已知x,y,z∈R,求证:x^2+y^2>=xy+x+y-1
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已知x,y,z∈R,求证:x^2+y^2>=xy+x+y-1已知x,y,z∈R,求证:x^2+y^2>=xy+x+y-1已知x,y,z∈R,求证:x^2+y^2>=xy+x+y-12(x^2+y^2)
已知x,y,z∈R,求证:x^2+y^2>=xy+x+y-1
已知x,y,z∈R,求证:x^2+y^2>=xy+x+y-1
已知x,y,z∈R,求证:x^2+y^2>=xy+x+y-1
2(x^2+y^2)-2(xy+x+y-1)
=2x^2+2y^2-2xy-2x-2y+2
=(x^2-2xy+y^2)+(x^2-2x+1)+(y^2-2y+1)
=(x-y)^2+(x-1)^2+(y-1)^2
平方相加大于等于0
(x-y)^2+(x-1)^2+(y-1)^2>=0
所以2(x^2+y^2)-2(xy+x+y-1)>=0
2(x^2+y^2)>=2(xy+x+y-1)
所以x^2+y^2>=xy+x+y-1
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