MathLabs

Problem 1

Let a1,…,ama_1,\ldots,a_m be distinct elements of {1,…,n}\{1,\ldots,n\} such that whenever ai+aj≤na_i+a_j\le n (with i≤ji\le j), the sum is also one of the aka_k. Prove a1+⋯+amm≥n+12\frac{a_1+\cdots+a_m}{m}\ge\frac{n+1}{2}.
Step 2 of 3: Prove the paired-sum inequality
In plain words

Prove the paired-sum inequality

ak+am−k+1≥n+1a_k+a_{m-k+1}\ge n+1
Detailed analysis

Suppose instead that ak+am−k+1≤na_k+a_{m-k+1}\le n. Then the kk distinct sums ai+am−k+1a_i+a_{m-k+1} for 1≤i≤k1\le i\le k are all at most nn, so each belongs to the set. Every one is strictly larger than am−k+1a_{m-k+1}, but there are only k−1k-1 elements with indices m−k+2,…,mm-k+2,\ldots,m, a contradiction.