令x1＜x2＜0
f(x2)-f(x1)
=x2^2+1-(x1^2+1)
=x2^2-x1^2
=(x2+x1)(x2-x1)
x2+x1＜0,x2-x1＞0
∴f(x）=(x2+x1)(x2-x1)＜0,得证