《数学通报》2004年第7期问题1504是:已知x,y,z∈(0,+∞),x+y+z=1,求1x2+y12+z82的最小值.我们将它一般化,得到定理设p,r,n∈N,n≥2,ai,xi∈(0,+∞),i=1,2,…,n,∑xip=1(以下总略去求和限),则(∑xarii)min=(∑aαi)1α,α=pp+r.证引入参数λ0,使如下平均不等式成立:aixir+…+x The “Mathematics Bulletin” 2004 issue 7, issue 1504 is: Know x, y, z ∈ (0, + ∞), x + y + z = 1, find the minimum value of 1x2 + y12 + z82. We will general it , get the theorem set p, r, n ∈ N, n ≥ 2, ai, xi ∈ (0, + ∞), i = 1, 2, ..., n, ∑ xip = 1 (total sum to the sum of ), then (∑xarii) min = (∑aαi) 1α, α = pp + r. The card introduces the parameter λ0 so that the following average inequality holds: aixir+...+x