设f(x)=x-lnx (x>1)
∵f'(x)=1-1/x=(x-1)/x>0 (x>1)
∴f(x)严格单调递增函数 ==>f(x)>f(1) (x>1)
==>x-lnx>1>0 (x>1)
==>x>lnx (x>1)
故若x>1时,x>lnx