/3}} ( {a,b} )$$

for all $$a > b > 0$$ with $$( {a - b} )/ ( {a + b} ) \in ( {0,{\delta _{2}}} )$$.

Case 3 $$p = 6 [ {\sqrt {2} {{\mathrm{e}} ^{ ( {\pi /4 - 1} )}} - \sqrt{2}} ]/ ( {7 - 6\sqrt{2}} )$$. Then, from (3.20) and (3.22) together with numerical computations, we get

$$G \bigl( {\sqrt{2}} \bigr) = 0, \qquad {G_{1}} \bigl( {\sqrt{2}} \bigr) = - 0.0033 \dots < 0.$$
(3.25)

Let $${\mu _{0}} = 1.0577 \dots$$ be the number given in Lemma 2.6(2). Then we divide the discussion into two subcases.

Subcase 1 $$x \in ( {1,{\mu _{0}}} ]$$. Then $${G_{1}} ( x ) > 0$$ for $$x \in ( {1,{\mu _{0}}} ]$$ follows easily from (3.22) and (3.23) together with Lemma 2.6(2).

Subcase 2 $$x \in ( {{\mu _{0}},\sqrt{2}} )$$. Then Lemma 2.6(2) and (3.23) lead to the conclusion that $${G_{1}} ( x )$$ is strictly decreasing on the interval $$[ {{\mu _{0}},\sqrt{2}} )$$. Then, from (3.25) and Subcase 1, we know that there exists $${\mu _{1}} \in ( {{\mu _{0}},\sqrt{2}} )$$ such that $${G_{1}} ( x ) > 0$$ for $$x \in [ {{\mu _{0}},{\mu _{1}}} )$$ and $${G_{1}} ( x ) < 0$$ for $$x \in ( {{\mu _{1}},\sqrt{2}} )$$.

It follows from Subcases 1 and 2 together with (3.21) that $$G ( x )$$ is strictly increasing on $$( {1,{\mu _{1}}} ]$$ and strictly decreasing on $$[ {{\mu _{1}},\sqrt{2}} )$$. Therefore,

$${R_{AQ}} ( {a,b} ) > p \biggl[ {\frac{1}{3}C ( {a,b} ) + \frac{2}{3}A ( {a,b} )} \biggr] + ( {1 - p} ){C^{1/3}} ( {a,b} ){A^{2/3}} ( {a,b} )$$

follows from (3.18)–(3.20) and (3.25) together with the piecewise monotonicity of $$G ( x )$$.

Case 4 $$6 [ {\sqrt {2} {{\mathrm{e}} ^{ ( {\pi /4 - 1} )}} - \sqrt{2}} ]/ ( {7 - 6\sqrt{2}} ) < p < 1$$. Then (3.21) leads to

$$G \bigl( {\sqrt{2}} \bigr) = \log ( {\sqrt {2} } ) + \frac{\pi }{4} - \log \biggl[ {\frac{7}{6}p + \sqrt{2} ( {1 - p} )} \biggr] - 1 < 0.$$
(3.26)

Equations (3.18) and (3.19) together with inequality (3.26) imply that there exists $$0 < \delta_{2}^{\ast}< 1$$ such that

$${R_{AQ}} ( {a,b} ) < p \biggl[ {\frac{1}{3}C ( {a,b} ) + \frac{2}{3}A ( {a,b} )} \biggr] + ( {1 - p} ){C^{1/3}} ( {a,b} ){A^{2/3}} ( {a,b} )$$

for all $$a > b > 0$$ with $$( {a - b} )/ ( {a + b} ) \in ( {1 - \delta_{2}^{\ast}, 1} )$$. □

## Results and discussion

In this paper, we provide the optimal upper and lower bounds for the Sándor–Yang means $$R_{QA}(a,b)$$ and $$R_{AQ}(a,b)$$ in terms of combinations of the arithmetic mean $$A(a,b)$$ and the contra-harmonic mean $$C(a,b)$$. Our approach may have further applications in the theory of bivariate means.

## Conclusion

In the article, we find several best possible bounds for the Sándor–Yang means $$R_{QA}(a,b)$$ and $$R_{AQ}(a,b)$$. These results are improvements and refinements of the previous results.

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## Funding

The research was supported by the Natural Science Foundation of China (Grants Nos. 61673169, 61374086, 11371125, 11401191), the Tianyuan Special Funds of the National Natural Science Foundation of China (Grant No. 11626101) and the Natural Science Foundation of the Department of Education of Zhejiang Province (Grant No. Y201635325).

## Author information

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### Contributions

All authors contributed equally to the writing of this paper. All authors read and approved the final manuscript.

### Corresponding author

Correspondence to Yu-Ming Chu.

## Ethics declarations

### Competing interests

The authors declare that they have no competing interests. 