已知$f\left( x \right)$是定义在$\left( {0 , + \infty } \right)$上的可导函数，满足$f\left( x \right) > 0$，且$f\left( x \right) + f'\left( x \right) < 0$．
(1) 讨论函数$F\left( x \right) = {{\rm{e}}^x}f\left( x \right)$的单调性；
(2)设$0 < x < 1$，比较$xf\left( x \right)$与$\dfrac{1}{x}f\left( {\dfrac{1}{x}} \right)$的大小．
分析与解 (1)$F'\left( x \right) = {{\rm{e}}^x}\left[ {f\left( x \right) + f'\left( x \right)} \right] < 0$，所以$F\left( x \right)$在$\left( {0 ,+ \infty } \right)$上单调递减；
(2)将比较的两个函数作商有$$\dfrac{{xf\left( x \right)}}{{\dfrac{1}{x}f\left( {\dfrac{1}{x}} \right)}} = {x^2} \cdot \dfrac{{f\left( x \right)}}{{f\left( {\dfrac{1}{x}} \right)}} = \dfrac{{{x^2} \cdot {{\rm{e}}^{\frac{1}{x}}}}}{{{{\rm{e}}^x}}} \cdot \dfrac{{{{\rm{e}}^x}f\left( x \right)}}{{{{\rm{e}}^{\frac{1}{x}}}f\left( {\dfrac{1}{x}} \right)}}.$$因为$x < \dfrac{1}{x}$，所以${{\rm{e}}^x}f\left( x \right) > {{\rm{e}}^{\frac{1}{x}}}f\left( {\dfrac{1}{x}} \right)$．
接下来试图证明当$0 < x < 1$时，${x^2} \cdot {{\rm{e}}^{\frac{1}{x} - x}} > 1$．
$${x^2} \cdot {{\rm{e}}^{\frac{1}{x} - x}} > 1 \Leftarrow 2\ln x + \dfrac{1}{x} - x > 0.$$记$g\left( x \right) = 2\ln x + \dfrac{1}{x} - x$，则$$g'\left( x \right) = \dfrac{2}{x} - \dfrac{1}{{{x^2}}} - 1 = \dfrac{{ - {{\left( {x - 1} \right)}^2}}}{{{x^2}}} < 0,$$所以$g\left( x \right)$是单调递减函数．
所以当$0 < x < 1$时，$g\left( x \right) > g\left( 1 \right) = 0$．
综上所述，$$\dfrac{{xf\left( x \right)}}{{\dfrac{1}{x}f\left( {\dfrac{1}{x}} \right)}} > 1,$$也即$$xf\left( x \right) > \dfrac{1}{x}f\left( x \right).$$