已知x.y.z∈r,求x/(2x+y+z)+y/(x+2y+z)+z/(x+y+2z)的最大值
1个回答
展开全部
x,y,z>0时x/(2x+y+z)+y/(x+2y+z)+z/(x+y+2z)-3/4
=（2x-y-z)/[4(2x+y+z)]+(2y-z-x)/[4(x+2y+z)]+(2z-x-y)/[4(x+y+2z)]
=(1/4){(x-y)[1/(2x+y+z)-1/(x+2y+z)]+(x-z)[1/(2x+y+z)-1/(x+y+2z)]+(y-z)[1/(x+2y+z)-1/(x+y+2z)]}
=(1/4){-(x-y)^2/[(2x+y+z)(x+2y+z)]-(x-z)^2/[(2x+y+z)(x+y+2z)]-(y-z)^2/[(x+2y+z)(x+y+2z)^2]}<=0,
∴x/(2x+y+z)+y/(x+2y+z)+z/(x+y+2z)<=3/4,
当x=y=z时取等号，
∴x/(2x+y+z)+y/(x+2y+z)+z/(x+y+2z)的最大值=3/4.
但是，x,y,z∈R时取x=1,2x+y+z→0+，最大值不存在.
=（2x-y-z)/[4(2x+y+z)]+(2y-z-x)/[4(x+2y+z)]+(2z-x-y)/[4(x+y+2z)]
=(1/4){(x-y)[1/(2x+y+z)-1/(x+2y+z)]+(x-z)[1/(2x+y+z)-1/(x+y+2z)]+(y-z)[1/(x+2y+z)-1/(x+y+2z)]}
=(1/4){-(x-y)^2/[(2x+y+z)(x+2y+z)]-(x-z)^2/[(2x+y+z)(x+y+2z)]-(y-z)^2/[(x+2y+z)(x+y+2z)^2]}<=0,
∴x/(2x+y+z)+y/(x+2y+z)+z/(x+y+2z)<=3/4,
当x=y=z时取等号，
∴x/(2x+y+z)+y/(x+2y+z)+z/(x+y+2z)的最大值=3/4.
但是，x,y,z∈R时取x=1,2x+y+z→0+，最大值不存在.
本回答由网友推荐
已赞过
已踩过<
评论
收起
你对这个回答的评价是？
推荐律师服务：
若未解决您的问题，请您详细描述您的问题，通过百度律临进行免费专业咨询
