Irish Mathematical Olympiad

Let $u_j = \binom{2j}{j}$, $j = 0, 1, 2, \dots$. For fixed $n, j$ we use the abbreviation $m = n-j$ to get $f(j) = u_j u_{n-j} = u_j u_m$ and $f(j+1) = u_{j+1} u_{n-j-1} = u_{j+1} u_{m-1}$.

a.
Since
$$u_{j+1} = \frac{2(2j+1)}{j+1} u_j \quad \text{and} \quad u_m = \frac{2(2m-1)}{m} u_{m-1}, \text{ we have}$$
$$\begin{align*} f(j) - f(j+1) &= u_j u_m - u_{j+1} u_{m-1} \\ &= 2 \left( \frac{2m-1}{m} - \frac{2j+1}{j+1} \right) u_{j+1} u_{m-1} \\ &= 2 \left( \frac{(2m-1)(j+1) - (2j+1)m}{m(j+1)} \right) u_{j+1} u_{m-1} \\ &= 2 \left( \frac{m-j-1}{m(j+1)} \right) u_{j+1} u_{m-1} \\ &\ge 0, \end{align*}$$
if $0 \le j < m = n-j$, i.e., $0 \le j < n/2$. This means that $f$ decreases steadily over the first $n/2$ terms and, because $f(n-j) = f(j)$, then increases, so that
$$\min\{f(j) : j = 0, 1, \dots, n\} = \begin{cases} u_{n/2}^2, & \text{if } n \text{ is even,} \\ u_{(n+1)/2}u_{(n-1)/2}, & \text{if } n \text{ is odd.} \end{cases}$$

b.
To establish (b), note, with the same notation, that
$$\begin{align*} f(j+1) + f(j-1) - 2f(j) &= u_{j+1}u_{m-1} + u_{j-1}u_{m+1} - 2u_j u_m \\ &= \frac{2(2j+1)}{j+1}u_{j-1}u_{m-1} + u_{j-1} \frac{2(2m+1)}{m+1}u_m - 2u_j u_m \\ &= \frac{4(4j^2-1)}{j(j+1)}u_{j-1}u_{m-1} + u_{j-1} \frac{4(4m^2-1)}{m(m+1)}u_{m-1} \\ &\quad - \frac{8(2j-1)(2m-1)}{j(m)}u_{j-1}u_{m-1} \\ &= 4u_{j-1}u_{m-1} \left( \frac{(4j^2-1)}{j(j+1)} + \frac{(4m^2-1)}{m(m+1)} - \frac{2(2j-1)(2m-1)}{jm} \right) \\ &= 4u_{j-1}u_{m-1} \frac{3m^2+m+3j^2+j-2jm-2}{j(j+1)m(m+1)} \\ &= 4u_{j-1}u_{m-1} \frac{(m+1)(2m-1)+(j+1)(2j-1)+(j-m)^2}{j(j+1)m(m+1)} \\ &> 0. \end{align*}$$
Hence (b) holds.