Let $f:\mathbb{R}\to\mathbb{R}$ a *convex decreasing* function. Let $x_0 < x_1 < x_2$.
Studying the behaviour of the difference quotient, it is clear that
$$f(x_0)-f(x_2) \leq M (f(x_0)-f(x_1))$$
with $M=\frac{x_2-x_0}{x_1-x_0}>0$.

Now take $F:\mathbb{R}^2\to\mathbb{R}$ *convex* and *decreasing* with respect to each variable. Let $x_0 < x_1 < x_2$ and $y_0 < y_1 < y_2$.
I ask if a similar condition holds, say for example
$$F(x_0,y_0)-F(x_2,y_2) \leq M (F(x_0,y_0)-F(x_1,y_1))$$
with $M=\max ( \frac{x_2-x_0}{x_1-x_0}, \frac{y_2-y_0}{y_1-y_0} )$ or $M=\frac{x_2-x_0}{x_1-x_0}+\frac{y_2-y_0}{y_1-y_0}$.

*Edit after Brian's answer*: you may add the hypothesis that $F$ is $C^2$ and also the mixed derivative $\frac{\partial^2F}{\partial x\partial y}$ is non-negative.

*Edit*: under the additional assumption the answer is yes with $M=\max(\frac{x_2-x_0}{x_1-x_0},\ \frac{x_2-x_0}{x_1-x_0})$. But is it possible to reach the same conclusion *without the additional assumption* on the mixed derivatives?