设a,b,c为正实数,并且满足abc=1证明:(a-1+1/b)(b-1+1/c)(c-1+1/a)≤1

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设a,b,c为正实数,并且满足abc=1证明:(a-1+1/b)(b-1+1/c)(c-1+1/a)≤1设a,b,c为正实数,并且满足abc=1证明:(a-1+1/b)(b-1+1/c)(c-1+1/

设a,b,c为正实数,并且满足abc=1证明:(a-1+1/b)(b-1+1/c)(c-1+1/a)≤1
设a,b,c为正实数,并且满足abc=1
证明:(a-1+1/b)(b-1+1/c)(c-1+1/a)≤1

设a,b,c为正实数,并且满足abc=1证明:(a-1+1/b)(b-1+1/c)(c-1+1/a)≤1
令a=x/y,b=y/z,c=z/x
那么原不等式等价于证(x+z-y)(y+z-x)(x+y-z)≤xyz
若x+z-y,y+z-x,x+y-z有一个不大于0,不妨设x+y≤z,那么y+z-x≥y+x+y-x=2y>0,同理x+z-y>0
于是(x+z-y)(y+z-x)(x+y-z)≤0<xyz,成立
若x+z-y,y+z-x,x+y-z全大于0
再令x+z-y=u,y+z-x=v,x+y-z=w,那么x=(u+w)/2,y=(v+w)/2,z=(u+v)/2
原不等式又变为uvw≤[(u+v)/2][(u+w)/2][(v+w)/2],注意到(u+v)/2≥√(uv),(u+w)/2≥√(uw),(v+w)/2≥√(vw),三式相乘即得uvw≤[(u+v)/2][(u+w)/2][(v+w)/2],原不等式得证