The Heilbronn triangle problem seems closely related, and the method of Roth for a power-saving upper bound goes through a similar area-covering argument. Let me sketch the basic idea Roth uses to show that for any $n$ points in a unit disk, some three of them form a triangle of area $n^{-1-\epsilon}$. All constants are missing intentionally.

Suppose that there are $n$ points in the unit disk $D$ no triple of whom form a triangle of area $<\Delta$, and $\Delta$ is large (i.e. $~1/n$ contrary to our expectations). Then, this means that for every pair of points $t = (p_1, p_2)\in D^2$, the infinite linear strip $H(t,\Delta/d(t))$ of width $\Delta/d(t)$ around the line $p_1 p_2$, where $d(t)$ is the distance between $p_1,p_2$, contains no other points of our set.

Roth then finishes by (A) if the pair $t$ is close together, as some pairs have to be by pigeonhole (~$1/\sqrt{n}$), this means that the corresponding linear strip $H(t,\Delta/d(t))$ contains much less than the expected number of points ($\approx \sqrt{n}$), and (B) if instead of considering all pairs $t$ we restrict to some subcollection of pairs with similar slopes, the corresponding regions essentially can't intersect (or else there would be a small-area triangle).

You might want to look for an analogue of (B): some subfamily of the all your pairs $f_{ij}$ have some structure making them repel one another such as having similar slopes. It might be possible to transform the problem so that if two pairs have similar slopes and the corresponding $f_{ij}$ intersect within the disk, among the four points involved some three will give you the triangle you want. I think the main area contribution of $f_{ij}$ looks like the pair of strips parallel to the line segment between the circle centers.

The Heilbronn triangle problem seems closely related, and the method of Roth for a power-saving upper bound goes through a similar area-covering argument. Let me sketch the basic idea Roth uses to show that for any $n$ points in a unit disk, some three of them form a triangle of area $<n^{-1-\epsilon}$. All constants are missing intentionally.

Suppose that there are $n$ points in the unit disk $D$ no triple of whom form a triangle of area $<\Delta$, and $\Delta$ is large (i.e. $~1/n$ contrary to our expectations). Then, this means that for every pair of points $t = (p_1, p_2)\in D^2$, the infinite linear strip $H(t,\Delta/d(t))$ of width $\Delta/d(t)$ around the line $p_1 p_2$, where $d(t)$ is the distance between $p_1,p_2$, contains no other points of our set.

Roth then finishes by (A) if the pair $t$ is close together, as some pairs have to be by pigeonhole (~$1/\sqrt{n}$), this means that the corresponding linear strip $H(t,\Delta/d(t))$ contains much less than the expected number of points ($\approx \sqrt{n}$), and (B) if instead of considering all pairs $t$ we restrict to some subcollection of pairs with similar slopes, the corresponding regions essentially can't intersect (or else there would be a small-area triangle).

You might want to look for an analogue of (B): some subfamily of the all your pairs $f_{ij}$ have some structure making them repel one another such as having similar slopes. It might be possible to transform the problem so that if two pairs have similar slopes and the corresponding $f_{ij}$ intersect within the disk, among the four points involved some three will give you the triangle you want.