设0<x1<x2
f(x1)-f(x2)=1/x1 -1/x2
=(x2-x1)/(x1 x2)
∵x2-x1>0,x1x2>0
∴f(x1)-f(x2)>0
即f(x1)>f(x2)
∴f(x)在(0,+∞)上单调减.