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2013-07-19
展开全部
根据题意,首先对原不等式进行变形有x+y+1xy≤1x+1y+xy�6�2xy(x+y)+1≤x+y+(xy)2;再用做差法,让右式-左式,通过变形、整理化简可得右式-左式=(xy-1)(x-1)(y-1),又由题意中x≥1,y≥1,判断可得右式-左式≥0,从而不等式得到证明.
证明:由于x≥1,y≥1;则x+y+1xy≤1x+1y+xy�6�2xy(x+y)+1≤x+y+(xy)2;
用作差法,右式-左式=(x+y+(xy)2)-(xy(x+y)+1)
=((xy)2-1)-(xy(x+y)-(x+y))
=(xy+1)(xy-1)-(x+y)(xy-1)
=(xy-1)(xy-x-y+1)
=(xy-1)(x-1)(y-1);
又由x≥1,y≥1,则xy≥1;即右式-左式≥0
证明:由于x≥1,y≥1;则x+y+1xy≤1x+1y+xy�6�2xy(x+y)+1≤x+y+(xy)2;
用作差法,右式-左式=(x+y+(xy)2)-(xy(x+y)+1)
=((xy)2-1)-(xy(x+y)-(x+y))
=(xy+1)(xy-1)-(x+y)(xy-1)
=(xy-1)(xy-x-y+1)
=(xy-1)(x-1)(y-1);
又由x≥1,y≥1,则xy≥1;即右式-左式≥0
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