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 February 17th, 2015, 07:13 AM #1 Member   Joined: Jan 2012 Posts: 51 Thanks: 1 induction a1
 February 17th, 2015, 07:28 AM #2 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,512 Thanks: 2514 Math Focus: Mainly analysis and algebra You can do the two inequalities separately: \begin{aligned} && a_1 &\lt {1 \over k}(a_1 + a_1 + \cdots + a_k) \\ && k a_1 &\lt a_1 + a_1 + \cdots + a_k \\ &\text{now add a_{k+1} to both sides} & ka_1 + a_{k+1} &\lt a_1 + a_1 + \cdots + a_k + a_{k+1} \\ &\text{since a_{k+1} \gt a_1} & ka_1 + a_1 &\lt a_1 + a_1 + \cdots + a_k + a_{k+1} \\ && (k+1)a_1 &\lt a_1 + a_1 + \cdots + a_k + a_{k+1} \\ && a_1 &\lt {1 \over k+1}(a_1 + a_1 + \cdots + a_k + a_{k+1}) \end{aligned} There should be a similar derivation to prove the other inequality.

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