Let $f$ be a function defined on the unit square $R = [0,1]^2 \subseteq \mathbf{R}^2$ satisfying

 - $f \geq 0$, $f(0,0) = 0$,
 - $\frac{\partial{f}}{\partial{x}} \geq 0$, $\frac{\partial{f}}{\partial{y}} \leq 0$,
 - $\frac{\partial^2{f} }{\partial{x}\partial{y}} \leq 0$.

 The last condition is equivalent to the inequality $f(x_1,y_1) + f(x_2,y_2) \geq f(\min\left(x_1,x_2\right), \min\left(y_1,y_2\right)) + f(\max\left(x_1,x_2\right), \max\left(y_1,y_2\right))$ on any rectangle, which can be obtained by integrating $\frac{\partial^2{f} }{\partial{x}\partial{y}}$ over the rectangle.

If we label the vertices of the rectangle counterclockwise $v_1, \dots, v_4$, starting at the upper right, this is saying that $f(v_2) + f(v_4) \geq f(v_1) + f(v_3)$. 

Does this property also hold for parallelograms inscribed in $R$ with wlog $v_1 = (1,1), v_3 = (0,0)$?

For $v_2 = (x_2,y_2)$, $v_4 = (x_1,y_1)$, the linear function mapping $R$ to the parallelogram $P$ is

\begin{pmatrix}
&x_1 & x_2 \\
&y_1 & y_2
\end{pmatrix}

After change of variables the differential condition on $P$ can then be written

$(x_1y_2 + y_1x_2)f_{xy} + x_1x_2f_{xx} + y_1y_2f_{yy}$, I don't know that that is necessarily nonpositive.

Note that without the first bullet condition, the answer is no <https://mathoverflow.net/questions/381405/planar-function-inequality-on-parallelograms>. 

I'm pretty sure the answer is yes with the additional condition - proof?