所求

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设AB=x,PA=y
则PD=y-1
所求为：x^2y/2的最小值.
易证明△PAB和△PCD相似
DC:AB=PD:PB
1/x=(y-1)/PB
1/x^2=(y-1)^2/PB^2
PB^2=x^2+y^2,代入
1/x^2=(y-1)^2/(x^2+y^2)
化简,得到：
y=2x^2/(x^2-1)
AB*S△PAB=x^2y/2=x^4/(x^2-1)
=[(x^2+1)(x^2-1)+1]/(x^2-1)
=x^2+1+1/(x^2-1)
=(x^2-1)+1/(x^2-1)+2
当x^2-1=1/(x^2-1)时,有最小值,
此时,(x^2-1)^2=1
x=√2,
AB*S△PAB有最小值：4