Optimal power mean bounds for Yang mean

Abstract

In this paper, we prove that the double inequality $M_p(a,b)<U(a,b)<M_q(a,b)$ holds for all $a,b>0$ with $a\ne b$ if and only if $p\le 2log2/(2log\pi -log2)=0.8684\cdots$ and $q\ge 4/3$, where $U(a,b)$ and $M_r(a,b)$ are the Yang and r th power means of a and b, respectively.

MSC:26E60.

1 Introduction

Let $p\in \mathbb{R}$ and $a,b>0$ with $a\ne b$. Then the p th power mean $M_p(a,b)$ of a and b is given by

$$M_p(a,b)={\left(\frac{a^p+b^p}{2}\right)}^{1/p}(p\ne 0),M_0(a,b)=\sqrt{ab}.$$

The main properties for the power mean are given in [1]. It is well known that $M_p(a,b)$ is strictly increasing with respect to $p\in \mathbb{R}$ for fixed $a,b>0$ with $a\ne b$. Many classical means are the 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.

Let $L(a,b)=(b-a)/(logb-loga)$, $P(a,b)=(a-b)/[2arcsin((a-b)/(a+b))]$, $M(a,b)=(a-b)/[2{sinh}^{-1}((a-b)/(a+b))]$, $I(a,b)={(a^a/b^b)}^{1/(a-b)}/e$ and $T(a,b)=(a-b)/[2arctan((a-b)/(a+b))]$ be the logarithmic, first Seiffert, Neuman-Sándor, identric, and second Seiffert means of two distinct positive real numbers a and b, respectively. Then it is well known that the inequalities

$$\begin{array}{rl}H(a,b)& <G(a,b)<L(a,b)<P(a,b)\\ <I(a,b)<A(a,b)<M(a,b)<T(a,b)<Q(a,b)\end{array}$$

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

Recently, the bounds for certain bivariate means in terms of the power mean have been the subject of intensive research. Seiffert [2] proved that the inequalities

$$\frac{2}{\pi }{M}_1(a,b)<P(a,b)<M_1(a,b)<T(a,b)<M_2(a,b)$$

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

Jagers [3] proved that the double inequality

$$M_{1/2}(a,b)<P(a,b)<M_{2/3}(a,b)$$

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

In [4, 5], Hästö established that

$$P(a,b)>M_{log2/log\pi }(a,b),P(a,b)>\frac{2\sqrt{2}}{\pi }{M}_{2/3}(a,b)$$

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

Witkowski [6] proved that the double inequality

$$\frac{2\sqrt{2}}{\pi }{M}_2(a,b)<T(a,b)<\frac{4}{\pi }{M}_1(a,b)$$

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

In [7], Costin and Toader presented the result that

$$M_{log2/(log\pi -log2)}(a,b)<T(a,b)<M_{5/3}(a,b)$$

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

Chu and Long [8] proved that the double inequality

$$M_p(a,b)<M(a,b)<M_q(a,b)$$

holds for all $a,b>0$ with $a\ne b$ if and only if $p\le log2/log[2log(1+\sqrt{2})]=1.224\cdots$ and $q\ge 4/3$.

The following sharp bounds for the logarithmic and identric means in terms of the power means can be found in the literature [9–16]:

$$\begin{array}{c}{M}_0(a,b)<L(a,b)<M_{1/3}(a,b),M_{2/3}(a,b)<I(a,b)<M_{log2}(a,b),\hfill \\ M_0(a,b)<L^{1/2}(a,b)I^{1/2}(a,b)<M_{1/2}(a,b),\hfill \\ M_{log2/(1+log2)}(a,b)<\frac{L(a,b)+I(a,b)}{2}<M_{1/2}(a,b)\hfill \end{array}$$

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

Recently, Yang [17] introduced the Yang mean $U(a,b)$ of two distinct positive real numbers a and b as follows:

$$U(a,b)=\frac{a-b}{\sqrt{2}arctan\frac{a-b}{\sqrt{2ab}}},$$

and he proved that the inequalities

$$\begin{array}{c}P(a,b)<U(a,b)<T(a,b),\frac{G(a,b)T(a,b)}{A(a,b)}<U(a,b)<\frac{P(a,b)Q(a,b)}{A(a,b)},\hfill \\ Q^{1/2}(a,b){\left[\frac{2G(a,b)+Q(a,b)}{3}\right]}^{1/2}<U(a,b)<Q^{2/3}(a,b){\left[\frac{G(a,b)+Q(a,b)}{2}\right]}^{1/3},\hfill \\ \frac{G(a,b)+Q(a,b)}{2}<U(a,b)<{[\frac{2}{3}{\left(\frac{G(a,b)+Q(a,b)}{2}\right)}^{1/2}+\frac{1}{3}{Q}^{1/2}(a,b)]}^2\hfill \end{array}$$

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

In [18], Yang et al. presented several sharp bounds for the Yang mean $U(a,b)$ in terms of the geometric mean $G(a,b)$ and quadratic mean $Q(a,b)$.

The main purpose of this article is to find the greatest value p and the least value q such that the double inequality

$$M_p(a,b)<U(a,b)<M_q(a,b)$$

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

2 Lemmas

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

Lemma 2.1 Let $f_1:(0,1)\times \mathbb{R}\to \mathbb{R}$ be defined by

$$f_1(x,p)=\frac{(1-x^2)(1+x^p)}{\sqrt{x}(1+x^2)(1+x^{p-1})}-\sqrt{2}arctan\frac{1-x}{\sqrt{2x}}.$$

(2.1)

Then

• (1) $f_1(x,p)$ is strictly decreasing with respect to x on $(0,1)$ if and only if $p\ge 4/3$;

• (2) $f_1(x,p)$ is strictly increasing with respect to x on $(0,1)$ if and only if $p\le 1/2$.

Proof It follows from (2.1) that

$$\frac{\partial f_1(x,p)}{\partial x}=\frac{(1-x)x^{p-1/2}}{2{(1+x^2)}^2{(x+x^p)}^2}{f}_2(x,p),$$

(2.2)

where

$$\begin{array}{rl}{f}_2(x,p)=& x^{1-p}(-1+x-5x^2-3x^3)+x^p(3+5x-x^2+x^3)-(2p-1)\\ +4x-4x^3+(2p-1)x^4.\end{array}$$

(2.3)

• (1) If $f_1(x,p)$ is strictly decreasing with respect to x on $(0,1)$, then (2.2) leads to the conclusion that $f_2(x,p)<0$ for all $x\in (0,1)$. In particular, from (2.3) we have

  $$\underset{x\to 1^{-}}{lim}\frac{f_2(x,p)}{1-x}=-24(p-\frac{4}{3})\le 0.$$

  (2.4)

Therefore, $p\ge 4/3$ follows from (2.4).

If $p\ge 4/3$, then it follows from (2.3) that

$$\begin{array}{rl}\frac{\partial f_2(x,p)}{\partial p}=& [x^p(x^3-x^2+5x+3)+x^{1-p}(3x^3+5x^2-x+1)]logx\\ -2(1-x^4)<0\end{array}$$

(2.5)

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

Equation (2.3) and inequality (2.5) lead to the conclusion that

$$\begin{array}{rl}{f}_2(x,p)\le & f_2(x,\frac{4}{3})=-\frac{x^{-1/3}}{3}{(1-x^{2/3})}^3\\ \times (3x^{8/3}+5x^{7/3}+9x^2+12x^{5/3}+6x^{4/3}+12x+9x^{2/3}+5x^{1/3}+3)<0\end{array}$$

(2.6)

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

Therefore, $f_1(x,p)$ is strictly decreasing with respect to x on $(0,1)$ follows from (2.2) and (2.6).

• (2) If $f_1(x,p)$ is strictly increasing with respect to x on $(0,1)$, then (2.2) leads to the conclusion that $f_2(x,p)>0$ for all $x\in (0,1)$. In particular, we have $f_2(0^{+},p)\ge 0$ and we assert that $p\le 1/2$. Indeed, from (2.3) we clearly see that $f_2(0^{+},p)=-\mathrm{\infty }$ for $p>1$, $f_2(0^{+},1)=-2$, $f_2(0^{+},0)=4$, $f_2(0^{+},p)=\mathrm{\infty }$ for $p<0$, and $f_2(0^{+},p)=1-2p$ for $0<p<1$.

If $p\le 1/2$, then inequality (2.5) holds again. It follows from (2.3) and (2.5) that

$$f_2(x,p)\ge f_2(x,\frac{1}{2})=2x^{1/2}(1-x)(x^2+2x^{3/2}+4x+2x^{1/2}+1)>0$$

(2.7)

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

Therefore, $f_1(x,p)$ is strictly increasing with respect to x on $(0,1)$ follows from (2.2) and (2.7). □

Lemma 2.2 Let $f_1:(0,1)\times \mathbb{R}\to \mathbb{R}$ be defined by (2.1). Then

• (1) $f_1(x,p)>0$ for all $x\in (0,1)$ if and only if $p\ge 4/3$;

• (2) $f_1(x,p)<0$ for all $x\in (0,1)$ if and only if $p\le 1/2$.

Proof (1) If $f_1(x,p)>0$ for all $x\in (0,1)$, then from (2.1) and the L’Hôpital rules we have

$$\underset{x\to 1^{-}}{lim}\frac{f_1(x,p)}{{(1-x)}^3}=\frac{1}{12}(3p-4)\ge 0$$

and $p\ge 4/3$.

If $p\ge 4/3$, then (2.1) and Lemma 2.1(1) lead to the conclusion that $f_1(x,p)>f_1(1,p)=0$ for all $x\in (0,1)$.

• (2) If $f_1(x,p)<0$ for all $x\in (0,1)$, then $f_1(0^{+},p)\le 0$. We claim that $p\le 1/2$. Indeed, it follows from (2.1) that $f_1(0^{+},p)=+\mathrm{\infty }$ if $p>1/2$.

If $p\le 1/2$, then (2.1) and Lemma 2.1(2) lead to the conclusion that $f_1(x,p)<f_1(1,p)=0$ for all $x\in (0,1)$. □

Lemma 2.3 Let $f_3:(0,1)\times \mathbb{R}\to \mathbb{R}$ be defined by

$$\begin{array}{rl}{f}_3(x,p)=& -x^{1-2p}+x^{2-2p}-5x^{3-2p}-3x^{4-2p}+3+5x-x^2+x^3\\ -(2p-1)x^{-p}+4x^{1-p}-4x^{3-p}+(2p-1)x^{4-p}.\end{array}$$

(2.8)

Then ${\partial }^4{f}_3(x,p)/\partial x^4<0$ for all $x\in (0,1)$ if $p\in (1,4/3)$.

Proof It follows (2.8) that

$$x^{p+4}\frac{{\partial }^4{f}_3(x,p)}{\partial x^4}=x^{1-p}(a_3{x}^3+a_2{x}^2+a_{1}x+a_0)+b_4{x}^4+b_3{x}^3+b_{1}x+b_0,$$

(2.9)

where

$$a_3=-3(2p-1)(2p-2)(2p-3)(2p-4)<0,$$

(2.10)

$$a_2=-10p(2p-1)(2p-2)(2p-3)>0,$$

(2.11)

$$a_1=2p(2p-1)(2p+1)(2p-2)>0,$$

(2.12)

$$a_0=-2p(2p-1)(2p+1)(2p+2)<0,$$

(2.13)

$$b_4=(2p-1)(p-1)(p-2)(p-3)(p-4)<0,$$

(2.14)

$$b_3=-4p(p-1)(p-2)(p-3)<0,$$

(2.15)

$$b_1=4p(p-1)(p+1)(p+2)>0,$$

(2.16)

$$b_0=-p(2p-1)(p+1)(p+2)(p+3)<0.$$

(2.17)

From (2.11)-(2.13) and (2.16) together with (2.17) we get

$$a_2{x}^2+a_{1}x+a_0<a_2+a_1+a_0=-4p(2p-1)(10p^2-21p+17)<0,$$

(2.18)

$$b_{1}x+b_0<b_1+b_0=-p(p+2)(p+1)(2p^2+p+1)<0$$

(2.19)

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

Therefore, Lemma 2.3 follows easily from (2.9), (2.10), (2.14), (2.15), (2.18), and (2.19). □

Lemma 2.4 Let $f_3:(0,1)\times \mathbb{R}\to \mathbb{R}$ be defined by (2.8). Then ${\partial }^2{f}_3(x,p)/\partial x^2<0$ for all $x\in (0,1)$ if $p\in (1/2,4/3)$.

Proof It follows from (2.8) that

$$\begin{array}{rl}{x}^{p+2}\frac{{\partial }^2{f}_3(x,p)}{\partial x^2}=& 6x^{p+3}+(2p-1)(p-3)(p-4)x^4-2x^{p+2}\\ -3(2p-3)(2p-4)x^{4-p}-5(2p-2)(2p-3)x^{3-p}\\ +(2p-1)(2p-2)x^{2-p}-2p(2p-1)x^{1-p}\\ -4(p-2)(p-3)x^3+4p(p-1)x-p(2p-1)(p+1),\end{array}$$

(2.20)

$$\frac{{\partial }^2{f}_3(x,p)}{\partial x^2}{|}_{x=1}=-48(\frac{4}{3}-p)(\frac{3}{2}-p)<0,$$

(2.21)

$$\frac{{\partial }^3{f}_3(x,p)}{\partial x^3}{|}_{x=1}=88p^3-300p^2+380p-144.$$

(2.22)

We divide the proof into two cases.

Case 1. $p\in (1/2,1]$. Then from

$$\begin{array}{c}(2p-1)(p-3)(p-4)>0,-3(2p-3)(2p-4)<0,-5(2p-2)(2p-3)\le 0,\hfill \\ (2p-1)(2p-2)\le 0,-2p(2p-1)<0,-4(p-2)(p-3)<0,\hfill \\ 4p(p-1)<0,-p(2p-1)(p+1)<0,\hfill \\ 0<x^4\le x^{p+3}<x^{4-p}\le x^3\le x^{p+2}<x^{3-p}\le x^2<x^{2-p}\le x<x^{1-p}\le 1\hfill \end{array}$$

and (2.20) we clearly see that

$$\begin{array}{rl}{x}^{p+2}\frac{{\partial }^2{f}_3(x,p)}{\partial x^2}<& [6+(2p-1)(p-3)(p-4)]x^{4-p}+[-2-3(2p-3)(2p-4)\\ -5(2p-2)(2p-3)+(2p-1)(2p-2)-2p(2p-1)\\ -4(p-2)(p-3)+4p(p-1)\\ -p(2p-1)(p+1)]x^{4-p}\\ =& -8(3p-4)(2p-3)x^{4-p}<0\end{array}$$

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

Case 2. $p\in (1,4/3]$. Then (2.22) leads to

$$\frac{{\partial }^3{f}_3(x,p)}{\partial x^3}{|}_{x=1}=88(p-1){(p-\frac{53}{44})}^2+\frac{887}{22}(p-1)+24>0.$$

(2.23)

It follows from Lemma 2.3 and (2.23) that ${\partial }^2{f}_3(x,p)/\partial x^2$ is strictly increasing with respect to x on $(0,1)$.

Therefore, ${\partial }^2{f}_3(x,p)/\partial x^2<0$ for all $x\in (0,1)$ follows from (2.21) and the monotonicity of the ${\partial }^2{f}_3(x,p)/\partial x^2$ with respect to x on the interval $(0,1)$. □

Lemma 2.5 Let $f_1:(0,1)\times \mathbb{R}\to \mathbb{R}$ be defined by (2.1). Then there exists $\lambda \in (0,1)$ such that $f_1(x,p)$ is strictly decreasing with respect to x on the interval $(0,\lambda ]$ and strictly increasing with respect to x on the interval $[\lambda ,1)$ if $p\in (1/2,4/3)$.

Proof Let $f_2(x,p)$ and $f_3(x,p)$ be defined by (2.3) and (2.8), respectively. Then from (2.8) we clearly see that

$$f_3(1,p)=0,f_3(0^{+},p)=-\mathrm{\infty },$$

(2.24)

$$\frac{\partial f_3(x,p)}{\partial x}{|}_{x=1}=8(3p-4)<0,\underset{x\to 0^{+}}{lim}\frac{\partial f_3(x,p)}{\partial x}=+\mathrm{\infty }.$$

(2.25)

It follows from Lemma 2.4 and (2.25) that there exists ${\lambda }_0\in (0,1)$ such that $f_3(x,p)$ is strictly increasing with respect to x on $(0,{\lambda }_0]$ and strictly decreasing with respect to x on $[{\lambda }_0,1)$. This in conjunction with (2.24) leads to the conclusion that there exists $\lambda \in (0,1)$ such that $f_3(x,p)<0$ for $x\in (0,\lambda )$ and $f_3(x,p)>0$ for $x\in (\lambda ,1)$.

Note that

$$f_2(x,p)=x^p{f}_3(x,p).$$

(2.26)

Therefore, Lemma 2.5 follows from (2.2) and (2.26) together with the piecewise positive and negative of $f_3(x,p)$ on $(0,1)$. □

Lemma 2.6 Let $f:(0,1)\times \mathbb{R}\to \mathbb{R}$ be defined by

$$f(x,p)=log\frac{U(1,x)}{M_p(1,x)}=log\frac{1-x}{\sqrt{2}arctan\frac{1-x}{\sqrt{2x}}}-\frac{1}{p}log\frac{1+x^p}{2}(p\ne 0),$$

(2.27)

$$f(x,0)=\underset{p\to 0}{lim}log\frac{U(1,x)}{M_p(1,x)}=log\frac{1-x}{\sqrt{2}arctan\frac{1-x}{\sqrt{2x}}}-\frac{1}{2}logx.$$

(2.28)

Then the following statements are true:

• (1) $f(x,p)$ is strictly increasing with respect to x on $(0,1)$ if and only if $p\ge 4/3$;

• (2) $f(x,p)$ is strictly decreasing with respect to x on $(0,1)$ if and only if $p\le 1/2$;

• (3) If $1/2<p<4/3$, then there exists $\mu \in (0,1)$ such that $f(x,p)$ is strictly increasing with respect to x on $(0,\mu ]$ and strictly decreasing with respect to x on $[\mu ,1)$.

Proof It follows from (2.27) and (2.28) that

$$\frac{\partial f(x,p)}{\partial x}=\frac{1+x^{p-1}}{\sqrt{2}(1-x)(1+x^p)arctan\frac{1-x}{\sqrt{2x}}}{f}_1(x,p),$$

(2.29)

where $f_1(x,p)$ is defined by (2.1).

Therefore, parts (1) and (2) follow from Lemma 2.2 and (2.29).

Next, we prove part (3). If $1/2<p<4/3$, then (2.1) leads to

$$f_1(0^{+},p)=+\mathrm{\infty },f_1(1,p)=0.$$

(2.30)

From Lemma 2.5 and (2.30) we clearly see that there exists $\mu \in (0,1)$ such that $f_1(x,p)>0$ for $x\in (0,\mu )$ and $f_1(x,p)<0$ for $x\in (\mu ,1)$.

Therefore, part (3) follows from (2.29) and the fact that $f_1(x,p)>0$ for $x\in (0,\mu )$ and $f_1(x,p)<0$ for $x\in (\mu ,1)$. □

3 Main results

Theorem 3.1 The double inequality

$$M_p(a,b)<U(a,b)<M_q(a,b)$$

holds for all $a,b>0$ with $a\ne b$ if and only if $p\le p_0=2log2/(2log\pi -log2)=0.8684\cdots$ and $q\ge 4/3$.

Proof Since both the Yang mean $U(a,b)$ and the r th power mean $M_r(a,b)$ are symmetric and homogeneous of degree 1, without loss of generality, we assume that $a=1$ and $b=x\in (0,1)$.

We first prove that the inequality $U(1,x)<M_q(1,x)$ holds for all $x\in (0,1)$ if and only if $q\ge 4/3$.

If $q=4/3$, then from (2.27) and Lemma 2.6(1) we get

$$log\frac{U(1,x)}{M_{4/3}(1,x)}=f(x,\frac{4}{3})<f(1^{-},\frac{4}{3})=0$$

(3.1)

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

Therefore, $U(1,x)<M_q(1,x)$ for all $x\in (0,1)$ and $q\ge 4/3$ follows from (3.1) and the monotonicity of the function $q\to M_q(1,x)$.

If $U(1,x)<M_q(1,x)$, then (2.27) and (2.28) lead to $f(x,q)<0$ for all $x\in (0,1)$. In particular, we have

$$\underset{x\to 1^{-}}{lim}\frac{f(x,q)}{{(1-x)}^2}=\frac{1}{8}(\frac{4}{3}-q)\le 0$$

and $q\ge 4/3$.

Next, we prove that the inequality $U(1,x)>M_p(1,x)$ holds for all $x\in (0,1)$ if and only if $p\le p_0$.

If $U(1,x)>M_p(1,x)$ holds for all $x\in (0,1)$, then (2.27) leads to $f(x,p)>0$ for all $x\in (0,1)$. In particular, we have

$$f(0^{+},p)=(\frac{1}{p}+\frac{1}{2})log2-log\pi \ge 0.$$

(3.2)

We claim that $p\le p_0$. Indeed, $p\le p_0$ follows from (3.2) if $p>0$, and $p<p_0$ is obvious if $p<0$.

If $p=p_0$, then (2.27) leads to

$$f(0^{+},p_0)=f(1,p_0)=0.$$

(3.3)

It follows from (2.27) and (3.3) together with Lemma 2.6(3) that

$$log\frac{U(1,x)}{M_{p_0}(1,x)}=f(x,p_0)>0$$

(3.4)

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

Therefore, $U(1,x)>M_p(1,x)$ for all $x\in (0,1)$ and $p\le p_0$ follows from (3.4) and the monotonicity of the function $p\to M_p(1,x)$. □

Theorem 3.2 Let $a,b>0$ with $a\ne b$. Then the double inequality

$$\frac{2^{5/4}}{\pi }{M}_{4/3}(a,b)<U(a,b)<\frac{2^{5/2}}{\pi }{M}_{1/2}(a,b)$$

holds with the best possible constants $2^{5/4}/\pi$ and $2^{5/2}/\pi$.

Proof It follows from Lemma 2.6(1) and (2) together with (2.27) that

$$log\frac{U(1,x)}{M_{1/2}(1,x)}=f(x,\frac{1}{2})<f(0^{+},\frac{1}{2})=log\frac{2^{5/2}}{\pi }$$

(3.5)

and

$$log\frac{U(1,x)}{M_{4/3}(1,x)}=f(x,\frac{4}{3})>f(0^{+},\frac{4}{3})=log\frac{2^{5/4}}{\pi }$$

(3.6)

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

Therefore, $2^{5/4}/\pi M_{4/3}(1,x)<U(1,x)<2^{5/2}/\pi M_{1/2}(1,x)$ for all $x\in (0,1)$ follows from (3.5) and (3.6), and the optimality of the parameters $2^{5/4}/\pi$ and $2^{5/2}/\pi$ follows from the monotonicity of the functions $f(x,1/2)$ and $f(x,4/3)$. □

Remark 3.1 For all $a_1,a_2,b_1,b_2>0$ with $a_1/b_1<a_2/b_2<1$. Then from Lemma 2.6(1) and (2) together with (2.27) we clearly see that the Ky Fan type inequalities

$$\frac{M_p(a_2,b_2)}{M_p(a_1,b_1)}<\frac{U(a_2,b_2)}{U(a_1,b_1)}<\frac{M_q(a_2,b_2)}{M_q(a_1,b_1)}$$

hold if and only if $p\ge 4/3$ and $q\le 1/2$.

Let $p\in \mathbb{R}$ and $L_p(a,b)=(a^{p+1}+b^{p+1})/(a^p+b^p)$ be the p th Lehmer mean of two positive real numbers and a and b. Then the function $f_1(x,p)$ defined by (2.1) can be rewritten as

$$f_1(x,p)=(1-x)[\frac{A(1,x)L_{p-1}(1,x)}{G(1,x)Q^2(1,x)}-\frac{1}{U(1,x)}].$$

(3.7)

From Lemma 2.2 and (3.7) we get Remark 3.2.

Remark 3.2 The double inequality

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

holds for all $a,b>0$ with $a\ne b$ if and only if $p\ge 4/3$ and $q\le 1/2$.