已知x,y∈R,求证:x+y+1≥x+y+xy

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已知x,y∈R,求证:x+y+1≥x+y+xy

已知x,y∈R,求证:x+y+1≥x+y+xy
已知x,y∈R,求证:x+y+1≥x+y+xy

已知x,y∈R,求证:x+y+1≥x+y+xy
2x+2y+2≥2x+2y+2xy x+x+y+y-2x-2y+1+1-2xy≥0 (x+1)+(y+1)+(x-y)≥0 平方都≥0所以成立

证明:∵x,y∈R ∴x+y+1-(x+y+xy) =(x-2x+1)+(y-2y+1)+x+y-xy-1 =(x-1)^2+(y-1)^2+(x-1)+y(1-x) =(x-1)^2+(y-1)^2-(x-1)(y-1) =(x-1)^2-(x-1)(y-1)+[(y-1)^2]/4+3[(y-1)^2]/4 =[(x-1)-(y-1)/2]^2+3[(y-1)^2]/4...

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证明:∵x,y∈R ∴x+y+1-(x+y+xy) =(x-2x+1)+(y-2y+1)+x+y-xy-1 =(x-1)^2+(y-1)^2+(x-1)+y(1-x) =(x-1)^2+(y-1)^2-(x-1)(y-1) =(x-1)^2-(x-1)(y-1)+[(y-1)^2]/4+3[(y-1)^2]/4 =[(x-1)-(y-1)/2]^2+3[(y-1)^2]/4 又∵[(x-1)-(y-1)/2]^2≥0;(y-1)^2≥0 ∴[(x-1)-(y-1)/2]^2+3[(y-1)^2]/4≥0 ∴x+y+1-(x+y+xy)≥0 ∴x+y+1≥ x+y+xy

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