Optimal bounds for a Gaussian Arithmetic-Geometric type mean by quadratic and contraharmonic means

TL;DR

This paper determines optimal bounds for a Gaussian arithmetic-geometric mean type using quadratic and contraharmonic means, providing new inequalities and bounds for elliptic integrals.

Contribution

It introduces the best possible parameters for inequalities involving the Gaussian arithmetic-geometric mean and classical means, extending bounds for elliptic integrals.

Findings

Established optimal parameters for inequalities involving AG_{Q,C}(a,b).

Derived new bounds for the complete elliptic integral of the first kind.

Extended classical mean inequalities to a Gaussian mean context.

Abstract

In this paper, we present the best possible parameters $\alpha_i, \beta_i\ (i=1,2,3)$ and $\alpha_4,\beta_4\in(1/2,1)$ such that the double inequalities \begin{align*} \alpha_1Q(a,b)+(1-\alpha_1)C(a,b)&<AG_{Q,C}(a,b)<\beta_1Q(a,b)+(1-\beta_1)C(a,b),\\ \qquad\ Q^{\alpha_2}(a,b)C^{1-\alpha_2}(a,b)&<AG_{Q,C}(a,b)<Q^{\beta_2}(a,b)C^{1-\beta_2}(a,b),\\ \frac{Q(a,b)C(a,b)}{\alpha_3Q(a,b)+(1-\alpha_3)C(a,b)}&<AG_{Q,C}(a,b)<\frac{Q(a,b)C(a,b)}{\beta_3Q(a,b)+(1-\beta_3)C(a,b)},\\ C\left(\sqrt{\alpha_4a^2+(1-\alpha_4)b^2},\sqrt{(1-\alpha_4)a^2+\alpha_4b^2}\right)&<AG_{Q,C}(a,b)<C\left(\sqrt{\beta_4a^2+(1-\beta_4)b^2},\sqrt{(1-\beta_4)a^2+\beta_4b^2}\right) \end{align*} hold for all $a, b>0$ with $a\neq b$, where $Q(a,b)$, $C(a,b)$ and $AG(a,b)$ are the quadratic, contraharmonic and Arithmetic-Geometric means, and $AG_{Q,C}(a,b)=AG[Q(a,b),C(a,b)]$. As consequences, we present new bounds for the complete elliptic integral of the first kind. Keywords: Arithmetic-Geometric mean, Complete elliptic integral, Quadratic mean, Contraharmonic mean

Equations

$${\alpha }_{1}Q(a,b)+(1-{\alpha }_1)C(a,b)$$

$$Q^{{\alpha }_2}(a,b)C^{1-{\alpha }_2}(a,b)$$

$$\frac{Q(a,b)C(a,b)}{{\alpha }_{3}Q(a,b)+(1-{\alpha }_3)C(a,b)}$$

$$C\left(\sqrt{{\alpha }_4{a}^2+(1-{\alpha }_4)b^2},\sqrt{(1-{\alpha }_4)a^2+{\alpha }_4{b}^2}\right)$$

$$a_{n+1}=A(a_n,b_n),b_{n+1}=G(a_n,b_n)$$

$$AG(1,r)\mathcal{K}(\sqrt{1-r^2})=\frac{\pi }{2}$$

$$AG(a,b)=\frac{\pi }{2{\int }_0^{\pi /2}dt/\sqrt{a^2{\cos }^{2}t+b^2{\sin }^{2}t}}.$$

$$\mathcal{K}\left(\frac{2\sqrt{r}}{1+r}\right)=(1+r)\mathcal{K}(r),$$

$$\frac{d\mathcal{K}(r)}{dr}=\frac{\mathcal{E}(r)-r^{\prime 2}\mathcal{K}(r)}{r{r}^{\prime 2}},$$

$$\frac{d\left[\mathcal{E}(r)-r^{\prime 2}\mathcal{K}(r)\right]}{dr}=r\mathcal{K}(r),$$

$$\mathcal{K}(\sqrt{2}/2)=\frac{{\Gamma }^2(1/4)}{4\sqrt{\pi }}=1.854\cdots ,\mathcal{E}(\sqrt{2}/2)=\frac{4{\Gamma }^2(3/4)+{\Gamma }^2(1/4)}{8\sqrt{\pi }}=1.350\cdots ,$$

$$A{G}_{X,Y}(a,b)=AG[X(a,b),Y(a,b)],$$

$$L_{-1}(a,b)=L(a,b)$$

$$\frac{1+\sqrt{r}}{2}AG(1,\sqrt{r})$$

$$\frac{1}{\lambda /L(a,b)+(1-\lambda )/A(a,b)}<AG(a,b)<\frac{1}{\mu /L(a,b)+(1-\mu )/A(a,b)}$$

$$S_p(a,b)<AG(a,b)<S_q(a,b),$$

$$\begin{array}{rl}{S}_{7/4,-1/4}(a,b)& <AG(a,b)<A^{1/4}(a,b)L^{3/4}(a,b),\\ AG(a,b)& <\sqrt{S_{p,1}(a,b)S_{1-p,1}(a,b)}\end{array}$$

$$Q^{{\alpha }_1}(a,b)A^{1-{\alpha }_1}(a,b)$$

$${\alpha }_{2}Q(a,b)+(1-{\alpha }_2)A(a,b)$$

$$Q\left[{\alpha }_{3}a+(1-{\alpha }_3)b,(1-{\alpha }_3)a+{\alpha }_{3}b\right]$$

$$\zeta (1/2)=Q(a,b)<\min \{Q(a,b),C(a,b)\}<A{G}_{Q,C}(a,b)<\max \{Q(a,b),C(a,b)\}=C(a,b)=\zeta (1).$$

$${\alpha }_{1}Q(a,b)+(1-{\alpha }_1)C(a,b)$$

$$Q^{{\alpha }_2}(a,b)C^{1-{\alpha }_2}(a,b)$$

$$\frac{Q(a,b)C(a,b)}{{\alpha }_{3}Q(a,b)+(1-{\alpha }_3)C(a,b)}$$

$$C\left(\sqrt{{\alpha }_4{a}^2+(1-{\alpha }_4)b^2},\sqrt{(1-{\alpha }_4)a^2+{\alpha }_4{b}^2}\right)$$

$$<C\left(\sqrt{{\beta }_4{a}^2+(1-{\beta }_4)b^2},\sqrt{(1-{\beta }_4)a^2+{\beta }_4{b}^2}\right)$$

$$\frac{f(x)-f(a)}{g(x)-g(a)}\text{and}\frac{f(x)-f(b)}{g(x)-g(b)}.$$

$$f(r)=\frac{1-\pi /[2\mathcal{K}(r)]}{1-\sqrt{1-r^2}},$$

$$\frac{f_1^{\prime }(r)}{f_2^{\prime }(r)}=\frac{\pi }{2}\cdot \frac{\mathcal{E}(r)-r^{\prime 2}\mathcal{K}(r)}{r^2}\cdot \frac{1}{{\left[r^{\prime 1/2}\mathcal{K}(r)\right]}^2}.$$

$$g(r)=\frac{\log \pi /2-\log \mathcal{K}(r)}{\log \sqrt{1-r^2}},$$

$$\frac{g_1^{\prime }(r)}{g_2^{\prime }(r)}=\frac{\mathcal{E}(r)-r^{\prime 2}\mathcal{K}(r)}{r^2\mathcal{K}(r)}.$$

$$h(r)=\frac{2\sqrt{1-r^2}\mathcal{K}(r)/\pi -1}{\sqrt{1-r^2}-1},$$

$$\frac{h_1^{\prime }(r)}{h_2^{\prime }(r)}=\frac{\mathcal{K}(r)-\mathcal{E}(r)}{r^2}.$$

$$\mathcal{K}(r)>\frac{\pi }{2}\left(1+\frac{r^2}{4}\right)$$

$$\frac{2\mathcal{K}(r)/\pi -1}{r^2}>\frac{1}{4}$$

$${\mu }_1(0)={\mu }_2(0)=0,\mu (r)=\frac{{\mu }_1(r)}{{\mu }_2(r)}.$$

Full text

[TABLE]

hold for all $a,b>0$ with $a\neq b$, where $Q(a,b)$, $C(a,b)$ and $AG(a,b)$ are the quadratic, contraharmonic and Arithmetic-Geometric means, and $AG_{Q,C}(a,b)=AG[Q(a,b),C(a,b)]$. As consequences, we present new bounds for the complete elliptic integral of the first kind.

1 Introduction

The classical arithmetic-geometric mean $AG(a,b)$ of two positive numbers $a$ and $b$ is defined by starting with $a_{0}=a,b_{0}=b$ and then iterating

[TABLE]

for $n\in\mathbb{N}$ until two sequences $\{a_{n}\}$ and $\{b_{n}\}$ converge to the same number, where $A(a,b)=(a+b)/2$ and $G(a,b)=\sqrt{ab}$ are arithmetic and geometric means, respectively.

The well-known Gauss identity AVV shows that

[TABLE]

for $r\in(0,1)$, where $\mathcal{K}(r)=\int^{\pi/2}_{0}(1-r^{2}\sin^{2}t)^{-1/2}dt$ ($r\in[0,1)$) is the complete elliptic integral of the first kind. By use of the homogeneity of (1.1) and (1.2), $AG(a,b)$ can be written explicitly as

[TABLE]

Let $r\in(0,1)$ and $r^{\prime}=\sqrt{1-r^{2}}$, then the complete elliptic integral of the second kind is given by $\mathcal{E}(r)=\int^{\pi/2}_{0}(1-r^{2}\sin^{2}t)^{1/2}dt$. We clearly see that $\mathcal{K}(r)$ is strictly increasing from $(0,1)$ onto $(\pi/2,+\infty)$ and $\mathcal{E}(r)$ is strictly decreasing from $(0,1)$ onto $(1,\pi/2)$. Moreover, $\mathcal{K}(r)$ and $\mathcal{E}(r)$ satisfy the following Landen identities and derivatives formulas (see (AVV, , Appendix E, p.474-475))

[TABLE]

Two special values $\mathcal{K}(\sqrt{2}/2)$ and $\mathcal{E}{\sqrt{2}/2}$ will be used later, which can be expressed as (see (AVV, , Theorem 1.7))

[TABLE]

where $\Gamma(x)=\int^{\infty}_{0}t^{x-1}e^{-t}dt$ is the classical Euler’s gamma function.

The special bivariate mean $AG_{X,Y}(a,b)$ derived from Arithmetric-Geometric mean for any bivariate means $X(a,b)$ and $Y(a,b)$ of positive numbers a, b is given by

[TABLE]

which is called a Arithmetric-Geometric type mean. We denote the pairs of means $\{X,Y\}$ the generating means of the Arithmetric-Geometric type mean defined in (1.4).

It is well known that the elliptic elliptic integrals $\mathcal{K}(r)$ and $\mathcal{E}(r)$ and the Gaussian arithmetic-geometric mean $AG(a,b)$ have many applications in mathematics, physics, mechanics, and engineering Ca ; EBJ ; Le ; Ho ; MF ; Ma . Recently, the Arithmetric-Geometric mean has been the subject of intensive research. The double inequalities

[TABLE]

hold for all $a,b>0$ with $a\neq b$, where $L_{p}(a,b)=\left[(b^{p+1}-a^{p+1})/((p+1)(b-a))\right]^{1/p}(p\neq-1,0)$, $L_{-1}(a,b)=L(a,b)=(b-a)/(\log b-\log a)$ and $L_{0}(a,b)=I(a,b)=(b^{b}/a^{a})^{1/(b-a)}/e$. The left inequality of (1.5) was first proposed by Carlson and Vuorinen CV and also was proved by different methods in Sa ; NS ; Ya . Vamanamurthy and Vuorinen VV proved that $AG(a,b)<(\pi/2)L(a,b)$ for all $a,b>0$ with $a\neq b$. The second inequality of (1.5) was proved by Borwein BB and Yang Ya . Very recently, Ding and Zhao DZ showed that $L_{-1}(a,b)<AG(a,b)<L_{-1/2}(a,b)$ for all $a,b>0$ with $a\neq b$, where $L_{-1}(a,b)$ and $L_{-1/2}(a,b)$ are the best possible lower and upper generalized logarithmic mean bounds, respectively.

In Ku , Kühnau refined the double inequality (1.6) and obtained the improved upper bound $\pi(1-r^{2}/9)/[2\log(4/r)]$. Qiu and Vamanamurthy QV presented the new lower and upper bounds for $AG(1,r)$ with $r\in(0,1)$, which are $4\pi/[(9-r^{2})\log(4/r)]$ and $(9-r^{2})\pi/[18.192\log(4/r)]$, respectively.

Alzer and Qiu AQ proved that the double inequality

[TABLE]

holds for all $a,b>0$ with $a\neq b$ if and only if $\lambda\geq 3/4$ and $\mu\leq 2/\pi$.

Chu and Wang CW proved that the double inequality

[TABLE]

for all $a,b>0$ with $a\neq b$ if and only if $p\leq 1/2,q\geq 1$, where $S_{p}(a,b)=[(a^{p-1}+b^{p-1})/(a+b)]^{1/(p-2)}$ $(p\neq 2)$ and $S_{2}(a,b)=(a^{a}b^{b})^{1/(a+b)}$ is the $p$th Gini mean of $a$ and $b$. In YSC , Yang et al. proved that inequalities

[TABLE]

hold for all $p\in(1/2,1)$ and $a,b>0$ with $a\neq b$, where $S_{p,q}(a,b)=\left[q(a^{p}-b^{p})/(p(a^{q}-b^{q}))\right]^{1/(p-q)}$ is the Stolarsky mean YCZ of $a$ and $b$.

Very recently, optimal bounds for $AG_{A,Q}(a,b)$ by several convex combinations of their generating means were established. Explicitly, Wang et al. WQC presented the best possible parameters $\alpha_{i},\beta_{i}\ (i=1,2,3)$ such that the double inequalities

[TABLE]

hold for all $a,b>0$ with $a\neq b$, where $Q(a,b)=\sqrt{(a^{2}+b^{2})/2}$ is the quadratic mean of $a$ and $b$.

Let $C(a,b)=(a^{2}+b^{2})/(a+b)$ is the contraharmonic mean, then it is easy to verify that the function $\zeta(x)=C\left[\sqrt{xa^{2}+(1-x)b^{2}},\sqrt{(1-x)a^{2}+xb^{2}}\right]$ is continuous and strictly increasing on $[1/2,1]$. Note that

[TABLE]

Motivated by inequality (1.10) and the results of WQC , it is natural to ask what are the best possible parameters $\alpha_{i},\beta_{i}\ (i=1,2,3)$ and $\alpha_{4},\beta_{4}\in(1/2,1)$ such that the double inequalities

[TABLE]

hold for all $a,b>0$ with $a\neq b$ The main purpose of this paper is to answer this question.

2 Lemmas

In order to prove the desired theorem, we need several lemmas which we present this section.

Lemma 2.1**.**

(see (AVV, , Theorem 1.25)) For $-\infty<a<b<\infty$, let $f,g:[a,b]\rightarrow\mathbb{R}$ be continuous on $[a,b]$, and be differetiable on $(a,b)$, let $g^{\prime}(x)\neq 0$ on (a,b). If $f^{\prime}(x)/g^{\prime}(x)$ is increasing (decreasing) on $(a,b)$, then so are

[TABLE]

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

Lemma 2.2**.**

1. (1)

$[\mathcal{E}(r)-r^{\prime 2}\mathcal{K}(r)]/r^{2}$* is strictly increasing from $(0,1)$ onto $(\pi/4,1)$;* 2. (2)

For each $c\in[1/2,\infty)$, the function $r^{\prime c}\mathcal{K}(r)$ is decreasing from $[0,1)$ onto $(0,\frac{\pi}{2}]$; 3. (3)

$[\mathcal{E}(r)-r^{\prime 2}\mathcal{K}(r)]/[r^{2}\mathcal{K}(r)]$* is strictly decreasing on $(0,1)$;* 4. (4)

$[\mathcal{K}(r)-\mathcal{E}(r)]/r^{2}$* is strictly increasing on $(0,1)$.*

Proof.

Parts (1)-(4) follow from (AVV, , Exercise 3.43 (11) and (46), Theorem 3.21 (1) and (7)). ∎

Lemma 2.3**.**

Let $\delta_{1}=(2+\sqrt{2})\left(1-\pi/[2\mathcal{K}(\sqrt{2}/2)]\right)=0.5216\cdots$ and

[TABLE]

then $f(r)$ is strictly increasing from $(0,\sqrt{2}/2)$ onto $(1/2,\delta_{1})$.

Proof.

Let $f_{1}(r)=1-\pi/[2\mathcal{K}(r)]$ and $f_{2}(r)=1-\sqrt{1-r^{2}}$, then we clearly see that $f_{1}(0)=f_{2}(0)=0,f(r)=f_{1}(r)/f_{2}(r)$ and

[TABLE]

It follows from Lemma 2.2 (1) and (2) that $[\mathcal{E}(r)-r^{\prime 2}\mathcal{K}(r)]/r^{2}$ and $1/\left[r^{\prime 1/2}\mathcal{K}(r)\right]^{2}$ are strictly increasing on $(0,1)$. This conjunction with (2.1) implies that $f^{\prime}_{1}(r)/f^{\prime}_{2}(r)$ is strictly increasing on $(0,1)$.

Therefore, Lemma 2.3 follows immediately from Lemma 2.1 and the limiting values $f(0^{+})=1/2$ and $f(\sqrt{2}/2^{-})=\delta_{1}$. ∎

Lemma 2.4**.**

Let $\delta_{2}=2\log[2\mathcal{K}(\sqrt{2}/2)/\pi]/\log 2=0.4784\cdots$ and

[TABLE]

then $g(r)$ is strictly decreasing from $(0,\sqrt{2}/2)$ onto $(\delta_{2},1/2)$.

Proof.

Let $g_{1}(r)=\log\pi/2-\log\mathcal{K}(r)$ and $g_{2}(r)=\log\sqrt{1-r^{2}}$, then we clearly see that $g_{1}(0)=g_{2}(0)=0,g(r)=g_{1}(r)/g_{2}(r)$ and

[TABLE]

It follows from Lemma 2.2 (3) and (2.2) that $g^{\prime}_{1}(r)/g^{\prime}_{2}(r)$ is strictly increasing on $(0,1)$.

Therefore, Lemma 2.4 follows from Lemma 2.1 and the limiting values $g(0^{+})=1/2$ and $g(\sqrt{2}/2^{-})=\delta_{2}$. ∎

Lemma 2.5**.**

Let $\delta_{3}=(\sqrt{2}+1)\left(\sqrt{2}-[2\mathcal{K}(\sqrt{2}/2)]/\pi\right)=0.5646\cdots$ and

[TABLE]

then $h(r)$ is strictly increasing from $(0,\sqrt{2}/2)$ onto $(1/2,\delta_{3})$.

Proof.

Let $h_{1}(r)=2\sqrt{1-r^{2}}\mathcal{K}(r)/\pi-1$ and $h_{2}(r)=\sqrt{1-r^{2}}-1$, then it is easy to see that $h_{1}(0)=h_{2}(0)=0,h(r)=h_{1}(r)/h_{2}(r)$ and

[TABLE]

Lemma 2.2 (4) and (2.3) lead to the conclusion that $h^{\prime}_{1}(r)/h^{\prime}_{2}(r)$ is strictly increasing on $(0,\sqrt{2}/2)$.

Therefore, Lemma 2.5 follows easily from Lemma 2.1 and the limiting values $h(0^{+})=1/2$ and $h(\sqrt{2}/2^{-})=\delta_{3}$. ∎

Lemma 2.6**.**

The inequality

[TABLE]

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

Proof.

In order to prove this lemma, it suffices to show the inequality

[TABLE]

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

Let $\mu(r)=[2\mathcal{K}(r)/\pi-1]/r^{2}$, $\mu_{1}(r)=2\mathcal{K}(r)/\pi-1$ and $\mu_{2}(r)=r^{2}$, then we clearly see that

[TABLE]

Taking the derivative of $\mu_{1}(r)$ and $\mu_{2}(r)$ yieds

[TABLE]

It follows from (2.6) and Lemma 2.2 (1) together with the monotonicity of $r^{\prime}=\sqrt{1-r^{2}}$ that $\mu^{\prime}_{1}(r)/\mu^{\prime}_{2}(r)$ is strictly increasing on $(0,1)$. This conjunction with (2.5) and Lemma 2.1 implies that $\mu(r)$ is strictly increasing on $(0,1)$.

Therefore, the desired inequality (2.4) follows from $\mu(0^{+})=1/4$ and the monotonicity of $\mu(r)$. ∎

Lemma 2.7**.**

The function $\eta(r)=\left[\pi/\left(2\mathcal{K}(r)\right)+r^{2}/4-1\right]/r^{4}$ is strictly decreasing from $(0,\sqrt{2}/2)$ onto $(-7/2+2\pi/\mathcal{K}(\sqrt{2}/2),-5/64)$.

Proof.

Let $\eta_{1}(r)=\pi/[2\mathcal{K}(r)]+r^{2}/4-1$ and $\eta_{2}(r)=r^{4}$, then it is easy to see that $\eta_{1}(0)=\eta_{2}(0)=0,\eta(r)=\eta_{1}(r)/\eta_{2}(r)$ and

[TABLE]

where $\eta_{11}(r)=r^{2}r^{\prime 2}\mathcal{K}^{2}(r)-\pi[\mathcal{E}(r)-r^{\prime 2}\mathcal{K}(r)]$ and $\eta_{22}(r)=8r^{4}r^{\prime 2}\mathcal{K}^{2}(r)$.

Observe that $\eta_{11}(0)=\eta_{22}(0)=0$. Taking the derivative of $\eta_{11}(r)$ and $\eta_{22}(r)$ yields

[TABLE]

where

[TABLE]

An easy computation leads to

[TABLE]

for $r\in(0,\sqrt{2}/2)$.

It follows from (JDZ, , Corollary 2.7) that $2\mathcal{E}(r)/\pi>1-r^{2}/4-r^{4}/8$ for $r\in(0,\sqrt{2}/2)$. This can be rewritten as

[TABLE]

for $r\in(0,\sqrt{2}/2)$.

Lemma 2.6 and (2.11) lead to the conclusion that

[TABLE]

for $r\in(0,\sqrt{2}/2)$. This conjunction with (2.10) implies that $\zeta_{1}(r)$ is strictly increasing on $(0,\sqrt{2}/2)$. Since $\zeta_{2}(r)$ can be rewritten as $\mathcal{E}(r)-\mathcal{K}(r)+2r^{\prime 2}\mathcal{K}(r)$, we conclude easily from Lemma 2.2 (2) and (4) that $\zeta_{2}(r)$ is strictly decreasing on $(0,\sqrt{2}/2)$.

Moreover, it follows easily from (2.9) that $\zeta_{1}(r)>0$ and $\zeta_{2}(r)>0$ for $r\in(0,\sqrt{2}/2)$. This conjunction with (2.8) together with the monotonicity of $\zeta_{1}(r)$ and $\zeta_{2}(r)$ implies that $\eta^{\prime}_{11}/(r)\eta^{\prime}_{22}(r)$ is strictly decreasing on $(0,\sqrt{2}/2)$.

Therefore, Lemma 2.4 follows immediately from (2.7) and Lemma 2.1 together with the limiting values $\eta(0^{+})=-5/64$ and $\eta(\sqrt{2}/2^{-})=-7/2+2\pi/\mathcal{K}(\sqrt{2}/2)$.

∎

Note that $-7/2+2\pi/\mathcal{K}(\sqrt{2}/2)=-0.111148<-7/64$, then the following corollary follows directly from Lemma 2.7.

Corollary 2.8**.**

The double inequality

[TABLE]

holds for $r\in(0,\sqrt{2}/2)$.

Lemma 2.9**.**

Let $\delta_{4}=4\sqrt{2\mathcal{K}^{2}(\sqrt{2}/2)\left[\pi^{2}-\mathcal{K}^{2}(\sqrt{2}/2)\right]}/\pi=1.8389\cdots$, $\lambda\in(0,2]$ and

[TABLE]

which is defined as in (JDZ, , Lemma 2.8), then the following statements are true:

1. (1)

$\Phi_{\delta_{4}}(r)>1-r^{2}/4$* holds for $r\in(0,33/50)$;* 2. (2)

$\mathcal{K}(r)/\sqrt{r}$* is strictly decreasing on $(0,\sqrt{2}/2)$;* 3. (3)

$\sqrt{r}\Phi_{\delta_{4}}(r)$* is strictly decreasing on $(33/50,\sqrt{2}/2)$.*

Proof.

(1) In order to prove that $\Phi_{\delta_{4}}(r)>1-r^{2}/4$ for $r\in(0,33/50)$, by squaring both sides of the inequality and simplifying, it suffices to show

[TABLE]

holds for $r\in(0,33/50)$.

The difference of both sides squares of (2.12) leads to

[TABLE]

where

[TABLE]

An easy calculation yields

[TABLE]

It follows from (2.14) and (2.15) that $\nu(r)>0$ for $r\in(0,33/50)$. This conjunction with (2.13) completes the proof of Lemma 2.9 (1).

(2) It suffices to determine the sign of the derivate of $\mathcal{K}(r)/\sqrt{r}$. An easy computation yields

[TABLE]

where

[TABLE]

It follows from Lemma 2.2 (1) and the monotonicity of $\mathcal{E}(r)$ that $\xi(r)$ is strictly increasing on $(0,\sqrt{2}/2)$. As a consequence, we obtain

[TABLE]

for $r\in(0,\sqrt{2}/2)$.

Therefore, we conclude from (2.16) and (2.17) that $\mathcal{K}(r)/\sqrt{r}$ is strictly decreasing on $(0,\sqrt{2}/2)$.

(3) Let $\omega_{1}(r)=\sqrt{1+\delta_{4}rr^{\prime}}$ and $\omega_{2}(r)=\sqrt{1-\delta_{4}rr^{\prime}}$, then we clearly see that $\omega_{1}(r)>\omega_{2}(r)>0$ for $r\in(0,\sqrt{2}/2)$ and $\Phi_{\delta_{4}}(r)=[\omega_{1}(r)-\omega_{2}(r)]/(\delta_{4}r)$.

Easy computations lead to

[TABLE]

for $r\in(0,\sqrt{2}/2)$. It follows from (2.18) and (2.19) that $\omega_{1}(r)-\omega_{2}(r)$ and $1/\omega_{1}(r)+1/\omega_{2}(r)$ are strictly increasing on $r\in(0,\sqrt{2}/2)$. Moreover, the monotonicity property of composite function leads to the conculsion that $(1-2r^{2})/\sqrt{1-r^{2}}=2\sqrt{1-r^{2}}-1/\sqrt{1-r^{2}}$ is strictly decreasing on $(0,\sqrt{2}/2)$. These properties imply that

[TABLE]

for $r\in(33/50,\sqrt{2}/2)$.

Therefore, Lemma 2.9 (3) follows directly from (2.20). ∎

3 Main results

Theorem 3.1**.**

The double inequality

[TABLE]

holds for all $a,b>0$ with $a\neq b$ if and only if $\alpha_{1}\geq\delta_{1}=0.5216\cdots$ and $\beta_{1}\leq 1/2$.

Proof.

Since $Q(a,b),C(a,b)$ and $AG(a,b)$ are symmetric and homogeneous of degree 1, without loss of generality, we may assume that $a>b>0$. Let $r=(a-b)/\sqrt{2(a^{2}+b^{2})}\in(0,\sqrt{2}/2)$, then we clearly see from (1.2) and (1.3) together with the definition of $Q(a,b)$ and $C(a,b)$ that

[TABLE]

It follows from (3.1) and (3.2) that

[TABLE]

where $f(r)$ is defined as in Lemma 2.3.

Therefore, Theorem 3.1 follows easily from (3.3) and Lemma 2.3 .

∎

Theorem 3.2**.**

The double inequality

[TABLE]

holds for all $a,b>0$ with $a\neq b$ if and only if $\alpha_{2}\geq 1/2$ and $\beta_{2}\leq\delta_{2}=0.4784\cdots$.

Proof.

Without loss of generality, we assume that $a>b>0$. Let $r=(a-b)/\sqrt{2(a^{2}+b^{2})}\in(0,\sqrt{2}/2)$, then from (3.1) and (3.2) we clearly see that

[TABLE]

where $g(r)$ is defined as in Lemma 2.4.

Therefore, Theorem 3.2 follows directly from (3.4) and Lemma 2.4.

∎

Theorem 3.3**.**

The double inequality

[TABLE]

holds for all $a,b>0$ with $a\neq b$ if and only if $\alpha_{3}\leq 1/2$ and $\beta_{3}\geq\delta_{3}=0.5646\cdots$.

Proof.

In order to prove the double inequality in Theorem 3.3, it suffices to find $\alpha_{3}$ and $\beta_{3}$ such that

[TABLE]

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

Without loss of generality, we assume that $a>b>0$. Let $r=(a-b)/\sqrt{2(a^{2}+b^{2})}\in(0,\sqrt{2}/2)$, then (3.1) and (3.2) lead to

[TABLE]

where $h(r)$ is defined as in Lemma 2.5.

Therefore, Theorem 3.3 follows directly from (3.5), (3.6) and Lemma 2.5. ∎

Theorem 3.4**.**

Let $\alpha_{4},\beta_{4}\in(1/2,1)$, then the double inequality

[TABLE]

holds for all $a,b>0$ with $a\neq b$ if and only if $\alpha_{4}\leq(\sqrt{2}+2)/4$ and $\beta_{4}\geq(\delta_{4}+2)/4=0.9597\cdots$.

Proof.

Since $AG(a,b)$ and $C(a,b)$ are symmetric and homogeneous of degree one, we assume that $a>b>0$. Let $r=(a-b)/\sqrt{2(a^{2}+b^{2})}\in(0,\sqrt{2}/2)$, then (3.2) and the definition of $C(a,b)$ lead to

[TABLE]

where $\Phi_{\lambda}(r)$ is defined as in Lemma 2.9.

It is easy to be verified that $C\left(\sqrt{pa^{2}+(1-p)b^{2}},\sqrt{pb^{2}+(1-p)a^{2}}\right)$ is continuous and strictly increasing on $[1/2,1]$ with respect to $p$ for fixed $a,b>0$ with $a\neq b$.

We divide the proof into three cases.

Case 1. $p_{1}=(\sqrt{2}+2)/4$. We clearly see from (JDZ, , Lemma 2.8 (1)) that

[TABLE]

for $r\in(0,\sqrt{2}/2)$.

It follows from Corollary 2.8 and (3.8) that

[TABLE]

for $r\in(0,\sqrt{2}/2)$.

Therefore, $AG_{Q,C}(a,b)>C\left(\sqrt{p_{1}a^{2}+(1-p_{1})b^{2}},\sqrt{(1-p_{1})a^{2}+p_{1}b^{2}}\right)$ follows from (3.7) and (3.9).

Case 2. $p_{2}=(\delta_{4}+2)/4$. Then from Corollary 2.8 and Lemma 2.9 (1) we clearly see that

[TABLE]

for $r\in(0,33/50)$.

Furthermore, it follows from Lemma 2.9 (2) and (3) that $\mathcal{K}(r)\Phi_{\delta_{4}}(r)=\left[\mathcal{K}(r)/\sqrt{r}\right]\cdot\left[\sqrt{r}\Phi_{\delta_{4}}(r)\right]$ is strictly decreasing on $(33/50,\sqrt{2}/2)$. As a consequence,

[TABLE]

for $r\in(0,33/50)$.

It follows from (3.11) that

[TABLE]

for $r\in(33/50,\sqrt{2}/2)$.

Therefore, $AG_{Q,C}(a,b)<C\left(\sqrt{p_{2}a^{2}+(1-p_{2})b^{2}},\sqrt{(1-p_{2})a^{2}+p_{2}b^{2}}\right)$ follows from (3.7), (3.10) and (3.12).

Case 3. $(\sqrt{2}+2)/4<p_{3}<(\delta_{4}+2)/4$. On the one hand, if $r\rightarrow 0$, then making use of Taylor series yields

[TABLE]

Equations (3.7) and (3.13) lead to the conclusion that there exists small enough $\tau_{1}\in(0,\sqrt{2}/2)$ such that $AG_{Q,C}(a,b)<C\left(\sqrt{p_{3}a^{2}+(1-p_{3})b^{2}},\sqrt{(1-p_{3})a^{2}+p_{3}b^{2}}\right)$ for all $a>b>0$ with $(a-b)/\sqrt{2(a^{2}+b^{2})}\in(0,\tau_{1})$.

On the other hand, it follows from

[TABLE]

for $p\in(1/2,1)$ that

[TABLE]

is strictly decreasing on $(1/2,1)$ with respect to $p$. This implies that

[TABLE]

Equations (3.7) and (3.14) lead to the conclusion that there exists small enough $\tau_{2}\in(0,\sqrt{2}/2)$ such that $AG_{Q,C}(a,b)>C\left(\sqrt{p_{3}a^{2}+(1-p_{3})b^{2}},\sqrt{(1-p_{3})a^{2}+p_{3}b^{2}}\right)$ for all $a>b>0$ with $(a-b)/\sqrt{2(a^{2}+b^{2})}\in(\sqrt{2}/2-\tau_{2},\sqrt{2}/2)$. ∎

4 Applications

In this section, we will present new bounds for the complete elliptic integrals $\mathcal{K}(r)$ and $\mathcal{E}(r)$ on $(0,\sqrt{2}/2)$.

Theorem 4.1 follows from Theorem 3.1, 3.2, 3.3 and 3.4 immediately.

Theorem 4.1**.**

Let $r^{\prime}=\sqrt{1-r^{2}}$ and

[TABLE]

where $\delta_{1},\delta_{2},\delta_{3}$ and $\delta_{4}$ are defined as in Lemmas 2.3, 2.4, 2.5 and 2.9, respectively. Then the double inequality

[TABLE]

holds for all $r\in(0,\sqrt{2}/2)$.

Observed that the double inequality

[TABLE]

for $0<r<1$ was presented in AVV . It follows easily from (4.1) that

[TABLE]

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

The following theorem is derived from Theorem 4.1 and (4.2) immediately.

Theorem 4.2**.**

Suppose that $m(r),M(r)$ are defined as in Theorem 4.1, then the double inequality

[TABLE]

holds for all $r\in(0,\sqrt{2}/2)$.

Acknowledgement

I am grateful to Professor Xingjiang Lu, Professor Shoufeng Shen, Doctor Zhengchao Ji for helpful conversations.

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