Let $A B C D$ be a unit square, and let $P$ be a point inside triangle $A B C$. Let $E$ and $F$ be the points on $A B$ and $A D$, respectively, such that $A E P F$ is a rectangle. Let $x$ and $y$ denote the lengths of $A E$ and $A F$, respectively.

Determine a two-variable polynomial $f(x, y)$ with the property that the area of $A E P F$ is greater than the area of each of the quadrilaterals $E B C P$ and $F P C D$ if and only if $f(x, y)>0$.
The diagram is as shown in Figure 1. The area of $A E P F$ is $x y$, the area of $E B C P$ is
$$
B E \cdot \frac{E P+B C}{2}=(1-x) \frac{y+1}{2}=\frac{(1-x)(1+y)}{2},
$$
and by symmetry (or similar calculations), the area of $F P C D$ is $\frac{(1-y)(1+x)}{2}$.
It is given that $P$ is inside triangle $A B C$, which implies $x>y$. The area of $F P C D$ minus the area of $E P C B$ is
$$
\begin{aligned}
\frac{(1-y)(1+x)}{2}-\frac{(1-x)(1+y)}{2} & =\frac{1+x-y-x y-(1+y-x-x y)}{2} \\
& =\frac{2 x-2 y}{2} \\
& =x-y>0,
\end{aligned}
$$
so this means that the area of $A E P F$ is greater than the area of each of the other two quadrilaterals if and only if it is greater than the area of $F P C D$. (Note that if we allow $P$ to lie on the boundary of $A B C$ then we have $x \geq y$, and we reach the same conclusion here that the area of $A E P F$ is greater than the area of each of the other two quadrilaterals if and only if it is greater than the area of $F P C D$.)

Now
$$
\begin{array}{l}
\operatorname{area}(A E P F)>\operatorname{area}(F P C D) \\
x y>\frac{(1-y)(1+x)}{2} \\
\Longleftrightarrow \quad 2 x y-(1-y)(1+x)>0 \\
\Longleftrightarrow \quad 2 x y-1-x+y+x y>0 \\
\Longleftrightarrow \quad 3 x y-x+y-1>0 \text {, } \\
\end{array}
$$
so we can take $f(x, y)=3 x y-x+y-1$.