Problem 2551

The maximum value of $f$ is $1$.
Firstly, we prove that $f \le 1$. It suffices to show that
$$n \sum_{i=1}^{n} (\sum_{j=1}^{m} a_{ij})^2 + m \sum_{j=1}^{m} (\sum_{i=1}^{n} a_{ij})^2 \le (\sum_{i=1}^{n} \sum_{j=1}^{m} a_{ij})^2 + mn \sum_{i=1}^{n} \sum_{j=1}^{m} a_{ij}^2,$$
$$\text{or} \quad \left(\sum_{i=1}^{n} \sum_{j=1}^{m} a_{ij}\right)^2 + mn \sum_{i=1}^{n} \sum_{j=1}^{m} a_{ij}^2 - n \sum_{i=1}^{n} \left(\sum_{j=1}^{m} a_{ij}\right)^2 - m \sum_{j=1}^{m} \left(\sum_{i=1}^{n} a_{ij}\right)^2 \ge 0,$$
$$\text{or} \quad \sum_{\substack{1 \le p < s \le n \\ 1 \le q < r \le m}} (a_{pq} + a_{sr} - a_{pr} - a_{sq})^2 \ge 0.$$
So $f \le 1$, and when all of $a_{ij}$ are equal to $1$, $f = 1$.

The minimum value of $f$ is $\frac{m+n}{mn + \min\{m, n\}}$.
To prove $f \ge \frac{m+n}{mn + \min\{m, n\}}$, without loss of generality, we assume $n \le m$. Hence it is sufficient to prove that
$$f \ge \frac{m+n}{mn+n} \qquad \textcircled{1}$$
Let
$$S = \frac{n^2(m+1)}{m+n} \sum_{i=1}^{n} r_i^2 + \frac{mn(m+1)}{m+n} \sum_{j=1}^{m} c_j^2$$
$$-(\sum_{i=1}^{n} \sum_{j=1}^{m} a_{ij})^2 - mn \sum_{i=1}^{n} \sum_{j=1}^{m} a_{ij}^2,$$
where $r_i = \sum_{j=1}^{m} a_{ij}$, $1 \le i \le n$, $c_j = \sum_{i=1}^{n} a_{ij}$, $1 \le j \le m$.

Now $\textcircled{1} \Leftrightarrow S \ge 0$. Consider Lagrange's equation.
$$(\sum_{i=1}^{n} a_i b_i)^2 = (\sum_{i=1}^{n} a_i^2)(\sum_{i=1}^{n} b_i^2) - \sum_{1 \le k < l \le n} (a_k b_l - a_l b_k)^2$$
Put $a_i = r_i$, $b_i = 1$, $1 \le i \le n$. Then
$$-(\sum_{i=1}^{n} \sum_{j=1}^{m} a_{ij})^2 = -n \sum_{i=1}^{n} r_i^2 + \sum_{1 \le k < l \le n} (r_k - r_l)^2,$$
and
$$\begin{aligned} S &= \frac{mn(n-1)}{m+n} \sum_{i=1}^{n} r_i^2 + \frac{mn(m+1)}{m+n} \sum_{j=1}^{m} c_j^2 \\ &= -mn \sum_{i=1}^{n} \sum_{j=1}^{m} a_{ij}^2 + \sum_{1 \le k < l \le n} (r_k - r_l)^2 \\ &= \frac{mn(n-1)}{m+n} \sum_{j=1}^{m} \sum_{i=1}^{n} a_{ij} (r_i - a_{ij})^2 \\ &\quad + \frac{mn(m+1)}{m+n} \sum_{i=1}^{n} \sum_{j=1}^{m} a_{ij} (c_j - a_{ij})^2 \\ &\quad + \sum_{1 \le k < l \le n} (r_k - r_l)^2. \end{aligned}$$
Since $a_{ij} \ge 0$, $r_i \ge a_{ij}$, $c_j \ge a_{ij}$, so $S \ge 0$.

When $a_{11} = a_{22} = \cdots = a_{mn} = 1$ and the other $a_{ij} = 0$, the minimum value of $f$ is $\frac{m+n}{mn+n}$.

With the above arguments, we conclude that the maximum value of $f$ is $1$ and the minimum value of $f$ is $\frac{m+n}{mn+\min\{m, n\}}$.