$0<f(x)<-f'(x)$,证明$xf(x)>\frac{1}{x}f(\frac{1}{x}),0<x<1$

hbghlyj

1+1=？

$0<f(x)<-f'(x)$等价于$e^xf(x)$单调递减,令$e^xf(x)=g(x)$，若$0<x<1$显然$g(x)>g(\frac{1}{x})$，要证明$xf(x)>\frac{1}{x}f(\frac{1}{x})$等价于证明$\frac{x}{e^x}>\frac{\frac{1}{x}}{e^{\frac{1}{x}}}$.