A Simple Bound on Super-Additive Functions If $f(x)$ is a positive super-additive function ($\sum f(x) \leq f(\sum(x) $), can we prove that:
$$I = \sum_i f\left(\sum_j x_{ij}\right) + \sum_j f\left(\sum_i x_{ij}\right) - 2 \sum_i \sum_j f(x_{ij}) \leq f\left(\sum_i \sum_j x_{ij}\right)$$
If not, what extra assumptions on function $f$ do we need to make, so that the inequality would hold, and is true for both $f(x)=x^2$ and $f(x)=(x+1)log(x+1)$. 
Note1: a larger upper bound $2 f(\sum_i \sum_j x_{ij})$ can be derived for $I$ simply, since from positivity:
$\sum_i f(\sum_j x_{ij}) + \sum_j f(\sum_i x_{ij}) - 2 \sum_i \sum_j f(x_{ij}) \leq \sum_i f(\sum_j x_{ij}) + \sum_j f(\sum_i x_{ij}) $ 
And from super-additivity we get:
$\sum_i f(\sum_j x_{ij}) + \sum_j f(\sum_i x_{ij}) \leq  f(\sum_i \sum_j x_{ij})+ f(\sum_i \sum_j x_{ij}) $ 
Note2: the following inequalities does not hold: 
$\sum_i f(\sum_j x_{ij})  -  \sum_i \sum_j f(x_{ij}) \nleq \frac{1}{2}  f(\sum_i \sum_j x_{ij}) $ 
nor
$ f(\sum_i \sum_j x_{ij}) \nleq  2\sum_i \sum_j f(x_{ij})$ 
Note 3: if consider Matrix $X$, we can re-write the problem as:
$$I = \mathbf{1} f(X \mathbf{1}^T )+ f(\mathbf{1} X) \mathbf{1}^T - 2 \mathbf{1} f(X) \mathbf{1}^T \leq f(\mathbf{1} X \mathbf{1}^T) $$
where $\mathbf{1}$ is a vector of ones with appropriate shape so that the matrix-vector product is valid.
 A: If we let $$X=\left[\begin{array}{c} 1 && 1\\ 1&&1\end{array}\right]$$
and chose some superadditve function such that $f(1)=1$, $f(2)=5$, and $f(4)=10$ - such as a piecewise linear function interpolating those points - we would have that the inequality would reduce to
$$4f(2)-8f(1)\leq f(4)$$
$$12\leq 10$$
which is false. More generally, if we had an $n\times n$ matrix of $1$s, the equation would be:
$$2nf(n)-2n^2f(1)<f(n^2)$$
where, if we set $f(n^2)=nf(n)$ and rearrange we get
$$nf(n)<2n^2f(1)$$
$$f(n)<2nf(1)$$
which is not necessarily true of a super-additive function. Further, the best condition it suggests is that we want
$$f\left(\sum x\right)\leq 2\left(\sum f(x)\right)$$
which would clearly suffice, but rules out $x^2$. However, the equations before this establish that this is the best usable bound of the form
$$f\left(\sum x\right)\leq c\left(\sum f(x)\right)$$
meaning anything that admits $x^2$ will have to be more complicated than that - meaning that it might be necessary to use a double summation in the requirement, which would be of similar complexity to the desired result (at which point, you might as well let the result be the condition).
I think a tempting idea would be to require something like $f'''$ being always non-negative (in addition to $f''$, $f'$, and $f$ which are positive from the original definitions), but I suspect that such an attempt would quickly turn into very unenjoyable algebra.
