IMO Shortlist 1999 problem A6


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April 2, 2012
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For n \geq 3 and a_{1} \leq a_{2} \leq \ldots \leq a_{n} given real numbers we have the following instructions:

- place out the numbers in some order in a ring;
- delete one of the numbers from the ring;
- if just two numbers are remaining in the ring: let S be the sum of these two numbers. Otherwise, if there are more the two numbers in the ring, replace

Afterwards start again with the step (2). Show that the largest sum S which can result in this way is given by the formula

S_{max}= \sum^n_{k=2} \begin{pmatrix} n -2 \\ [\frac{k}{2}] - 1\end{pmatrix}a_{k}.
Source: Međunarodna matematička olimpijada, shortlist 1999