Is there a fractal like probability distribution inside an equilateral triangle such that $$F(x)+G(y)+H(z)=3/2$$ for every point inside the fractal and $$F(x)+G(y)+H(z)\le3/2$$ for every point outside it?
Here $(x,y,z)$ are the perpendiculars to the equilateral triangle, with $x+y+z=1$, and $F(x),G(y),H(z)$ are the cumulative probability functions of the three coordinates
We can construct three random variables whose support is a fractal subset of the unit simplex, and whose marginals are uniform, as follows. Let $\pi^1,\pi^2,\pi^3,\dots$ be an iid sequence of permutations of $\{0,1,2\}$ (e.g. $\pi^1=\{0\to 1,1\to0,2\to 2\}$), and let $$ X = \frac23\sum_{n\ge 1} \pi^n(0)\cdot 3^{-n},\\ Y = \frac23\sum_{n\ge 1} \pi^n(1)\cdot 3^{-n},\\ Z = \frac23\sum_{n\ge 1} \pi^n(2)\cdot 3^{-n},\\ $$ Then $X+Y+Z=1$ always, and $X\sim \text{Unif}(0,\frac23)$, since $\pi^n(0)$ is a just a sequence of iid ternary digits.
This means that $F(x)=G(x)=H(x)=\min(\frac32 x, 1)$ for all $x\ge 0$. It follows that $F(x)+G(y)+H(z)\le \frac32(x+y+z)=\frac32$, and as long as $(x,y,z)$ is in the fractal support, we have $\max(x,y,z)\le \frac23$, so equality is attained.
The fractal looks like this. Inscribe the largest possible hexagon in the unit simplex, and then inscribe six congruent hexagons in that hexagon so the smaller hexagons fit snuggly in the larger hexagon and the smaller hexagons touch corner to corner. Delete everything except these six smaller hexagons, and apply the same procedure to each of the six hexagons, repeated infinitely. We will find that $F(x)+G(y)+H(z)=\frac32$ for all points on the larger original hexagon, which certainly includes the whole fractal.
Here is a picture of the resulting fractal. As you can see, the complement is made of (whole and partial) Koch snowflakes.