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Combinatorics Difficulty 6.5 National olympiad Prove it Estonia

Call a tuple (a1,,an)(a_1, \dots, a_n) of real numbers stable if the sums a1+a2++aka_1 + a_2 + \dots + a_k, as well as the sums ak+ak+1++ana_k + a_{k+1} + \dots + a_n, where in both cases 0<kn0 < k \le n, are either all negative or all non-negative.
For instance, the tuple (3,1,2)(3, -1, 2) is stable, since:
30,20,3+(1)0,(1)+20,3+(1)+20;3+(1)+20. \begin{array}{lclcl} 3 & \ge & 0, & 2 & \ge 0, \\ 3 + (-1) & \ge & 0, & (-1) + 2 & \ge 0, \\ 3 + (-1) + 2 & \ge & 0; & 3 + (-1) + 2 & \ge 0. \end{array}
Prove that in any stable tuple with at least 3 terms where all terms are alternately negative and non-negative (it is unknown whether the first term is negative or non-negative), there exist 3 consecutive terms that together (without reordering) form a stable tuple on their own.

Solution

Consider terms whose absolute value is minimal in the tuple. If there exists a negative such element, denote it aia_i, then the sum of aia_i and its any neighbour is non-negative. Thus aia_i is neither the first nor the last in the tuple because of stability of the tuple. But then both ai1+aia_{i-1} + a_i and ai+ai+1a_i + a_{i+1} are non-negative, as well as ai1+ai+ai+1a_{i-1} + a_i + a_{i+1}, hence ai1a_{i-1}, aia_i, ai+1a_{i+1} together form a stable subtuple. If all elements with minimal absolute value are non-negative then let aia_i be any of them. Analogously to the previous case, both ai1+aia_{i-1} + a_i and ai+ai+1a_i + a_{i+1} are negative, as well as ai1+ai+ai+1a_{i-1} + a_i + a_{i+1}, hence ai1a_{i-1}, aia_i, ai+1a_{i+1} together form a stable tuple.

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