判断并证明f(x)=x/x²+1在(0,﹢∞)上的单调性

f(x)在(0,1]上是增函数,在[1,+∞)是减函数.证明如下：
f(x1)-f(x2)=x1/(x1^2+1)-x2/(x2^2+1)
=[x1(x2^2+1)-x2(x1^2+1)]/[(x1^2+1)(x2^2+1)]
=[x1x2(x2-x1)+x1-x2]/[(x1^2+1)(x2^2+1)]
=(x2-x1)(x1x2-1)/[(x1^2+1)(x2^2+1)],
设0