已知X,y是正实数,且xy-x-y=1,求证x+y>/2+2√2

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也可以先从x+y考虑xy-(x+y)=1≤(x+y)^2/4-(x+y)∴(x+y)^2/4-(x+y)≥1∴(x+y)^2-4(x+y)-4≥0解得:x+y≥2+2√2 综上所述x+y的取值范围是:x+y≥2+2√2