Question 157
A sequence of real numbers $(x_n)_{n=0}^{\infty}$ is *concave* if
$$
x_n \ge \frac{x_{n-1} + x_{n+1}}{2}, \quad \text{for all } n \in \mathbb{N}.
$$
Given two sequences of real numbers $(a_n)_{n=0}^{\infty}$ and $(b_n)_{n=0}^{\infty}$, we define a new sequence $(c_n)_{n=0}^{\infty}$ by:
$$
c_n = \max\{a_i + b_{n-i} \mid 0 \le i \le n\}, \quad n \ge 0.
$$
Show that if $(a_n)_{n=0}^{\infty}$ and $(b_n)_{n=0}^{\infty}$ are concave, then so is $(c_n)_{n=0}^{\infty}$.
$$
x_n \ge \frac{x_{n-1} + x_{n+1}}{2}, \quad \text{for all } n \in \mathbb{N}.
$$
Given two sequences of real numbers $(a_n)_{n=0}^{\infty}$ and $(b_n)_{n=0}^{\infty}$, we define a new sequence $(c_n)_{n=0}^{\infty}$ by:
$$
c_n = \max\{a_i + b_{n-i} \mid 0 \le i \le n\}, \quad n \ge 0.
$$
Show that if $(a_n)_{n=0}^{\infty}$ and $(b_n)_{n=0}^{\infty}$ are concave, then so is $(c_n)_{n=0}^{\infty}$.