x>=1,y>=1证明:x+y+1\xy
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x>=1,y>=1证明:x+y+1\xy<=1\x+1\y+xy
证明:由于x≥1,y≥1;则x+y+1xy≤1x+1y+xy?xy(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
再问: 能不能说一下解题思路
再答: 由题设可知: (xy)≥1 原不等式两边同乘以xy, 可得: x²y+xy²+1≤x+y+(xy)² 移项,因式分解可得: (xy-1)(xy+1-x-y)≥0. (xy-1)(x-1)(y-1)≥0 由题设,显然成立。 证明: 因为x≥1,y≥1, 所以(xy-1)(x-1)(y-1)≥0, 展开得x^2y^2-x^2y-xy^2+x+y-1≥0,移项得:x^2y+xy^2+1≤x^2y^2+x+y。 两边同除以xy得x+y+1/xy≤1/x+1/y+xy
再问: Tank you
再答: thank you 哦 少个H
再问: .......
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
再问: 能不能说一下解题思路
再答: 由题设可知: (xy)≥1 原不等式两边同乘以xy, 可得: x²y+xy²+1≤x+y+(xy)² 移项,因式分解可得: (xy-1)(xy+1-x-y)≥0. (xy-1)(x-1)(y-1)≥0 由题设,显然成立。 证明: 因为x≥1,y≥1, 所以(xy-1)(x-1)(y-1)≥0, 展开得x^2y^2-x^2y-xy^2+x+y-1≥0,移项得:x^2y+xy^2+1≤x^2y^2+x+y。 两边同除以xy得x+y+1/xy≤1/x+1/y+xy
再问: Tank you
再答: thank you 哦 少个H
再问: .......
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