Question

一条不等式的证明题证明：x^2+y^>=xy+x+y-1

Answer 1

(x^2+y^2)-(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

(x^2+y^2)-(xy+x+y-1)≥0

x^2+y^2≥xy+x+y-1