Toán Lớp 8: Chứng minh rằng: `x^2 +y^2 +1 ≥xy+x+y`

Question

Toán Lớp 8:  Chứng minh rằng: x^2 +y^2 +1 ≥xy+x+y

baoquyen · Answer

x&sup2; + y&sup2; + 1 &ge; xy + x + y   &hArr;2(x&sup2; + y&sup2; + 1) &ge; 2(xy + x + y)   &hArr;2x&sup2; + 2y&sup2; + 2 &ge; 2xy + 2x + 2y   &hArr;2x&sup2; + 2y&sup2; + 2 - 2xy - 2x - 2y &ge;0   &hArr;x&sup2;+ x&sup2; + y&sup2; + y&sup2;+ 1 +1 - 2xy - 2x - 2y &ge;0   &hArr; (x&sup2; -2xy + y&sup2;) + (x&sup2; -2x + 1) + (y&sup2; -2y + 1) &ge;0   &hArr; (x-y)&sup2; + (x-1)&sup2; + (y-1)&sup2; &ge;0 ( lu&ocirc;n đ&uacute;ng v&igrave; (x-y)&sup2;&ge;0; (x-1)&sup2;&ge;0; (y-1)&sup2;&ge;0)   Vậy x&sup2; + y&sup2; + 1 &ge; xy + x + y

truonganhthi · Answer

x^2+y^2+1>=xy+x+y <=>2(x^2+y^2+1)>=2(xy+x+y) <=>2x^2+2y^2+2>=2xy+2x+2y <=>2x^2+2y^2+2-2xy-2x-2y>=0 <=>(x^2-2xy+y^2)+(x^2-2x+1)+(y^2-2y+1)>=0 <=>(x-y)^2+(x-1)^2+(y-1)^2>=0(luôn đúng) =>Điều phải chứng minh