x>0,y>0根据基本不等式:x+y≥2√(xy)∴xy-x-y=xy-(x+y)=1≤xy-2√(xy)∴xy-2√(xy)≥1xy-2√(xy)-1≥0令√(xy)=t （t≥0）解得:√(xy)≤1-√2(舍去)√(xy)≥1+√2∴xy≥(1+√2)^2=3+2√2∵x+y=xy-1∴x+y≥2+2√2也可...