x,y,z属于实数,u=x/(2x+y+z)+y/(x+2y+z)+z/(x+y+2z)的最大值,竞赛题急用!
1个回答

只考虑x,y,z>0的情景

x/[(x+y)+(x+z)]+y/[(x+y)+(y+z)]+z/[(x+z)+(y+z)]

设x≥y≥z则1/[(x+y)+(x+z)]≤1/[(x+y)+(y+z)]≤1/[(x+z)+(y+z)]

4倒序和≤(2x+y+z)/(2x+y+z)+(x+2y+z)/(x+2y+z)+(x+y+2z)/(x+y+2z)=3

所以原式的最大值为3/4