Take for instance $f(x,y)=x^2$, that satisfies the assumptions. Since one has $f(x)+f(1-x)= 1-2x+2x^2 < 1=f(0)+f(1)$ for all $0<x<1$, it verifies the stated inequality for no parallelogram (inscribed in $R$, with vertices in $(0,0)$ and $(1,1)$), but $R$ itself.

Take for instance $f(x,y)=x^2$, that satisfies the assumptions. Since one has $f(x,y)+f(1-x,1-y)= 1-2x+2x^2 < 1=f(0,0)+f(1,1)$ for all $0<x<1$, it verifies the stated inequality for no parallelogram (inscribed in $R$, with vertices in $(0,0)$ and $(1,1)$) with non-vertical edges.