Sharp bounds for the Neuman mean in terms of the quadratic and second Seiffert means

Abstract

In this paper, we prove that $\alpha =0$ and $\beta =\frac{\sqrt{3}\pi -4log(2+\sqrt{3})}{(\sqrt{2}\pi -4)log(2+\sqrt{3})}=0.29758\cdots$ are the best possible constants such that the double inequality

$$\alpha Q(a,b)+(1-\alpha )T(a,b)<S_{CA}(a,b)<\beta Q(a,b)+(1-\beta )T(a,b)$$

holds for all $a,b>0$ with $a\ne b$, where $Q(a,b)=\sqrt{(a^2+b^2)/2}$,

$$S_{CA}(a,b)=\frac{(a-b)\sqrt{3(a^2+b^2)+2ab}}{2(a+b){sinh}^{-1}(\frac{(a-b)\sqrt{3(a^2+b^2)+2ab}}{{(a+b)}^2})}$$

and $T(a,b)=(a-b)/[2arctan((a-b)/(a+b))]$ are the quadratic, Neuman and second Seiffert means of a and b, respectively.

MSC:26E60.

1 Introduction

For $a,b>0$ with $a\ne b$, the Neuman mean $S_{CA}(a,b)$ [1, 2] derived from the Schwab-Borchardt mean [3, 4], the quadratic mean $Q(a,b)$ and the second Seiffert mean $T(a,b)$ [5] are given by

$$S_{CA}(a,b)=\frac{(a-b)\sqrt{3(a^2+b^2)+2ab}}{2(a+b){sinh}^{-1}(\frac{(a-b)\sqrt{3(a^2+b^2)+2ab}}{{(a+b)}^2})},$$

(1.1)

$$Q(a,b)=\sqrt{\frac{a^2+b^2}{2}}$$

(1.2)

and

$$T(a,b)=\frac{a-b}{2arctan(\frac{a-b}{a+b})},$$

(1.3)

respectively, where ${sinh}^{-1}(x)=log(x+\sqrt{1+x^2})$ is the inverse hyperbolic sine function. Recently, the Neuman, quadratic and second Seiffert means have been the subject of intensive research. In particular, many remarkable inequalities for these means can be found in the literature [1–4, 6–15].

Let $A(a,b)=(a+b)/2$ and $C(a,b)=(a^2+b^2)/(a+b)$ be the arithmetic and contraharmonic means of a and b, respectively. Then Neuman [1] proved that the inequalities

$$A(a,b)<T(a,b)<S_{CA}(a,b)<Q(a,b)<C(a,b)$$

(1.4)

hold for any $a,b>0$ with $a\ne b$.

In [1, 2], Neuman found that ${\alpha }_1=[\sqrt{3}-log(2+\sqrt{3})]/log(2+\sqrt{3})=0.315\cdots$, ${\beta }_1=1/3$, ${\alpha }_2=1/3$, ${\beta }_2=[log3-2log(log(2+\sqrt{3}))]/(2log2)=0.395\cdots$, ${\alpha }_3=2log(2+\sqrt{3})/3-1=0.520\cdots$ and ${\beta }_3=2/3$ are the best possible constants such that the double inequalities

$$\begin{array}{c}{\alpha }_{1}C(a,b)+(1-{\alpha }_1)A(a,b)<S_{CA}(a,b)<{\beta }_{1}C(a,b)+(1-{\beta }_1)A(a,b),\hfill \\ C^{{\alpha }_2}(a,b)A^{1-{\alpha }_2}(a,b)<S_{CA}(a,b)<C^{{\beta }_2}(a,b)A^{1-{\beta }_2}(a,b)\hfill \end{array}$$

and

$$\frac{{\alpha }_3}{A(a,b)}+\frac{1-{\alpha }_3}{C(a,b)}<\frac{1}{S_{CA}(a,b)}<\frac{{\beta }_3}{A(a,b)}+\frac{1-{\beta }_3}{C(a,b)}$$

hold for any $a,b>0$ with $a\ne b$.

He et al. [16] proved that $\alpha =1/2+\sqrt{\sqrt{3}/log(2+\sqrt{3})-1}/2$ and $\beta =1/2+\sqrt{3}/6$ are the best possible constants in $[1/2,1]$ such that the double inequality

$$C[\alpha a+(1-\alpha )b,\alpha b+(1-\alpha )a]<S_{CA}(a,b)<C[\beta a+(1-\beta )b,\beta b+(1-\beta )a]$$

holds for any $a,b>0$ with $a\ne b$.

In [17, 18], the authors proved that the double inequalities

$$\begin{array}{r}\alpha [\frac{1}{3}C(a,b)+\frac{2}{3}A(a,b)]+(1-\alpha )C^{1/3}(a,b)A^{2/3}(a,b)\\ <S_{CA}(a,b)<\beta [\frac{1}{3}C(a,b)+\frac{2}{3}A(a,b)]+(1-\beta )C^{1/3}(a,b)A^{2/3}(a,b)\end{array}$$

and

$$\lambda A(a,b)+(1-\lambda )Q(a,b)<S_{CA}(a,b)<\mu A(a,b)+(1-\mu )Q(a,b)$$

hold for any $a,b>0$ with $a\ne b$ if and only if $\alpha \le \frac{3[\sqrt[3]{2}log(2+\sqrt{3})-\sqrt{3}]}{(3\sqrt[3]{2}-4)log(2+\sqrt{3})}=0.7528\cdots$, $\beta \ge 4/5$, $\lambda \ge 1/3$ and $\mu \le \frac{\sqrt{2}log(2+\sqrt{3})-\sqrt{3}}{(\sqrt{2}-1)log(2+\sqrt{3})}=0.2390\cdots$.

The main purpose of this paper is to present the best possible constants α and β such that the double inequality

$$\alpha Q(a,b)+(1-\alpha )T(a,b)<S_{CA}(a,b)<\beta Q(a,b)+(1-\beta )T(a,b)$$

holds for any $a,b>0$ with $a\ne b$. All numerical computations are carried out using MATHEMATICA software.

2 Lemmas

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

Lemma 2.1 The double inequality

$$-\frac{2x}{3}+\frac{16x^3}{45}-\frac{2x^5}{7}<\frac{x}{(1+x^2){arctan}^{2}x}-\frac{1}{arctanx}<-\frac{2x}{3}+\frac{16x^3}{45}$$

(2.1)

holds for $x\in (0,0.6)$.

Proof Let

$${\varphi }_1(x)=x-(1+x^2)arctanx+(\frac{2x}{3}-\frac{16x^3}{45}+\frac{2x^5}{7})(1+x^2){arctan}^{2}x,$$

(2.2)

$${\varphi }_2(x)=x-(1+x^2)arctanx+(\frac{2x}{3}-\frac{16x^3}{45})(1+x^2){arctan}^{2}x.$$

(2.3)

Then we only need to show that ${\varphi }_1(x)>0$ and ${\varphi }_2(x)<0$ for $x\in (0,0.6)$.

Taking the differentiation of ${\varphi }_1(x)$ yields

$${\varphi }_1(0)=0,$$

(2.4)

$${\varphi }_1^{\prime }(x)=\frac{2arctanx}{315}{\varphi }_1^{\ast }(x),$$

(2.5)

where

$${\varphi }_1^{\ast }(x)=(105+147x^2-55x^4+315x^6)arctanx-x(105+112x^2-90x^4),$$

(2.6)

$${\varphi }_1^{\ast }(0)=0,$$

(2.7)

$${{\varphi }^{\ast }}_1^{\prime }(x)=\frac{x}{1+x^2}{\varphi }_1^{\ast \ast }(x),$$

(2.8)

where

$${\varphi }_1^{\ast \ast }(x)=2(147+37x^2+835x^4+945x^6)arctanx-x(294-59x^2-765x^4).$$

(2.9)

It is well known that the inequality

$$arctanx>x-\frac{x^3}{3}$$

(2.10)

holds for all $x\in (0,1)$.

Equation (2.9) and inequality (2.10) lead to the conclusion that

$$\begin{array}{rl}{\varphi }_1^{\ast \ast }(x)& >2(147+37x^2+835x^4+945x^6)(x-\frac{x^3}{3})-x(294-59x^2-765x^4)\\ =\frac{x^3}{3}[105+7,231x^2+2,110x^4+1,890x^4(1-x^2)]>0\end{array}$$

(2.11)

for $x\in (0,0.6)$.

Therefore, ${\varphi }_1(x)>0$ for $x\in (0,0.6)$ follows easily from (2.4)-(2.8) and (2.11).

Differentiating ${\varphi }_2(x)$ leads to

$${\varphi }_2(0)=0,$$

(2.12)

$${\varphi }_2^{\prime }(x)=-\frac{2arctanx}{45}{\varphi }_2^{\ast }(x),$$

(2.13)

where

$${\varphi }_2^{\ast }(x)=(15x+16x^3)-(15+21x^2-40x^4)arctanx.$$

(2.14)

It is well known that the inequality

$$arctanx<x-\frac{x^3}{3}+\frac{x^5}{5}$$

(2.15)

holds for all $x\in (0,1)$.

Equation (2.14) and inequality (2.15) lead to the conclusion that

$$\begin{array}{rl}{\varphi }_2^{\ast }(x)& >(15x+16x^3)-(15+21x^2-40x^4)(x-\frac{x^3}{3}+\frac{x^5}{5})\\ =\frac{x^5}{15}(660-263x^2+120x^4)>0\end{array}$$

(2.16)

for $x\in (0,0.6)$.

Therefore, ${\varphi }_2(x)<0$ for $x\in (0,0.6)$ follows from (2.12) and (2.13) together with (2.16). □

Lemma 2.2 The double inequality

$$\frac{x}{\sqrt{1+x^2}}+\frac{x}{(1+x^2){arctan}^{2}x}-\frac{1}{arctanx}>\frac{x}{3}-\frac{x^3}{6}$$

(2.17)

holds for $x\in (0,0.6)$.

Proof A simple computation leads to

$$\begin{array}{r}{(1-\frac{x^2}{2}+\frac{x^4}{4})}^2(1+x^2)\\ =1-\frac{x^4}{16}[8(\frac{\sqrt{2}}{2}+x)(\frac{\sqrt{2}}{2}-x)+2x^4+x^4(1-x^2)]<1\end{array}$$

for $x\in (0,0.6)$. This implies

$$\frac{x}{\sqrt{1+x^2}}>x-\frac{x^3}{2}+\frac{x^5}{4}$$

(2.18)

for $x\in (0,0.6)$.

From Lemma 2.1 and (2.18) we clearly see that

$$\begin{array}{r}\frac{x}{\sqrt{1+x^2}}+\frac{x}{(1+x^2){arctan}^{2}x}-\frac{1}{arctanx}\\ >(x-\frac{x^3}{2}+\frac{x^5}{4})+(-\frac{2x}{3}+\frac{16x^3}{45}-\frac{2x^5}{7})\\ =\frac{x}{3}-\frac{13x^3}{90}-\frac{x^5}{28}=\frac{x}{3}-\frac{x^3}{6}+\frac{x^3}{28}(\sqrt{\frac{28}{45}}+x)(\sqrt{\frac{28}{45}}-x)>\frac{x}{3}-\frac{x^3}{6}\end{array}$$

for $x\in (0,0.6)$. □

Lemma 2.3 The inequality

$$\frac{x}{{[{sinh}^{-1}(x\sqrt{2+x^2})]}^2}-\frac{1+x^2}{\sqrt{2+x^2}{sinh}^{-1}(x\sqrt{2+x^2})}>-\frac{x}{3}+\frac{2x^3}{45}-\frac{x^5}{63}$$

(2.19)

holds for $x\in (0,1)$.

Proof Let

$$\begin{array}{rl}\phi (x)=& x\sqrt{2+x^2}-(1+x^2){sinh}^{-1}\left(x\sqrt{2+x^2}\right)\\ +(\frac{x}{3}-\frac{2x^3}{45}+\frac{x^5}{63}){\left[{sinh}^{-1}\left(x\sqrt{2+x^2}\right)\right]}^2\sqrt{2+x^2}.\end{array}$$

(2.20)

Then we only need to show that $\phi (x)>0$ for $x\in (0,1)$.

Differentiating (2.20) leads to

$$\phi (0)=0,$$

(2.21)

$${\phi }^{\prime }(x)=\frac{2x{sinh}^{-1}(x\sqrt{2+x^2})}{315(1+x^2)}{\phi }_1(x),$$

(2.22)

where

$$\begin{array}{rl}{\phi }_1(x)=& -105-133x^2-18x^4+10x^6\\ +3(35+56x^2+20x^4+4x^6+5x^8)\frac{{sinh}^{-1}(x\sqrt{2+x^2})}{x\sqrt{2+x^2}}.\end{array}$$

(2.23)

We claim that

$$\frac{{sinh}^{-1}(x\sqrt{2+x^2})}{x\sqrt{2+x^2}}>1-\frac{x^2}{3}+\frac{2x^4}{15}-\frac{2x^6}{35}$$

(2.24)

for $x\in (0,1)$. Indeed, let

$$\omega (x)={sinh}^{-1}\left(x\sqrt{2+x^2}\right)-x\sqrt{2+x^2}(1-\frac{x^2}{3}+\frac{2x^4}{15}-\frac{2x^6}{35}),$$

then $\omega (x)>0$ for $x\in (0,1)$ follows from the fact that

$$\omega (0)=0,{\omega }^{\prime }(x)=\frac{16x^8}{35\sqrt{2+x^2}}>0.$$

It follows from (2.23) and (2.24) that

$$\begin{array}{rl}{\phi }_1(x)>& -105-133x^2-18x^4+10x^6\\ +3(35+56x^2+20x^4+4x^6+5x^8)(1-\frac{x^2}{3}+\frac{2x^4}{15}-\frac{2x^6}{35})\\ =& \frac{x^6}{35}[644+90x^2+16x^6+(1-x^2)(239x^2+30x^6)]>0\end{array}$$

(2.25)

for $x\in (0,1)$.

Therefore, $\phi (x)>0$ for $x\in (0,1)$ follows from (2.21) and (2.22) together with (2.25). □

Lemma 2.4 The inequality

$$arctanx>\frac{\pi }{4}+\frac{x-1}{2}-\frac{2{(x-1)}^2}{7}>\frac{\pi }{4}+\frac{3(x-1)}{4}$$

(2.26)

holds for $x\in [0.55,1)$.

Proof Let

$$\nu (x)=arctanx-[\frac{\pi }{4}+\frac{x-1}{2}-\frac{2{(x-1)}^2}{7}].$$

(2.27)

Then simple computations lead to

$$\nu (0.55)=0.00030219\cdots ,\nu (1)=0,$$

(2.28)

$${\nu }^{\prime }(x)=\frac{{\nu }_1(x)}{14(1+x^2)},$$

(2.29)

$${\nu }_1(x)=-1+8x-15x^2+8x^3,$$

(2.30)

$${\nu }_1(0.55)=0.1935,{\nu }_1(1)=0,$$

(2.31)

$${\nu }_1^{\prime }(x)=24(x-\frac{15-\sqrt{33}}{24})(x-\frac{15+\sqrt{33}}{24}).$$

(2.32)

From (2.32) and $(15-\sqrt{33})/24=0.385643\cdots <0.55$ together with $0.55<(15+\sqrt{33})/24=0.864357\cdots <1$, we clearly see that ${\nu }_1(x)$ is strictly decreasing on $[0.55,(15+\sqrt{33})/24]$ and strictly increasing on $[(15+\sqrt{33})/24,1)$. This in conjunction with (2.31) implies that there exists $x_1\in (0.55,1)$ such that ${\nu }_1(x)>0$ for $x\in [0.55,x_1)$ and ${\nu }_1(x)<0$ for $x\in (x_1,1)$. Then equation (2.29) leads to the conclusion that $\nu (x)$ is strictly increasing on $[0.55,x_1]$ and strictly decreasing on $[x_1,1]$.

Therefore, $\nu (x)>0$ for $x\in [0.55,1)$ follows from (2.28) and the piecewise monotonicity of $\nu (x)$. Moreover, the second inequality in (2.26) follows from

$$\frac{x-1}{2}-\frac{2{(x-1)}^2}{7}>\frac{3(x-1)}{4}+\frac{(1-x)(8x-1)}{28}>\frac{3(x-1)}{4}.$$

 □

Lemma 2.5 The inequality

$$x-arctanx<\frac{7}{20}x{arctan}^{2}x$$

(2.33)

holds for $x\in [0.55,1)$.

Proof Let

$$\mu (x)=x-arctanx-\frac{7}{20}x{arctan}^{2}x.$$

(2.34)

Then it suffices to show $\mu (x)<0$ for $x\in [0.55,1)$.

Differentiating $\mu (x)$ yields

$${\mu }^{\prime }(x)=\frac{{\mu }_1(x)}{20(1+x^2)},$$

(2.35)

where

$${\mu }_1(x)=20x^2-14xarctanx-7{arctan}^{2}x-7x^2{arctan}^{2}x.$$

(2.36)

It is well known that

$$arctanx>x-\frac{x^3}{3}+\frac{x^5}{5}-\frac{x^7}{7}$$

(2.37)

for $x\in (0,1)$.

For $x\in [0.55,0.7]$, it follows from (2.36) and (2.37) that

$$\begin{array}{rl}{\mu }_1(x)<& 20x^2-14x(x-\frac{x^3}{3}+\frac{x^5}{5}-\frac{x^7}{7})-7{(x-\frac{x^3}{3}+\frac{x^5}{5}-\frac{x^7}{7})}^2\\ -7x^2{(x-\frac{x^3}{3}+\frac{x^5}{5}-\frac{x^7}{7})}^2=\frac{x^2}{1,575}{\mu }^{\ast }\left(x^2\right),\end{array}$$

(2.38)

where

$$\begin{array}{rl}{\mu }^{\ast }(x)=& -1,575+3,675x-2,695x^2+2,135x^3\\ +3,129x^4-861x^5+405x^6-225x^7,\end{array}$$

(2.39)

$${\mu }^{\ast }(0.49)=-9.99966\cdots .$$

(2.40)

Differentiating ${\mu }^{\ast }(x)$ yields

$$\begin{array}{rl}{{\mu }^{\ast }}^{\prime }(x)=& (3,675-5,390x+6,405x^2)+(12,516x^3-4,305x^4)\\ +(2,430x^5-1,575x^6)>0\end{array}$$

(2.41)

for $x\in [0.3025,0.49]$.

Therefore, ${\mu }^{\ast }(x)<0$ for $x\in [0.3025,0.49]$ follows from (2.40) and (2.41). This in conjunction with (2.35) and (2.38) implies that $\mu (x)$ is strictly decreasing on $[0.55,0.7]$. Therefore, we get $\mu (x)\le \mu (0.55)=-0.00151709\cdots <0$ for $x\in [0.55,0.7]$.

It follows from Lemma 2.4 that

$$\mu (x)<x-[\frac{\pi }{4}+\frac{x-1}{2}-\frac{2{(x-1)}^2}{7}]-\frac{7}{20}{[\frac{\pi }{4}+\frac{x-1}{2}-\frac{2{(x-1)}^2}{7}]}^2=\frac{{\mu }_2(x)}{2,240}$$

(2.42)

for $x\in (0.7,1)$, where

$$\begin{array}{rl}{\mu }_2(x)=& (1,760-560\pi )+(308\pi -49{\pi }^2-644)x\\ +(1,960-420\pi )x^2+(112\pi -1,252)x^3+480x^4-64x^5.\end{array}$$

(2.43)

Differentiating ${\mu }_2(x)$ yields

$${\mu }_2(0.7)=-1.68877\cdots ,{\mu }_2(1)=-2.9025\cdots ,$$

(2.44)

$$\begin{array}{rl}{\mu }_2^{\prime }(x)=& (-644+308\pi -49{\pi }^2)+(3,920-840\pi )x+(336\pi -3,756)x^2\\ +1,920x^3-320x^4,\end{array}$$

(2.45)

$${\mu }_2^{\prime }(0.7)=-4.73674\cdots ,{\mu }_2^{\prime }(1)=20.6372\cdots ,$$

(2.46)

$${\mu }_2^{″}(x)=8(490-105\pi -939x+84\pi x+720x^2-160x^3),$$

(2.47)

$${\mu }_2^{″}(0.7)=-116.173\cdots ,{\mu }_2^{″}(1)=360.212\cdots ,$$

(2.48)

$$\begin{array}{rl}{\mu }_2^{‴}(x)& =24(28\pi -313+480x-160x^2)\\ >24(28\pi -313+480\times 0.7-160\times {(0.7)}^2)=781.55\cdots >0.\end{array}$$

(2.49)

It follows from (2.48) and (2.49) that there exists $x_2\in (0.7,1)$ such that ${\mu }_2^{\prime }(x)$ is strictly decreasing on $(0.7,x_2]$ and strictly increasing on $[x_2,1)$. This in conjunction with (2.46) implies that there exists $x_3\in (0.7,1)$ such that ${\mu }_2(x)$ is strictly decreasing on $(0.7,x_3]$ and strictly increasing on $[x_3,1)$. From (2.44) and the piecewise monotonicity of ${\mu }_2(x)$, we know that ${\mu }_2(x)<0$ for $x\in (0.7,1)$; this in conjunction with (2.42) implies $\mu (x)<0$ for $x\in (0.7,1)$. □

Lemma 2.6 The function

$$\sigma (x)=\frac{\sqrt{1+x^2}{arctan}^{3}x-2(x-arctanx)}{{(1+x^2)}^2{arctan}^{3}x}$$

is strictly decreasing on $[0.55,1)$. Moreover, $\sigma (x)<0.236$ for $x\in [0.55,1)$.

Proof Differentiating $\sigma (x)$ yields

$${\sigma }^{\prime }(x)=\frac{{\sigma }_1(x)}{{(1+x^2)}^3{arctan}^{4}x},$$

(2.50)

where

$${\sigma }_1(x)=6(x-arctanx)+6x^{2}arctanx-8x{arctan}^{2}x-3x\sqrt{1+x^2}{arctan}^{4}x.$$

(2.51)

From Lemma 2.5 and (2.51) we clearly see that

$${\sigma }_1(x)<6x^{2}arctanx-\frac{59}{10}x{arctan}^{2}x-3x{arctan}^{4}x=xarctanx{\sigma }_2(x)$$

(2.52)

for $x\in [0.55,1)$, where

$${\sigma }_2(x)=6x-\frac{59}{10}arctanx-3{arctan}^{3}x.$$

(2.53)

Differentiating ${\sigma }_2(x)$ leads to

$${\sigma }_2(0.55)=-0.0482086\cdots ,{\sigma }_2(1)=-0.0872684\cdots ,$$

(2.54)

$${\sigma }_2^{\prime }(x)=\frac{{\sigma }_3(x)}{10(1+x^2)},$$

(2.55)

$${\sigma }_3(x)=1+60x^2-90{arctan}^{2}x,$$

(2.56)

$${\sigma }_3(0.55)=-3.60662\cdots ,{\sigma }_3(1)=5.48348\cdots ,$$

(2.57)

$${\sigma }_3^{\prime }(x)=\frac{60{\sigma }_4(x)}{1+x^2},$$

(2.58)

$${\sigma }_4(x)=2x+2x^3-3arctanx,$$

(2.59)

$${\sigma }_4(0.55)=-0.0757796\cdots ,{\sigma }_4(1)=1.64381\cdots ,$$

(2.60)

$${\sigma }_4^{\prime }(x)=\frac{-1+8x^2+6x^4}{1+x^2}>0.$$

(2.61)

It follows from (2.58)-(2.61) that there exists $x_4\in (0.55,1)$ such that ${\sigma }_3(x)$ is strictly decreasing on $(0.55,x_4]$ and strictly increasing on $[x_4,1)$. This in conjunction with (2.55)-(2.57) implies that there exists $x_5\in (0.55,1)$ such that ${\sigma }_2(x)$ is strictly decreasing on $(0.55,x_5]$ and strictly increasing on $[x_5,1)$. Then from (2.54) we clearly see that ${\sigma }_2(x)<0$ for $x\in (0.55,1)$.

Therefore, it follows from (2.50) and (2.52) that $\sigma (x)$ is strictly decreasing on $[0.55,1)$. Moreover, $\sigma (x)\le \sigma (0.55)=0.235477\cdots <0.236$ for $x\in [0.55,1)$. □

Lemma 2.7 The function

$$\kappa (x)=\frac{2(4+3x^2){sinh}^{-1}(x\sqrt{2+x^2})-8x\sqrt{2+x^2}}{(2+x^2){[{sinh}^{-1}(x\sqrt{2+x^2})]}^3}$$

is strictly decreasing on $[0.55,1)$. Moreover, $\kappa (x)<0.771$ for $x\in [0.55,1)$.

Proof Simple computations lead to

$$\kappa (0.55)=0.770758\cdots ,$$

(2.62)

$${\kappa }^{\prime }(x)=\frac{8{\kappa }_1(x)}{{(2+x^2)}^2{[{sinh}^{-1}(x\sqrt{2+x^2})]}^4},$$

(2.63)

where

$${\kappa }_1(x)=6x(2+x^2)-3{(2+x^2)}^{3/2}{sinh}^{-1}\left(x\sqrt{2+x^2}\right)+x{\left[{sinh}^{-1}\left(x\sqrt{2+x^2}\right)\right]}^2.$$

(2.64)

We claim that

$$\sqrt{2}x-\frac{x^3}{6\sqrt{2}}<{sinh}^{-1}\left(x\sqrt{2+x^2}\right)<\sqrt{2}x$$

(2.65)

for $x\in (0,1)$. Indeed, let

$${\eta }_1(x)={sinh}^{-1}\left(x\sqrt{2+x^2}\right)-\sqrt{2}x+\frac{x^3}{6\sqrt{2}},$$

(2.66)

$${\eta }_2(x)={sinh}^{-1}\left(x\sqrt{2+x^2}\right)-\sqrt{2}x.$$

(2.67)

Then we clearly see that

$${\eta }_1(0)={\eta }_2(0)=0,$$

(2.68)

$${\eta }_1^{\prime }(x)=\frac{2}{\sqrt{2+x^2}}+\frac{\sqrt{2}}{4}{x}^2-\sqrt{2},$$

(2.69)

$${\eta }_2^{\prime }(x)=\frac{2}{\sqrt{2+x^2}}-\sqrt{2}<0,$$

(2.70)

$${\eta }_1^{\prime }(0)=0,$$

(2.71)

$${\eta }_1^{″}(x)=x(\frac{1}{\sqrt{2}}-\frac{2}{{(2+x^2)}^{3/2}})>0.$$

(2.72)

Therefore, the double inequality (2.65) follows easily from (2.68)-(2.72).

Equation (2.64) and inequality (2.65) imply that

$${\kappa }_1(x)<6x(2+x^2)-3{(2+x^2)}^{3/2}(\sqrt{2}x-\frac{x^3}{6\sqrt{2}})+x{(\sqrt{2}x)}^2=\frac{x}{4}{\kappa }_2(x),$$

(2.73)

where

$${\kappa }_2(x)=16(3+2x^2)-\sqrt{2}(12-x^2){(2+x^2)}^{3/2}.$$

(2.74)

Let $u=\sqrt{2+x^2}$, then $x^2=u^2-2$, $\sqrt{2}<u<\sqrt{3}$ and ${\kappa }_2(x)$ becomes

$$\tilde{\kappa }(u)=-16+32u^2-14\sqrt{2}{u}^3+\sqrt{2}{u}^5.$$

(2.75)

Equation (2.75) leads to

$$\tilde{\kappa }(\sqrt{2})=0,$$

(2.76)

$${\tilde{\kappa }}^{\prime }(u)=u(64-42\sqrt{2}u+5\sqrt{2}{u}^3)=u{\tilde{\kappa }}_1(u),$$

(2.77)

$${\tilde{\kappa }}_1(u)=64-42\sqrt{2}u+5\sqrt{2}{u}^3,{\tilde{\kappa }}_1(\sqrt{2})=0,{\tilde{\kappa }}_1(\sqrt{3})=-2.1362\cdots ,$$

(2.78)

$${\tilde{\kappa }}_1^{\prime }(u)=15\sqrt{2}(u-\sqrt{\frac{14}{5}})(u+\sqrt{\frac{14}{5}}).$$

(2.79)

From (2.79) we clearly see that ${\tilde{\kappa }}_1^{\prime }(u)<0$ for $u\in (\sqrt{2},\sqrt{14/5})$ and ${\tilde{\kappa }}_1^{\prime }(u)>0$ for $u\in (\sqrt{14/5},\sqrt{3})$. This in conjunction with (2.77) implies that ${\tilde{\kappa }}^{\prime }(u)$ is strictly decreasing on $(\sqrt{2},\sqrt{14/5}]$ and strictly increasing on $[\sqrt{14/5},\sqrt{3})$. Thus ${\tilde{\kappa }}^{\prime }(u)<0$ for $u\in (\sqrt{2},\sqrt{3})$ follows from (2.78) and the piecewise monotonicity of ${\tilde{\kappa }}^{\prime }(u)$.

Therefore, ${\kappa }_2(x)=\tilde{\kappa }(u)<0$ follows from (2.76). This in conjunction with (2.63) and (2.73) implies that $\kappa (x)$ is strictly decreasing on $[0.55,1)$. Moreover, it follows from (2.62) that $\kappa (x)\le \kappa (0.55)=0.770758\cdots <0.771$ for $x\in [0.55,1)$. □

Lemma 2.8 The function

$$\tau (x)=\frac{2(x-arctanx)}{{(1+x^2)}^2{arctan}^{3}x}-\frac{2x(3+x^2)}{{(2+x^2)}^{3/2}{sinh}^{-1}(x\sqrt{2+x^2})}<-0.88$$

for $x\in [0.55,1)$.

Proof We first prove

$$\sqrt{2+x^2}{sinh}^{-1}\left(x\sqrt{2+x^2}\right)<2x+\frac{x^3}{3}$$

(2.80)

for $x\in (0,1)$. Let

$$\epsilon (x)=\sqrt{2+x^2}{sinh}^{-1}\left(x\sqrt{2+x^2}\right)-(2x+\frac{x^3}{3}).$$

Then $\epsilon (x)<0$ follows from $\epsilon (0)=0$ and the fact that

$${\epsilon }^{\prime }(x)=\frac{x}{\sqrt{2+x^2}}({sinh}^{-1}\left(x\sqrt{2+x^2}\right)-x\sqrt{2+x^2})<\frac{x}{\sqrt{2+x^2}}(\sqrt{2}x-x\sqrt{2+x^2})<0,$$

where the second term follows from (2.65).

From Lemma 2.5 and (2.10) we clearly see that

$$\frac{x-arctanx}{{arctan}^{3}x}<\frac{7x}{20arctanx}<\frac{21}{20(3-x^2)}$$

(2.81)

for $x\in [0.55,1)$.

It follows from (2.80) and (2.81) that

$$\tau (x)<\frac{21}{10{(1+x^2)}^2(3-x^2)}-\frac{6(3+x^2)}{(2+x^2)(6+x^2)}=:{\tau }_1(x)$$

(2.82)

for $x\in [0.55,1)$.

Simple computation yields

$${\tau }_1(0.55)=-0.906585\cdots ,{\tau }_1(1)=-0.880357\cdots ,$$

(2.83)

$${\tau }_1^{\prime }(x)=\frac{3x}{5{(x^2-3)}^2{(1+x^2)}^3{(2+x^2)}^2{(6+x^2)}^2}\tilde{\tau }(x),$$

(2.84)

where

$$\begin{array}{rl}\tilde{\tau }(x)=& -2,880+2,424x^2+6,052x^4+1,468x^6\\ -939x^8-219x^{10}+60x^{12}+20x^{14},\end{array}$$

(2.85)

$$\tilde{\tau }(0.55)=-1,560.68\cdots ,\tilde{\tau }(1)=5,986,$$

(2.86)

$$\begin{array}{rl}{\tilde{\tau }}^{\prime }(x)=& 2x(2,424+12,104x^2+4,404x^4-3,756x^6\\ -1,095x^8+360x^{10}+140x^{12})>0.\end{array}$$

(2.87)

From (2.85)-(2.87) we know that there exists $x_6\in (0.55,1)$ such that $\tilde{\tau }(x)<0$ for $x\in (0.55,x_6)$ and $\tilde{\tau }(x)>0$ for $x\in (x_6,1)$. This in conjunction with (2.84) implies that ${\tau }_1(x)$ is strictly decreasing on $[0.55,x_6)$ and strictly increasing on $[x_6,1)$.

Therefore, $\tau (x)<{\tau }_1(x)\le max\{{\tau }_1(0.55),{\tau }_1(1)\}=-0.880357\cdots <-0.88$ follows from (2.83) and the piecewise monotonicity of ${\tau }_1(x)$. □

3 Main result

Theorem 3.1 The double inequality

$$\alpha Q(a,b)+(1-\alpha )T(a,b)<S_{CA}(a,b)<\beta Q(a,b)+(1-\beta )T(a,b)$$

(3.1)

holds for all $a,b>0$ with $a\ne b$ if and only if $\alpha \le 0$ and $\beta \ge {\beta }_0=\frac{\sqrt{3}\pi -4log(2+\sqrt{3})}{(\sqrt{2}\pi -4)log(2+\sqrt{3})}=0.29758\cdots$.

Proof Since the Neuman mean $S_{CA}(a,b)$, the quadratic mean $Q(a,b)$ and the second Seiffert mean $T(a,b)$ are symmetric and homogeneous of degree 1, without loss of generality, we assume that $a>b$. Let $v=(a-b)/(a+b)\in (0,1)$, then from (1.1)-(1.3) one has

$$S_{CA}(a,b)=A(a,b)\frac{v\sqrt{2+v^2}}{{sinh}^{-1}(v\sqrt{2+v^2})},$$

(3.2)

$$T(a,b)=A(a,b)\frac{v}{arctan(v)},Q(a,b)=A(a,b)\sqrt{1+v^2}.$$

(3.3)

Equations (3.2) and (3.3) lead to

$$\frac{S_{CA}(a,b)-T(a,b)}{Q(a,b)-T(a,b)}=\frac{\frac{v\sqrt{2+v^2}}{{sinh}^{-1}(v\sqrt{2+v^2})}-\frac{v}{arctan(v)}}{\sqrt{1+v^2}-\frac{v}{arctan(v)}}.$$

(3.4)

It is easy to find that

$$\underset{v\to 0^{+}}{lim}\frac{\frac{v\sqrt{2+v^2}}{{sinh}^{-1}(v\sqrt{2+v^2})}-\frac{v}{arctan(v)}}{\sqrt{1+v^2}-\frac{v}{arctan(v)}}=0,$$

(3.5)

$$\underset{v\to 1^{-}}{lim}\frac{\frac{v\sqrt{2+v^2}}{{sinh}^{-1}(v\sqrt{2+v^2})}-\frac{v}{arctan(v)}}{\sqrt{1+v^2}-\frac{v}{arctan(v)}}={\beta }_0.$$

(3.6)

We investigate the difference between the convex combination of $Q(a,b)$, $T(a,b)$ and $S_{CA}(a,b)$ as follows:

$$\begin{array}{r}pQ(a,b)+(1-p)T(a,b)-S_{CA}(a,b)\\ =A(a,b)[p\sqrt{1+v^2}+(1-p)\frac{v}{arctan(v)}-\frac{v\sqrt{2+v^2}}{{sinh}^{-1}(v\sqrt{2+v^2})}].\end{array}$$

(3.7)

Let

$$D_p(v)=p\sqrt{1+v^2}+(1-p)\frac{v}{arctan(v)}-\frac{v\sqrt{2+v^2}}{{sinh}^{-1}(v\sqrt{2+v^2})}.$$

(3.8)

Then simple computations lead to

$$D_p\left(0^{+}\right)=0,D_p\left(1^{-}\right)=p(\sqrt{2}-\frac{4}{\pi })+\frac{4}{\pi }-\frac{\sqrt{3}}{log(2+\sqrt{3})},D_{{\beta }_0}\left(1^{-}\right)=0,$$

(3.9)

$$\begin{array}{rl}{D}_p^{\prime }(v)=& p[\frac{v}{\sqrt{1+v^2}}+\frac{v}{(1+v^2){arctan}^{2}v}-\frac{1}{arctanv}]+\frac{v}{{({sinh}^{-1}(v\sqrt{2+v^2}))}^2}\\ -\frac{1+v^2}{\sqrt{2+v^2}{sinh}^{-1}(v\sqrt{2+v^2})}-\frac{v}{(1+v^2){arctan}^{2}v}+\frac{1}{arctanv},\end{array}$$

(3.10)

$$\begin{array}{rl}{D}_p^{″}(v)=& p\frac{\sqrt{1+v^2}{arctan}^{3}v-2(v-arctanv)}{{(1+v^2)}^2{arctan}^{3}v}\\ +\frac{2(4+3v^2){sinh}^{-1}(v\sqrt{2+v^2})-8v\sqrt{2+v^2}}{(2+v^2){({sinh}^{-1}(v\sqrt{2+v^2}))}^3}\\ +\frac{2(v-arctanv)}{{(1+v^2)}^2{arctan}^{3}v}-\frac{2v(3+v^2)}{{(2+v^2)}^{3/2}{sinh}^{-1}(v\sqrt{2+v^2})}\\ =& p\sigma (v)+\kappa (v)+\tau (v),\end{array}$$

(3.11)

where $\sigma (x)$, $\kappa (x)$ and $\tau (x)$ are defined as in Lemmas 2.6, 2.7 and 2.8, respectively.

From Lemmas 2.1-2.3 and (3.10) we clearly see that

$$\begin{array}{rl}{D}_{{\beta }_0}^{\prime }(v)& >{\beta }_0(\frac{v}{3}-\frac{v^3}{6})-\frac{v}{3}+\frac{2v^3}{45}-\frac{v^5}{63}+\frac{2v}{3}-\frac{16v^3}{45}\\ =\frac{v}{630}[210(1+{\beta }_0)-7(28+15{\beta }_0)v^2-10v^4]\\ >\frac{v}{630}[210(1+0.29758)-7(28+15\times 0.29759)\times {(0.55)}^2-10\times {(0.55)}^4]\\ =\frac{v}{630}\times 202.83\cdots >0\end{array}$$

(3.12)

for $v\in (0,0.55]$.

It follows from Lemmas 2.6-2.8 and (3.11) that

$$D_{{\beta }_0}^{″}(v)={\beta }_0\sigma (v)+\kappa (v)+\tau (v)<0.236{\beta }_0+0.771-0.88=-0.0387709\cdots$$

(3.13)

for $v\in [0.55,1)$. Then from $D_{{\beta }_0}^{\prime }(0.55)=0.0139552\cdots$ and $D_{{\beta }_0}^{\prime }(1)=-0.0650268\cdots$ we know that there exists $v_0\in (0.55,1)$ such that $D_{{\beta }_0}^{\prime }(v)>0$ for $v\in [0.55,v_0)$ and $D_{{\beta }_0}^{\prime }(x)<0$ for $v\in (v_0,1)$. This in conjunction with (3.13) leads to the conclusion that $D_{{\beta }_0}(v)$ is strictly increasing on $[0.55,v_0]$ and strictly decreasing on $[v_0,1)$.

Therefore, $D_{{\beta }_0}(v)>0$ for $v\in (0,1)$ follows from (3.9) and the monotonicity of $D_{{\beta }_0}(v)$. In other words, we obtain

$${\beta }_{0}Q(a,b)+(1-{\beta }_0)T(a,b)>S_{CA}(a,b)$$

(3.14)

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

Obviously, if $\alpha =0$, then (1.4) gives

$$T(a,b)<S_{CA}(a,b)$$

(3.15)

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

Therefore, Theorem 3.1 follows from (3.14) and (3.15) together with the following statements:

• If $\alpha >0$, then (3.4) and (3.5) imply that there exists ${\delta }_1\in (0,1)$ such that $S_{CA}(a,b)<\alpha Q(a,b)+(1-\alpha )T(a,b)$ for all $a,b>0$ with $(a-b)/(a+b)\in (0,{\delta }_1)$.

• If $\beta <{\beta }_0$, then (3.4) and (3.6) imply that there exists ${\delta }_2\in (0,1)$ such that $S_{CA}(a,b)>\beta Q(a,b)+(1-\beta )T(a,b)$ for all $a,b>0$ with $(a-b)/(a+b)\in (1-{\delta }_2,1)$.

 □