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  浙江大学学报(理学版)  2018, Vol. 45 Issue (6): 665-672  DOI:10.3785/j.issn.1008-9497.2018.06.004
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引用本文 [复制中英文]

张帆, 杨月英, 钱伟茂. Sándor-Yang平均关于经典平均凸组合的确界[J]. 浙江大学学报(理学版), 2018, 45(6): 665-672. DOI: 10.3785/j.issn.1008-9497.2018.06.004.
[复制中文]
ZHANG Fan, YANG Yueying, QIAN Weimao. Sharp bounds for Sándor-Yang means in terms of the convex combination of classical bivariate means[J]. Journal of Zhejiang University(Science Edition), 2018, 45(6): 665-672. DOI: 10.3785/j.issn.1008-9497.2018.06.004.
[复制英文]

基金项目

浙江省自然科学基金资助项目(LY13A010004);浙江广播电视大学科学研究课题(XKT-17G26);湖州职业技术学院校教改课题(2016xj26)

作者简介

张帆(1977-), ORCID:http://orcid.org/0000-0003-0689-946X, 男, 硕士, 讲师, 主要从事解析不等式研究

通信作者

杨月英, ORCID:http://orcid.org/0000-0003-3644-460X, E-mail: yyy1008hz@163.com

文章历史

收稿日期:2018-02-28
Sándor-Yang平均关于经典平均凸组合的确界
张帆1 , 杨月英2 , 钱伟茂3     
1. 湖州职业技术学院 建筑工程学院, 浙江 湖州 313000;
2. 湖州职业技术学院 机电与汽车工程学院, 浙江 湖州 313000;
3. 湖州广播电视大学 远程教育学院, 浙江 湖州 313000
摘要: 应用实分析方法,研究Sándor-Yang平均RGQ关于算术平均A与几何平均G(或调和平均H)凸组合和Sándor-Yang平均RQG与算术平均A与二次平均Q(或反调和平均C)凸组合的序关系,以及两Sándor-Yang平均RGQRQG与几何平均G、算术平均A、二次平均Q的序关系,得到了4个精确双向不等式和一个新的不等式链.
关键词: Sándor-Yang平均    经典平均    不等式    
Sharp bounds for Sándor-Yang means in terms of the convex combination of classical bivariate means
ZHANG Fan1, YANG Yueying2, QIAN Weimao3     
1. School of Architecture Engineering, Huzhou Vocational & Technical College, Huzhou 313000, Zhejiang Province, China;
2. Mechanic Electronic and Automobile Egineering College, Huzhou Vocational & Technical College, Huzhou 313000, Zhejiang Province, China;
3. School of Distance Education, Huzhou Broadcast and TV University, Huzhou 313000, Zhejiang Province, China
Abstract: This article presents several sharp bounds for the Sándor-Yang mean RGQ in terms of the convex combination of arithmetic mean A and geometric mean G(arithmetic mean A and harmonic mean H), the Sándor-Yang mean RQG in terms of the convex combination of quadratic mean Q and arithmetic mean A(contra-harmonic mean C and arithmetic mean A). A new chain of inequalities for the geometric mean G, arithmetic mean A, quadratic mean Q and two Sándor-Yang means RGQ and RQG are then derived.
Key Words: Sándor-Yang mean    classical mean    inequality    
0 引言

2013年, YANG[1]介绍了4个Sándor-Yang平均:

$ \left\{ \begin{array}{l} {R_{AQ}}\left( {a,b} \right) = Q\left( {a,b} \right){{\rm{e}}^{A\left( {a,b} \right)/T\left( {a,b} \right) - 1}},\\ {R_{QA}}\left( {a,b} \right) = A\left( {a,b} \right){{\rm{e}}^{Q\left( {a,b} \right)/\text{NS}\left( {a,b} \right) - 1}},\\ {R_{GQ}}\left( {a,b} \right) = Q\left( {a,b} \right){{\rm{e}}^{G\left( {a,b} \right)/U\left( {a,b} \right) - 1}},\\ {R_{QG}}\left( {a,b} \right) = G\left( {a,b} \right){{\rm{e}}^{Q\left( {a,b} \right)/V\left( {a,b} \right) - 1}}, \end{array} \right. $ (1)

其中,

$ G\left( {a,b} \right) = \sqrt {ab} ,A\left( {a,b} \right) = \frac{{a + b}}{2},Q\left( {a,b} \right) = \sqrt {\frac{{{a^2} + {b^2}}}{2}} , $ (2)
$ \left\{ \begin{array}{l} {\rm{NS}}\left( {a,b} \right) = \frac{{a - b}}{{2\sin \;{h^{ - 1}}\left[ {\left( {a - b} \right)/\left( {a + b} \right)} \right]}},\\ T\left( {a,b} \right) = \frac{{a - b}}{{2{{\tan }^{ - 1}}\left[ {\left( {a - b} \right)/\left( {a + b} \right)} \right]}},\\ U\left( {a,b} \right) = \frac{{a - b}}{{\sqrt 2 {{\tan }^{ - 1}}\left[ {\left( {a - b} \right)/\sqrt {2ab} } \right]}},\\ V\left( {a,b} \right) = \frac{{a - b}}{{\sqrt 2 \sin \;{h^{ - 1}}\left[ {\left( {a - b} \right)/\sqrt {2ab} } \right]}}, \end{array} \right. $ (3)

分别为2个正数ab的几何平均、算术平均、二次平均, Neuman-Sándor平均[2-3]和第二类Seiffert平均[4], 第一类Yang平均和第二类Yang平均[5-7].

pRa, b > 0, 则调和平均H(a, b), 反调和平均C(a, b)和p阶幂平均Mp(a, b)[8]分别定义为

$ \begin{array}{*{20}{c}} {H\left( {a,b} \right) = \frac{{2ab}}{{a + b}},C\left( {a,b} \right) = \frac{{{a^2} + {b^2}}}{{a + b}},}\\ {{M_p}\left( {a,b} \right) = {{\left( {\frac{{{a^p} + {b^p}}}{2}} \right)}^{1/p}},p \ne 0,}\\ {{M_0}\left( {a,b} \right) = \sqrt {ab} ,} \end{array} $ (4)

并且p阶幂平均Mp(a, b)对于固定的a, b > 0和ab关于pR是连续和严格单调上升的.

不等式

$ \begin{array}{l} H\left( {a,b} \right) = {M_{ - 1}}\left( {a,b} \right) < G\left( {a,b} \right) = {M_0}\left( {a,b} \right) < \\ \;\;\;\;\;\;\;L\left( {a,b} \right) < P\left( {a,b} \right) < A\left( {a,b} \right) = {M_1}\left( {a,b} \right) < \\ \;\;\;\;\;\;\;{\rm{NS}}\left( {a,b} \right) < T\left( {a,b} \right) < Q\left( {a,b} \right) = {M_2}\left( {a,b} \right) < \\ \;\;\;\;\;\;\;C\left( {a,b} \right) \end{array} $

$ \begin{array}{l} L\left( {a,b} \right) < V\left( {a,b} \right) < P\left( {a,b} \right) < U\left( {a,b} \right) < \\ \;\;\;\;\;\;{\rm{NS}}\left( {a,b} \right) < T\left( {a,b} \right) \end{array} $

对所有a, b > 0且ab均成立, 其中,P(a, b)=(a-b)/[2sin-1((a-b)/(a+b))]和L(a, b)=(b-a)/(ln b-ln a)分别为2个正数ab的第一类Seiffert平均和对数平均.

最近, Sándor-Yang平均关于其他二元平均或其组合的比较研究取得了一定进展.国内外学者证明了在特殊情形下一些涉及Sándor-Yang平均的重要不等式[1, 9-14].

ZHAO等[11]证明了α≤ln 2/(1+ln 2-$ \sqrt 2$ln(1+$ \sqrt 2$)), β≥5/3, λ≤4ln 2/(4+2ln 2-π)和μ≥4/3是使得双向不等式

$ {M_a}\left( {a,b} \right) < {R_{QA}}\left( {a,b} \right) < {M_\beta }\left( {a,b} \right), $
$ {M_\lambda }\left( {a,b} \right) < {R_{AQ}}\left( {a,b} \right) < {M_\mu }\left( {a,b} \right) $

对所有a, b > 0且ab成立的最佳参数.

徐会作[13]证明了双向不等式

$ \begin{array}{*{20}{c}} {{\alpha _1}Q\left( {a,b} \right) + \left( {1 - {\alpha _1}} \right)A\left( {a,b} \right) < {R_{AQ}}\left( {a,b} \right) < }\\ {{\beta _1}Q\left( {a,b} \right) + \left( {1 - {\beta _1}} \right)A\left( {a,b} \right),} \end{array} $
$ \begin{array}{*{20}{c}} {{\alpha _2}Q\left( {a,b} \right) + \left( {1 - {\alpha _2}} \right)A\left( {a,b} \right) < {R_{QA}}\left( {a,b} \right) < }\\ {{\beta _2}Q\left( {a,b} \right) + \left( {1 - {\beta _2}} \right)A\left( {a,b} \right),} \end{array} $
$ \begin{array}{*{20}{c}} {{\alpha _3}C\left( {a,b} \right) + \left( {1 - {\alpha _3}} \right)A\left( {a,b} \right) < {R_{AQ}}\left( {a,b} \right) < }\\ {{\beta _3}C\left( {a,b} \right) + \left( {1 - {\beta _3}} \right)A\left( {a,b} \right),} \end{array} $
$ \begin{array}{*{20}{c}} {{\alpha _4}C\left( {a,b} \right) + \left( {1 - {\alpha _4}} \right)A\left( {a,b} \right) < {R_{QA}}\left( {a,b} \right) < }\\ {{\beta _4}C\left( {a,b} \right) + \left( {1 - {\beta _4}} \right)A\left( {a,b} \right)} \end{array} $

对所有a, b > 0且ab成立当且仅当α1≤1/3,

$ \begin{array}{*{20}{c}} {{\beta _1} \ge \left( {\sqrt 2 {{\rm{e}}^{\frac{{\rm{\pi }}}{4} - 1}} - 1} \right)/\left( {\sqrt 2 - 1} \right) = 0.340\;5 \cdots ,}\\ {{\alpha _2} \le 2/3,} \end{array} $
$ {\beta _2} \ge \left( {1 + \sqrt 2 } \right)\left[ {{{\left( {1 + \sqrt 2 } \right)}^{\sqrt 2 }} - {\rm{e}}} \right]/{\rm{e = 0}}{\rm{.674}}\;{\rm{7}} \cdots , $
$ {\alpha _3} \le \sqrt 2 {{\rm{e}}^{\frac{{\rm{\pi }}}{4} - 1}} - 1 = 0.141\;0 \cdots ,{\beta _3} \ge 1/6, $
$ {\alpha _4} \le {\left( {1 + \sqrt 2 } \right)^{\sqrt 2 }}/{\rm{e}} - 1 = 0.279\;4 \cdots ,{\beta _4} \ge 1/3. $

本文发现并证明了最佳参数α1, α2, α3, α4, β1, β2, β3, β4∈(0, 1), 双向不等式

$ \begin{array}{*{20}{c}} {{\alpha _1}A\left( {a,b} \right) + \left( {1 - {\alpha _1}} \right)H\left( {a,b} \right) < {R_{GQ}}\left( {a,b} \right) < }\\ {{\beta _1}A\left( {a,b} \right) + \left( {1 - {\beta _1}} \right)H\left( {a,b} \right),} \end{array} $
$ \begin{array}{*{20}{c}} {{\alpha _2}A\left( {a,b} \right) + \left( {1 - {\alpha _2}} \right)G\left( {a,b} \right) < {R_{GQ}}\left( {a,b} \right) < }\\ {{\beta _2}A\left( {a,b} \right) + \left( {1 - {\beta _2}} \right)G\left( {a,b} \right),} \end{array} $
$ \begin{array}{*{20}{c}} {{\alpha _3}Q\left( {a,b} \right) + \left( {1 - {\alpha _3}} \right)A\left( {a,b} \right) < {R_{QG}}\left( {a,b} \right) < }\\ {{\beta _3}Q\left( {a,b} \right) + \left( {1 - {\beta _3}} \right)A\left( {a,b} \right),} \end{array} $
$ \begin{array}{*{20}{c}} {{\alpha _4}C\left( {a,b} \right) + \left( {1 - {\alpha _4}} \right)A\left( {a,b} \right) < {R_{QG}}\left( {a,b} \right) < }\\ {{\beta _4}C\left( {a,b} \right) + \left( {1 - {\beta _4}} \right)A\left( {a,b} \right)} \end{array} $

对所有a, b > 0且ab均成立, 并且推得一个新的不等式链:

$ G\left( {a,b} \right) < {R_{GQ}}\left( {a,b} \right) < A\left( {a,b} \right) < {R_{QG}}\left( {a,b} \right) < Q\left( {a,b} \right). $
1 引理

为证明本文的主要结论, 需以下引理:

引理1  设p∈(0, 1),

$ \begin{array}{l} f\left( x \right) = \frac{{\sqrt {2\left( {{x^2} - 1} \right)} \left[ {\left( {2 - p} \right){x^2} + p - 1} \right]}}{{{x^2}\left( {p{x^2} + 1 - p} \right)}} - \\ \;\;\;\;\;\;\;\;\;{\tan ^{ - 1}}\left( {\sqrt {2\left( {{x^2} - 1} \right)} } \right). \end{array} $

(1) 若p=5/6, 则对所有x∈(1, +∞)有f(x) < 0;

(2) 若p=$ \sqrt 2 $/e, 则存在λ∈(1, +∞)使得当x∈(1, λ)时有f(x) > 0,当x∈(λ, +∞)时有f(x) < 0.

证明  经简单计算得

$ f\left( {{1^ + }} \right) = 0,\mathop {\lim }\limits_{x \to + \infty } f\left( x \right) = - \frac{{\rm{\pi }}}{2}, $ (5)
$ f'\left( x \right) = - \frac{{\sqrt {2\left( {{x^2} - 1} \right)} }}{{{x^3}\left( {2{x^2} - 1} \right){{\left( {p{x^2} + 1 - p} \right)}^2}}}{f_1}\left( x \right), $ (6)

其中,

$ \begin{array}{l} {f_1}\left( x \right) = p\left( {4 - p} \right){x^6} + 2\left( {3{p^2} - 2p - 2} \right){x^4} - \\ \;\;\;\;\;\;\;\;\;\;\;\left( {7{p^2} - 10p + 3} \right){x^2} + 2{\left( {1 - p} \right)^2}. \end{array} $ (7)

下面分2种情形讨论:

(1) 当p=5/6时, 式(7)变为

$ \begin{array}{l} {f_1}\left( x \right) = \frac{1}{{36}}\left( {{x^2} - 1} \right)\left( {95{x^4} - 19{x^2} - 2} \right) > \\ \frac{{37}}{{18}}{x^4}\left( {{x^2} - 1} \right) > 0 \end{array} $ (8)

对所有x∈(1, +∞)成立.

所以, 由式(5)、(6)和(8)易得对所有x∈(1, +∞)有f(x) < 0.

(2) 当p=$ \sqrt 2 $/e时, 由式(7)和数值计算得到

$ 6p - 5 = - 1.878\;4 \cdots < 0, $ (9)
$ 2{p^2} + 14p - 11 = - 3.175\;0 \cdots < 0, $ (10)
$ 3{p^2} + 8p - 4 = 0.974\;0 \cdots > 0, $ (11)

式(7)和式(9)~(11)经简单计算可得

$ {f_1}\left( 1 \right) = 6p - 5 < 0,\mathop {\lim }\limits_{x \to + \infty } {f_1}\left( x \right) = + \infty , $ (12)
$ {{f'}_1}\left( x \right) = 2x{f_2}\left( x \right), $ (13)

其中,对所有x∈(0, +∞),

$ \begin{array}{l} {f_2}\left( x \right) = 3p\left( {4 - p} \right){x^4} + 4\left( {3{p^2} - 2p - 2} \right){x^2} - \\ \;\;\;\;\;\;\;\;\;\;\;\;\left( {7{p^2} - 10p + 3} \right),\\ \;\;\;\;\;\;\;\;\;\;\;\;{f_2}\left( 1 \right) = 2{p^2} + 14p - 11 < 0,\\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\mathop {\lim }\limits_{x \to + \infty } {f_2}\left( x \right) = + \infty , \end{array} $ (14)
$ \begin{array}{l} {{f'}_2}\left( x \right) = 12p\left( {4 - p} \right){x^3} + 8\left( {3{p^2} - 2p - 2} \right)x > \\ \;\;\;\;\;\;\;\;\;\;\;\;4\left( {3{p^2} + 8p - 4} \right)x > 0. \end{array} $ (15)

由式(14)和(15)可得,存在λ0∈(1, +∞),使得当x∈(1, λ0)时f2(x) < 0,当x∈(λ0, +∞)时f2(x) > 0.

由式(12)和(13)及当x∈(1, λ0)时f2(x) < 0可知,f1(x) < 0.由式(13)和当x∈(λ0, +∞)时f2(x) > 0得到函数f1(x)在区间(λ0, +∞)上严格单调上升;则由式(12)和f1(λ0) < 0及函数f1(x)在区间(λ0, +∞)上的单调性, 可得存在λ1 > λ0,使得当x∈(λ0, λ1)时f1(x) < 0,当x∈(λ1, +∞)时f1(x) > 0.

下面分2种情形讨论:

情形1  x∈(1, λ1].易由式(5)和(6)及在区间(1, λ1)上f1(x) < 0得到f(x) > 0.

情形2  x∈(λ1, +∞).由式(6)和f1(x) > 0,得到函数f(x)在区间(λ1, +∞)上严格单调下降.由式(5)和f(λ1) > 0及函数f(x)在区间(λ1, +∞)上的单调性,可得存在λ > λ1 > λ0∈(1, +∞),使得当x∈(λ1, λ)时f(x) > 0,当x∈(λ, +∞)时f(x) < 0.

引理2  设p∈(0, 1),

$ \begin{array}{*{20}{c}} {g\left( x \right) = \frac{{\sqrt {2\left( {{x^2} - 1} \right)} \left[ {\left( {1 - p} \right)x + p} \right]}}{{x\left( {px + 1 - p} \right)}} - }\\ {{{\tan }^{ - 1}}\left( {\sqrt {2\left( {{x^2} - 1} \right)} } \right).} \end{array} $

(1) 若p=2/3, 则对所有x∈(1, +∞)有g(x) < 0;

(2) 若p=$ \sqrt 2 $/e, 则存在μ∈(1, +∞),使得当x∈(1, μ)时有g(x) > 0,当x∈(μ, +∞)时有g(x) < 0.

证明  经简单计算可得

$ g\left( {{1^ + }} \right) = 0,\mathop {\lim }\limits_{x \to + \infty } g\left( x \right) = \frac{{\sqrt 2 }}{p} - \sqrt 2 - \frac{{\rm{\pi }}}{2}, $ (16)
$ g'\left( x \right) = \frac{{\sqrt 2 \left( {{x^2} - 1} \right)}}{{{x^2}\sqrt {{x^2} - 1} \left( {2{x^2} - 1} \right){{\left( {px + 1 - p} \right)}^2}}}{g_1}\left( x \right), $ (17)

其中,

$ {g_1}\left( x \right) = - \left( {{p^2} + 4p - 2} \right){x^3} + 2{p^2}x + p\left( {1 - p} \right). $ (18)

下面分2种情形讨论:

(1) 当p=2/3时, 式(18)变为

$ {g_1}\left( x \right) = - \frac{2}{9}\left( {x - 1} \right)\left( {5{x^2} + 5x + 1} \right) < 0, $ (19)

且对所有x∈(1, +∞)成立.

所以, 由式(16)、(17)和(19)易得, 对所有x∈(1, +∞)有g(x) < 0.

(2) 当p=$ \sqrt 2 $/e时, 则由式(18)可得

$ \begin{array}{*{20}{c}} {{g_1}\left( 1 \right) = 2 - 3p = 0.439\;2 \cdots > 0,}\\ {\mathop {\lim }\limits_{x \to + \infty } {g_1}\left( x \right) = - \infty ,} \end{array} $ (20)
$ \begin{array}{l} {{g'}_1}\left( x \right) = - 3\left( {{p^2} + 4p - 2} \right){x^2} + 2{p^2} < \\ \;\;\;\;\;\;\; - {p^2} - 12p + 6 = - 0.513\;7 \cdots < 0 \end{array} $ (21)

对所有x∈(1, +∞)…成立.

由式(20)和(21)可知,存在μ0∈(1, +∞),使得当x∈(1, μ0)时有g1(x) > 0,当x∈(μ0, +∞)时有g1(x) < 0.

下面分2种情形证明:

情形1  x∈(1, μ0].根据式(16)和(17)及在区间(1, μ0)上g1(x) > 0,可知g(x) > 0.

情形2  x∈(μ0, +∞).由式(17)和在区间(μ0, +∞)上g1(x) < 0得到函数g(x)在区间[μ0, +∞)上严格单调下降.注意到式(16)变成

$ \mathop {\lim }\limits_{x \to + \infty } g\left( x \right) = {\rm{e}} - \sqrt 2 - \frac{{\rm{\pi }}}{2} < 0. $ (22)

由不等式(22)和g(μ0) > 0及函数g(x)在区间[μ0, +∞)上的单调性, 可得存在μ∈(μ0, +∞)$ \subset $(1, +∞),使得当x∈(μ0, μ)时有g(x) > 0,当x∈(μ, +∞)时有g(x) < 0.

证毕.

引理3  设p∈(0, 1),

$ \begin{array}{l} h\left( x \right) = \frac{{\sqrt {2\left( {{x^2} - 1} \right)} \left[ {\left( {1 - p} \right)x + p} \right]}}{{px + 1 - p}} - \\ \;\;\;\;\;\;\sin \;{h^{ - 1}}\left( {\sqrt {\frac{{2\left( {{x^2} - 1} \right)}}{{2 - {x^2}}}} } \right). \end{array} $

(1) 若p=1/3, 则对所有x∈(1, $ \sqrt 2 $)有h(x) < 0;

(2) 若p=(2$ \sqrt 2 $/e-1)/($ \sqrt 2 $-1), 则存在σ∈(1, $ \sqrt 2 $),使得当x∈(1, σ)时有h(x) > 0,当x∈(σ, $ \sqrt 2 $)时有h(x) < 0.

证明  经简单计算可得

$ h\left( {{1^ + }} \right) = 0,h\left( {\sqrt 2 } \right) = - \infty , $ (23)
$ h'\left( x \right) = \frac{{\sqrt {2\left( {{x^2} - 1} \right)} }}{{\left( {2 - {x^2}} \right){{\left( {px + 1 - p} \right)}^2}}}{h_1}\left( x \right), $ (24)

其中,

$ \begin{array}{l} {h_1}\left( x \right) = - p\left( {1 - p} \right){x^3} - 2{\left( {1 - p} \right)^2}{x^2} + \\ \;\;\;\;\;\;\;\;\;\;\;{p^2} - 6p + 3. \end{array} $ (25)

下面分2种情形讨论:

(1) 当p=1/3时, 式(25)变为

$ {h_1}\left( x \right) = - \frac{2}{9}\left( {x - 1} \right)\left( {{x^2} + 5x + 5} \right) < 0, $ (26)

且对所有x∈(1, $ \sqrt 2 $)成立.

所以, 由式(23)、(24)和(26)易得对所有x∈(1, $ \sqrt 2 $)有h(x) < 0.

(2) 当p=(2$ \sqrt 2 $/e-1)/($ \sqrt 2 $-1)时, 由式(25)可得

$ \begin{array}{*{20}{c}} {{h_1}\left( 1 \right) = 1 - 3p = 0.706\;5 \cdots > 0,}\\ {{h_1}\left( {\sqrt 2 } \right) = - 8/{{\rm{e}}^2} < 0,} \end{array} $ (27)
$ {{h'}_1}\left( x \right) = - \left( {1 - p} \right)x\left[ {3px + 4\left( {1 - p} \right)} \right] < 0 $ (28)

对所有x∈(0, +∞)成立.

由不等式(27)和(28), 可得存在σ0∈(1, $ \sqrt 2 $),使得当x∈(1, σ0)时有h1(x) > 0,当x∈(σ0, $ \sqrt 2 $)时有h1(x) < 0.

下面分2种情形证明:

情形1  x∈(1, σ0].由式(23)和(24)及在区间(1, σ0)上h1(x) > 0,可知h(x) > 0.

情形2  x∈(σ0, $ \sqrt 2 $).由式(24)和在区间(σ0, $ \sqrt 2 $)上h1(x) < 0可知,函数h(x)在区间[σ0, $ \sqrt 2 $)上严格单调下降.

由式(23)和h(σ0) > 0及函数h(x)在区间[σ0, $ \sqrt 2 $)上的单调性, 可知存在σ∈(σ0, $ \sqrt 2 $)$ \subset $(1, $ \sqrt 2 $),使得当x∈(σ0, σ)时有h(x) > 0,当x∈(σ, $ \sqrt 2 $)时有h(x) < 0.

引理4  设p∈(0, 1),

$ \begin{array}{l} k\left( x \right) = \frac{{x\sqrt {2\left( {{x^2} - 1} \right)} \left( { - p{x^2} + p + 1} \right)}}{{p{x^2} - p + 1}} - \\ \;\;\;\;\;\;\sin \;{h^{ - 1}}\left( {\sqrt {\frac{{2\left( {{x^2} - 1} \right)}}{{2 - {x^2}}}} } \right). \end{array} $

(1) 若p=1/6, 则对所有x∈(1, $ \sqrt 2 $)有

$ k\left( x \right) < 0; $

(2) 若p=2$ \sqrt 2 $/e-1, 则存在τ∈(1, $ \sqrt 2 $),使得当x∈(1, τ)时有k(x) > 0,当x∈(τ, $ \sqrt 2 $)时有k(x) < 0.

证明  经简单计算可得

$ k\left( {{1^ + }} \right) = 0,k\left( {\sqrt 2 } \right) = - \infty , $ (29)
$ k'\left( x \right) = \frac{{\sqrt {2\left( {{x^2} - 1} \right)} }}{{\left( {2 - {x^2}} \right){{\left( {p{x^2} - p + 1} \right)}^2}}}{k_1}\left( x \right), $ (30)

其中,

$ \begin{array}{l} {k_1}\left( x \right) = 2{p^2}{x^6} + p\left( {4 - 7p} \right){x^4} + \\ \;\;\;\;\;\;2\left( {3{p^2} - 4p - 1} \right){x^2} - \left( {{p^2} + 2p - 3} \right). \end{array} $ (31)

下面分2种情形讨论:

(1) 当p=1/6时, 式(31)变为

$ \begin{array}{l} {k_1}\left( x \right) = \frac{1}{{36}}\left( {{x^2} - 1} \right)\left( {2{x^4} + 19{x^2} - 95} \right) < \\ \;\;\;\;\;\;\;\;\;\;\;\; - \frac{{49}}{{36}}\left( {{x^2} - 1} \right) < 0, \end{array} $ (32)

且对所有x∈(1, $ \sqrt 2 $)成立.

所以, 由式(29)、(30)和(32)易得对所有x∈(1, $ \sqrt 2 $)有k(x) < 0.

(2) 当p=2$ \sqrt 2 $/e-1时, 由式(31)并经数值计算可得

$ 4 - 7p = 3.716\;3 \cdots > 0, $ (33)
$ {p^2} + 4p - 1 = - 0.8362 \cdots < 0, $ (34)
$ \begin{array}{*{20}{c}} {{k_1}\left( 1 \right) = 1 - 6p = 0.756\;8 \cdots > 0,}\\ {{k_1}\left( {\sqrt 2 } \right) = - {{\left( {1 + p} \right)}^2} < 0,} \end{array} $ (35)
$ \begin{array}{*{20}{c}} {{{k'}_1}\left( x \right) = 12{p^2}{x^5} + 4p\left( {4 - 7p} \right){x^3} + }\\ {4\left( {3{p^2} - 4p - 1} \right)x,} \end{array} $ (36)

由不等式(33)、(34)和等式(36)可知

$ {{k'}_1}\left( x \right) < 4\left( {{p^2} + 4p - 1} \right)x < 0. $ (37)

由不等式(35)和(37)可知,存在τ0∈(1, $ \sqrt 2 $),使得当x∈(1, τ0)时有k1(x) > 0,当x∈(τ0, $ \sqrt 2 $)时有k1(x) < 0.

下面分2种情形证明:

情形1  x∈(1, τ0].由式(29)和(30)及在区间(1, τ0)上k1(x) > 0,可知k(x) > 0.

情形2  x∈(τ0, $ \sqrt 2 $).由式(30)和在区间(τ0, $ \sqrt 2 $)上k1(x) < 0可得函数k(x)在区间[τ0, $ \sqrt 2 $)上严格单调下降.

由式(29)和k(τ0) > 0及函数k(x)在区间[τ0, $ \sqrt 2 $)上单调下降可知,存在τ>τ0∈(1, $ \sqrt 2 $),使得当x∈(τ0, τ)时有k(x) > 0,当x∈(τ, $ \sqrt 2 $)时有k(x) < 0.

2 主要结果

定理1  双向不等式

$ \begin{array}{l} {\alpha _1}A\left( {a,b} \right) + \left( {1 - {\alpha _1}} \right)H\left( {a,b} \right) < {R_{GQ}}\left( {a,b} \right) < \\ \;\;\;\;\;\;\;{\beta _1}A\left( {a,b} \right) + \left( {1 - {\beta _1}} \right)H\left( {a,b} \right) \end{array} $

对所有a, b > 0且ab成立当且仅当α1$ \sqrt 2 $/e=0.520 2…,β1≥5/6.

证明  由调和平均H(a, b)、算术平均A(a, b)和Sándor-Yang平均RGQ(a, b)是对称的且齐次系数为1, 不失一般性, 假设a > b > 0.设p∈(0, 1),x=(a+b)/(2$ \sqrt {ab} $)∈(1, +∞).则由式(1)~(3)可得

$ \begin{array}{*{20}{c}} {H\left( {a,b} \right) = G\left( {a,b} \right)/x,}\\ {Q\left( {a,b} \right) = G\left( {a,b} \right)\sqrt {2{x^2} - 1} ,} \end{array} $ (38)
$ {R_{GQ}}\left( {a,b} \right) = G\left( {a,b} \right)\sqrt {2{x^2} - 1} {{\rm{e}}^{{{\tan }^{ - 1}}}}^{\left( {\sqrt {2\left( {{x^2} - 1} \right)} } \right)/\sqrt {2\left( {{x^2} - 1} \right)} - 1}. $ (39)

由式(38)和(39)有

$ \begin{array}{l} \frac{{{R_{GQ}}\left( {a,b} \right) - H\left( {a,b} \right)}}{{A\left( {a,b} \right) - H\left( {a,b} \right)}} = \\ \;\;\;\;\;\;\frac{{\sqrt {2{x^2} - 1} {{\rm{e}}^{{{\tan }^{ - 1}}}}^{\left( {\sqrt {2\left( {{x^2} - 1} \right)} } \right)/\sqrt {2\left( {{x^2} - 1} \right)} - 1} - 1/x}}{{x - 1/x}}, \end{array} $ (40)
$ \mathop {\lim }\limits_{x \to {1^ + }} \frac{{\sqrt {2{x^2} - 1} {{\rm{e}}^{{{\tan }^{ - 1}}}}^{\left( {\sqrt {2\left( {{x^2} - 1} \right)} } \right)/\sqrt {2\left( {{x^2} - 1} \right)} - 1} - 1/x}}{{x - 1/x}} = \frac{5}{6}, $ (41)
$ \begin{array}{l} \mathop {\lim }\limits_{x \to + \infty } \frac{{\sqrt {2{x^2} - 1} {{\rm{e}}^{{{\tan }^{ - 1}}}}^{\left( {\sqrt {2\left( {{x^2} - 1} \right)} } \right)/\sqrt {2\left( {{x^2} - 1} \right)} - 1} - 1/x}}{{x - 1/x}} = \\ \;\;\;\;\;\;\frac{{\sqrt 2 }}{{\rm{e}}} = 0.520\;2 \cdots , \end{array} $ (42)
$ \begin{array}{l} \log \left[ {{R_{GQ}}\left( {a,b} \right)} \right] - \log \left[ {pA\left( {a,b} \right) + \left( {1 - p} \right)H\left( {a,b} \right)} \right] = \\ \;\;\;\;\;\;\frac{1}{2}\ln \left( {2{x^2} - 1} \right) + \frac{{{{\tan }^{ - 1}}\left( {\sqrt {2\left( {{x^2} - 1} \right)} } \right)}}{{\sqrt {2\left( {{x^2} - 1} \right)} }} - \\ \;\;\;\;\;\;\ln \left( {\frac{{p{x^2} + 1 - p}}{x}} \right) - 1. \end{array} $ (43)

$ \begin{array}{l} F\left( x \right) = \frac{1}{2}\ln \left( {2{x^2} - 1} \right) + \frac{{{{\tan }^{ - 1}}\left( {\sqrt {2\left( {{x^2} - 1} \right)} } \right)}}{{\sqrt {2\left( {{x^2} - 1} \right)} }} - \\ \;\;\;\;\;\;\ln \left( {\frac{{p{x^2} + 1 - p}}{x}} \right) - 1, \end{array} $ (44)

经简单计算可得

$ F\left( {{1^ + }} \right) = 0, $ (45)
$ \mathop {\lim }\limits_{x \to + \infty } F\left( x \right) = \frac{1}{2}\ln 2 - \ln p - 1, $ (46)
$ F'\left( x \right) = \frac{x}{{\sqrt 2 {{\left( {{x^2} - 1} \right)}^{3/2}}}}f\left( x \right), $ (47)

其中f(x)的定义见引理1.

下面分2种情形证明:

情形1  p=5/6.由式(43)~(45)、(47)及引理1(1)得

$ {R_{GQ}}\left( {a,b} \right) < \frac{5}{6}A\left( {a,b} \right) + \frac{1}{6}H\left( {a,b} \right). $ (48)

情形2  p=$ \sqrt 2 $/e.由式(47)和引理1(2)可得,存在λ∈(1, +∞),使得函数F(x)在区间(1, λ]严格单调上升,在区间[λ, +∞)严格单调下降.注意到式(46)变为

$ \mathop {\lim }\limits_{x \to + \infty } F\left( x \right) = 0, $ (49)

由式(43)~(45)、(49)及函数F(x)的分段单调性可知

$ {R_{GQ}}\left( {a,b} \right) > \frac{{\sqrt 2 }}{{\rm{e}}}A\left( {a,b} \right) + \left( {1 - \frac{{\sqrt 2 }}{{\rm{e}}}} \right)H\left( {a,b} \right). $ (50)

所以, 定理1可由等式(48)和(50)结合下面的陈述得到.

(1) 若p < 5/6, 则由式(40)和(41)可得存在足够小的δ1 > 0,对所有a > b > 0且(a+b)/(2$ \sqrt {ab} $)∈(1, 1+δ1),有

$ {R_{GQ}}\left( {a,b} \right) > pA\left( {a,b} \right) + \left( {1 - p} \right)H\left( {a,b} \right). $

(2) 若p > $ \sqrt 2 $/e, 则由式(40)和(42)可得,存在足够大的M1 > 1,对所有a > b > 0且(a+b)/(2$ \sqrt {ab} $)∈(M1, +∞),有

$ {R_{GQ}}\left( {a,b} \right) < pQ\left( {a,b} \right) + \left( {1 - p} \right)H\left( {a,b} \right). $

定理2  双向不等式

$ \begin{array}{*{20}{c}} {{\alpha _2}A\left( {a,b} \right) + \left( {1 - {\alpha _2}} \right)G\left( {a,b} \right) < {R_{GQ}}\left( {a,b} \right) < }\\ {{\beta _2}A\left( {a,b} \right) + \left( {1 - {\beta _2}} \right)G\left( {a,b} \right)} \end{array} $

对所有a, b > 0且ab成立当且仅当α2$ \sqrt 2 $/e=0.520 2…和β2≥2/3.

证明  根据几何平均G(a, b)、算术平均A(a, b)和Sándor-Yang平均RGQ(a, b)是对称的且齐次系数为1, 不失一般性, 假设a > b > 0.设p∈(0, 1),x=(a+b)/(2$ \sqrt {ab} $)∈(1, +∞),则由式(38)和(39)可得

$ \begin{array}{l} \frac{{{R_{GQ}}\left( {a,b} \right) - G\left( {a,b} \right)}}{{A\left( {a,b} \right) - G\left( {a,b} \right)}} = \\ \;\;\;\;\;\frac{\sqrt{2{{x}^{2}}-1}{{\text{e}}^{{{\tan }^{-1}}}}\left( ^{\sqrt{2\left( {{x}^{2}}-1 \right)}} \right){{/}^{\sqrt{2\left( {{x}^{2}}-1 \right)}-1}}-1}{x-1}, \end{array} $ (51)
$ \mathop {\lim }\limits_{x \to {1^ + }} \frac{{\sqrt {2{x^2} - 1} {{\rm{e}}^{{{\tan }^{ - 1}}}}\left( {^{\sqrt {2\left( {{x^2} - 1} \right)} }} \right){/^{\sqrt {2\left( {{x^2} - 1} \right)} - 1}} - 1}}{{x - 1}} = \frac{2}{3}, $ (52)
$ \begin{array}{l} \mathop {\lim }\limits_{x \to + \infty } \frac{{\sqrt {2{x^2} - 1} {{\rm{e}}^{{{\tan }^{ - 1}}}}\left( {^{\sqrt {2\left( {{x^2} - 1} \right)} }} \right){/^{\sqrt {2\left( {{x^2} - 1} \right)} - 1}} - 1}}{{x - 1}} = \\ \;\;\;\;\frac{{\sqrt 2 }}{{\rm{e}}} = 0.520\;2 \cdots , \end{array} $ (53)
$ \begin{array}{l} \ln \left[ {{R_{GQ}}\left( {a,b} \right)} \right] - \ln \left[ {pA\left( {a,b} \right) + \left( {1 - p} \right)G\left( {a,b} \right)} \right] = \\ \;\;\;\;\;\;\frac{1}{2}\ln \left( {2{x^2} - 1} \right) + \frac{{{{\tan }^{ - 1}}\left( {\sqrt {2\left( {{x^2} - 1} \right)} } \right)}}{{\sqrt {2\left( {{x^2} - 1} \right)} }} - \\ \;\;\;\;\;\;\ln \left( {px + 1 - p} \right) - 1. \end{array} $ (54)

$ \begin{array}{l} G\left( x \right) = \frac{1}{2}\ln \left( {2{x^2} - 1} \right) + \frac{{{{\tan }^{ - 1}}\left( {\sqrt {2\left( {{x^2} - 1} \right)} } \right)}}{{\sqrt {2\left( {{x^2} - 1} \right)} }} - \\ \;\;\;\;\;\;\;\ln \left( {px + 1 - p} \right) - 1, \end{array} $ (55)

经简单计算得

$ G\left( {{1^ + }} \right) = 0, $ (56)
$ \mathop {\lim }\limits_{x \to + \infty } G\left( x \right) = \frac{1}{2}\ln 2 - \ln p - 1, $ (57)
$ G'\left( x \right) = \frac{x}{{\sqrt 2 {{\left( {{x^2} - 1} \right)}^{3/2}}}}g\left( x \right), $ (58)

其中g(x)的定义见引理2.

下面, 分2种情形证明:

情形1  p=2/3.由式(54)~(56)、(58)及引理2(1)得

$ {R_{GQ}}\left( {a,b} \right) < \frac{2}{3}A\left( {a,b} \right) + \frac{1}{3}G\left( {a,b} \right). $ (59)

情形2  p=$ \sqrt 2 $/e.由式(58)和引理2(2)可得存在μ∈(1, +∞)使得函数G(x)在区间(1, μ]严格单调上升,在区间[μ, +∞)严格单调下降.注意到式(57)变为

$ \mathop {\lim }\limits_{x \to + \infty } G\left( x \right) = 0, $ (60)

由式(54)~(56)、(60)及函数G(x)的分段单调性知

$ {R_{GQ}}\left( {a,b} \right) > \frac{{\sqrt 2 }}{{\rm{e}}}A\left( {a,b} \right) + \left( {1 - \frac{{\sqrt 2 }}{{\rm{e}}}} \right)G\left( {a,b} \right). $ (61)

所以, 由等式(51)~(53)和不等式(59)、(61)易得定理2.

定理3  双向不等式

$ \begin{array}{*{20}{c}} {{\alpha _3}Q\left( {a,b} \right) + \left( {1 - {\alpha _3}} \right)A\left( {a,b} \right) < {R_{GQ}}\left( {a,b} \right) < }\\ {{\beta _3}Q\left( {a,b} \right) + \left( {1 - {\beta _3}} \right)A\left( {a,b} \right)} \end{array} $

对所有a, b > 0且ab成立当且仅当α3≤(2$ \sqrt 2 $/e-1)/($ \sqrt 2 $-1)=0.0978…和β3≥1/3.

证明  由算术平均A(a, b)、二次平均Q(a, b)和Sándor-Yang平均RQG(a, b)是对称的且齐次系数为1, 不失一般性, 假设a > b > 0.若$ p \in \left( {0, 1} \right), x = \sqrt {2({a^2} + {b^2})} $/(a+b)∈(1, $ \sqrt 2 $),则由式(1)~(3)得

$ \begin{array}{*{20}{c}} {Q\left( {a,b} \right) = A\left( {a,b} \right)x,}\\ {G\left( {a,b} \right) = A\left( {a,b} \right)\sqrt {2 - {x^2}} ,} \end{array} $ (62)
$ \begin{array}{l} {R_{GQ}}\left( {a,b} \right) = \\ A\left( {a,b} \right)\sqrt {2 - {x^2}} {{\rm{e}}^{x\sin \;{h^{ - 1}}\left( {\sqrt {2\left( {{x^2} - 1} \right)/\left( {2 - {x^2}} \right)} } \right)/\sqrt {2\left( {{x^2} - 1} \right)} - 1}}. \end{array} $ (63)

由式(62)和(63), 得到

$ \begin{array}{l} \frac{{{R_{QG}}\left( {a,b} \right) - A\left( {a,b} \right)}}{{Q\left( {a,b} \right) - A\left( {a,b} \right)}} = \\ \frac{{\sqrt {2 - {x^2}} {{\rm{e}}^{x\sin \;{h^{ - 1}}\left( {\sqrt {2\left( {{x^2} - 1} \right)/\left( {2 - {x^2}} \right)} } \right)/\sqrt {2\left( {{x^2} - 1} \right)} - 1}} - 1}}{{x - 1}}, \end{array} $ (64)
$ \mathop {\lim }\limits_{x \to {1^ + }} \frac{{\sqrt {2 - {x^2}} {{\rm{e}}^{x\sin \;{h^{ - 1}}\left( {\sqrt {2\left( {{x^2} - 1} \right)/\left( {2 - {x^2}} \right)} } \right)/\sqrt {2\left( {{x^2} - 1} \right)} - 1}} - 1}}{{x - 1}} = \frac{1}{3}, $ (65)
$ \begin{array}{l} \mathop {\lim }\limits_{x \to {{\sqrt 2 }^ - }} \frac{{\sqrt {2 - {x^2}} {{\rm{e}}^{x\sin \;{h^{ - 1}}\left( {\sqrt {2\left( {{x^2} - 1} \right)/\left( {2 - {x^2}} \right)} } \right)/\sqrt {2\left( {{x^2} - 1} \right)} - 1}} - 1}}{{x - 1}} = \\ \;\;\;\;\;\;\;\frac{{4/{\rm{e}} - \sqrt 2 }}{{2 - \sqrt 2 }} = 0.097\;8 \cdots , \end{array} $ (66)
$ \begin{array}{l} \ln \left[ {{R_{QG}}\left( {a,b} \right)} \right] - \ln \left[ {pQ\left( {a,b} \right) + \left( {1 - p} \right)A\left( {a,b} \right)} \right] = \\ \frac{1}{2}\ln \left( {2 - {x^2}} \right) + \frac{{x\sin \;{h^{ - 1}}\left[ {\sqrt {2\left( {{x^2} - 1} \right)/\left( {2 - {x^2}} \right)} } \right]}}{{\sqrt {2\left( {{x^2} - 1} \right)} }} - \\ \ln \left( {px + 1 - p} \right) - 1. \end{array} $ (67)

$ \begin{array}{l} H\left( x \right) = \frac{1}{2}\ln \left( {2 - {x^2}} \right) + \\ \;\;\;\;\;\;\;\;\;\;\;\frac{{x\sin \;{h^{ - 1}}\left[ {\sqrt {2\left( {{x^2} - 1} \right)/\left( {2 - {x^2}} \right)} } \right]}}{{\sqrt {2\left( {{x^2} - 1} \right)} }} - \\ \;\;\;\;\;\;\;\;\;\;\;\ln \left( {px + 1 - p} \right) - 1. \end{array} $ (68)

经简单计算得

$ H\left( {{1^ + }} \right) = 0, $ (69)
$ \begin{array}{l} H\left( {{{\sqrt 2 }^ - }} \right) = \frac{3}{2}\ln 2 + \ln \left( {1 + \sqrt 2 } \right) - \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\ln \left( {1 + \sqrt 2 + p} \right) - 1, \end{array} $ (70)
$ H'\left( x \right) = \frac{1}{{\sqrt 2 {{\left( {{x^2} - 1} \right)}^{3/2}}}}h\left( x \right), $ (71)

其中h(x)的定义见引理3.

下面, 分2种情形证明:

情形1  p=1/3.由式(67)~(69)、(71)及引理3(1)得

$ {R_{QG}}\left( {a,b} \right) < \frac{1}{3}Q\left( {a,b} \right) + \frac{2}{3}A\left( {a,b} \right), $ (72)

情形2  p=(2$ \sqrt 2 $/e-1)/($ \sqrt 2 $-1).由式(71)和引理3(2)可得,存在σ∈(1, $ \sqrt 2 $),使得函数H(x)在区间(1, σ]严格单调上升,在区间[σ, $ \sqrt 2 $)严格单调下降.注意到式(70)变为

$ H\left( {{{\sqrt 2 }^ - }} \right) = 0, $ (73)

由式(67)~(69)、(73)及函数H(x)的分段单调性知

$ \begin{array}{*{20}{c}} {{R_{QG}}\left( {a,b} \right) > \frac{{2\sqrt 2 /{\rm{e}} - {\rm{1}}}}{{\sqrt 2 - 1}}Q\left( {a,b} \right) + }\\ {\frac{{\sqrt 2 \left( {1 - 2/{\rm{e}}} \right)}}{{\sqrt 2 - 1}}A\left( {a,b} \right).} \end{array} $ (74)

所以, 定理3可由式(72)和(74)结合下面陈述得到.

(1) 若p < 1/3, 则由式(64)和(65)得存在足够小的δ2 > 0对所有a > b > 0且$\sqrt {2({a^2} + {b^2})} {\rm{ }}/(a + b) \in (1, 1 + {\delta _2})$,有

$ {R_{QG}}\left( {a,b} \right) > pQ\left( {a,b} \right) + \left( {1 - p} \right)A\left( {a,b} \right). $

(2) 若p > (2$ \sqrt 2 $/e-1)/($ \sqrt 2 $-1), 则由式(64)和(66)得存在足够大的1 < M2 < $ \sqrt 2 $使对所有a > b > 0且$ \sqrt {2({a^2} + {b^2})} $ /(a+b)∈(M2, $ \sqrt 2 $),有

$ {R_{QG}}\left( {a,b} \right) < pQ\left( {a,b} \right) + \left( {1 - p} \right)A\left( {a,b} \right). $

定理4  双向不等式

$ \begin{array}{*{20}{c}} {{\alpha _4}C\left( {a,b} \right) + \left( {1 - {\alpha _4}} \right)A\left( {a,b} \right) < {R_{QG}}\left( {a,b} \right) < }\\ {{\beta _4}C\left( {a,b} \right) + \left( {1 - {\beta _4}} \right)A\left( {a,b} \right)} \end{array} $

对所有a, b > 0且ab成立当且仅当α4≤2$ \sqrt 2 $/e-1=0.040 5…,β4≥1/6.

证明  不失一般性, 假设a > b > 0.设p∈(0, 1), x= $ \sqrt {2({a^2} + {b^2})} $ /(a+b)∈(1, $ \sqrt 2 $), 则由式(62),(63)和C(a, b)=A(a, b)x2可得

$ \begin{array}{l} \frac{{{R_{QG}}\left( {a,b} \right) - A\left( {a,b} \right)}}{{C\left( {a,b} \right) - A\left( {a,b} \right)}} = \\ \;\;\;\;\;\;\;\;\;\frac{{\sqrt {2 - {x^2}} {{\rm{e}}^{x\sin \;{h^{ - 1}}\left( {\sqrt {2\left( {{x^2} - 1} \right)/\left( {2 - {x^2}} \right)} } \right)/\sqrt {2\left( {{x^2} - 1} \right)} - 1}} - 1}}{{{x^2} - 1}}, \end{array} $ (75)
$ \mathop {\lim }\limits_{x \to {1^ + }} \frac{{\sqrt {2 - {x^2}} {{\rm{e}}^{x\sin \;{h^{ - 1}}\left( {\sqrt {2\left( {{x^2} - 1} \right)/\left( {2 - {x^2}} \right)} } \right)/\sqrt {2\left( {{x^2} - 1} \right)} - 1}} - 1}}{{{x^2} - 1}} = \frac{1}{6}, $ (76)
$ \begin{array}{l} \mathop {\lim }\limits_{x \to {{\sqrt 2 }^ - }} \frac{{\sqrt {2 - {x^2}} {{\rm{e}}^{x\sin \;{h^{ - 1}}\left( {\sqrt {2\left( {{x^2} - 1} \right)/\left( {2 - {x^2}} \right)} } \right)/\sqrt {2\left( {{x^2} - 1} \right)} - 1}} - 1}}{{{x^2} - 1}} = \\ \;\;\;\;\;\;\;2\sqrt 2 /{\rm{e}} - 1, \end{array} $ (77)
$ \begin{array}{l} \ln \left[ {{R_{QG}}\left( {a,b} \right)} \right] - \ln \left[ {pC\left( {a,b} \right) + \left( {1 - p} \right)A\left( {a,b} \right)} \right] = \\ \frac{1}{2}\ln \left( {2 - {x^2}} \right) + \frac{{x\sin \;{h^{ - 1}}\left[ {\sqrt {2\left( {{x^2} - 1} \right)/\left( {2 - {x^2}} \right)} } \right]}}{{\sqrt {2\left( {{x^2} - 1} \right)} }} - \\ \ln \left( {p{x^2} + 1 - p} \right) - 1. \end{array} $ (78)

$ \begin{array}{l} H\left( x \right) = \frac{1}{2}\ln \left( {2 - {x^2}} \right) + \\ \;\;\;\;\;\;\;\;\;\;\;\frac{{x\sin \;{h^{ - 1}}\left[ {\sqrt {2\left( {{x^2} - 1} \right)/\left( {2 - {x^2}} \right)} } \right]}}{{\sqrt {2\left( {{x^2} - 1} \right)} }} - \\ \;\;\;\;\;\;\;\;\;\;\;\ln \left( {p{x^2} + 1 - p} \right) - 1. \end{array} $ (79)

经简单计算得

$ K\left( {{1^ + }} \right) = 0, $ (80)
$ K\left( {{{\sqrt 2 }^ - }} \right) = \frac{3}{2}\ln 2 - \ln \left( {1 + \sqrt 2 } \right) - 1, $ (81)
$ K'\left( x \right) = \frac{1}{{\sqrt 2 {{\left( {{x^2} - 1} \right)}^{3/2}}}}k\left( x \right), $ (82)

其中k(x)的定义见引理4.

下面, 分2种情形证明:

情形1  p=1/6.由式(78)~(80)、(82)及引理4(1)得

$ {R_{QG}}\left( {a,b} \right) < \frac{1}{6}C\left( {a,b} \right) + \frac{5}{6}A\left( {a,b} \right). $ (83)

情形2  p=2$ \sqrt 2 $/e-1.由式(82)和引理4(2)可得,存在τ∈(1, $ \sqrt 2 $),使得函数K(x)在区间(1, τ]严格单调上升,在区间[τ, $ \sqrt 2 $)严格单调下降.注意到式(81)可变为

$ K\left( {{{\sqrt 2 }^ - }} \right) = 0, $ (84)

由式(78)~(80)、(84)及函数K(x)的分段单调性知

$ \begin{array}{*{20}{c}} {{R_{QG}}\left( {a,b} \right) > \left( {2\sqrt 2 /{\rm{e}} - {\rm{1}}} \right)C\left( {a,b} \right) + }\\ {2\left( {1 - \sqrt 2 /{\rm{e}}} \right)A\left( {a,b} \right).} \end{array} $ (85)

所以, 由等式(75)、(77)和不等式(83)、(85)易得定理4.

注记  由定理2和定理3, 不难得到对所有a, b > 0且ab,不等式链G(a, b) < RGQ(a, b) < A(a, b) < RQG(a, b) < Q(a, b)成立.

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