由 (n+1)an+1−(a0+⋯+an+an+1)=−dn+1(n+1)a_{n+1}-(a_0+\cdots+a_n+a_{n+1})=-d_{n+1}(n+1)an+1−(a0+⋯+an+an+1)=−dn+1 得 nan+1−(a0+⋯+an)=−dn+1na_{n+1}-(a_0+\cdots+a_n)=-d_{n+1}nan+1−(a0+⋯+an)=−dn+1。因此不等式 a0+⋯+ann≤an+1\frac{a_0+\cdots+a_n}{n}\le a_{n+1}na0+⋯+an≤an+1 等价于 a0+⋯+an≤nan+1a_0+\cdots+a_n\le na_{n+1}a0+⋯+an≤nan+1,即 −dn+1≥0-d_{n+1}\ge0−dn+1≥0,即 dn+1≤0d_{n+1}\le0dn+1≤0。于是问题归结为寻找唯一的 n≥1n\ge1n≥1 使 dn>0≥dn+1d_n>0\ge d_{n+1}dn>0≥dn+1。