x、y、z∈R+,依基本不等式得xy^2z^2＝16·x·(y/2)·(y/2)·(z/2)·(z/2)≤16[(x+y/2+y/2+z/2+z/2)/5]^5＝16[(x+y+z)/5]^5＝16/3125.∴x＝y/2＝z/2且x+y+z＝1,即x＝1/5,y＝2/5,z＝2/5时,所求最大值为：16/3125....