Question #1 is quite obvious after rewritting the inequalities.

Rewrite a_n <(a_0 + a_1 + ... + a_n)/n <= a_{n+1} as a_0 + a_1 + ... + a_{n-1} > (n-1)a_n and a_0 + a_1 + ... + a_{n-1} + a_n <= na_{n+1}.
It is natural to consider the sequences A_i = a_0 + a_1 + ... + a_{i-1} and B_i = (i-1)a_i. Both are increasing sequences.
A_1 > B_1, and for large enough i, as long as a_i >= a_0 + a_1, B_i >= A_i. n is the smallest i such that B_i >= A_i.

所有跟帖： 

• B_{i+1}-A_{i+1}=i(a_{i+1}-a_i)+(B_i-A_i) implies -dr_yin- ♂ (135 bytes) () 07/12/2014 postreply 19:49:27