MathLabs

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 1 of 4: Normalize the summands
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bi=aii,b1+⋯+bn≥anb_i=\frac{a_i}{i},\qquad b_1+\cdots+b_n\ge a_n
Detailed analysis

Put bi=ai/i. The desired inequality becomes b1+...+bn >= an. It is true for n=1, so we use induction.