直线与圆相切，<==>圆心(1,1)到直线的距离=|m+n|/√[(m+1)^2+(n+1)^2]=1(半径),设m+n=u,则u^2=m^2+n^2+2u+2,m^2+n^2>=u^2/2,∴u^2>=u^2/2+2u+2,∴u^2-4u-4>=0,∴"u<=2-2√2或u>=2+2√2，即m+n<=2-2√2,或m+n>=2+2√2,为所求.
