【若实数x,y满足x^2+y^2=1时,2xy/(x+y-1)>=m恒成立,求m范围】

孙业国回答：

　　1.若x+y+1不等于0,2xy/(x+y-1)=2xy(x+y+1)/[(x+y-1)(x+y+1)]=x+y+1>=m,此时x+y的最小值为-根号2（x+y)^2=1+2xy1-根号2;

　　综上述,得证