Let $f\colon \mathbf{R} \times \mathbf{R}^+ \rightarrow \mathbf{R}$ be defined by $f(x,y) = \frac{x^2}{y}$. Let $X = \left\lbrace x_1, \dots, x_n\right\rbrace \subseteq \mathbf{R}$, $Y = \left\lbrace y_1, \dots, y_n\right\rbrace \subseteq \mathbf{R}^+$ be ordered so that $\frac{x_1}{y_1} \leq \dots \leq \frac{x_n}{y_n}$. Define the set function $F\colon 2^n \rightarrow \mathbf{R}$ by $F(S) = \frac{(\sum_{i \in S}x_i)^2}{\sum_{i \in S}y_i}$ for $S \subseteq \left\lbrace 1, \dots n\right\rbrace$

$F$ may fail to be submodular, even for $X$ positive - for $X = \left\lbrace0, 7, 8, 9\right\rbrace$, $Y = \left\lbrace4, 7, 1, 1\right\rbrace$ take $$ \begin{align} S &= \left\lbrace 1, 3\right\rbrace \\ T &= \left\lbrace0, 2, 3\right\rbrace \\ S \cap T &= \left\lbrace 3\right\rbrace\\ S \cup T &= \left\lbrace 0, 1, 2, 3\right\rbrace \\ \end{align} $$ and $$ F(S) + F(T) \approx 80.1667 \\ F(S \cup T) + F(S \cap T) \approx 125.3077 $$

I think $F$ is submodular for intervals, however, in other words $$ F(S) + F(T) \geq F(S \cup T) + F(S \cap T) $$

for $S$, $T$ intervals of the form $\left\lbrace j, j+1, \dots k\right\rbrace$, for $j \leq k$, for any specification of $X$, $Y$. I haven't been able to prove this - can anyone prove or provide a counterexample?