f(x+t)=(x+t)^2+2(x+t)+1
=x^2+2xt+t^2+2x+2t+1
≤x
x^2+(2t+1)x+(t+1)^2≤0
令g(x)=x^2+(2t+1)x+(t+1)^2,方程x^2+(2t+1)x+(t+1)^2=0的两个根为x1,x2(假设x1
f(x+t)=(x+t)^2+2(x+t)+1
=x^2+2xt+t^2+2x+2t+1
≤x
x^2+(2t+1)x+(t+1)^2≤0
令g(x)=x^2+(2t+1)x+(t+1)^2,方程x^2+(2t+1)x+(t+1)^2=0的两个根为x1,x2(假设x1