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设x≥1,y≥1,求证x^2y^2+x+y≥xy(x+y)+1
人气:274 ℃ 时间:2020-06-21 08:38:14
解答
因为:x≥1,y≥1x-1>=0,y-1>=0,xy-1>=0x^2y^2+x+y-xy(x+y)-1=(x^2y^2-1)-(x+y)(xy-1)=(xy-1)(xy+1)-(x+y)(xy-1)=(xy-1)(xy-x-y+1)=(xy-1)(x-1)(y-1)>=0所以:x^2y^2+x+y≥xy(x+y)+1
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