amparvardi wrote: 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 \[0 \leq (n + 1)(a_n - a_{n+1}) \leq 2 \qquad \text{ for } n = 1, 2, 3, \ldots\] for $n = 1, 2, 3, \ldots$, let $b_n = a_n - a_{n + 1} $, then $0 \le a_n \le 1 \Rightarrow - 1 \le b_n \le 1,\;a_n - 2a_{n + 1} + a_{n + 2} \ge 0 \Leftrightarrow b_n \ge b_{n + 1} .$ LHS: if exist $k\in N^+$ such that $b_k < 0$, then for any $n\in N^+$, we have $a_k - a_{k + n} = b_k + b_{k + 1} + ... + b_{k + n - 1} \le nb_k .$ but $n \to + \infty $ give $a_k - a_{k + n} \to - \infty $, obvious contradiction to $0 \le a_n \le 1$. so it must have $b_k\ge0$ for any $k\in N^+$, i.e. $\left( {n + 1} \right)\left( {a_n - a_{n + 1} } \right) \ge 0$; RHS: $\left( {n + 1} \right)b_n \le b_1 + b_2 + ... + b_n + b_n = a_1 + a_n - 2a_{n + 1} \le 2$ $ \Rightarrow \left( {n + 1} \right)\left( {a_n - a_{n + 1} } \right) \le 2.$