设f(x)=e^x-xe(x>=1),则f(1)=0;
f(x)=f(x)-f(1)=f'(x')(x-1)
x'属于[1,x]
f'(x)=e^x-e,x>1时,f'(x)>0,
于是f(x)=f'(x')(x-1)>0
于是e^x>xe