Optimal power mean bounds for the second Yang mean

Abstract

In this paper, we present the best possible parameters p and q such that the double inequality

$$ M_{p}(a,b)< V(a,b)< M_{q}(a,b) $$

holds for all $a, b>0$ with $a\neq b$, where $M_{r}(a,b)=[(a^{r}+b^{r})/2]^{1/r}$ ($r\neq0$) and $M_{0}(a,b)= \sqrt {ab}$ is the rth power mean and $V(a,b)=(a-b)/[\sqrt{2}\sinh^{-1}((a-b)/\sqrt{2ab})]$ is the second Yang mean.

1 Introduction

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

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

(1.1)

It is well known that $M_{r}(a,b)$ 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 special cases of the power mean, for example, $M_{-1}(a,b)=2ab/(a+b)=H(a,b)$ is the harmonic mean, $M_{0}(a,b)=\sqrt{ab}=G(a,b)$ is the geometric mean, $M_{1}(a,b)=(a+b)/2=A(a,b)$ is the arithmetic mean, and $M_{2}(a,b)=\sqrt{(a^{2}+b^{2})/2}=Q(a,b)$ is the quadratic mean. The main properties for the power mean are given in [1].

Let

$$\begin{aligned}& L(a,b)=\frac{a-b}{\log a-\log b},\quad\quad I(a,b)=\frac{1}{e} \biggl(\frac{a^{a}}{b^{b}} \biggr)^{1/(a-b)},\quad\quad P(a,b)=\frac{a-b}{2\arcsin(\frac{a-b}{a+b} )}, \\& U(a,b)=\frac{a-b}{\sqrt{2}\arctan(\frac{a-b}{\sqrt{2ab}} )},\quad\quad T^{\ast }(a,b)=\frac{2}{\pi} \int_{0}^{\pi/2}\sqrt{a^{2}\cos^{2} \theta+b^{2}\sin^{2}\theta}d\theta, \\& NS(a,b)=\frac{a-b}{2\sinh^{-1} (\frac{a-b}{a+b} )},\quad\quad X(a,b)=A(a,b)e^{G(a,b)/P(a,b)-1}, \\& T(a,b)=\frac{a-b}{2\arctan(\frac{a-b}{a+b} )},\quad\quad B(a,b)=Q(a,b)e^{A(a,b)/T(a,b)-1}, \end{aligned}$$

and

$$ V(a,b)=\frac{a-b}{\sqrt{2}\sinh^{-1} (\frac{a-b}{\sqrt{2ab}} )} $$

(1.2)

be, respectively, the logarithmic mean, identric mean, first Seiffert mean [2], first Yang mean [3], Toader mean [4], Neuman-Sándor mean [5, 6], Sándor mean [7], second Seiffert mean [8], Sándor-Yang mean [3], and second Yang mean [3] of two distinct positive real numbers a and b, where $\sinh^{-1}(x)=\log(x+\sqrt {x^{2}+1})$ is the inverse hyperbolic sine function.

Recently, the bounds for certain bivariate means in terms of the power mean have attracted the attention of many mathematicians. Radó [9] (see also [10–12]) proved that the double inequalities

$$ \begin{aligned} &M_{p}(a,b)< L(a,b)< M_{q}(a,b), \\ &M_{\lambda}(a,b)< I(a,b)< M_{\mu}(a,b) \end{aligned} $$

(1.3)

hold for all $a, b>0$ with $a\neq b$ if and only if $p\leq0$, $q\geq 1/3$, $\lambda\leq2/3$, and $\mu\geq\log2$.

In [13–16], the authors proved that the double inequality

$$ M_{p}(a,b)< T^{\ast}(a,b)< M_{q}(a,b) $$

holds for all $a, b>0$ with $a\neq b$ if and only if $p\leq3/2$ and $q\geq\log2/(\log\pi-\log2)$.

Jagers [17], Hästö [18, 19], Yang [20], and Costin and Toader [21] proved that $p_{1}=\log2/\log\pi$, $q_{1}=2/3$, $p_{2}=\log 2/(\log\pi-\log2)$, and $q_{2}=5/3$ are the best possible parameters such that the double inequalities

$$ \begin{aligned} &M_{p_{1}}(a,b)< P(a,b)< M_{q_{1}}(a,b), \\ &M_{p_{2}}(a,b)< T(a,b)< M_{q_{2}}(a,b) \end{aligned} $$

(1.4)

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

In [21–26], the authors proved that the double inequalities

$$\begin{aligned}& M_{\lambda_{1}}(a,b)< NS(a,b)< M_{\mu_{1}}(a,b), \\& M_{\lambda_{2}}(a,b)< U(a,b)< M_{\mu_{2}}(a,b), \\& M_{\lambda_{3}}(a,b)< X(a,b)< M_{\mu_{3}}(a,b), \end{aligned}$$

hold for all $a, b>0$ with $a\neq b$ if and only if $\lambda_{1}\leq \log2/\log[2\log(1+\sqrt{2})]$, $\mu_{1}\geq4/3$, $\lambda_{2}\leq 2\log2/(2\log\pi-\log2)$, $\mu_{2}\geq4/3$, $\lambda_{3}\leq1/3$, and $\mu_{3}\geq\log2/(1+\log2)$.

Very recently, Yang and Chu [27] showed that $p=4\log2/(4+2\log2-\pi )$ and $q=4/3$ are the best possible parameters such that the double inequality

$$ M_{p}(a,b)< B(a,b)< M_{q}(a,b) $$

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

The main purpose of this paper is to present the best possible parameters p and q such that the double inequality

$$ M_{p}(a,b)< V(a,b)< M_{q}(a,b) $$

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

2 Lemmas

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

Lemma 2.1

Let $t>0$, $p\in\mathbb{R}$, and

$$\begin{aligned} f(t, p) =& 2\sinh\bigl[2(p-1)t\bigr] +\sinh\bigl[2(p+1)t\bigr]+\sinh\bigl[2(p-2)t \bigr] \\ &{} +p\sinh(4t)-\sinh(2t). \end{aligned}$$

(2.1)

Then the following statements are true:

• (i) $f(t,p)>0$ for all $t>0$ if and only if $p\geq2/3$;

• (ii) $f(t,p)<0$ for all $t>0$ if and only if $p\leq0$.

Proof

It follows from (2.1) that

$$\begin{aligned} \frac{\partial f(t,p)}{\partial t} =&\sinh(4t)+4t\cosh\bigl[2(p-1)t\bigr] +2t\cosh\bigl[2(p+1)t\bigr]+2t\cosh\bigl[2(p-2)t\bigr] \\ >&0 \end{aligned}$$

(2.2)

for all $t>0$ and $p\in\mathbb{R}$.

(i) If $f(t,p)>0$ for all $t>0$, then (2.1) leads to

$$ \lim_{t\rightarrow0^{+}}\frac{f(t, p)}{t}=12 \biggl(p-\frac{2}{3} \biggr)\geq0, $$

which gives $p\geq2/3$.

If $p\geq2/3$, then (2.1) and (2.2) lead to the conclusion that

$$\begin{aligned} f(t,p) \geq& f \biggl(t, \frac{2}{3} \biggr)=\frac{2}{3}\sinh(4t)- \sinh(2t)-2\sinh\biggl(\frac{2}{3}t \biggr) -\sinh\biggl( \frac{8}{3}t \biggr)+\sinh\biggl(\frac{10}{3}t \biggr) \\ =&\frac{8}{3}\sinh^{3} \biggl(\frac{2}{3}t \biggr) \cosh\biggl(\frac{2}{3}t \biggr) \biggl[8\cosh^{2} \biggl( \frac{2}{3}t \biggr) +6\cosh\biggl(\frac{2}{3}t \biggr)-3 \biggr]>0 \end{aligned}$$

for all $t>0$.

(ii) If $f(t, p)<0$ for all $t>0$, then from part (i) we know that $p<2/3$. We assert that $p\leq0$, otherwise $0< p<2/3$ and (2.1) leads to

$$\begin{aligned}& \lim_{t\rightarrow+\infty}\frac{f(t, p)}{e^{4t}} \\& \quad=\lim_{t\rightarrow+\infty}\frac{-2\sinh[2(1-p)t]+\sinh[2(1+p)t]-\sinh [2(2-p)t]+p\sinh(4t)-\sinh(2t)}{e^{4t}} \\& \quad=\frac{p}{2}>0, \end{aligned}$$

which contradicts with $f(t, p)<0$ for all $t>0$. 

If $p\leq0$, then from (2.1) and (2.2) we have

$$ f(t,p)\leq f(t, 0)=-2\sinh(2t)-\sinh(4t)< 0 $$

for all $t>0$. □

Lemma 2.2

The double inequality

$$ \bigl[\cosh(pt)\bigr]^{1/p}< \frac{\sqrt{2}\sinh(t)}{\sinh^{-1}[\sqrt {2}\sinh(t)]}< \bigl[\cosh(qt) \bigr]^{1/q} $$

(2.3)

holds for all $t>0$ if and only if $p\leq0$ and $q\geq2/3$. Here

$$\bigl[\cosh(pt)\bigr]^{1/p}\big|_{p=0}:=\lim_{p\rightarrow0}\bigl[\cosh(pt)\bigr]^{1/p}. $$

Proof

Let $t>0$, $p\in\mathbb{R}$ and $F(t, p)$ be defined by

$$ F(t, p)=\log\biggl[\frac{\sqrt{2}\sinh(t)}{\sinh^{-1} (\sqrt{2}\sinh (t) )} \biggr]-\frac{1}{p}\log\bigl[ \cosh(pt)\bigr]. $$

(2.4)

Then making use of the power series formulas

$$\begin{aligned}& \sinh(t)=t+\frac{t^{3}}{3!}+\frac{t^{5}}{5!}+\frac{t^{7}}{7!}+\cdots =\sum _{n=0}^{\infty}\frac{t^{2n+1}}{(2n+1)!}, \\& \cosh(t)=1+\frac{t^{2}}{2!}+\frac{t^{4}}{4!}+\frac{t^{6}}{6!}+\cdots =\sum _{n=0}^{\infty}\frac{t^{2n}}{(2n)!}, \\& \sinh^{-1}(t)=t-\frac{1}{2}\times\frac{t^{3}}{3}+ \frac{1\times3}{2\times4}\times\frac{t^{5}}{5}-\frac{1\times3\times 5}{2\times4\times6}\times \frac{t^{7}}{7}+\cdots \\& \hphantom{\sinh^{-1}(t)}=\sum_{n=0}^{\infty}\frac{(-1)^{n}(2n)!t^{2n+1}}{2^{2n}(n!)^{2}(2n+1)} \end{aligned}$$

we get

$$ \log\biggl[\frac{\sqrt{2}\sinh(t)}{\sinh^{-1} (\sqrt{2}\sinh(t) )} \biggr]=\frac{t^{2}}{3}+o \bigl(t^{2} \bigr),\quad\quad \frac{1}{p}\log\bigl[\cosh(pt)\bigr]=- \frac{1}{2}pt^{2}+o\bigl(t^{2}\bigr) $$

(2.5)

for $t\rightarrow0^{+}$.

It follows from (2.4) and (2.5) that

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

(2.6)

$$\begin{aligned}& \frac{\partial F(t, p)}{\partial t}=\frac{\cosh[(p-1)t]}{\sinh(t)\cosh (pt)\sinh^{-1}[\sqrt{2}\sinh(t)]}f_{1}(t, p), \end{aligned}$$

(2.7)

where

$$\begin{aligned}& f_{1}(t, p)=\sinh^{-1}\bigl[\sqrt{2}\sinh(t)\bigr]- \frac{\sqrt{2}\sinh(t)\cosh(pt)\cosh(t)}{\sqrt{\cosh(2t)}\cosh[(p-1)t]}, \end{aligned}$$

(2.8)

$$\begin{aligned}& f_{1}(0, p)=0, \end{aligned}$$

(2.9)

$$\begin{aligned}& \frac{\partial f_{1}(t, p)}{\partial t}=-\frac{\sqrt{2}\sinh(t)}{4[\cosh (2t)]^{3/2}\cosh^{2}[(p-1)t]}f(t,p), \end{aligned}$$

(2.10)

where $f(t,p)$ is defined by Lemma 2.1.

$$ \lim_{t\rightarrow0}\frac{F(t,p)}{t^{2}}=-\frac{1}{2} \biggl(p- \frac{2}{3} \biggr) $$

(2.11)

and

$$ \lim_{t\rightarrow+\infty}F(t,p)=-\infty $$

(2.12)

if $p>0$.

We first prove that the inequality

$$ \frac{\sqrt{2}\sinh(t)}{\sinh^{-1}[\sqrt{2}\sinh(t)]}< \bigl[\cosh (pt)\bigr]^{1/p} $$

(2.13)

holds for all $t>0$ if and only if $p\geq2/3$.

If $p\geq2/3$, then inequality (2.13) holds for all $t>0$ follows easily from Lemma 2.1(i), (2.4), (2.6), (2.7), (2.9), and (2.10).

If inequality (2.13) holds for all $t>0$, then (2.4) and (2.11) lead to $p\geq2/3$.

Next, we prove that the inequality

$$ \frac{\sqrt{2}\sinh(t)}{\sinh^{-1}[\sqrt{2}\sinh(t)]}>\bigl[\cosh (pt)\bigr]^{1/p} $$

(2.14)

holds for all $t>0$ if and only if $p\leq0$.

If $p\leq0$, then that inequality (2.14) holds for all $t>0$ follows easily from Lemma 2.1(ii), (2.4), (2.6), (2.7), (2.9), and (2.10).

If inequality (2.14) holds for all $t>0$, then (2.4) leads to $F(t,p)>0$. We assert that $p\leq0$, otherwise $p>0$ and (2.12) implies that there exists large enough $T_{0}>0$ such that $F(t, p)<0$ for $t\in(T_{0}, \infty)$. □

Lemma 2.3

Let $t>0$, $p\in\mathbb{R}$, and $f_{1}(t,p)$ be defined by (2.8). Then the following statements are true:

• (i) $f_{1}(t,p)<0$ for all $t>0$ if and only if $p\geq2/3$;

• (ii) $f(t,p)>0$ for all $t>0$ if and only if $p\leq0$.

Proof

(i) If $p\geq2/3$, then $f_{1}(t,p)<0$ for all $t>0$ follows easily from (2.9) and (2.10) together with Lemma 2.1(i).

If $f_{1}(t,p)<0$ for all $t>0$, then (2.8) leads to

$$ \lim_{t\rightarrow0}\frac{f_{1}(t,p)}{t^{3}}=\frac{-\sqrt{2} (p-\frac {2}{3} )t^{3}+o(t^{3})}{t^{3}}=-\sqrt{2} \biggl(p-\frac{2}{3} \biggr)\leq0, $$

which gives $p\geq2/3$.

(ii) If $p\leq0$, then $f_{1}(t,p)>0$ for all $t>0$ follows easily from (2.9) and (2.10) together with Lemma 2.1(ii).

Note that

$$\begin{aligned}& \frac{f_{1}(t,p)}{e^{(|p|-|p-1|)t}\sinh(t)} \\& \quad=\frac{\sinh^{-1}[\sqrt{2}\sinh(t)]}{e^{(|p|-|p-1|)t}\sinh(t)}-\frac {\sqrt{2}\cosh(t)\cosh(pt)}{e^{(|p|-|p-1|)t}\cosh[(p-1)t]\sqrt{\cosh(2t)}} \\& \quad=\frac{\log[\sqrt{2}\sinh(t)+\sqrt{\cosh(2t)} ]}{e^{(|p|-|p-1|)t}\sinh (t)}-\frac{\sqrt{2} (1+e^{-2|p|t} )\cosh(t)}{ (1+e^{-2|p-1|t} )\sqrt{\cosh(2t)}}. \end{aligned}$$

(2.15)

If $f_{1}(t,p)>0$ for all $t>0$, then

$$ \lim_{t\rightarrow+\infty}\frac{f_{1}(t,p)}{e^{(|p|-|p-1|)t}\sinh (t)}\geq0 $$

and we assert that $p\leq0$. Otherwise, equation (2.15) leads to

$$ \lim_{t\rightarrow+\infty}\frac{f_{1}(t,p)}{e^{(|p|-|p-1|)t}\sinh (t)}=-\frac{\sqrt{2}}{2}< 0 $$

if $p=1$ and

$$ \lim_{t\rightarrow+\infty}\frac{f_{1}(t,p)}{e^{(|p|-|p-1|)t}\sinh (t)}=-\sqrt{2}< 0 $$

if $p\in(0, 1)\cup(1, \infty)$. □

3 Main results

Theorem 3.1

The double inequality

$$ M_{p}(a,b)< V(a,b)< M_{q}(a,b) $$

holds for all $a, b>0$ with $a\neq b$ if and only if $p\leq0$ and $q\geq2/3$.

Proof

Since both $M_{r}(a,b)$ and $V(a,b)$ are symmetric and homogeneous of degree 1, without loss of generality, we assume that $a>b>0$. Let $t=\frac{1}{2}\log(a/b)>0$ and $r\in\mathbb{R}$, then (1.1) and (1.2) lead to

$$ V(a,b)=\sqrt{ab}V \biggl(\sqrt{\frac{a}{b}}, \sqrt{\frac{b}{a}} \biggr)=\sqrt{ab}V \bigl(e^{t}, e^{-t} \bigr) = \frac{\sqrt{2ab}\sinh(t)}{\sinh^{-1}[\sqrt{2}\sinh(t)]} $$

(3.1)

and

$$ M_{r}(a,b)=\sqrt{ab}M_{r} \biggl(\sqrt{\frac{a}{b}}, \sqrt{\frac{b}{a}} \biggr)=\sqrt{ab}M_{r} \bigl(e^{t}, e^{-t} \bigr) =\sqrt{ab}\bigl[\cosh(rt)\bigr]^{1/r}. $$

(3.2)

Therefore, Theorem 3.1 follows easily from (3.1) and (3.2) together with Lemma 2.2. □

Theorem 3.2

The double inequality

$$ \frac{a^{p-1}+b^{p-1}}{a^{p}+b^{p}}\frac{ab\sqrt{2 (a^{2}+b^{2} )}}{a+b}< V(a,b) < \frac{a^{q-1}+b^{q-1}}{a^{q}+b^{q}} \frac{ab\sqrt{2 (a^{2}+b^{2} )}}{a+b} $$

holds for all $a, b>0$ with $a\neq b$ if and only if $p\geq2/3$ and $q\leq0$.

Proof

Without loss of generality, we assume that $a>b>0$. Let $t=\frac{1}{2}\log(a/b)>0$ and $r\in\mathbb{R}$, then

$$ \frac{a^{r-1}+b^{r-1}}{a^{r}+b^{r}}\frac{ab\sqrt{2 (a^{2}+b^{2} )}}{a+b}=\frac{\sqrt{ab}\cosh[(r-1)t]\sqrt{\cosh(2t)}}{\cosh(t)\cosh(rt)}. $$

(3.3)

Therefore, Theorem 3.2 follows easily from (3.1) and (3.3) together with Lemma 2.3. □

Let $p\in\mathbb{R}$ and $a, b>0$. Then the pth Lehmer mean [28] $L_{p}(a,b)=\frac{a^{p+1}+b^{p+1}}{a^{p}+b^{p}}$ is strictly increasing with respect to $p\in\mathbb{R}$ for fixed $a, b>0$ with $a\neq b$. From Theorem 3.2 we get Corollary 3.3 immediately.

Corollary 3.3

The double inequality

$$ \frac{Q(a,b)G^{2}(a,b)}{A(a,b)L_{p-1}(a,b)}< V(a,b)< \frac {Q(a,b)G^{2}(a,b)}{A(a,b)L_{q-1}(a,b)} $$

holds for all $a, b>0$ with $a\neq b$ if and only if $p\geq2/3$ and $q\leq0$.

Let $p=2/3, 1, 2, +\infty$ and $q=0, -1/2, -1, -2, -\infty$. Then Corollary 3.3 leads to

Corollary 3.4

The inequalities

$$\begin{aligned} \min\{a, b\}\frac{Q(a,b)}{A(a,b)} < &\frac{G^{2}(a,b)}{Q(a,b)}< \frac {G^{2}(a,b)Q(a,b)}{A^{2}(a,b)} \\ < &\frac {Q(a,b)G^{4/3}(a,b)M^{1/3}_{1/3}(a,b)}{A(a,b)M^{2/3}_{2/3}(a,b)}< V(a,b) < Q(a,b) \\ < &\frac{Q(a,b)[2A(a,b)-G(a,b)]}{A(a,b)} \\ < &\frac{Q^{3}(a,b)}{A^{2}(a,b)} < \frac {2Q^{2}(a,b)-G^{2}(a,b)}{Q(a,b)}< \max\{a, b\}\frac{Q(a,b)}{A(a,b)} \end{aligned}$$

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

From (1.3), (1.4), and Theorem 3.1 we clearly see that $M_{2/3}(a,b)$ is the sharp upper power mean bound for the 2-order generalized logarithmic mean $L^{1/2}(a^{2}, b^{2})$, the first Seiffert mean $P(a,b)$, and the second Yang mean $V(a,b)$. In [29], Theorem 3, Yang and Chu proved that the inequality

$$ P(a,b)>L^{1/r} \bigl(a^{r}, b^{r} \bigr) $$

(3.4)

holds for all $a, b>0$ with $a\neq b$ if and only if $r\leq2$.

As a result of comparing $V(a,b)$ with $L^{1/2} (a^{2}, b^{2} )$, we have the following.

Theorem 3.5

The inequality

$$ V(a,b)< L^{1/2} \bigl(a^{2}, b^{2} \bigr) $$

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

Proof

We assume that $a>b$. Let $t=\frac{1}{2}\log(a/b)>0$, then

$$ L^{1/2} \bigl(a^{2}, b^{2} \bigr)= \biggl( \frac{a^{2}-b^{2}}{2(\log a-\log b)} \biggr)^{1/2}=\sqrt{ab}\sqrt{\frac {\sinh(2t)}{2t}}. $$

(3.5)

It follows from (3.1) and (3.5) that

$$\begin{aligned}& L^{1/2} \bigl(a^{2}, b^{2} \bigr)-V(a,b) \\& \quad=\frac{\sqrt{ab}\sqrt{\sinh(2t)}}{\sqrt{2t}\sinh^{-1} (\sqrt{2}\sinh(t) )} \bigl[\sinh^{-1} \bigl(\sqrt{2}\sinh(t) \bigr)- \sqrt{2t}\tanh(t) \bigr]. \end{aligned}$$

(3.6)

Let

$$ g(t)=\sinh^{-1} \bigl(\sqrt{2}\sinh(t) \bigr)-\sqrt{2t}\tanh(t). $$

(3.7)

Then simple computation leads to

$$\begin{aligned}& g(0)=0, \end{aligned}$$

(3.8)

$$\begin{aligned}& g^{\prime}(t)=\sqrt{2} \biggl(\frac{\cosh(t)}{\sqrt{\cosh(2t)}}-\frac {t+\sinh(t)\cosh(t)}{2\cosh^{2}(t)\sqrt{t\tanh(t)}} \biggr), \end{aligned}$$

(3.9)

$$\begin{aligned}& \biggl(\frac{\cosh(t)}{\sqrt{\cosh(2t)}} \biggr)^{2}- \biggl( \frac{t+\sinh(t)\cosh(t)}{2\cosh^{2}(t)\sqrt{t\tanh(t)}} \biggr)^{2} \\& \quad=\frac{\cosh^{2}(t)}{\cosh(2t)}-\frac{(t+\sinh(t)\cosh(t))^{2}}{4t\sinh (t)\cosh^{3}(t)} \\& \quad=\frac{(2t\cosh(2t)-\sinh(2t))(\sinh(2t)\cosh(2t)-2t)}{16t\sinh(t)\cosh (2t)\cosh^{3}(t)} \\& \quad =\frac{\sinh(4t)-4t}{16t\sinh(t)\cosh(2t)\cosh^{3}(t)} \biggl(\cosh(2t)- \frac{\sinh(2t)}{2t} \biggr)>0 \end{aligned}$$

(3.10)

for $t>0$.

Therefore, Theorem 3.5 follows easily from (3.6)-(3.10). □

Remark 3.6

From (1.4), (3.4), Theorems 3.1, and 3.5 we get the inequalities

$$ M_{0}(a,b)< V(a,b)< L^{1/2} \bigl(a^{2}, b^{2} \bigr)< P(a,b)< M_{2/3}(a,b) $$

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