设x≥1,y≥1,求证x^2y^2+x+y≥xy(x+y)+1
问题描述：
设x≥1,y≥1,求证x^2y^2+x+y≥xy(x+y)+1
答
因为：x≥1,y≥1x-1>=0,y-1>=0,xy-1>=0x^2y^2+x+y-xy(x+y)-1=(x^2y^2-1)-(x+y)(xy-1)=(xy-1)(xy+1)-(x+y)(xy-1)=(xy-1)(xy-x-y+1)=(xy-1)(x-1)(y-1)>=0所以：x^2y^2+x+y≥xy(x+y)+1