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?



A consequence of convexity
You can use the triangle inequality to solve this by looking at each coordinate separately.


$F(x\_0,y\_0)-F(x\_2,y\_2)=F(x\_0,y\_0)-F(x\_2,y\_0)+F(x\_2,y\_0)-F(x\_2,y\_2)$ $\leq M\_1(F(x\_0,y\_0)-F(x\_1,y\_0))+M\_2(F(x\_2,y\_0)-F(x\_2,y\_1))$. Replacing $F(x\_1,y\_0)$ with $F(x\_1,y\_1)$ only increases the right side, and replacing both $x\_2$'s on the far right side with $x\_0$ only makes the right hand side larger by convexity. Finally, replace the last $x\_0$ that we just added with $x\_1$ without ncreasing the right hand side. This gives us $F(x\_0,y\_0)-F(x\_2,y\_2)\leq M(F(x\_0,y\_0)-F(x\_1,y\_1))$, where $M$ is $2\max(\frac{x\_2-x\_0}{x\_1-x\_0},\frac{y\_2-y\_0}{y\_1-y\_0})$. $M$ can also be the sum instead of the max, in which case we can drop the 2.


**Edit:** This argument doesn't work without additional assumptions; see the comments.


