# Note on certain inequalities for Neuman means

## Abstract

In this paper, we give the explicit formulas for the Neuman means ${N}_{AH}$, ${N}_{HA}$, ${N}_{AC}$, and ${N}_{CA}$, and present the best possible upper and lower bounds for these means in terms of the combinations of harmonic mean H, arithmetic mean A, and contraharmonic mean C.

MSC:26E60.

## 1 Introduction

Let $a,b,c\ge 0$ with $ab+ac+bc\ne 0$. Then the symmetric integral ${R}_{F}\left(a,b,c\right)$  of the first kind is defined as

${R}_{F}\left(a,b,c\right)=\frac{1}{2}{\int }_{0}^{\mathrm{\infty }}{\left[\left(t+a\right)\left(t+b\right)\left(t+c\right)\right]}^{-1/2}\phantom{\rule{0.2em}{0ex}}dt.$

The degenerate case of ${R}_{F}$, denoted by ${R}_{C}$, plays an important role in the theory of special functions [1, 2], which is given by

${R}_{C}\left(a,b\right)={R}_{F}\left(a,b,b\right).$

For $a,b>0$ with $a\ne b$, the Schwab-Borchardt mean $SB\left(a,b\right)$  of a and b is given by

$SB\left(a,b\right)=\left\{\begin{array}{cc}\frac{\sqrt{{b}^{2}-{a}^{2}}}{{cos}^{-1}\left(a/b\right)},\hfill & ab,\hfill \end{array}$

where ${cos}^{-1}\left(x\right)$ and ${cosh}^{-1}\left(x\right)=log\left(x+\sqrt{{x}^{2}-1}\right)$ are the inverse cosine and inverse hyperbolic cosine functions, respectively.

$SB\left(a,b\right)={\left[{R}_{C}\left({a}^{2},{b}^{2}\right)\right]}^{-1}.$

Recently, the Schwab-Borchardt mean has been the subject of intensive research. In particular, many remarkable inequalities for the Schwab-Borchardt mean and its generated means can be found in the literature [35, 811].

Let $a>b>0$, $v=\left(a-b\right)/\left(a+b\right)\in \left(0,1\right)$, $p\in \left(0,\mathrm{\infty }\right)$, $q\in \left(0,\pi /2\right)$, $r\in \left(0,log\left(2+\sqrt{3}\right)\right)$, and $s\in \left(0,\pi /3\right)$ be the parameters such that $1/cosh\left(p\right)=cos\left(q\right)=1-{v}^{2}$, $cosh\left(r\right)=sec\left(s\right)=1+{v}^{2}$, $H\left(a,b\right)=2ab/\left(a+b\right)$, $G\left(a,b\right)=\sqrt{ab}$, $A\left(a,b\right)=\left(a+b\right)/2$, $Q\left(a,b\right)=\sqrt{\left({a}^{2}+{b}^{2}\right)/2}$, and $C\left(a,b\right)=\left({a}^{2}+{b}^{2}\right)/\left(a+b\right)$ be, respectively, the harmonic, geometric, arithmetic, quadratic, and contraharmonic means of a and b, ${S}_{AH}\left(a,b\right)=SB\left[A\left(a,b\right),H\left(a,b\right)\right]$, ${S}_{HA}\left(a,b\right)=SB\left[H\left(a,b\right),A\left(a,b\right)\right]$, ${S}_{AC}\left(a,b\right)=SB\left[A\left(a,b\right),C\left(a,b\right)\right]$, ${S}_{CA}\left(a,b\right)=SB\left[C\left(a,b\right),A\left(a,b\right)\right]$. Then Neuman  gave the explicit formulas

${S}_{AH}\left(a,b\right)=A\left(a,b\right)\frac{tanh\left(p\right)}{p},\phantom{\rule{2em}{0ex}}{S}_{HA}\left(a,b\right)=A\left(a,b\right)\frac{sinq}{q},$
(1.1)
${S}_{CA}\left(a,b\right)=A\left(a,b\right)\frac{sinh\left(r\right)}{r},\phantom{\rule{2em}{0ex}}{S}_{AC}\left(a,b\right)=A\left(a,b\right)\frac{tans}{s}.$
(1.2)

Very recently, Neuman  found a new mean $N\left(a,b\right)$ derived from the Schwab-Borchardt mean as follows:

$N\left(a,b\right)=\frac{1}{2}\left[a+\frac{{b}^{2}}{SB\left(a,b\right)}\right].$
(1.3)

Let ${N}_{AH}\left(a,b\right)=N\left[A\left(a,b\right),H\left(a,b\right)\right]$, ${N}_{HA}\left(a,b\right)=N\left[H\left(a,b\right),A\left(a,b\right)\right]$, ${N}_{AG}\left(a,b\right)=N\left[A\left(a,b\right),G\left(a,b\right)\right]$, ${N}_{GA}\left(a,b\right)=N\left[G\left(a,b\right),A\left(a,b\right)\right]$, ${N}_{AC}\left(a,b\right)=N\left[A\left(a,b\right),C\left(a,b\right)\right]$, ${N}_{CA}\left(a,b\right)=N\left[C\left(a,b\right),A\left(a,b\right)\right]$, ${N}_{AQ}\left(a,b\right)=N\left[A\left(a,b\right),Q\left(a,b\right)\right]$, and ${N}_{QA}\left(a,b\right)=N\left[Q\left(a,b\right),A\left(a,b\right)\right]$ be the Neuman means. Then Neuman  proved that

$G\left(a,b\right)<{N}_{AG}\left(a,b\right)<{N}_{GA}\left(a,b\right)

for all $a,b>0$ with $a\ne b$, and the double inequalities

$\begin{array}{r}{\alpha }_{1}A\left(a,b\right)+\left(1-{\alpha }_{1}\right)G\left(a,b\right)<{N}_{GA}\left(a,b\right)<{\beta }_{1}A\left(a,b\right)+\left(1-{\beta }_{1}\right)G\left(a,b\right),\\ {\alpha }_{2}Q\left(a,b\right)+\left(1-{\alpha }_{2}\right)A\left(a,b\right)<{N}_{AQ}\left(a,b\right)<{\beta }_{2}Q\left(a,b\right)+\left(1-{\beta }_{2}\right)A\left(a,b\right),\\ {\alpha }_{3}A\left(a,b\right)+\left(1-{\alpha }_{3}\right)G\left(a,b\right)<{N}_{AG}\left(a,b\right)<{\beta }_{3}A\left(a,b\right)+\left(1-{\beta }_{3}\right)G\left(a,b\right),\\ {\alpha }_{4}Q\left(a,b\right)+\left(1-{\alpha }_{4}\right)A\left(a,b\right)<{N}_{QA}\left(a,b\right)<{\beta }_{4}Q\left(a,b\right)+\left(1-{\beta }_{4}\right)A\left(a,b\right)\end{array}$

hold for all $a,b>0$ with $a\ne b$ if and only if ${\alpha }_{1}\le 2/3$, ${\beta }_{1}\ge \pi /4$, ${\alpha }_{2}\le 2/3$, ${\beta }_{2}\ge \left(\pi -2\right)/\left[4\left(\sqrt{2}-1\right)\right]=0.689\dots$ , ${\alpha }_{3}\le 1/3$, ${\beta }_{3}\ge 1/2$, ${\alpha }_{4}\le 1/3$, and ${\beta }_{4}\ge \left[log\left(1+\sqrt{2}\right)+\sqrt{2}-2\right]/\left[2\left(\sqrt{2}-1\right)\right]=0.356\dots$ .

Zhang et al.  presented the best possible parameters ${\alpha }_{1},{\alpha }_{2},{\beta }_{1},{\beta }_{2}\in \left[0,1/2\right]$ and ${\alpha }_{3},{\alpha }_{4},{\beta }_{3},{\beta }_{4}\in \left[1/2,1\right]$ such that the double inequalities

$\begin{array}{c}G\left({\alpha }_{1}a+\left(1-{\alpha }_{1}\right)b,{\alpha }_{1}b+\left(1-{\alpha }_{1}\right)a\right)<{N}_{AG}\left(a,b\right)

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

In , the authors found the greatest values ${\alpha }_{1}$, ${\alpha }_{2}$, ${\alpha }_{3}$, ${\alpha }_{4}$, ${\alpha }_{5}$, ${\alpha }_{6}$, ${\alpha }_{7}$, ${\alpha }_{8}$, and the least values ${\beta }_{1}$, ${\beta }_{2}$, ${\beta }_{3}$, ${\beta }_{4}$, ${\beta }_{5}$, ${\beta }_{6}$, ${\beta }_{7}$, ${\beta }_{8}$ such that the double inequalities

$\begin{array}{c}{A}^{{\alpha }_{1}}\left(a,b\right){G}^{1-{\alpha }_{1}}\left(a,b\right)<{N}_{GA}\left(a,b\right)<{A}^{{\beta }_{1}}\left(a,b\right){G}^{1-{\beta }_{1}}\left(a,b\right),\hfill \\ \frac{{\alpha }_{2}}{G\left(a,b\right)}+\frac{1-{\alpha }_{2}}{A\left(a,b\right)}<\frac{1}{{N}_{GA}\left(a,b\right)}<\frac{{\beta }_{2}}{G\left(a,b\right)}+\frac{1-{\beta }_{2}}{A\left(a,b\right)},\hfill \\ {A}^{{\alpha }_{3}}\left(a,b\right){G}^{1-{\alpha }_{3}}\left(a,b\right)<{N}_{AG}\left(a,b\right)<{A}^{{\beta }_{3}}\left(a,b\right){G}^{1-{\beta }_{3}}\left(a,b\right),\hfill \\ \frac{{\alpha }_{4}}{G\left(a,b\right)}+\frac{1-{\alpha }_{4}}{A\left(a,b\right)}<\frac{1}{{N}_{AG}\left(a,b\right)}<\frac{{\beta }_{4}}{G\left(a,b\right)}+\frac{1-{\beta }_{4}}{A\left(a,b\right)},\hfill \\ {Q}^{{\alpha }_{5}}\left(a,b\right){A}^{1-{\alpha }_{5}}\left(a,b\right)<{N}_{AQ}\left(a,b\right)<{Q}^{{\beta }_{5}}\left(a,b\right){A}^{1-{\beta }_{5}}\left(a,b\right),\hfill \\ \frac{{\alpha }_{6}}{A\left(a,b\right)}+\frac{1-{\alpha }_{6}}{Q\left(a,b\right)}<\frac{1}{{N}_{AQ}\left(a,b\right)}<\frac{{\beta }_{6}}{A\left(a,b\right)}+\frac{1-{\beta }_{6}}{Q\left(a,b\right)},\hfill \\ {Q}^{{\alpha }_{7}}\left(a,b\right){A}^{1-{\alpha }_{7}}\left(a,b\right)<{N}_{QA}\left(a,b\right)<{Q}^{{\beta }_{7}}\left(a,b\right){A}^{1-{\beta }_{7}}\left(a,b\right),\hfill \\ \frac{{\alpha }_{8}}{A\left(a,b\right)}+\frac{1-{\alpha }_{8}}{Q\left(a,b\right)}<\frac{1}{{N}_{QA}\left(a,b\right)}<\frac{{\beta }_{8}}{A\left(a,b\right)}+\frac{1-{\beta }_{8}}{Q\left(a,b\right)}\hfill \end{array}$

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

The main purpose of this paper is to give the explicit formulas for the Neuman means ${N}_{AH}$, ${N}_{HA}$, ${N}_{AC}$, and ${N}_{CA}$, and to present the best possible upper and lower bounds for these means in terms of the combinations of harmonic, arithmetic, and contraharmonic means. Our main results are Theorems 1.1-1.3.

Theorem 1.1 Let $a>b>0$, $v=\left(a-b\right)/\left(a+b\right)\in \left(0,1\right)$, $p\in \left(0,\mathrm{\infty }\right)$, $q\in \left(0,\pi /2\right)$, $r\in \left(0,log\left(2+\sqrt{3}\right)\right)$, and $s\in \left(0,\pi /3\right)$ be the parameters such that $1/cosh\left(p\right)=cos\left(q\right)=1-{v}^{2}$, $cosh\left(r\right)=sec\left(s\right)=1+{v}^{2}$. Then we have

${N}_{AH}\left(a,b\right)=\frac{1}{2}A\left(a,b\right)\left[1+\frac{2p}{sinh\left(2p\right)}\right],$
(1.4)
${N}_{HA}\left(a,b\right)=\frac{1}{2}A\left(a,b\right)\left[cos\left(q\right)+\frac{q}{sin\left(q\right)}\right],$
(1.5)
${N}_{CA}\left(a,b\right)=\frac{1}{2}A\left(a,b\right)\left[cosh\left(r\right)+\frac{r}{sinh\left(r\right)}\right],$
(1.6)
${N}_{AC}\left(a,b\right)=\frac{1}{2}A\left(a,b\right)\left[1+\frac{2s}{sin\left(2s\right)}\right],$
(1.7)

and

$\begin{array}{rl}H\left(a,b\right)& <{N}_{AH}\left(a,b\right)<{N}_{HA}\left(a,b\right)
(1.8)

Theorem 1.2 The double inequalities

${\alpha }_{1}A\left(a,b\right)+\left(1-{\alpha }_{1}\right)H\left(a,b\right)<{N}_{AH}\left(a,b\right)<{\beta }_{1}A\left(a,b\right)+\left(1-{\beta }_{1}\right)H\left(a,b\right),$
(1.9)
${\alpha }_{2}A\left(a,b\right)+\left(1-{\alpha }_{2}\right)H\left(a,b\right)<{N}_{HA}\left(a,b\right)<{\beta }_{2}A\left(a,b\right)+\left(1-{\beta }_{2}\right)H\left(a,b\right),$
(1.10)
${\alpha }_{3}C\left(a,b\right)+\left(1-{\alpha }_{3}\right)A\left(a,b\right)<{N}_{CA}\left(a,b\right)<{\beta }_{3}C\left(a,b\right)+\left(1-{\beta }_{3}\right)A\left(a,b\right),$
(1.11)
${\alpha }_{4}C\left(a,b\right)+\left(1-{\alpha }_{4}\right)A\left(a,b\right)<{N}_{AC}\left(a,b\right)<{\beta }_{4}C\left(a,b\right)+\left(1-{\beta }_{4}\right)A\left(a,b\right)$
(1.12)

hold for all $a,b>0$ with $a\ne b$ if and only if ${\alpha }_{1}\le 1/3$, ${\beta }_{1}\ge 1/2$, ${\alpha }_{2}\le 2/3$, ${\beta }_{2}\ge \pi /4=0.7853\dots$ , ${\alpha }_{3}\le 1/3$, ${\beta }_{3}\ge \sqrt{3}log\left(2+\sqrt{3}\right)/6=0.3801\dots$ , ${\alpha }_{4}\le 2/3$, and ${\beta }_{4}\ge \left(4\sqrt{3}\pi -9\right)/18=0.7901\dots$ .

Theorem 1.3 The double inequalities

$\frac{{\alpha }_{5}}{H\left(a,b\right)}+\frac{1-{\alpha }_{5}}{A\left(a,b\right)}<\frac{1}{{N}_{AH}\left(a,b\right)}<\frac{{\beta }_{5}}{H\left(a,b\right)}+\frac{1-{\beta }_{5}}{A\left(a,b\right)},$
(1.13)
$\frac{{\alpha }_{6}}{H\left(a,b\right)}+\frac{1-{\alpha }_{6}}{A\left(a,b\right)}<\frac{1}{{N}_{HA}\left(a,b\right)}<\frac{{\beta }_{6}}{H\left(a,b\right)}+\frac{1-{\beta }_{6}}{A\left(a,b\right)},$
(1.14)
$\frac{{\alpha }_{7}}{A\left(a,b\right)}+\frac{1-{\alpha }_{7}}{C\left(a,b\right)}<\frac{1}{{N}_{CA}\left(a,b\right)}<\frac{{\beta }_{7}}{A\left(a,b\right)}+\frac{1-{\beta }_{7}}{C\left(a,b\right)},$
(1.15)
$\frac{{\alpha }_{8}}{A\left(a,b\right)}+\frac{1-{\alpha }_{8}}{C\left(a,b\right)}<\frac{1}{{N}_{AC}\left(a,b\right)}<\frac{{\beta }_{8}}{A\left(a,b\right)}+\frac{1-{\beta }_{8}}{C\left(a,b\right)},$
(1.16)

hold for all $a,b>0$ with $a\ne b$ if and only if ${\alpha }_{5}\le 0$, ${\beta }_{5}\ge 2/3$, ${\alpha }_{6}\le 0$, ${\beta }_{6}\ge 1/3$, ${\alpha }_{7}\le \left[2\sqrt{3}-log\left(2+\sqrt{3}\right)\right]/\left[2\sqrt{3}+log\left(2+\sqrt{3}\right)\right]=0.4490\dots$ , ${\beta }_{7}\ge 2/3$, ${\alpha }_{8}\le \left(9\sqrt{3}-4\pi \right)/\left(3\sqrt{3}+4\pi \right)=0.1701\dots$ , and ${\beta }_{8}\ge 1/3$.

## 2 Lemmas

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

Lemma 2.1 (See [, Theorem 1.25])

For $-\mathrm{\infty }, let $f,g:\left[a,b\right]\to \mathbb{R}$ be continuous on $\left[a,b\right]$, and be differentiable on $\left(a,b\right)$, let ${g}^{\prime }\left(x\right)\ne 0$ on $\left(a,b\right)$. If ${f}^{\prime }\left(x\right)/{g}^{\prime }\left(x\right)$ is increasing (decreasing) on $\left(a,b\right)$, then so are

$\frac{f\left(x\right)-f\left(a\right)}{g\left(x\right)-g\left(a\right)}\phantom{\rule{1em}{0ex}}\mathit{\text{and}}\phantom{\rule{1em}{0ex}}\frac{f\left(x\right)-f\left(b\right)}{g\left(x\right)-g\left(b\right)}.$

If ${f}^{\prime }\left(x\right)/{g}^{\prime }\left(x\right)$ is strictly monotone, then the monotonicity in the conclusion is also strict.

Lemma 2.2 (See [, Lemma 1.1])

Suppose that the power series $f\left(x\right)={\sum }_{n=0}^{\mathrm{\infty }}{a}_{n}{x}^{n}$ and $g\left(x\right)={\sum }_{n=0}^{\mathrm{\infty }}{b}_{n}{x}^{n}$ have the radius of convergence $r>0$ and ${a}_{n},{b}_{n}>0$ for all $n\ge 0$. If the sequence $\left\{{a}_{n}/{b}_{n}\right\}$ is (strictly) increasing (decreasing) for all $n\ge 0$, then the function $f\left(x\right)/g\left(x\right)$ is also (strictly) increasing (decreasing) on $\left(0,r\right)$.

Lemma 2.3 (See [, Theorem 4.1])

If $a>b$, then

$N\left(b,a\right)>N\left(a,b\right).$

Lemma 2.4 The function

${\phi }_{1}\left(t\right)=\frac{sinh\left(2t\right)-4sinh\left(t\right)+2t}{sinh\left(2t\right)-2sinh\left(t\right)}$

is strictly increasing from $\left(0,\mathrm{\infty }\right)$ onto $\left(2/3,1\right)$.

Proof Making use of power series expansion we get

${\phi }_{1}\left(t\right)=\frac{{\sum }_{n=1}^{\mathrm{\infty }}\frac{{2}^{2n+1}-4}{\left(2n+1\right)!}{t}^{2n+1}}{{\sum }_{n=1}^{\mathrm{\infty }}\frac{{2}^{2n+1}-2}{\left(2n+1\right)!}{t}^{2n+1}}=\frac{{\sum }_{n=0}^{\mathrm{\infty }}\frac{{2}^{2n+3}-4}{\left(2n+3\right)!}{t}^{2n}}{{\sum }_{n=0}^{\mathrm{\infty }}\frac{{2}^{2n+3}-2}{\left(2n+3\right)!}{t}^{2n}}.$
(2.1)

Let

${a}_{n}=\frac{{2}^{2n+3}-4}{\left(2n+3\right)!},\phantom{\rule{2em}{0ex}}{b}_{n}=\frac{{2}^{2n+3}-2}{\left(2n+3\right)!}.$
(2.2)

Then

${a}_{n}>0,\phantom{\rule{2em}{0ex}}{b}_{n}>0,$
(2.3)

and ${a}_{n}/{b}_{n}=1-1/\left({2}^{2n+2}-1\right)$ is strictly increasing for all $n\ge 0$.

Note that

${\phi }_{1}\left({0}^{+}\right)=\frac{{a}_{0}}{{b}_{0}}=\frac{2}{3},\phantom{\rule{2em}{0ex}}{\phi }_{1}\left(\mathrm{\infty }\right)=\underset{n\to \mathrm{\infty }}{lim}\frac{{a}_{n}}{{b}_{n}}=1.$
(2.4)

Therefore, Lemma 2.4 follows easily from Lemma 2.2 and (2.1)-(2.4) together with the monotonicity of the sequence $\left\{{a}_{n}/{b}_{n}\right\}$. □

Lemma 2.5 The function

${\phi }_{2}\left(t\right)=\frac{2t-sin\left(2t\right)}{sint\left(1-cost\right)}$

is strictly increasing from $\left(0,\pi /2\right)$ onto $\left(8/3,\pi \right)$.

Proof Let ${f}_{1}\left(t\right)=2t-sin\left(2t\right)$ and ${g}_{1}\left(t\right)=sint\left(1-cost\right)$. Then simple computations lead to

${\phi }_{2}\left(t\right)=\frac{{f}_{1}\left(t\right)-{f}_{1}\left(0\right)}{{g}_{1}\left(t\right)-{g}_{1}\left(0\right)}$
(2.5)

and ${f}_{1}^{\prime }\left(t\right)/{g}_{1}^{\prime }\left(t\right)=4\left[1-1/\left(2+1/cost\right)\right]$ is strictly increasing on $\left(0,\pi /2\right)$.

Note that

${\phi }_{2}\left({0}^{+}\right)=\underset{t\to {0}^{+}}{lim}\frac{{f}_{1}^{\prime }\left(t\right)}{{g}_{1}^{\prime }\left(t\right)}=\frac{8}{3},\phantom{\rule{2em}{0ex}}{\phi }_{2}\left(\pi /2\right)=\pi .$
(2.6)

Therefore, Lemma 2.5 follows from Lemma 2.1, (2.5), (2.6), and the monotonicity of ${f}_{1}^{\prime }\left(t\right)/{g}_{1}^{\prime }\left(t\right)$. □

Lemma 2.6 The function

${\phi }_{3}\left(t\right)=\frac{sinh\left(t\right)cosh\left(t\right)-t}{\left[sinh\left(t\right)cosh\left(t\right)+t\right]\left(cosh\left(t\right)-1\right)}$

is strictly decreasing from $\left(0,\mathrm{\infty }\right)$ onto $\left(0,2/3\right)$.

$\begin{array}{rl}{\phi }_{3}\left(t\right)& =\frac{2sinh\left(2t\right)-4t}{sinh\left(3t\right)+4tcosh\left(t\right)+sinh\left(t\right)-2sinh\left(2t\right)-4t}\\ =\frac{{\sum }_{n=0}^{\mathrm{\infty }}\frac{{2}^{2n+4}}{\left(2n+3\right)!}{t}^{2n}}{{\sum }_{n=0}^{\mathrm{\infty }}\frac{{3}^{2n+3}-{2}^{2n+4}+8n+13}{\left(2n+3\right)!}{t}^{2n}}.\end{array}$
(2.7)

Let

${a}_{n}=\frac{{2}^{2n+4}}{\left(2n+3\right)!},\phantom{\rule{2em}{0ex}}{b}_{n}=\frac{{3}^{2n+3}-{2}^{2n+4}+8n+13}{\left(2n+3\right)!}.$
(2.8)

Then

${a}_{n}>0,\phantom{\rule{2em}{0ex}}{b}_{n}>0,$
(2.9)

and

$\frac{{a}_{n+1}}{{b}_{n+1}}-\frac{{a}_{n}}{{b}_{n}}=-\frac{{2}^{2n+4}\left(5×{3}^{2n+3}-24n-31\right)}{\left({3}^{2n+5}-{2}^{2n+6}+8n+21\right)\left({3}^{2n+3}-{2}^{2n+4}+8n+13\right)}<0$
(2.10)

for all $n\ge 0$.

Note that

${\phi }_{3}\left({0}^{+}\right)=\frac{{a}_{0}}{{b}_{0}}=\frac{2}{3},\phantom{\rule{2em}{0ex}}{\phi }_{3}\left(\mathrm{\infty }\right)=\underset{n\to \mathrm{\infty }}{lim}\frac{{a}_{n}}{{b}_{n}}=0.$
(2.11)

Therefore, Lemma 2.6 follows easily from (2.7)-(2.11) and Lemma 2.2. □

Lemma 2.7 The function

$f\left(t\right)=9cost+\frac{t}{sint}$

is strictly decreasing on the interval $\left(0,\pi /2\right)$.

Proof Let ${f}_{2}\left(t\right)=9sintcost+t$ and ${g}_{2}\left(t\right)=sint$. Then simple computations lead to

$\begin{array}{r}f\left(t\right)=\frac{{f}_{2}\left(t\right)-{f}_{2}\left(0\right)}{{g}_{2}\left(t\right)-{g}_{2}\left(0\right)},\\ \frac{{f}_{2}^{\prime }\left(t\right)}{{g}_{2}^{\prime }\left(t\right)}=\frac{18{cos}^{2}t-8}{cost},\end{array}$
(2.12)

and

${\left[\frac{{f}_{2}^{\prime }\left(t\right)}{{g}_{2}^{\prime }\left(t\right)}\right]}^{\prime }=-\frac{2sint\left(9{cos}^{2}t+4\right)}{{cos}^{2}t}<0$
(2.13)

for $t\in \left(0,\pi /2\right)$.

Therefore, Lemma 2.7 follows easily from (2.12) and (2.13) together with Lemma 2.1. □

Lemma 2.8 The function

${\phi }_{4}\left(t\right)=\frac{sintcost-t}{\left(t+sintcost\right)\left(1-cost\right)}$

is strictly decreasing from $\left(0,\pi /2\right)$ onto $\left(-1,-2/3\right)$.

Proof Let ${f}_{3}\left(t\right)=sintcost-t$ and ${g}_{3}\left(t\right)=\left(t+sintcost\right)\left(1-cost\right)$. Then simple computations lead to

${\phi }_{4}\left(t\right)=\frac{{f}_{3}\left(t\right)}{{g}_{3}\left(t\right)}=\frac{{f}_{3}\left(t\right)-{f}_{3}\left(0\right)}{{g}_{3}\left(t\right)-{g}_{3}\left(0\right)},$
(2.14)
$\frac{{f}_{3}^{\prime }\left(t\right)}{{g}_{3}^{\prime }\left(t\right)}=\frac{{f}_{3}^{\prime }\left(t\right)-{f}_{3}^{\prime }\left(0\right)}{{g}_{3}^{\prime }\left(t\right)-{g}_{3}^{\prime }\left(0\right)},$
(2.15)

and

$\frac{{f}_{3}^{″}\left(t\right)}{{g}_{3}^{″}\left(t\right)}=\frac{4}{4-\left(9cost+\frac{t}{sint}\right)}.$
(2.16)

Note that

${\phi }_{4}\left({0}^{+}\right)=\underset{t\to {0}^{+}}{lim}\frac{{f}_{3}^{″}\left(t\right)}{{g}_{3}^{″}\left(t\right)}=-\frac{2}{3},\phantom{\rule{2em}{0ex}}{\phi }_{4}\left(\frac{\pi }{2}\right)=-1.$
(2.17)

Therefore, Lemma 2.8 follows from Lemma 2.1 and Lemma 2.7 together with (2.14)-(2.17). □

## 3 Proofs of Theorems 1.1-1.3

Proof of Theorem 1.1 It follows from (1.1)-(1.3) as we clearly see that

$\begin{array}{c}\begin{array}{rl}{N}_{AH}\left(a,b\right)& =\frac{1}{2}\left[A\left(a,b\right)+\frac{{H}^{2}\left(a,b\right)}{{S}_{AH}\left(a,b\right)}\right]=\frac{1}{2}A\left(a,b\right)\left[1+{\left(1-{v}^{2}\right)}^{2}\frac{p}{tanh\left(p\right)}\right]\\ =\frac{1}{2}A\left(a,b\right)\left[1+\frac{p}{tanh\left(p\right){cosh}^{2}\left(p\right)}\right]=\frac{1}{2}A\left(a,b\right)\left[1+\frac{2p}{sinh\left(2p\right)}\right],\end{array}\hfill \\ \begin{array}{rl}{N}_{HA}\left(a,b\right)& =\frac{1}{2}\left[H\left(a,b\right)+\frac{{A}^{2}\left(a,b\right)}{{S}_{HA}\left(a,b\right)}\right]=\frac{1}{2}A\left(a,b\right)\left[\left(1-{v}^{2}\right)+\frac{q}{sinq}\right]\\ =\frac{1}{2}A\left(a,b\right)\left[cosq+\frac{q}{sinq}\right],\end{array}\hfill \\ \begin{array}{rl}{N}_{CA}\left(a,b\right)& =\frac{1}{2}\left[C\left(a,b\right)+\frac{{A}^{2}\left(a,b\right)}{{S}_{CA}\left(a,b\right)}\right]=\frac{1}{2}A\left(a,b\right)\left[\left(1+{v}^{2}\right)+\frac{r}{sinh\left(r\right)}\right]\\ =\frac{1}{2}A\left(a,b\right)\left[cosh\left(r\right)+\frac{r}{sinh\left(r\right)}\right],\end{array}\hfill \\ \begin{array}{rl}{N}_{AC}\left(a,b\right)& =\frac{1}{2}\left[A\left(a,b\right)+\frac{{C}^{2}\left(a,b\right)}{{S}_{AC}\left(a,b\right)}\right]=\frac{1}{2}A\left(a,b\right)\left[1+{\left(1+{v}^{2}\right)}^{2}\frac{s}{tan\left(s\right)}\right]\\ =\frac{1}{2}A\left(a,b\right)\left[1+\frac{s}{tan\left(s\right){cos}^{2}s}\right]=\frac{1}{2}A\left(a,b\right)\left[1+\frac{2s}{sin\left(2s\right)}\right].\end{array}\hfill \end{array}$

Inequalities (1.8) follow easily from $H\left(a,b\right) and Lemma 2.3 together with the fact that ${N}_{KL}\left(a,b\right)$ is a mean of $K\left(a,b\right)$ and $L\left(a,b\right)$ for $K\left(a,b\right),L\left(a,b\right)\in \left\{H\left(a,b\right),A\left(a,b\right),C\left(a,b\right)\right\}$. □

Proof of Theorem 1.2 Without loss of generality, we assume that $a>b$. Let $v=\left(a-b\right)/\left(a+b\right)\in \left(0,1\right)$, $p\in \left(0,\mathrm{\infty }\right)$, $q\in \left(0,\pi /2\right)$, $r\in \left(0,log\left(2+\sqrt{3}\right)\right)$, and $s\in \left(0,\pi /3\right)$ be the parameters such that $1/cosh\left(p\right)=cos\left(q\right)=1-{v}^{2}$, $cosh\left(r\right)=sec\left(s\right)=1+{v}^{2}$. Then from (1.4)-(1.7) we have

$\begin{array}{rl}\frac{{N}_{AH}\left(a,b\right)-H\left(a,b\right)}{A\left(a,b\right)-H\left(a,b\right)}& =\frac{\left[1+2p/sinh\left(2p\right)\right]/2-\left(1-{v}^{2}\right)}{{v}^{2}}\\ =\frac{\left[1+2p/sinh\left(2p\right)\right]/2-1/cosh\left(p\right)}{1-1/cosh\left(p\right)}={\phi }_{1}\left(p\right),\end{array}$
(3.1)
$\begin{array}{rl}\frac{{N}_{HA}\left(a,b\right)-H\left(a,b\right)}{A\left(a,b\right)-H\left(a,b\right)}& =\frac{\left[cosq+q/sinq\right]/2-\left(1-{v}^{2}\right)}{{v}^{2}}\\ =\frac{\left[cosq+q/sinq\right]/2-cosq}{1-cosq}=\frac{1}{4}{\phi }_{2}\left(q\right),\end{array}$
(3.2)
$\begin{array}{rl}\frac{{N}_{CA}\left(a,b\right)-A\left(a,b\right)}{C\left(a,b\right)-A\left(a,b\right)}& =\frac{\left[cosh\left(r\right)+r/sinh\left(r\right)\right]/2-1}{{v}^{2}}\\ =\frac{\left[cosh\left(r\right)+r/sinh\left(r\right)\right]/2-1}{cosh\left(r\right)-1}=\frac{1}{2}{\phi }_{1}\left(r\right),\end{array}$
(3.3)
$\begin{array}{rl}\frac{{N}_{AC}\left(a,b\right)-A\left(a,b\right)}{C\left(a,b\right)-A\left(a,b\right)}& =\frac{\left[1+2s/sin\left(2s\right)\right]/2-1}{{v}^{2}}\\ =\frac{\left[1+2s/sin\left(2s\right)\right]/2-1}{sec\left(s\right)-1}=\frac{1}{4}{\phi }_{2}\left(s\right),\end{array}$
(3.4)

where the functions ${\phi }_{1}$ and ${\phi }_{2}$ are defined as in Lemmas 2.4 and 2.5, respectively.

Note that

${\phi }_{1}\left[log\left(2+\sqrt{3}\right)\right]=\sqrt{3}log\left(2+\sqrt{3}\right)/6$
(3.5)

and

${\phi }_{2}\left(\frac{\pi }{3}\right)=\frac{8\sqrt{3}\pi -18}{9}.$
(3.6)

Therefore, inequality (1.9) holds for all $a,b>0$ with $a\ne b$ if and only if ${\alpha }_{1}\le 1/3$ and ${\beta }_{1}\ge 1/2$ follows from (3.1) and Lemma 2.4, inequality (1.10) holds for all $a,b>0$ with $a\ne b$ if and only if ${\alpha }_{2}\le 2/3$ and ${\beta }_{2}\ge \pi /4$ follows from (3.2) and Lemma 2.5, inequality (1.11) holds for all $a,b>0$ with $a\ne b$ if and only if ${\alpha }_{3}\le 1/3$ and ${\beta }_{3}\ge \sqrt{3}log\left(2+\sqrt{3}\right)/6$ follows from (3.3) and (3.5) together with Lemma 2.4, and inequality (1.12) holds for all $a,b>0$ with $a\ne b$ if and only if ${\alpha }_{4}\le 2/3$ and ${\beta }_{4}\ge \left(4\sqrt{3}\pi -9\right)/18$ follows from (3.4) and (3.6) together with Lemma 2.5. □

Proof of Theorem 1.3 Without loss of generality, we assume that $a>b$. Let $v=\left(a-b\right)/\left(a+b\right)\in \left(0,1\right)$, $p\in \left(0,\mathrm{\infty }\right)$, $q\in \left(0,\pi /2\right)$, $r\in \left(0,log\left(2+\sqrt{3}\right)\right)$, and $s\in \left(0,\pi /3\right)$ be the parameters such that $1/cosh\left(p\right)=cos\left(q\right)=1-{v}^{2}$, $cosh\left(r\right)=sec\left(s\right)=1+{v}^{2}$. Then from (1.4)-(1.7) we have

$\frac{1/{N}_{AH}\left(a,b\right)-1/A\left(a,b\right)}{1/H\left(a,b\right)-1/A\left(a,b\right)}=\frac{\frac{2}{1+2p/sinh\left(2p\right)}-1}{\frac{1}{1-{v}^{2}}-1}=\frac{\frac{2sinh\left(2p\right)}{2p+sinh\left(2p\right)}-1}{cosh\left(p\right)-1}={\phi }_{3}\left(p\right),$
(3.7)
$\frac{1/{N}_{HA}\left(a,b\right)-1/A\left(a,b\right)}{1/H\left(a,b\right)-1/A\left(a,b\right)}=\frac{\frac{2}{cos\left(q\right)+q/sin\left(q\right)}-1}{\frac{1}{1-{v}^{2}}-1}=\frac{\frac{2sin\left(q\right)-sin\left(q\right)cos\left(q\right)-q}{sin\left(q\right)cos\left(q\right)+q}}{\frac{1-cos\left(q\right)}{cos\left(q\right)}}=1+{\phi }_{4}\left(q\right),$
(3.8)
$\frac{1/{N}_{CA}\left(a,b\right)-1/C\left(a,b\right)}{1/A\left(a,b\right)-1/C\left(a,b\right)}=\frac{\frac{2}{cosh\left(r\right)+r/sinh\left(r\right)}-\frac{1}{1+{v}^{2}}}{1-\frac{1}{1+{v}^{2}}}={\phi }_{3}\left(r\right),$
(3.9)

and

$\frac{1/{N}_{AC}\left(a,b\right)-1/C\left(a,b\right)}{1/A\left(a,b\right)-1/C\left(a,b\right)}=1+{\phi }_{4}\left(s\right),$
(3.10)

where the functions ${\phi }_{3}$ and ${\phi }_{4}$ are defined as in Lemmas 2.6 and 2.8, respectively.

Note that

${\phi }_{3}\left[log\left(2+\sqrt{3}\right)\right]=\frac{2\sqrt{3}-log\left(2+\sqrt{3}\right)}{2\sqrt{3}+log\left(2+\sqrt{3}\right)}$
(3.11)

and

${\phi }_{4}\left(\frac{\pi }{3}\right)=-\frac{8\pi -6\sqrt{3}}{4\pi +3\sqrt{3}}.$
(3.12)

Therefore, Theorem 1.3 follows easily from (3.7)-(3.12) together with Lemmas 2.6 and 2.8. □

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## Acknowledgements

This research was supported by the Natural Science Foundation of China under Grants 61374086, 11371125, and 11401192, the Natural Science Foundation of Hunan Province under Grant 12C0577, and the Research Foundation of Education Bureau of Hunan Province under Grant 14A026.

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