(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]
因为(x-y)^2≥0,(x-1)^2≥0,(y-1)^2≥0
(三项都取=号,有解x=y=1)
所以
(x2+y2)-(xy+x+y-1)≥0
x^2+y^2≥xy+x+y-1
已知x,y∈R,求证:x^2+y^2≥xy+x+y-1
(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]
因为(x-y)^2≥0,(x-1)^2≥0,(y-1)^2≥0
(三项都取=号,有解x=y=1)
所以
(x2+y2)-(xy+x+y-1)≥0
x^2+y^2≥xy+x+y-1