Functional properties for $f_{1}$ and $f_{2}$ Can we say anything about the functions $f_{1}$ and $f_{2}$ using the inequalities?
$c \in [0,1]$ is a constant and both functions are $f_{i}: [0,1]^2\rightarrow[0,1]$, $i=1,2$
$c(f_{1}(x,y)+f_{2}(x,y))-x\times f_{2}(x,y)\geq c(f_{1}(x',y')+f_{2}(x',y'))-x\times f_{2}(x',y')$
$c(f_{1}(x,y)+f_{2}(x,y))-y\times f_{1}(x,y)\geq c(f_{1}(x',y')+f_{2}(x',y'))-y\times f_{1}(x',y')$
These inequalities can be written in this form as well:
$f_{1}(x,y)-f_{1}(x',y') \geq (\frac{x}{c}-1)(f_{2}(x,y)-f_{2}(x',y')) $
$f_{2}(x,y)-f_{2}(x',y') \geq (\frac{y}{c}-1)(f_{1}(x,y)-f_{1}(x',y')) $
 A: This is not an actual answer, but maybe you'll find some elements that will give you an idea.
Starting from your third and fourth inequalities (combining them):
$$f_{1}(x,y)-f_{1}(x',y') \ge \left(\frac{x}{c}-1\right)(f_{2}(x,y)-f_{2}(x',y'))$$
$$\Rightarrow f_{1}(x,y)-f_{1}(x',y') \geq \left(\frac{x}{c}-1\right)\left(\dfrac{y}{c} - 1\right)(f_{1}(x,y)-f_{1}(x',y'))$$
Both hands involve $f_{1}(x,y)-f_{1}(x',y')$, so that teach us that:
$$\left( \dfrac{x}{c} - 1 \right)\left( \dfrac{y}{c} - 1\right) = \left( \dfrac{x - c}{c} \right)\left( \dfrac{y - c}{c} \right) \le 1$$
Thus $(x - c)(y - c) = xy - cy - cx + c^2 \le c^2 \Leftrightarrow xy \le c(x+y)$.
Or, we know by AM-GM inequalities that $\dfrac{x + y}{2} \ge \sqrt{xy}$.
But since we have $(x,y) \in [0,1]^2$, we have $\sqrt{x} \ge x$.
From there we get that $\dfrac{x+y}{2} \ge \sqrt{xy} \ge xy$. But since $xy \le c(x+y)$, we get that $c \ge \dfrac{1}{2}$, otherwise that would contradict the AM-GM inequality (because we want it to work for any $x, y$).
That's the only info I could get from the statements, but I thought it would still be better to publish it than to keep it for me (and too big for a comment).
