Problem 2

Let a1,a2,... be a sequence of real numbers satisfying a(i+j) <= ai+aj for all positive integers i,j. Prove that a1+a2/2+...+an/n >= an for every positive integer n.

Step 3 of 4: Pair terms using subadditivity

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2(a_1+\cdots+a_n)=(a_1+a_n)+(a_2+a_{n-1})+\cdots+(a_n+a_1)\ge n a_{n+1}

Detailed analysis

The given condition gives ai+a(n+1-i) >= a(n+1) for each i. Pairing all terms therefore yields 2(a1+...+an) >= n a(n+1). Combining this with the preceding line gives (n+1) times the partial sum of the bi at least n a(n+1).