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−  Let <math>a_1, a_2, \dots a_m</math> satisfy the given conditions. We will prove that for all <math>j, 1 \le j \le m,</math>
 
   
−  <cmath>a_j+a_{mj+1} \ge n+1</cmath>
 
− 
 
−  WLOG, let <math>a_1 < a_2 < \dots < a_m</math>. Assume that for some <math>j, 1 \le j \le m,</math>
 
− 
 
−  <cmath> a_j + a_{mj+1} \le n</cmath>
 
− 
 
−  This implies, for each <math>i, 1 \le i \le j,</math>
 
−  <cmath> a_i + a_{mj+1} \le n </cmath>
 
−  because <math>a_i \le a_j</math>
 
− 
 
−  For each of these values of i, we must have <math>a_i + a_{mj+1} = a_{k_i}</math> such that <math>a_{k_i}</math> is a member of the sequence for each <math>i</math>. Because <math>a_i > 0, a_{k_i} > a_{mj+1}</math>.
 
−  Combining all of our conditions we have that each of <math>k_i</math> must be distinct integers such that
 
−  <cmath>mj+1 < k_i \le m</cmath>
 
− 
 
−  However, there are <math>j</math> distinct <math>k_i</math>, but only <math>j1</math> integers satisfying the above inequality, so we have a contradiction. Our assumption that <math>a_j + a_{mj+1} \le n</math> was false, so <math>a_j + a_{mj+1} \ge n+1</math> for all <math>j</math> such that <math>1 \le j \le m</math> Summing these inequalities together for <math>1 \le j \le m</math> gives
 
−  <cmath> 2(a_1+a_2+ \dots a_m) \ge m(n+1) </cmath>
 
−  which rearranges to
 
−  <cmath> \frac{a_1+a_2+ \dots a_m}{m} \ge \frac{n+1}{2} </cmath>
 