设A(x1,y1),B(x2,y2).
0A*OB=x1x2+y1y2
设x1+y1=s1 x1-y1=t1
x2+y2=s2 x2-y2=t2
因为(x+y)(x-y)=1 s1t1=1 s2t2=1
则OA*OB=(s1+t1)(s2+t2)/4+(s1-t1)(s2-t2)
=(s1s2+t1t2)/2
>=根号下s1t1s2t2=1
总上,最小值是1
设A(x1,y1),B(x2,y2).
0A*OB=x1x2+y1y2
设x1+y1=s1 x1-y1=t1
x2+y2=s2 x2-y2=t2
因为(x+y)(x-y)=1 s1t1=1 s2t2=1
则OA*OB=(s1+t1)(s2+t2)/4+(s1-t1)(s2-t2)
=(s1s2+t1t2)/2
>=根号下s1t1s2t2=1
总上,最小值是1