P(x,x/(x+1))
AP^2=x^2+[x/(x+1)-2]^2=x^2+[-1/(x+1)-1]^2=x^2+1/(x+1)^2+1+2/(x+1)
令t=1/(x+1),则
AP^2=(1/t-1)^2+t^2+2t+1=1/t^2+t^2-2/t+2t+2
=(t-1/t)^2+2(t-1/t)+4,令 r=t-1/t
=r^2+2r+4
=(r+1)^2+3
所以当r=-1,即t-1/t=-1,t=(-1±√5)/2时
最小值AP=√3