China National Team Selection Test

The maximum value of $f$ is $1$. Firstly, we prove that $f\le 1$. It suffices to show that $$n\sum _{i=1}^n(\sum _{j=1}^m{a}_{ij}{)}^2+m\sum _{j=1}^m(\sum _{i=1}^n{a}_{ij}{)}^2\le (\sum _{i=1}^n\sum _{j=1}^m{a}_{ij}{)}^2+mn\sum _{i=1}^n\sum _{j=1}^m{a}_{ij}^2,$$ $$\text{or}{\left(\sum _{i=1}^n\sum _{j=1}^m{a}_{ij}\right)}^2+mn\sum _{i=1}^n\sum _{j=1}^m{a}_{ij}^2-n\sum _{i=1}^n{\left(\sum _{j=1}^m{a}_{ij}\right)}^2-m\sum _{j=1}^m{\left(\sum _{i=1}^n{a}_{ij}\right)}^2\ge 0,$$ $$\text{or}\sum _{\begin{array}{c}1\le p<s\le n\\ 1\le q<r\le m\end{array}}(a_{pq}+a_{sr}-a_{pr}-a_{sq}{)}^2\ge 0.$$ 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 $$f\ge \frac{m+n}{mn+n}\stackrel{◯}{1}$$ 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 $\stackrel{◯}{1}\iff S\ge 0$. 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 $$\begin{array}{rl}S& =\frac{mn(n-1)}{m+n}\sum _{i=1}^n{r}_i^2+\frac{mn(m+1)}{m+n}\sum _{j=1}^m{c}_j^2\\ & =-mn\sum _{i=1}^n\sum _{j=1}^m{a}_{ij}^2+\sum _{1\le k<l\le n}(r_k-r_l{)}^2\\ & =\frac{mn(n-1)}{m+n}\sum _{j=1}^m\sum _{i=1}^n{a}_{ij}(r_i-a_{ij}{)}^2\\ & +\frac{mn(m+1)}{m+n}\sum _{i=1}^n\sum _{j=1}^m{a}_{ij}(c_j-a_{ij}{)}^2\\ & +\sum _{1\le k<l\le n}(r_k-r_l{)}^2.\end{array}$$ 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 $\frac{m+n}{mn+\min \{m,n\}}$.