0 引言

设a, b>0和a≠b, 则几何平均G(a, b)、指数平均I(a, b)、对数平均L(a, b)、算术平均A(a, b)、二次平均Q(a, b)和Toader-Qi平均TQ(a, b)^[1-3]分别为

$ \begin{array}{l} G\left( {a,b} \right) = \sqrt {ab} ,I\left( {a,b} \right) = {\left( {{a^a}/{b^b}} \right)^{1/\left( {a - b} \right)}}/e,\\ L\left( {a,b} \right) = \frac{{a - b}}{{\log a - \log b}},A\left( {a,b} \right) = \frac{{a + b}}{2},\\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;Q\left( {a,b} \right) = \sqrt {\frac{{{a^2} + {b^2}}}{2}} , \end{array} $

(1)

$ {\rm{TQ}}\left( {a,b} \right)\frac{2}{{\rm{\pi }}}\int_0^{{\rm{\pi }}/2} {{a^{{{\cos }^2}\theta }}{b^{{{\sin }^2}\theta }}{\rm{d}}\theta } . $

(2)

近年来, 关于若干经典平均与Toader-Qi平均的比较受到了关注, 得到了一定的研究, 有关Toader-Qi平均的重要不等式可参阅文献[2-7].QI等^[2]证明了

$ {\rm{TQ}}\left( {a,b} \right) = \sqrt {ab} {I_0}\left( {\frac{1}{2}\log \frac{b}{a}} \right) $

(3)

和不等式

$ \begin{array}{*{20}{c}} {L\left( {a,b} \right) < {\rm{TQ}}\left( {a,b} \right) < \frac{{A\left( {a,b} \right) + G\left( {a,b} \right)}}{2} < }\\ {\frac{{2A\left( {a,b} \right) + G\left( {a,b} \right)}}{3} < I\left( {a,b} \right)} \end{array} $

对所有a, b>0和a≠b成立, 并且

$ {I_0}\left( t \right) = \sum\limits_{n = 0}^\infty {\frac{{{t^{2n}}}}{{{2^{2n}}{{\left( {n!} \right)}^2}}}} $

(4)

是第1类改进Bessel函数^[8].

YANG等^[4]证明了

$ \frac{a}{{1 + \log \left( {a/b} \right)}} < {\rm{TQ}}\left( {a,b} \right) < \frac{a}{{\sqrt {1 + \log \left( {a/b} \right)} }} $

对所有a>b>0成立, 并且

$ \begin{array}{l} {\alpha _1}\sqrt {L\left( {a,b} \right)A\left( {a,b} \right)} < {\rm{TQ}}\left( {a,b} \right) < \\ \;\;\;\;\;\;\;{\beta _1}\sqrt {L\left( {a,b} \right)A\left( {a,b} \right)} , \end{array} $

$ \begin{array}{l} {L^{{\alpha _2}}}\left( {a,b} \right){A^{1 - {\alpha _2}}}\left( {a,b} \right) < {\rm{TQ}}\left( {a,b} \right) < \\ \;\;\;\;\;{\beta _2}L\left( {a,b} \right) + \left( {1 - {\beta _2}} \right)A\left( {a,b} \right) \end{array} $

对所有a, b>0和a≠b成立当且仅当α₁≤$\sqrt{2/\text{ }\!\!\pi\!\!\text{ }}$, β₁≥1, α₂≥3/4和β₂≤3/4.

由文献[6], 推得精确不等式

$ {\rm{TQ}}\left( {a,b} \right) > {L_{3/2}}\left( {a,b} \right) $

对所有a>b>0成立.其中L_p(a, b)=[(a^p-b^p)/p(log a-log b)]^1/p(p≠0) 和L₀(a, b)=$\mathop {\lim }\limits_{p \to 0} {\mkern 1mu} $ ^L_p(a, b)=$\sqrt{ab}$是p阶对数平均.

设a>b>0，t=(log a-log b)/2>0, 则由式(1) 可推得

$ \left\{ \begin{array}{l} L\left( {a,b} \right) = \sqrt {ab} \frac{{\sinh \left( t \right)}}{t},\\ A\left( {a,b} \right) = \sqrt {ab} \cosh \left( t \right),\\ Q\left( {a,b} \right) = \sqrt {ab} {\cosh ^{1/2}}\left( {2t} \right). \end{array} \right. $

(5)

由式(2) 和(3) 可推得

$ {\rm{TQ}}\left( {a,b} \right) = \sqrt {ab} {I_0}\left( t \right). $

(6)

本文的主要目的是给出以下几个关于Toader-Qi平均与几何平均G(a, b)、对数平均L(a, b)、算术平均A(a, b)和二次平均Q(a, b)特殊组合的精确不等式:

$ \begin{array}{l} \frac{{{\alpha _1}{Q^2}\left( {a,b} \right) + \left( {1 - {\alpha _1}} \right){G^2}\left( {a,b} \right)}}{{G\left( {a,b} \right)}} < {\rm{TQ}}\left( {a,b} \right) < \\ \;\;\;\;\;\;\;\frac{{{\beta _1}{Q^2}\left( {a,b} \right) + \left( {1 - {\beta _1}} \right){G^2}\left( {a,b} \right)}}{{G\left( {a,b} \right)}}, \end{array} $

$ \begin{array}{l} \frac{{{\alpha _2}{Q^2}\left( {a,b} \right) + \left( {1 - {\alpha _2}} \right){G^2}\left( {a,b} \right)}}{{L\left( {a,b} \right)}} < {\rm{TQ}}\left( {a,b} \right) < \\ \;\;\;\;\;\;\;\frac{{{\beta _2}{Q^2}\left( {a,b} \right) + \left( {1 - {\beta _2}} \right){G^2}\left( {a,b} \right)}}{{L\left( {a,b} \right)}}, \end{array} $

$ \begin{array}{l} \frac{{{\alpha _3}{Q^2}\left( {a,b} \right) + \left( {1 - {\alpha _3}} \right){G^2}\left( {a,b} \right)}}{{A\left( {a,b} \right)}} < {\rm{TQ}}\left( {a,b} \right) < \\ \;\;\;\;\;\;\;\frac{{{\beta _3}{Q^2}\left( {a,b} \right) + \left( {1 - {\beta _3}} \right){G^2}\left( {a,b} \right)}}{{A\left( {a,b} \right)}}, \end{array} $

$ \begin{array}{l} \sqrt {{\alpha _4}{Q^2}\left( {a,b} \right) + \left( {1 - {\alpha _4}} \right){G^2}\left( {a,b} \right)} < {\rm{TQ}}\left( {a,b} \right) < \\ \;\;\;\;\;\;\;\;\sqrt {{\beta _4}{Q^2}\left( {a,b} \right) + \left( {1 - {\beta _4}} \right){G^2}\left( {a,b} \right)} \end{array} $

对所有a, b>0和a≠b成立.

1 预备知识和引理

为了证明主要结果, 本节给出一些经典Gamma函数、第1类Bessel函数的基本知识和相关引理.

设x>0, $\Gamma \left( x \right)=\int_{0}^{\infty }{{{\text{e}}^{-t}}{{t}^{x-1}}\text{d}t}$是经典Gamma函数.则Γ(x)满足下列公式(见文献[9], p33引理2.4 (1) (3), p46练习2.46，4 (b)):

$ \Gamma \left( {1/2} \right) = \sqrt {\rm{\pi }} ,\Gamma \left( {3/2} \right) = \sqrt {\rm{\pi }} /2, $

$ \Gamma \left( {n + 1} \right) = n\Gamma \left( n \right) = n!, $

$ \Gamma \left( {n + \frac{1}{2}} \right) = \frac{{\left( {2n} \right)!}}{{{2^{2n}}n!}}\sqrt {\rm{\pi }} , $

$ \Gamma \left( {n + \frac{3}{2}} \right) = \frac{{\left( {2n + 2} \right)!}}{{{2^{2n + 2}}\left( {n + 1} \right)!}}\sqrt {\rm{\pi }} , $

对所有正整数n成立.

设v>-1, 第1类修正Bessel函数I_v(t)定义(见文献[10], p77) 为

$ {I_v}\left( t \right) = \sum\limits_{n = 0}^\infty {\frac{1}{{n!\Gamma \left( {v + n + 1} \right)}}{{\left( {\frac{t}{2}} \right)}^{2n + v}}} . $

(7)

注意到在特殊情形下有：

$ cosh\left( t \right) = \sqrt {\rm{\pi }} {\left( {t/2} \right)^{1/2}}{I_{ - 1/2}}\left( t \right), $

(8)

$ \sinh \left( t \right) = \sqrt {\rm{\pi }} {\left( {t/2} \right)^{1/2}}{I_{1/2}}\left( t \right). $

(9)

引理1^[11]  假设幂级数$f\left( x \right)=\sum\limits_{n=0}^{\infty }{{{a}_{n}}{{x}^{n}}}$和$g\left( x \right)=\sum\limits_{n=0}^{\infty }{{{b}_{n}}{{x}^{n}}}$对所有的n∈{0, 1, 2, …}的收敛半径r>0且a_n, b_n>0.设h(x)=f(x)/g(x), 如果序列{a_n/b_n}_n=0^∞是(严格)上升(下降)的, 则h(x)在(0, r)内也(严格)上升(下降).

引理2(见文献[3], 引理2.8)  设I₀(t)定义为式(4).则等式

$ I_0^2\left( t \right) = \sum\limits_{n = 0}^\infty {\frac{{\left( {2n} \right)!}}{{{2^{2n}}{{\left( {n!} \right)}^4}}}{t^{2n}}} $

对所有t∈R成立.

引理3  等式

$ \cosh \left( t \right){I_0}\left( t \right) = \sum\limits_{n = 0}^\infty {\frac{{\left( {4n} \right)!}}{{{2^{2n}}{{\left[ {\left( {2n} \right)!} \right]}^3}}}{t^{2n}}} $

(A)

与

$ \sinh \left( t \right){I_0}\left( t \right) = \sum\limits_{n = 0}^\infty {\frac{{\left( {4n + 2} \right)!}}{{{2^{2n + 1}}{{\left[ {\left( {2n + 1} \right)!} \right]}^3}}}{t^{2n + 1}}} $

(B)

对所有t∈R成立.

证明   (A)由式(8) 和Cauchy乘积公式(见文献[12], (3.5))

$ \begin{array}{l} {I_\lambda }\left( t \right){I_\mu }\left( t \right) = \sum\limits_{n = 0}^\infty {\left( {\begin{array}{*{20}{c}} {2n + \lambda + \mu }\\ n \end{array}} \right){C_{\lambda ,n}}{C_{\mu ,n}}{{\left( {t/2} \right)}^{2n + \lambda + \mu }}} ,\\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;{C_{\lambda ,n}} = \frac{1}{{\Gamma \left( {n + \lambda + 1} \right)}}, \end{array} $

对所有t∈R, 有

$ \begin{array}{l} cosh\left( t \right){I_0}\left( t \right) = \sqrt {\rm{\pi }} {\left( {t/2} \right)^{1/2}}{I_{ - 1/2}}\left( t \right){I_0}\left( t \right) = \\ \sqrt {\rm{\pi }} {\left( {t/2} \right)^{1/2}}\sum\limits_{n = 0}^\infty {\frac{{\Gamma \left( {2n + 1/2} \right)}}{{n!{\Gamma ^2}\left( {n + 1/2} \right)\Gamma \left( {n + 1} \right)}}{{\left( {t/2} \right)}^{2n - 1/2}}} = \\ \sqrt {\rm{\pi }} \sum\limits_{n = 0}^\infty {\frac{{\Gamma \left( {2n + 1/2} \right)}}{{{{\left( {n!} \right)}^2}{\Gamma ^2}\left( {n + 1/2} \right)}}{{\left( {t/2} \right)}^{2n}}} = \\ \sum\limits_{n = 0}^\infty {\frac{{\left( {4n} \right)!}}{{{2^{2n}}{{\left[ {\left( {2n} \right)!} \right]}^3}}}{t^{2n}}} . \end{array} $

(B)由式(9) 和Cauchy乘积公式(见文献[12], (3.5))

$ \begin{array}{*{20}{c}} {{I_\lambda }\left( t \right){I_\mu }\left( t \right) = \sum\limits_{n = 0}^\infty {\left( {\begin{array}{*{20}{c}} {2n + \lambda + \mu }\\ n \end{array}} \right){C_{\lambda ,n}}{C_{\mu ,n}}{{\left( {t/2} \right)}^{2n + \lambda + \mu }}} ,}\\ {{C_{\lambda ,n}} = \frac{1}{{\Gamma \left( {n + \lambda + 1} \right)}},} \end{array} $

对所有t∈R, 有

$ \begin{array}{l} \sinh \left( t \right){I_0}\left( t \right) = \sqrt {\rm{\pi }} {\left( {t/2} \right)^{1/2}}{I_{1/2}}\left( t \right){I_0}\left( t \right) = \\ \sqrt {\rm{\pi }} {\left( {t/2} \right)^{1/2}}\sum\limits_{n = 0}^\infty {\frac{{\Gamma \left( {2n + 3/2} \right)}}{{n!{\Gamma ^2}\left( {n + 3/2} \right)\Gamma \left( {n + 1} \right)}}{{\left( {t/2} \right)}^{2n + 1/2}}} = \\ \sqrt {\rm{\pi }} \sum\limits_{n = 0}^\infty {\frac{{\Gamma \left( {2n + 3/2} \right)}}{{{{\left( {n!} \right)}^2}{\Gamma ^2}\left( {n + 3/2} \right)}}{{\left( {t/2} \right)}^{2n + 1}}} = \\ \sum\limits_{n = 0}^\infty {\frac{{\left( {4n + 2} \right)!}}{{{2^{2n + 1}}{{\left[ {\left( {2n + 1} \right)!} \right]}^3}}}{t^{2n + 1}}} . \end{array} $

引理4  函数$f\left( t \right)=\frac{{{I}_{0}}\left( t \right)-1}{\cosh \left( 2t \right)-1}$在(0, ∞)内严格单调下降.

证明  由式(4) 得

$ \begin{array}{l} f\left( t \right) = \frac{{\sum\limits_{n = 0}^\infty {\frac{{{t^{2n}}}}{{{2^{2n}}{{\left( {n!} \right)}^2}}} - 1} }}{{\sum\limits_{n = 0}^\infty {\frac{{{2^{2n}}}}{{\left( {2n} \right)!}}{t^{2n}} - 1} }} = \frac{{\sum\limits_{n = 1}^\infty {\frac{1}{{{2^{2n}}{{\left( {n!} \right)}^2}}}{t^{2n}}} }}{{\sum\limits_{n = 1}^\infty {\frac{{{2^{2n}}}}{{\left( {2n} \right)!}}{t^{2n}}} }} = \\ \;\;\;\;\;\;\;\;\;\;\;\frac{{\sum\limits_{n = 0}^\infty {\frac{1}{{{2^{2n + 2}}\left[ {\left( {n + 1} \right)!} \right]}}{t^{2n + 2}}} }}{{\sum\limits_{n = 0}^\infty {\frac{{{2^{2n + 2}}}}{{\left( {2n + 2} \right)!}}} {t^{2n + 2}}}} = \frac{{\sum\limits_{n = 0}^\infty {{a_n}{t^{2n}}} }}{{\sum\limits_{n = 0}^\infty {{b_n}{t^{2n}}} }}, \end{array} $

(10)

其中，

$ {a_n} = \frac{1}{{{2^{2n + 2}}{{\left[ {\left( {n + 1} \right)!} \right]}^2}}} > 0,{b_n} = \frac{{{2^{2n + 2}}}}{{\left( {2n + 2} \right)!}} > 0. $

(11)

经简单计算得到

$ \begin{array}{l} \frac{{{a_{n + 1}}}}{{{b_{n + 1}}}} - \frac{{{a_n}}}{{{b_n}}} = \frac{{\left( {2n + 4} \right)!}}{{{2^{4n + 8}}{{\left[ {\left( {n + 2} \right)!} \right]}^2}}} - \frac{{\left( {2n + 2} \right)!}}{{{2^{4n + 4}}{{\left[ {\left( {n + 1} \right)!} \right]}^2}}} = \\ \;\;\;\;\;\;\; - \frac{{\left( {n + 2} \right)\left( {6n + 13} \right)\left( {2n + 2} \right)!}}{{{2^{4n + 7}}{{\left[ {\left( {n + 2} \right)!} \right]}^2}}} < 0. \end{array} $

(12)

从不等式(11) 和(12) 中可清楚地看到序列{a_n/b_n}₀^∞对所有n≥0是严格单调下降的.所以, 由引理1和式(10) 协同序列{a_n/b_n}₀^∞的单调性可容易地得到引理4.

引理5  函数$g\left( t \right)=\frac{\sinh \left( t \right){{I}_{0}}\left( t \right)-1}{t\left[\cosh \left( 2t \right)-1 \right]}$在(0, ∞)内严格单调下降.

证明  由引理3(B)得到

$ \begin{array}{l} g\left( t \right) = \frac{{\sum\limits_{n = 0}^\infty {\frac{{\left( {4n + 2} \right)!}}{{{2^{2n + 1}}{{\left[ {\left( {2n + 1} \right)!} \right]}^3}}}{t^{2n + 1}} - t} }}{{t\sum\limits_{n = 0}^\infty {\frac{{{2^{2n}}}}{{\left( {2n} \right)!}}{t^{2n}} - t} }} = \\ \;\;\;\;\;\;\;\;\;\;\frac{{\sum\limits_{n = 1}^\infty {\frac{{\left( {4n + 2} \right)!}}{{{2^{2n + 1}}{{\left[ {\left( {2n + 1} \right)!} \right]}^3}}}{t^{2n + 1}}} }}{{\sum\limits_{n = 1}^\infty {\frac{{{2^{2n}}}}{{\left( {2n} \right)!}}{t^{2n + 1}}} }} = \\ \;\;\;\;\;\;\;\;\;\;\frac{{\sum\limits_{n = 0}^\infty {\frac{{\left( {4n + 6} \right)!}}{{{2^{2n + 3}}{{\left[ {\left( {2n + 3} \right)!} \right]}^3}}}{t^{2n + 3}}} }}{{\sum\limits_{n = 0}^\infty {\frac{{{2^{2n + 2}}}}{{\left( {2n + 2} \right)!}}{t^{2n + 3}}} }} = \frac{{\sum\limits_{n = 0}^\infty {{c_n}{t^{2n}}} }}{{\sum\limits_{n = 0}^\infty {{d_n}{t^{2n}}} }}, \end{array} $

(13)

其中，

$ {c_n} = \frac{{\left( {4n + 6} \right)!}}{{{2^{2n + 3}}{{\left[ {\left( {2n + 3} \right)!} \right]}^3}}} > 0,{d_n} = \frac{{{2^{2n + 2}}}}{{\left( {2n + 2} \right)!}} > 0. $

(14)

经简单计算得到

$ \begin{array}{l} \frac{{{c_{n + 1}}}}{{{d_{n + 1}}}} - \frac{{{c_n}}}{{{d_n}}} = \frac{{\left( {4n + 10} \right)!}}{{{2^{4n + 9}}{{\left[ {\left( {2n + 5} \right)!} \right]}^2}}} - \frac{{\left( {4n + 6} \right)!}}{{{2^{4n + 5}}{{\left[ {\left( {2n + 3} \right)!} \right]}^2}}} = \\ \;\;\;\;\;\; - \frac{{\left( {n + 2} \right)\left( {48{n^2} + 202n + 211} \right)\left( {4n + 5} \right)!}}{{{2^{4n + 5}}{{\left[ {\left( {2n + 5} \right)!} \right]}^2}}} < 0. \end{array} $

(15)

从不等式(14) 和(15) 中可清楚地看到序列{c_n/d_n}_n=0^∞对所有n≥0是严格单调下降的.所以, 由引理1和式(13) 协同序列{c_n/d_n}_n=0^∞的单调性可容易地得到引理5.

引理6  函数$h\left( t \right)=\frac{\cosh \left( t \right){{I}_{0}}\left( t \right)-1}{t\left[\cosh \left( 2t \right)-1 \right]}$在(0, ∞)内严格单调下降.

证明  由引理3(A)得到

$ \begin{array}{l} h\left( t \right) = \frac{{\sum\limits_{n = 0}^\infty {\frac{{\left( {4n} \right)!}}{{{2^{2n}}{{\left[ {\left( {2n} \right)!} \right]}^3}}}{t^{2n}} - 1} }}{{\sum\limits_{n = 0}^\infty {\frac{{{2^{2n}}}}{{\left( {2n} \right)!}}{t^{2n}} - 1} }} = \frac{{\sum\limits_{n = 1}^\infty {\frac{{\left( {4n} \right)!}}{{{2^{2n}}{{\left[ {\left( {2n} \right)!} \right]}^3}}}{t^{2n}}} }}{{\sum\limits_{n = 1}^\infty {\frac{{{2^{2n}}}}{{\left( {2n} \right)!}}{t^{2n}}} }} = \\ \;\;\;\;\;\;\;\;\;\frac{{\sum\limits_{n = 0}^\infty {\frac{{\left( {4n + 4} \right)!}}{{{2^{2n + 2}}{{\left[ {\left( {2n + 2} \right)!} \right]}^3}}}{t^{2n + 2}}} }}{{\sum\limits_{n = 0}^\infty {\frac{{{2^{2n + 2}}}}{{\left( {2n + 2} \right)!}}{t^{2n + 2}}} }} = \frac{{\sum\limits_{n = 0}^\infty {{u_n}{t^{2n}}} }}{{\sum\limits_{n = 0}^\infty {{v_n}{t^{2n}}} }}, \end{array} $

(16)

其中，

$ {u_n} = \frac{{\left( {4n + 4} \right)!}}{{{2^{2n + 2}}{{\left[ {\left( {2n + 2} \right)!} \right]}^3}}} > 0,{v_n} = \frac{{{2^{2n + 2}}}}{{\left( {2n + 2} \right)!}} > 0. $

(17)

经简单计算得到

$ \begin{array}{l} \frac{{{u_{n + 1}}}}{{{v_{n + 1}}}} - \frac{{{u_n}}}{{{v_n}}} = \frac{{\left( {4n + 8} \right)!}}{{{2^{4n + 8}}{{\left[ {\left( {2n + 4} \right)!} \right]}^2}}} - \frac{{\left( {4n + 4} \right)!}}{{{2^{4n + 4}}{{\left[ {\left( {2n + 2} \right)!} \right]}^2}}} = \\ - \frac{{\left( {n + 2} \right)\left( {2n + 3} \right)\left( {8n + 13} \right)\left( {4n + 4} \right)!}}{{{2^{4n + 5}}{{\left[ {\left( {2n + 4} \right)!} \right]}^2}}} < 0. \end{array} $

(18)

从不等式(17) 和(18) 中可清楚地看到序列{u_n/v_n}₀^∞对所有n≥0是严格单调下降的.所以，由引理1和式(16) 协同序列{u_n/v_n}₀^∞的单调性可容易地得到引理6.

引理7  函数$k\left( t \right)=\frac{I_{0}^{2}\left( t \right)-1}{\cosh \left( 2t \right)-1}$在(0, ∞)内严格单调下降.

证明  由引理2得到

$ \begin{array}{l} k\left( t \right) = \frac{{\sum\limits_{n = 0}^\infty {\frac{{\left( {2n} \right)}}{{{2^{2n}}{{\left( {n!} \right)}^4}}}{t^{2n}} - 1} }}{{\sum\limits_{n = 0}^\infty {\frac{{{2^{2n}}}}{{\left( {2n} \right)!}}{t^{2n}} - 1} }} = \frac{{\sum\limits_{n = 1}^\infty {\frac{{\left( {2n} \right)!}}{{{2^{2n}}{{\left( {n!} \right)}^4}}}{t^{2n}}} }}{{\sum\limits_{n = 1}^\infty {\frac{{{2^{2n}}}}{{\left( {2n} \right)!}}{t^{2n}}} }} = \\ \;\;\;\;\;\;\;\;\;\;\frac{{\sum\limits_{n = 0}^\infty {\frac{{\left( {2n + 2} \right)}}{{{2^{2n + 2}}{{\left[ {\left( {n + 1} \right)!} \right]}^4}}}{t^{2n + 2}}} }}{{\sum\limits_{n = 0}^\infty {\frac{{{2^{2n + 2}}}}{{\left( {2n + 2} \right)!}}{t^{2n + 2}}} }} = \frac{{\sum\limits_{n = 0}^\infty {{x_n}{t^{2n}}} }}{{\sum\limits_{n = 0}^\infty {{y_n}{t^{2n}}} }}, \end{array} $

(19)

其中，

$ {x_n} = \frac{{\left( {2n + 2} \right)!}}{{{2^{2n + 2}}{{\left[ {\left( {n + 1} \right)!} \right]}^4}}} > 0,{y_n} = \frac{{{2^{2n + 2}}}}{{\left( {2n + 2} \right)!}} > 0. $

(20)

经简单计算得到

$ \begin{array}{l} \frac{{{x_{n + 1}}}}{{{y_{n + 1}}}} - \frac{{{x_n}}}{{{y_n}}} = \frac{{{{\left[ {\left( {2n + 4} \right)!} \right]}^2}}}{{{2^{4n + 8}}{{\left[ {\left( {n + 2} \right)!} \right]}^4}}} - \frac{{{{\left[ {\left( {2n + 2} \right)!} \right]}^2}}}{{{2^{4n + 4}}{{\left[ {\left( {n + 1} \right)!} \right]}^4}}} = \\ \;\;\;\;\;\;\; - \frac{{{{\left( {n + 2} \right)}^2}\left( {4n + 7} \right){{\left[ {\left( {2n + 2} \right)!} \right]}^2}}}{{{2^{4n + 6}}{{\left[ {\left( {n + 2} \right)!} \right]}^4}}} < 0. \end{array} $

(21)

从不等式(20) 和(21) 中可清楚地看到序列{x_n/y_n}₀^∞对所有n≥0是严格单调下降的.所以，由引理1和式(19) 协同序列{x_n/y_n}₀^∞的单调性可容易地得到引理7.

2 主要结果及证明

定理1  双向不等式

$ \begin{array}{l} \frac{{{\alpha _1}{Q^2}\left( {a,b} \right) + \left( {1 + {\alpha _1}} \right){G^2}\left( {a,b} \right)}}{{G\left( {a,b} \right)}} < {\rm{TQ}}\left( {a,b} \right) < \\ \;\;\;\;\;\;\;\;\frac{{{\beta _1}{Q^2}\left( {a,b} \right) + \left( {1 - {\beta _1}} \right){G^2}\left( {a,b} \right)}}{{G\left( {a,b} \right)}} \end{array} $

(22)

对所有a, b>0和a≠b成立当且仅当α₁≤0且β₁≥1/8.

证明  不等式(22) 可写成如下形式:

$ {\alpha _1}\frac{{G\left( {a,b} \right){\rm{TQ}}\left( {a,b} \right) - {G^2}\left( {a,b} \right)}}{{{Q^2}\left( {a,b} \right) - {G^2}\left( {a,b} \right)}} < {\beta _1}. $

(23)

因G(a, b), Q(a, b)和TQ(a, b)是对称和一阶齐次的, 不失一般性, 假设a>b>0，t=(log a-log b)/2>0, 则由式(5) 和(6), 不等式(23) 变成

$ {\alpha _1} < f\left( t \right) < {\beta _1}, $

(24)

其中函数f(t)的定义见引理4.

注意到

$ \mathop {\lim }\limits_{t \to {0^ + }} f\left( t \right) = \frac{1}{8},\;\;\;\mathop {\lim }\limits_{t \to \infty } f\left( t \right) = 0, $

(25)

由不等式(23)(24) 和等式(25) 协同引理4, 则不等式(22) 对所有a, b>0和a≠b成立当且仅当α₁≤0且β₁≥1/8.

推论1  不等式I₀(t) < $\frac{1}{4}$sinh²(t)+1对所有t∈(0, +∞)成立.

定理2  双向不等式

$ \begin{array}{l} \frac{{{\alpha _2}{Q^2}\left( {a,b} \right) + \left( {1 - {\alpha _2}} \right){G^2}\left( {a,b} \right)}}{{L\left( {a,b} \right)}} < {\rm{TQ}}\left( {a,b} \right) < \\ \;\;\;\;\;\;\;\;\frac{{{\beta _2}{Q^2}\left( {a,b} \right) + \left( {1 - {\beta _2}} \right){G^2}\left( {a,b} \right)}}{{L\left( {a,b} \right)}} \end{array} $

(26)

对所有a, b>0和a≠b成立当且仅当α₂≤0且β₂≥5/24.

证明  不等式(26) 可改写成

$ {\alpha _2}\frac{{L\left( {a,b} \right){\rm{TQ}}\left( {a,b} \right) - {G^2}\left( {a,b} \right)}}{{{Q^2}\left( {a,b} \right) - {G^2}\left( {a,b} \right)}} < {\beta _2}. $

(27)

因G(a, b), L(a, b), Q(a, b)和TQ(a, b)是对称和一阶齐次的, 不失一般性, 假设a>b>0，t=(log a-log b)/2>0, 则由式(5) 和(6)，不等式(27) 变成

$ {\alpha _2} < g\left( t \right) < {\beta _2}, $

(28)

其中函数g(t)的定义见引理5.

注意到

$ \mathop {\lim }\limits_{t \to {0^ + }} g\left( t \right) = \frac{4}{{25}},\;\;\;\mathop {\lim }\limits_{t \to \infty } g\left( t \right) = 0, $

(29)

由不等式(27)(28) 和等式(29) 协同引理5, 则不等式(26) 对所有a, b>0和a≠b成立当且仅当α₂≤0且β₂≥5/24.

推论2  双向不等式

$ tcsch\left( t \right) < {I_0}\left( t \right) < \frac{5}{{12}}t\sinh \left( t \right) + tcsch\left( t \right) $

对所有t∈(0, +∞)成立.

定理3  双向不等式

$ \begin{array}{l} \frac{{{\alpha _3}{Q^2}\left( {a,b} \right) + \left( {1 - {\alpha _3}} \right){G^2}\left( {a,b} \right)}}{{A\left( {a,b} \right)}} < {\rm{TQ}}\left( {a,b} \right) < \\ \;\;\;\;\;\;\;\;\frac{{{\beta _3}{Q^2}\left( {a,b} \right) + \left( {1 - {\beta _3}} \right){G^2}\left( {a,b} \right)}}{{A\left( {a,b} \right)}} \end{array} $

(30)

对所有a, b>0和a≠b成立当且仅当α₃≤0且β₃≥3/8.

证明  不等式(30) 可改写为

$ {\alpha _3}\frac{{A\left( {a,b} \right){\rm{TQ}}\left( {a,b} \right) - {G^2}\left( {a,b} \right)}}{{{Q^2}\left( {a,b} \right) - {G^2}\left( {a,b} \right)}} < {\beta _3}. $

(31)

因G(a, b), A(a, b), Q(a, b)和TQ(a, b)是对称和一阶齐次的, 不失一般性, 假设a>b>0，t=(log a-log b)/2>0, 则由式(5) 和(6)，不等式(31) 变成

$ {\alpha _3} < h\left( t \right) < {\beta _3}, $

(32)

其中函数h(t)的定义见引理6.

注意到

$ \mathop {\lim }\limits_{t \to {0^ + }} h\left( t \right) = \frac{3}{8},\;\;\;\mathop {\lim }\limits_{t \to \infty } h\left( t \right) = 0, $

(33)

由不等式(31)(32) 和等式(33) 协同引理6, 则不等式(30) 对所有a, b>0和a≠b成立当且仅当α₃≤0且β₃≥3/8.

推论3  双向不等式

$ sech\left( t \right) < {I_0}\left( t \right) < \frac{3}{4}\cosh \left( t \right) + \frac{1}{4}{\rm{sech}}\left( t \right) $

对所有t∈(0, +∞)成立.

定理4  双向不等式

$ \begin{array}{l} \sqrt {{\alpha _4}{Q^2}\left( {a,b} \right) + \left( {1 - {\alpha _4}} \right){G^2}\left( {a,b} \right)} < {\rm{TQ}}\left( {a,b} \right) < \\ \;\;\;\;\;\sqrt {{\beta _4}{Q^2}\left( {a,b} \right) + \left( {1 - {\beta _4}} \right){G^2}\left( {a,b} \right)} \end{array} $

(34)

对所有a, b>0和a≠b成立当且仅当α₄≤0且β₄≥1/4.

证明  不等式(30) 可改写为

$ {\alpha _4} < \frac{{{\rm{T}}{{\rm{Q}}^2}\left( {a,b} \right) - {G^2}\left( {a,b} \right)}}{{{{\rm{Q}}^2}\left( {a,b} \right) - {G^2}\left( {a,b} \right)}} < {\beta _4}. $

(35)

不失一般性, 假设a>b>0，t=(log a-log b)/2>0, 则由式(5) 和(6)，不等式(34) 变成

$ {\alpha _4} < k\left( t \right) < {\beta _4}, $

(36)

其中函数k(t)的定义见引理7.

注意到

$ \mathop {\lim }\limits_{t \to {0^ + }} k\left( t \right) = \frac{1}{4},\;\;\;\mathop {\lim }\limits_{t \to \infty } k\left( t \right) = 0, $

(37)

由不等式(35)(36) 和式(37) 协同引理7, 则不等式(34) 对所有a, b>0和a≠b成立当且仅当α₄≤0且β₄≥1/4.

推论4  不等式${{I}_{0}}\left( t \right)<\frac{1}{2}\sqrt{\cosh \left( 2t \right)+3}$对所有t∈(0, +∞)成立.