(x2+y2)-(xy+x+y-1)
=(1/2)*[(x^2-2xy+y^2)+(x^2-2x+1)+(y^2-2y+1)]
=(1/2)*[(x-y)^2+(x-1)^2+(y-1)^2]
(x-y)^2≥0,(x-1)^2≥0,(y-1)^2≥0
所以
(x2+y2)-(xy+x+y-1)≥0
x^2+y^2+1≥xy+x+y
已知x.y∈R,求证x2+y2+1≥x+y+xy
(x2+y2)-(xy+x+y-1)
=(1/2)*[(x^2-2xy+y^2)+(x^2-2x+1)+(y^2-2y+1)]
=(1/2)*[(x-y)^2+(x-1)^2+(y-1)^2]
(x-y)^2≥0,(x-1)^2≥0,(y-1)^2≥0
所以
(x2+y2)-(xy+x+y-1)≥0
x^2+y^2+1≥xy+x+y