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a和b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].

设p∈R，a, b > 0, 则调和平均H(a, b), 反调和平均C(a, b)和p阶幂平均M_p(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M_p(a, b)对于固定的a, b > 0和a≠b关于p∈R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} $

和

$ \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a≠b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a和b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a≠b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a≠b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a≠b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a≠b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R_GQ(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)

设

$ \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a≠b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R_GQ(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)

设

$ \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a≠b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R_QG(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)

设

$ \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a≠b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)

设

$ \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a≠b，不等式链G(a, b) < R_GQ(a, b) < A(a, b) < R_QG(a, b) < Q(a, b)成立.