China National Team Selection Test

The problem conditions imply that
$$X_{i,j} \ge (i + j - 1) \cdot a_{i,j}, \quad Y_{i,j} \le (m + n - i - j + 1) \cdot a_{i,j}.$$
Hence, we have
$$\begin{align*} \prod_{i=1}^{m} \prod_{j=1}^{n} X_{i,j} &\ge \prod_{i=1}^{m} \prod_{j=1}^{n} (i+j-1) \cdot a_{i,j} \\ &= \prod_{i=1}^{m} \prod_{j=1}^{n} (m+n-i-j+1) \cdot a_{i,j} \\ &\ge \prod_{i=1}^{m} \prod_{j=1}^{n} Y_{i,j}, \end{align*}$$
where the equality on the second line is due to the one-to-one correspondence $(i, j) \leftrightarrow (m + 1 - i, n + 1 - j)$, and $i + j - 1$ corresponds to $m + n - i - j + 1$.

We shall use the following inequality:
For $0 \le c_1 \le \cdots \le c_t$, $0 \le d_1 \le \cdots \le d_t$,
$$(c_1+d_1)(c_2+d_2)\cdots(c_t+d_t) \le (c_1+d_t)(c_2+d_{t-1})\cdots(c_t+d_1). \quad \textcircled{1}$$
This is because, for each $i$,
$$(c_i+d_i)(c_{t+1-i}+d_{t+1-i}) \le (c_i+d_{t+1-i})(c_{t+1-i}+d_i).$$
Multiplying the above inequalities for $i = 1, \dots, t$ gives
$$\prod_{i=1}^{t} ((c_i+d_i)(c_{t+1-i}+d_{t+1-i})) \le \prod_{i=1}^{t} ((c_i+d_{t+1-i})(c_{t+1-i}+d_i)),$$
and taking the square root on both sides of the above inequality gives (1).
For the original problem, let $L_{i,j} = \sum_{q=1}^{j} a_{i,q}$, $R_{i,j} = \sum_{q=j}^{n} a_{i,q}$,
$$U_{i,j} = \sum_{p=1}^{i} a_{p,j}, \quad D_{i,j} = \sum_{p=i}^{m} a_{p,j}.$$
Then
$$X_{i,j} = L_{i,j} + U_{i,j} - a_{i,j}, \quad Y_{i,j} = D_{i,j} + R_{i,j} - a_{i,j}.$$
Apply (1) to derive
$$\begin{align*} \prod_{i=1}^{m} \prod_{j=1}^{n} X_{i,j} &= \prod_{i=1}^{m} \prod_{j=1}^{n} (L_{i,j} + (U_{i,j} - a_{i,j})) \\ &\ge \prod_{i=1}^{m} \prod_{j=1}^{n} (L_{i,n+1-j} + (U_{i,j} - a_{i,j})) \\ &\quad \text{(since } \{L_{i,j}\}_{j=1}^{n} \text{ increases while } \{U_{i,j} - a_{i,j}\}_{j=1}^{n} \text{ decreases)} \\ &\ge \prod_{i=1}^{m} \prod_{j=1}^{n} (R_{i,j} + (U_{i,j} - a_{i,j})) \quad \text{(since } L_{i,n+1-j} \ge R_{i,j}) \\ &= \prod_{j=1}^{n} \prod_{i=1}^{m} ((R_{i,j} - a_{i,j}) + U_{i,j}) \\ &\ge \prod_{j=1}^{n} \prod_{i=1}^{m} ((R_{i,j} - a_{i,j}) + U_{m+1-i,j}) \\ &\quad \text{(since } \{R_{i,j} - a_{i,j}\}_{i=1}^{m} \text{ decreases while } \{U_{i,j}\}_{i=1}^{m} \text{ increases)} \end{align*}$$
$$\begin{align*} &\ge \prod_{j=1}^{n} \prod_{i=1}^{m} ((R_{i,j} - a_{i,j}) + D_{i,j}) \quad (\text{since } U_{m+1-i,j} \ge D_{i,j}) \\ &= \prod_{i=1}^{m} \prod_{j=1}^{n} Y_{i,j}. \end{align*}$$