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若x+y+z=xyz(x,z,y>0)求1/(1+xy)+1/(1+xz)+1/(1+yz)最大值
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若x+y+z=xyz (x,z,y>0) 求1/(1+xy)+1/(1+xz)+1/(1+yz)最大值
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答案和解析
显然xy>1.xz>1.yz>1
原式<=1/(1+x*x)+1(1+y*y)+1(1+z*z)(做差可证)
当且仅当x=y=z=根号3 取等
故max=3/4
原式<=1/(1+x*x)+1(1+y*y)+1(1+z*z)(做差可证)
当且仅当x=y=z=根号3 取等
故max=3/4
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