### 1975 IMO SL #4

Let $a_1, a_2, \ldots , a_n, \ldots$ be a sequence of real numbers such that $0 \leq a_n \leq 1$ and $a_n - 2a_{n+1} + a_{n+2} \geq 0$ for $n = 1, 2, 3, \ldots$. Prove that $0 \leq (n + 1)(a_n - a_{n+1}) \leq 2 \qquad \text{ for } n = 1, 2, 3, \ldots$

Let $\triangle a_n = a_n - a_{n-1}$. Due to the given condition, we have $\triangle a_n > \triangle a_{n+1}$. Suppose that for some $n$, we have $\triangle a_n<0$. Then, for each $k \ge n$, we get $\triangle a_k < \triangle a_n$. Hence $$a_n-a_{n-m}=\triangle a_n + \cdots + \triangle a_{n-m+1}<m \triangle a_n.$$ So, for sufficiently large $m$, it holds that $a_n-a_{n-m}<-1$, which is impossible. This shows the first part of the inequality.

Next, observe that $$n \ge \sum_{k=1}^{n} a_k = na_{n+1} + \sum_{k=1}^{n} k \triangle a_k \ge (1+2+\cdots+n)\triangle a_n = \frac{n(n+1)}{2} \triangle a_n.$$ Hence, we also have $(n+1)\triangle a_n \le 2$.

We are done. $\square$

1. Anonymous2/22/2022

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