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Algebra Difficulty 4.5 AIME Prove it Soviet Union

Problem:

Given a0a_0, a1a_1, ..., ana_n, satisfying a0=an=0a_0 = a_n = 0, and ak12ak+ak+10a_{k - 1} - 2a_k + a_{k + 1} \geq 0 for k=1,2,...,n1k = 1, 2, ..., n-1. Prove that all the numbers are negative or zero.

Solution

Solution:

The essential point is that if we plot the values ara_r against rr, then the curve formed by joining the points is cup shaped. Its two endpoints are on the axis, so the other points cannot be above it. There are many ways of turning this insight into a formal proof. Barry Paul's was neater than mine: ar+1ararar1a_{r + 1} - a_r \geq a_r - a_{r - 1}. Hence (easy induction) if asas1>0a_s - a_{s - 1} > 0, then an>asa_n > a_s. Take asa_s to be the first positive, then certainly as>as1a_s > a_{s - 1}, so an>0a_n > 0. Contradiction.

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