令x1>x2>=1
f(x1)-f(x2)
=x1²+2x1+1-x2²-2x2-1
=(x1+x2)(x1-x2)-2(x1-x2)
=(x1-x2)(x1+x2-2)
x1>x2>=1
所以x1-x2>0,x1+x2-2>0
所以x1>x2>=1设,f(x1)>f(x2)
递增
证明y=x^2=2x+1在[1,+∞)的单调性
令x1>x2>=1
f(x1)-f(x2)
=x1²+2x1+1-x2²-2x2-1
=(x1+x2)(x1-x2)-2(x1-x2)
=(x1-x2)(x1+x2-2)
x1>x2>=1
所以x1-x2>0,x1+x2-2>0
所以x1>x2>=1设,f(x1)>f(x2)
递增