February 17th, 2015, 06:13 AM  #1 
Member Joined: Jan 2012 Posts: 51 Thanks: 1  induction
a1<a2...<an prove by induction that a1<(a1+a2+...+an)/n<an My proof: I have proved it for n=2: Now assuming that is works for n=k we have: a1< (a1+a2+...+ak)/k < ak but how to make a(k+1) appear in this inequality? I have tried to sum a(k+1) on both sides, but it didn't work 
February 17th, 2015, 06:28 AM  #2 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,355 Thanks: 2469 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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