Inequalities for certain means in two arguments

Abstract

In this paper, we present the sharp bounds of the ratios $U(a,b)/L_{4}(a,b)$, $P_{2}(a,b)/U(a,b)$, $NS(a,b)/P_{2}(a,b)$ and $B(a,b)/NS(a,b)$ for all $a, b>0$ with $a\neq b$, where $L_{4}(a,b)=[(b^{4}-a^{4})/(4(\log b-\log a))]^{1/4}$, $U(a,b)=(b-a)/[\sqrt{2}\arctan((b-a)/\sqrt{2ab})]$, $P_{2}(a,b)=[(b^{2}-a^{2})/(2\arcsin ((b^{2}-a^{2})/(b^{2}+a^{2})))]^{1/2}$, $NS(a,b)=(b-a)/[2\sinh ^{-1}((b-a)/(b+a))]$, $B(a,b)=Q(a,b)e^{A(a,b)/T(a,b)-1}$, $A(a,b)=(a+b)/2$, $Q(a,b)=\sqrt{(a^{2}+b^{2})/2}$, and $T(a,b)=(a-b)/[2\arctan((a-b)/(a+b))]$.

1 Introduction

For $r\in\mathbb{R}$, the rth power mean $M(a,b; r)$ of two distinct positive real numbers a and b is defined by

$$ M(a,b; r)= \textstyle\begin{cases} (\frac{a^{r}+b^{r}}{2} )^{1/r},& r\neq0, \\ \sqrt{ab}, & r=0. \end{cases} $$

(1.1)

It is well known that $M(a,b; r)$ is continuous and strictly increasing with respect to $r\in\mathbb{R}$ for fixed $a, b>0$ with $a\neq b$. Many classical means are the special cases of the power mean, for example, $M(a,b; -1)=2ab/(a+b)=H(a,b)$ is the harmonic mean, $M(a,b; 0)=\sqrt{ab}=G(a,b)$ is the geometric mean, $M(a,b; 1)=(a+b)/2=A(a,b)$ is the arithmetic mean, and $M(a,b; 2)=\sqrt {(a^{2}+b^{2})/2}=Q(a,b)$ is the quadratic mean. The main properties for the power mean are given in [1].

Let

$$\begin{aligned}& L(a,b)=\frac{a-b}{\log a-\log b}, \qquad P(a,b)=\frac{a-b}{2\arcsin (\frac{a-b}{a+b} )}, \\& U(a,b)=\frac{a-b}{\sqrt{2}\arctan (\frac{a-b}{\sqrt {2ab}} )},\qquad NS(a,b)=\frac{a-b}{2\sinh^{-1} (\frac {a-b}{a+b} )}, \end{aligned}$$

(1.2)

$$\begin{aligned}& T(a,b)=\frac{a-b}{2\arctan (\frac{a-b}{a+b} )}, \qquad B(a,b)=Q(a,b)e^{A(a,b)/T(a,b)-1} \end{aligned}$$

(1.3)

be, respectively, the logarithmic mean, first Seiffert mean [2], Yang mean [3], Neuman-Sándor mean [4, 5], second Seiffert mean [6], Sándor-Yang mean [3, 7] of two distinct positive real numbers a and b.

Recently, the sharp bounds for certain bivariate means in terms of the power mean have attracted the attention of many mathematicians.

Radó [8] and Lin [9], Jagers [10] and Hästö [11, 12] proved that the double inequalities

$$\begin{aligned}& M(a,b; 0)< L(a,b)< M(a,b; 1/3), \end{aligned}$$

(1.4)

$$\begin{aligned}& M(a,b; \log2/\log\pi)< P(a,b)< M(a,b; 2/3) \end{aligned}$$

(1.5)

hold for all $a, b>0$ and $a\neq b$ with the best possible parameters 0, $1/3$, $\log2/\log\pi$, and $2/3$.

In [13–17], the authors proved that the double inequalities

$$\begin{aligned}& M(a,b; \alpha)< NS(a,b)< M(a,b; \beta), \end{aligned}$$

(1.6)

$$\begin{aligned}& M(a,b; \lambda)< U(a,b)< M(a,b; \mu) \end{aligned}$$

(1.7)

hold for all $a, b>0$ and $a\neq b$ if and only if $\alpha\leq\log 2/\log[2\log(1+\sqrt{2})]$, $\beta\geq4/3$, $\lambda\leq2\log 2/(2\log\pi-\log2)$ and $\mu\geq4/3$.

Very recently, Yang and Chu [18] presented that $p=4\log2/(4+2\log 2-\pi)$ and $q=4/3$ are the best possible parameters such that the double inequality

$$ M(a,b; p)< B(a,b)< M(a,b; q) $$

(1.8)

holds for all $a, b>0$ and $a\neq b$.

Let

$$ L_{4}(a,b)=L^{1/4}\bigl(a^{4}, b^{4}\bigr)= \biggl(\frac{b^{4}-a^{4}}{4(\log b-\log a)} \biggr)^{1/4} $$

(1.9)

and

$$ P_{2}(a,b)=P^{1/2}\bigl(a^{2}, b^{2}\bigr)= \biggl(\frac{b^{2}-a^{2}}{2\arcsin (\frac{b^{2}-a^{2}}{b^{2}+a^{2}} )} \biggr)^{1/2} $$

(1.10)

be, respectively, the fourth-order logarithmic and second-order first Seiffert means of a and b.

Then from (1.4)-(1.10) we clearly see that $M(a, b; 4/3)$ is the common sharp upper power mean bound for $L_{4}(a,b)$, $U(a,b)$, $P_{2}(a,b)$, $NS(a,b)$, and $B(a,b)$. Therefore, it is natural to ask what are the size relationships among these means? The main purpose of this paper is to answer this question.

2 Lemmas

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

Lemma 2.1

(See Lemma 7 of [19])

Let $\{a_{k}\}_{k=0}^{\infty}$ be a nonnegative real sequence with $a_{m}>0$ and $\sum_{k=m+1}^{\infty }a_{k}>0$, and

$$ P(t)=-\sum_{k=0}^{m}a_{k}t^{k}+ \sum_{k=m+1}^{\infty}a_{k}t^{k} $$

be a convergent power series on the interval $(0, \infty)$. Then there exists $t_{m+1}\in(0, \infty)$ such that $P(t_{m+1})=0$, $P(t)<0$ for $t\in(0, t_{m+1})$ and $P(t)>0$ for $t\in(t_{m+1}, \infty)$.

Lemma 2.2

Let $n\in\mathbb{N}$. Then

$$ 9(n-3)4^{2n}-8n(4n-11)3^{2n}+72n(n-1)2^{2n}-72n(20n-13)>0 $$

for all $n\geq6$.

Proof

Let

$$\begin{aligned}& v_{n}=9(n-3)4^{2n}-8n(4n-11)3^{2n}+72n(n-1)2^{2n}-72n(20n-13), \\& v^{\ast}_{n}=9\times \biggl(\frac{4}{3} \biggr)^{2n}-\frac{8n(4n-11)}{n-3}. \end{aligned}$$

(2.1)

Then we clearly see that

$$\begin{aligned}& v^{\ast}_{6}=\frac{4\text{,}495\text{,}024}{59\text{,}049}>0, \end{aligned}$$

(2.2)

$$\begin{aligned}& v_{n}\geq9(n-3)4^{2n}-8n(4n-11)3^{2n}+72n(n-1)2^{12}-72n(20n-13) \\& \hphantom{v_{n}}=9(n-3)4^{2n}-8n(4n-11)3^{2n}+72n(4 \text{,}076n-4\text{,}083) \\& \hphantom{v_{n}}>9(n-3)4^{2n}-8n(4n-11)3^{2n} \\& \hphantom{v_{n}}=(n-3)3^{2n} \times v^{\ast}_{n}, \end{aligned}$$

(2.3)

$$\begin{aligned}& v^{\ast}_{n+1}- \biggl(\frac{4}{3} \biggr)^{2}v^{\ast}_{n}= \frac {8(28n^{3}-169n^{2}+334n-189)}{9(n-2)(n-3)}>0 \end{aligned}$$

(2.4)

for $n\geq6$.

It follows from (2.2) and (2.4) that

$$ v^{\ast}_{n}>0 $$

(2.5)

for $n\geq6$.

Therefore, Lemma 2.2 follows easily from (2.1), (2.3), and (2.5). □

Lemma 2.3

Let $t>0$ and

$$ g_{1}(t)=\frac{\sqrt{2}}{2}\arctan \bigl(\sqrt{2} \sinh(t) \bigr)-\frac{4t\sinh^{2}(2t)}{\sinh(4t)\sinh(t)+4t\sinh(3t)}. $$

(2.6)

Then there exists a unique $t_{0}\in(0, \infty)$ such that $g_{1}(t)<0$ for $t\in(0, t_{0})$, $g_{1}(t_{0})=0$, and $g_{1}(t)>0$ for $t\in(t_{0}, \infty)$.

Proof

It follows from (2.6) that

$$\begin{aligned}& g_{1}\bigl(0^{+}\bigr)=0,\qquad \lim_{t\rightarrow\infty}g_{1}(t)= \frac{\sqrt {2}}{4}\pi>0, \end{aligned}$$

(2.7)

$$\begin{aligned}& g_{1}(t)=\frac{\sqrt{2}}{2}\arctan \bigl(\sqrt{2}\sinh(t) \bigr)- \frac{16t\sinh(t)\cosh^{2}(t)}{\sinh(4t)+16t\cosh^{2}(t)-4t}, \\& g^{\prime}_{1}(t)=\frac{\cosh(t)}{ (1+2\sinh^{2}(t) ) (\sinh(4t)+16t\cosh^{2}(t)-4t )^{2}}g_{2}(t), \end{aligned}$$

(2.8)

where

$$\begin{aligned} g_{2}(t) =&t^{2}\bigl[128\cosh^{2}(t) \sinh^{2}(t)-512\cosh^{4}(t)\sinh ^{2}(t)-64 \cosh^{2}(t)+256\sinh^{4}(t) \\ &{}+128\sinh^{2}(t)+16\bigr]+t\bigl[16 \sinh(4t)\cosh ^{2}(t)-32\sinh(4t)\cosh^{2}(t) \sinh^{2}(t) \\ &{}+128\cosh(4t)\cosh(t)\sinh^{3}(t)+64\cosh(4t)\cosh(t)\sinh (t) \\ &{}-64 \sinh(4t)\sinh^{4}(t)-32\sinh(4t)\sinh^{2}(t)-8\sinh(4t)\bigr] \\ &{}+\sinh^{2}(4t)-32 \cosh(t)\sinh (4t)\sinh^{3}(t)-16\cosh(t)\sinh(4t)\sinh(t) \\ =&-\frac{3}{2}\cosh(8t)+2t\sinh(8t)-16t^{2}\cosh(6t)+12t\sinh (6t)+16t^{2}\cosh(4t) \\ &{}-4t\sinh(4t)-80t^{2}\cosh(2t)+12t\sinh(2t)+32t^{2}+ \frac{3}{2}. \end{aligned}$$

(2.9)

Making use of power series formulas, (2.9) gives

$$ g_{2}(t)=\sum_{n=2}^{\infty} \frac{v_{n}}{18\times(2n)!}(2t)^{2n}, $$

(2.10)

where $v_{n}$ is defined by (2.1).

Note that

$$ v_{2}=v_{3}=0, \qquad v_{4}=-258 \text{,}048,\qquad v_{5}=-940\text{,}032. $$

(2.11)

From Lemma 2.1, (2.8), (2.10), and (2.11) we know that there exists $t_{1}\in(0, \infty)$ such that $g_{1}(t)$ is strictly decreasing on $(0, t_{1}]$ and strictly increasing on $[t_{1}, \infty)$.

Therefore, Lemma 2.3 follows easily from (2.7) and the piecewise monotonicity of $g_{1}(t)$. □

Lemma 2.4

The inequality

$$ -2x^{2}\cos x+\sin^{2}x\cos x+2x^{2} \cos^{2}x+x\sin x+x^{2}-3x\cos x\sin x>0 $$

holds for all $x\in(0, \pi/2)$.

Proof

Simple computations lead to

$$\begin{aligned}& -2x^{2}\cos x+\sin^{2}x\cos x+2x^{2} \cos^{2}x+x\sin x+x^{2}-3x\cos x\sin x \\& \quad =x^{2}\cos(2x)-2x^{2}\cos x+\frac{1}{4}\cos x- \frac{1}{4}\cos (3x)+x\sin x-\frac{3}{2}x\sin(2x)+2x^{2} \\& \quad =\sum_{n=2}^{\infty}(-1)^{n-1} \frac {3^{2n}+4n(n-2)2^{2n}-32n^{2}+24n-1}{4\times(2n)!}x^{2n}. \end{aligned}$$

(2.12)

Let

$$\begin{aligned}& \omega_{n}=\frac{3^{2n}+4n(n-2)2^{2n}-32n^{2}+24n-1}{4\times(2n)!}x^{2n}, \end{aligned}$$

(2.13)

$$\begin{aligned}& \omega^{\ast}_{n}=3^{2n}+4n(n-2)2^{2n}-32n^{2}+24n-1. \end{aligned}$$

(2.14)

Then

$$\begin{aligned}& \omega_{2}=0, \qquad \omega_{3}=\frac{4x^{6}}{9}>0, \end{aligned}$$

(2.15)

$$\begin{aligned}& \omega^{\ast}_{n}> 4n(n-2)2^{6}-32n^{2}+24n=8n(28n-61)>0 \quad (n\geq3), \end{aligned}$$

(2.16)

$$\begin{aligned}& \omega^{\ast}_{n+1}-9\omega^{\ast }_{n}=- \bigl(5n^{2}-18n+4\bigr)4^{n+1}+256n(n-1)< 0 \quad (n \geq4), \end{aligned}$$

(2.17)

$$\begin{aligned}& \frac{\omega_{4}}{\omega_{3}}=\frac{x^{2}}{56}\frac{\omega^{\ast }_{4}}{\omega^{\ast}_{3}}=\frac{x^{2}}{56}\times \frac{14\text{,}336}{1\text{,}280} =\frac{x^{2}}{5}< \frac{\pi^{2}}{20}. \end{aligned}$$

(2.18)

It follows from (2.13), (2.14), (2.16), and (2.17) that

$$\begin{aligned}& \omega_{n}>0 \quad (n\geq3), \end{aligned}$$

(2.19)

$$\begin{aligned}& \frac{\omega_{n+1}}{\omega_{n}}=\frac{x^{2}}{(2n+1)(2n+2)}\frac {\omega^{\ast}_{n+1}}{\omega^{\ast}_{n}} < \frac{9x^{2}}{(2n+1)(2n+2)}< \frac{\pi^{2}}{40} \quad (n\geq4). \end{aligned}$$

(2.20)

Inequalities (2.18)-(2.20) imply that the sequence $\{\omega_{n}\}$ is strictly decreasing for $n\geq3$, $\lim_{n\rightarrow\infty}\omega _{n}=0$ and $\sum_{n=2}^{\infty}(-1)^{n-1}\omega_{n}$ is a Leibniz series. Therefore, Lemma 2.4 follows from (2.12), (2.13), and (2.15). □

Lemma 2.5

The inequality

$$ \frac{\sqrt{2}\sinh(2t)\cosh(t)\arcsin(\tanh(2t))}{\sinh (2t)+\cosh(2t)\arcsin(\tanh(2t))}-\arctan \bigl(\sqrt{2}\sinh (t) \bigr)>0 $$

hold for all $t\in(0, \infty)$.

Proof

Let $x=\arcsin(\tanh(2t))\in(0, \pi/2)$ and

$$ h_{1}(t)=\frac{\sqrt{2}\sinh(2t)\cosh(t)\arcsin(\tanh(2t))}{\sinh (2t)+\cosh(2t)\arcsin(\tanh(2t))}-\arctan \bigl(\sqrt{2}\sinh (t) \bigr). $$

(2.21)

Then

$$\begin{aligned}& \sinh(2t)=\tan x, \qquad \cosh(2t)=\frac{1}{\cos x},\qquad \tanh(t)= \frac {1-\cos x}{\sin x}, \\& h_{1}\bigl(0^{+}\bigr)=0, \end{aligned}$$

(2.22)

$$\begin{aligned}& h_{1}(t)=\frac{\sqrt{2}x\sin x}{x+\sin x}\cosh(t)-\arctan \bigl(\sqrt{2}\sinh(t) \bigr), \\& h^{\prime}_{1}(t)=\frac{d}{dx} \biggl(\frac{\sqrt{2}x\sin x}{x+\sin x} \biggr)\frac{d[\arcsin(\tanh(2t))]}{dt}\cosh(t) +\frac{\sqrt{2}x\sin x}{x+\sin x}\sinh(t)-\frac{\sqrt{2}\cosh (t)}{\cosh(2t)} \\& \hphantom{h^{\prime}_{1}(t)}=\frac{\sqrt{2}\cosh(t) [2x^{2}\cos x-2x\sin x-x^{2}+\sin ^{2}x ]}{(x+\sin x)^{2}\cosh(2t)} +\frac{\sqrt{2}x\sin x}{x+\sin x}\sinh(t) \\& \hphantom{h^{\prime}_{1}(t)}=\sqrt{2}\cosh(t) \biggl[\frac{\cos x (2x^{2}\cos x-2x\sin x-x^{2}+\sin^{2}x )}{(x+\sin x)^{2}}+\frac{x\sin x(1-\cos x)}{\sin x(x+\sin x)} \biggr] \\& \hphantom{h^{\prime}_{1}(t)}=\frac{\sqrt{2}\cosh(t)[-2x^{2}\cos x+\sin^{2}x\cos x+2x^{2}\cos ^{2}x+x\sin x+x^{2}-3x\cos x\sin x]}{(x+\sin x)^{2}}. \end{aligned}$$

(2.23)

Therefore, Lemma 2.5 follows easily from (2.21)-(2.23) and Lemma 2.4. □

3 Main results

Theorem 3.1

The double inequality

$$ \lambda_{1}L_{4}(a,b)\leq U(a,b)< \mu_{1}L_{4}(a,b) $$

holds for all $a, b>0$ with $a\neq b$ if and only if $\lambda_{1}\leq c_{0}$ and $\mu_{1}=\infty$, where

$$ c_{0}=e^{\log(\sinh(t_{0}))-\log(\arctan(\sqrt{2}\sinh (t_{0})))-\log(\sinh(4t_{0})/t_{0})/4+\log2} $$

and $t_{0}\in(0, \infty)$ is defined by Lemma  2.3. Moreover, numerical computations show that $t_{0}=1.1336\ldots$ and $c_{0}=0.9991\ldots$ .

Proof

Since $U(a,b)$ and $L_{4}(a,b)$ are symmetric and homogeneous of degree 1, without loss of generality, we assume that $b>a>0$. Let $t=\log\sqrt{b/a}>0$, then (1.2) and (1.9) lead to

$$\begin{aligned}& U(a,b)=\frac{\sqrt{2ab}\sinh(t)}{\arctan (\sqrt{2}\sinh (t) )}, \qquad L_{4}(a,b)=\sqrt{ab} \biggl(\frac{\sinh (4t)}{4t} \biggr)^{1/4}, \end{aligned}$$

(3.1)

$$\begin{aligned}& \log\frac{U(a,b)}{L_{4}(a,b)}=\log\bigl(\sinh(t)\bigr)-\log \bigl(\arctan \bigl( \sqrt{2}\sinh(t) \bigr) \bigr) \\& \hphantom{\log\frac{U(a,b)}{L_{4}(a,b)}={}}{}-\frac{1}{4}\log\bigl(\sinh(4t)\bigr) + \frac{1}{4}\log t+\log2. \end{aligned}$$

(3.2)

Let

$$\begin{aligned} g(t) =&\log\bigl(\sinh(t)\bigr)-\log \bigl(\arctan \bigl(\sqrt{2} \sinh (t) \bigr) \bigr) \\ &{}-\frac{1}{4}\log\bigl(\sinh(4t)\bigr) + \frac{1}{4}\log t+\log2. \end{aligned}$$

(3.3)

Then

$$\begin{aligned}& g\bigl(0^{+}\bigr)=0, \qquad \lim_{t\rightarrow\infty}g(t)= \infty, \end{aligned}$$

(3.4)

$$\begin{aligned}& g^{\prime}(t)=\frac{\cosh(t)}{\sinh(t)}-\frac{\sqrt{2}\cosh (t)}{\arctan (\sqrt{2}\sinh(t) )\cosh(2t)}-\frac{\cosh (4t)}{\sinh(4t)}+ \frac{1}{4t} \\& \hphantom{g^{\prime}(t)}=\frac{\sinh(4t)\sinh(t)+4t\sinh(3t)}{4t\sinh(4t)\sinh(t)}-\frac {\sqrt{2}\cosh(t)}{\arctan (\sqrt{2}\sinh(t) )\cosh(2t)} \\& \hphantom{g^{\prime}(t)}=\frac{\sqrt{2}(\sinh(4t)\sinh(t)+4t\sinh(3t))}{4t\sinh(4t)\sinh (t)\arctan (\sqrt{2}\sinh(t) )}g_{1}(t), \end{aligned}$$

(3.5)

where $g_{1}(t)$ is defined by (2.6).

It follows from Lemma 2.3 and (3.5) that there exists a unique $t_{0}\in(0, \infty)$ such that $g_{1}(t_{0})=0$, $g(t)$ is strictly decreasing on $(0, t_{0}]$ and strictly increasing on $[t_{0}, \infty)$.

Therefore, Theorem 3.1 follows from (3.2)-(3.4) and the piecewise monotonicity of $g(t)$. □

Theorem 3.2

The double inequality

$$ \lambda_{2}U(a,b)< P_{2}(a,b)< \mu_{2}U(a,b) $$

holds for all $a, b>0$ with $a\neq b$ if and only if $\lambda_{2}\leq 1$ and $\mu_{2}\geq\sqrt{\pi/2}=1.2533\ldots$ .

Proof

Since $U(a,b)$ and $P_{2}(a,b)$ are symmetric and homogeneous of degree 1, without loss of generality, we assume that $b>a>0$. Let $t=\log\sqrt{b/a}>0$, then (1.10) and (3.1) lead to

$$\begin{aligned}& P_{2}(a,b)=\sqrt{ab} \biggl(\frac{\sinh(2t)}{\arcsin(\tanh (2t))} \biggr)^{1/2}, \end{aligned}$$

(3.6)

$$\begin{aligned}& \log\frac{P_{2}(a,b)}{U(a,b)}=\log \bigl(\arctan \bigl(\sqrt {2}\sinh(t) \bigr) \bigr) \\& \hphantom{\log\frac{P_{2}(a,b)}{U(a,b)}={}}{}- \frac{1}{2}\log\bigl(\arcsin\bigl(\tanh (2t)\bigr)\bigr)-\frac{1}{2} \log\bigl(\tanh(t)\bigr). \end{aligned}$$

(3.7)

Let

$$ h(t)=\log \bigl(\arctan \bigl(\sqrt{2}\sinh(t) \bigr) \bigr)- \frac{1}{2}\log\bigl(\arcsin\bigl(\tanh(2t)\bigr)\bigr)-\frac{1}{2} \log\bigl(\tanh(t)\bigr). $$

(3.8)

Then simple computations lead to

$$\begin{aligned}& h\bigl(0^{+}\bigr)=0,\qquad \lim_{t\rightarrow\infty}h(t)= \frac{1}{2}(\log\pi -\log2), \end{aligned}$$

(3.9)

$$\begin{aligned}& h^{\prime}(t)=\frac{\sqrt{2}\cosh(t)}{\cosh(2t)\arctan (\sqrt{2}\sinh(t) )}-\frac{1}{\cosh(2t)\arcsin(\tanh (2t))}-\frac{1}{\sinh(2t)} \\& \hphantom{h^{\prime}(t)}=\frac{\sinh(2t)+\cosh(2t)\arcsin(\tanh(2t))}{\sinh(2t)\cosh (2t)\arcsin(\tanh(2t))\arctan (\sqrt{2}\sinh(t) )} \\& \hphantom{h^{\prime}(t)={}}{}\times \biggl[\frac{\sqrt{2}\sinh(2t)\cosh(t)\arcsin(\tanh (2t))}{\sinh(2t)+\cosh(2t)\arcsin(\tanh(2t))}-\arctan \bigl(\sqrt {2} \sinh(t) \bigr) \biggr]. \end{aligned}$$

(3.10)

It follows from Lemma 2.5 and (3.10) that $h(t)$ is strictly increasing on $(0, \infty)$. Therefore, Theorem 3.2 follows easily from (3.7)-(3.9) and the monotonicity of $h(t)$. □

Remark 3.1

Let $b>a>0$ and $t=\log\sqrt{b/a}>0$. Then

$$ A(a,b)=\sqrt{ab}\cosh(t), \qquad Q(a,b)=\sqrt{ab} \cosh^{1/2}(2t). $$

(3.11)

It follows from Lemma 2.5 that

$$ \frac{\sqrt{2}\sinh(t)}{\arctan (\sqrt{2}\sinh(t) )}>\frac{\frac{\sinh(2t)}{\arcsin(\tanh(2t))}+\cosh(2t)}{2\cosh^{2}(t)}. $$

(3.12)

Equations (3.1), (3.6), and (3.11) together with inequality (3.12) lead to the conclusion that the inequality

$$ U(a,b)>\frac{P^{2}_{2}(a,b)+Q^{2}(a,b)}{2A^{2}(a,b)}G(a,b) $$

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

Theorem 3.3

The double inequality

$$ \lambda_{3}P_{2}(a,b)< NS(a,b) (a,b)< \mu_{3}P_{2}(a,b) $$

holds for all $a, b>0$ with $a\neq b$ if and only if $\lambda_{3}\leq 1$ and $\mu_{3}\geq\sqrt{\pi}/[2\log(1+\log2)]=1.0055\ldots$ .

Proof

Since $NS(a,b)$ and $P_{2}(a,b)$ are symmetric and homogeneous of degree 1, without loss of generality, we assume that $b>a>0$. Let $t=\log\sqrt{b/a}>0$, then (1.2) and (3.6) lead to

$$\begin{aligned}& NS(a,b)=\sqrt{ab}\frac{\sinh(t)}{\sinh^{-1}(\tanh(t))}, \end{aligned}$$

(3.13)

$$\begin{aligned}& \log\frac{NS(a,b)}{P_{2}(a,b)}=\frac{1}{2}\log\bigl(\tanh(t)\bigr)-\log \bigl( \sinh^{-1}\bigl(\tanh(t)\bigr)\bigr) \\& \hphantom{\log\frac{NS(a,b)}{P_{2}(a,b)}={}}{}+\frac{1}{2}\log\bigl(\arcsin \bigl(\tanh(2t)\bigr)\bigr)-\frac {1}{2}\log2. \end{aligned}$$

(3.14)

Let

$$\begin{aligned} h_{2}(t) =&\frac{1}{2}\log\bigl(\tanh(t)\bigr)-\log \bigl(\sinh^{-1}\bigl(\tanh (t)\bigr)\bigr) \\ &{}+\frac{1}{2}\log\bigl( \arcsin\bigl(\tanh(2t)\bigr)\bigr)-\frac{1}{2}\log2. \end{aligned}$$

(3.15)

Then simple computations lead to

$$\begin{aligned}& h_{2}\bigl(0^{+}\bigr)=0, \qquad \lim _{t\rightarrow\infty}h_{2}(t)=\log \biggl(\frac{\sqrt{\pi}}{2\log(1+\sqrt{2})} \biggr), \end{aligned}$$

(3.16)

$$\begin{aligned}& h^{\prime}_{2}(t)=\frac{1}{\sinh(2t)}-\frac{1}{\cosh(t)\sqrt {\cosh(2t)}\sinh^{-1}(\tanh(t))}+ \frac{1}{\cosh(2t)\arcsin(\tanh(2t))} \\& \hphantom{h^{\prime}_{2}(t)}=\frac{\sinh(2t)\cosh(t)+\cosh(t)\cosh(2t)\arcsin(\tanh (2t))}{\sinh(2t)\cosh(2t)\cosh(t)\sinh^{-1}(\tanh(t))\arcsin (\tanh(2t))}h_{3}(t), \end{aligned}$$

(3.17)

where

$$\begin{aligned}& h_{3}(t)=\sinh^{-1}\bigl(\tanh(t)\bigr)- \frac{2\sqrt{\cosh(2t)}\sinh (t)\arcsin(\tanh(2t))}{\sinh(2t)+\cosh(2t)\arcsin(\tanh(2t))}, \end{aligned}$$

(3.18)

$$\begin{aligned}& h_{3}\bigl(0^{+}\bigr)=0, \end{aligned}$$

(3.19)

$$\begin{aligned}& h^{\prime}_{3}(t)=\frac{\sqrt{\cosh(2t)}[\arcsin(\tanh(2t))-\tanh (2t)][\sinh(2t)-\arcsin(\tanh(2t))]}{ \cosh(t)[\sinh(2t)+\cosh(2t)\arcsin(\tanh(2t))]^{2}}. \end{aligned}$$

(3.20)

Let $x=\arcsin(\tanh(2t))\in(0, \pi/2)$. Then

$$\begin{aligned}& \arcsin\bigl(\tanh(2t)\bigr)-\tanh(2t)=x-\sin x>0, \end{aligned}$$

(3.21)

$$\begin{aligned}& \sinh(2t)-\arcsin\bigl(\tanh(2t)\bigr)=\tan x-x>0. \end{aligned}$$

(3.22)

From (3.17)-(3.22) we clearly see that $h_{2}(t)$ is strictly increasing on $(0, \infty)$. Therefore, Theorem 3.3 follows from (3.14)-(3.16) and the monotonicity of $h_{2}(t)$. □

Remark 3.2

From the proof of Theorem 3.2 we know that

$$ h_{3}(t)=\sinh^{-1}\bigl(\tanh(t)\bigr)- \frac{2\sqrt{\cosh(2t)}\sinh (t)\arcsin(\tanh(2t))}{\sinh(2t)+\cosh(2t)\arcsin(\tanh(2t))}>0, $$

which is equivalent to

$$ \frac{\frac{\sinh(2t)}{\arcsin(\tanh(2t))}+\cosh(2t)}{2\sqrt {\cosh(2t)}}>\frac{\sinh(t)}{\sinh^{-1}(\tanh(t))}. $$

(3.23)

Equations (3.6), (3.11), and (3.13) together with inequality (3.23) lead to the conclusion that the inequality

$$ NS(a,b)< \frac{P^{2}_{2}(a,b)+Q^{2}(a,b)}{2Q(a,b)} $$

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

Theorem 3.4

The double inequality

$$ \lambda_{4}NS(a,b)< B(a,b)< \mu_{4}NS(a,b) $$

holds for all $a, b>0$ with $a\neq b$ if and only if $\lambda_{4}\leq 1$ and $\mu_{4}\geq\sqrt{2}e^{\pi/4-1}\log(1+\sqrt {2})=1.0057\ldots$ .

Proof

Since $NS(a,b)$ and $B(a,b)$ are symmetric and homogeneous of degree 1, without loss of generality, we assume that $b>a>0$. Let $t=\log\sqrt{b/a}>0$, then (1.3) and (3.13) lead to

$$ \begin{aligned} &B(a,b)=\sqrt{ab}\cosh^{1/2}(2t)e^{\arctan(\tanh(t))/\tanh(t)-1}, \\ &\log\frac{B(a,b)}{NS(a,b)}=\frac{1}{2}\log\bigl(\cosh(2t)\bigr)+ \frac {\arctan(\tanh(t))}{\tanh(t)}-\log \biggl(\frac{\sinh(t)}{\sinh ^{-1}(\tanh(t))} \biggr)-1. \end{aligned} $$

(3.24)

Let

$$ f(t)=\frac{1}{2}\log\bigl(\cosh(2t)\bigr)+ \frac{\arctan(\tanh(t))}{\tanh (t)}-\log \biggl(\frac{\sinh(t)}{\sinh^{-1}(\tanh(t))} \biggr)-1. $$

(3.25)

Then simple computations lead to

$$\begin{aligned}& f\bigl(0^{+}\bigr)=0, \qquad \lim_{t\rightarrow\infty}f(t)= \frac{\pi}{4}-1+\frac {1}{2}\log2+\log\bigl[\log(1+\sqrt{2}) \bigr], \end{aligned}$$

(3.26)

$$\begin{aligned}& f^{\prime}(t)=\frac{f_{1}(t)}{\sinh^{2}(t)\sinh^{-1}(\tanh(t))}, \end{aligned}$$

(3.27)

where

$$ f_{1}(t)=\frac{\sinh^{2}(t)}{\sqrt{\cosh(2t)}\cosh(t)}-\sinh ^{-1}\bigl(\tanh(t)\bigr) \arctan\bigl(\tanh(t)\bigr). $$

Let $x=\tanh(t)\in(0, 1)$. Then

$$\begin{aligned}& f_{1}(t)=\frac{x^{2}}{\sqrt{1+x^{2}}}-\sinh^{-1}(x) \arctan(x):=f_{2}(x), \end{aligned}$$

(3.28)

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

(3.29)

$$\begin{aligned}& f^{\prime}_{2}(x)=\frac{1}{\sqrt{x^{2}+1}} \biggl[x\frac {x^{2}+2}{x^{2}+1}- \frac{\sinh^{-1}(x)}{\sqrt{x^{2}+1}}-\arctan (x) \biggr] :=\frac{f_{3}(x)}{\sqrt{x^{2}+1}}, \end{aligned}$$

(3.30)

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

(3.31)

$$\begin{aligned}& f^{\prime}_{3}(x)=\frac{x}{(x^{2}+1)^{3/2}} \biggl[\sinh ^{-1}(x)-\frac{x-x^{3}}{\sqrt{x^{2}+1}} \biggr]:=\frac {x}{(x^{2}+1)^{3/2}}f_{4}(x), \end{aligned}$$

(3.32)

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

(3.33)

$$\begin{aligned}& f^{\prime}_{4}(x)=\frac{2x^{2}(x^{2}+2)}{(x^{2}+1)^{3/2}}>0 \end{aligned}$$

(3.34)

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

It follows from (3.27)-(3.34) that $f(t)$ is strictly increasing on $(0, \infty)$. Therefore, Theorem 3.4 follows easily from (3.24)-(3.26) and the monotonicity of $f(t)$. □

Remark 3.3

From the proof of Theorem 3.4 we know that the inequalities

$$\begin{aligned}& \frac{x^{2}}{\sqrt{1+x^{2}}}>\sinh^{-1}(x)\arctan(x), \end{aligned}$$

(3.35)

$$\begin{aligned}& x\frac{x^{2}+2}{x^{2}+1}>\frac{\sinh^{-1}(x)}{\sqrt {x^{2}+1}}+\arctan(x), \end{aligned}$$

(3.36)

$$\begin{aligned}& \sinh^{-1}(x)>\frac{x-x^{3}}{\sqrt{x^{2}+1}} \end{aligned}$$

(3.37)

hold for all $x\in(0, \infty)$. Inequalities (3.35)-(3.37) lead to the conclusion that the inequalities

$$\begin{aligned}& NS(a,b)T(a,b)>A(a,b)Q(a,b), \\& \frac{A^{2}(a,b)}{G^{2}(a,b)}>\frac {NS(a,b)}{Q(a,b)}, \\& \frac{A(a,b)}{Q(a,b)}+\frac{Q(a,b)}{A(a,b)}>\frac {A(a,b)}{NS(a,b)}+\frac{Q(a,b)}{T(a,b)} \end{aligned}$$

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

Remark 3.4

Let $I(a,b)=(b^{b}/a^{a})^{1/(b-a)}/e$ be the identric mean of two distinct positive real numbers a and b, and $I_{2}(a,b)=I^{1/2} (a^{2}, b^{2} )$ be the second-order identric mean. Then from Theorems 3.1-3.4 and the inequalities $M(a,b;2/3)< I(a,b)< M(a, b; \log2)$ [20, 21] and $P_{2}(a,b)>L_{4}(a,b)$ [22] we get two inequalities chains as follows:

$$\begin{aligned} \frac{999}{1\text{,}000}L_{4}(a,b) < &U(a,b)< P_{2}(a,b)< NS(a,b) \\ < &B(a,b)< M(a, b; 4/3)< I_{2}(a,b) \\ < &M(a, b; 2\log2) \end{aligned}$$

and

$$\begin{aligned} L_{4}(a,b) < &P_{2}(a,b)< NS(a,b)< B(a,b) \\ < &M(a, b; 4/3)< I_{2}(a,b) \\ < &M(a, b; 2\log2) \end{aligned}$$

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