Let $P_1,P_2,P_3,P_4$ be four distinct, cyclically ordered points on a circle.
Write $S_{ijk}$ for the area of $\triangle P_iP_jP_k$. Then $(S_{123},S_{124},S_{134},S_{234})$ lies in a 3-dimensional semialgebraic region contained in the 3-dimensional vector subspace $S_{234}-S_{134}+S_{124}-S_{123}=0$ of $\mathbb{R}^4$.
It is a positive cone ($rx ∈ C$ for all real $r > 0$ and all $x ∈ C$.)
Parametrize by consecutive central gap angles:
$$ a=\angle P_1OP_2,\quad b=\angle P_2OP_3,\quad c=\angle P_3OP_4,\quad d=\angle P_4OP_1,\qquad a,b,c,d\in(0,2\pi),\ a+b+c+d=2\pi, $$
Question: How can one describe the image of the map
$$ (a,b,c,d)\mapsto (S_{123},S_{124},S_{134},S_{234}) $$
as a semialgebraic set? Its boundaries should be defined by homogeneous polynomials in $S_{123},S_{124},S_{134},S_{234}$.
The following conditions are necessary:
$$S_{234}-S_{134}+S_{124}-S_{123}=0$$
$$S_{ijk}>0$$
$$(\sqrt{S_{123}S_{134}}-\sqrt{S_{234}S_{124}})^2\;\le\; S_{123}S_{234}+S_{134}S_{124}.$$ equality holds if and only if $P_1P_2\parallel P_3P_4$
$$ (\sqrt{S_{123}S_{124}}-\sqrt{S_{234}S_{134}})^2\ \le\ S_{123}S_{234}+S_{124}S_{134}. $$ equality holds if and only if $P_2P_3\parallel P_4P_1$
Proof of necessity of condition 3./4.
Scale the radius of the circle to $1$. Then
$$ \begin{aligned} 4S_{123}&=s_{12}s_{23}d_{13},& 4S_{234}&=s_{23}s_{34}d_{24},\\ 4S_{134}&=s_{34}s_{41}d_{13},& 4S_{124}&=s_{41}s_{12}d_{24}, \end{aligned} $$
and Ptolemy’s polynomial for a cyclic quadrilateral:
$$ \mathrm{Pt}:=d_{13}d_{24}-s_{12}s_{34}-s_{23}s_{41}=0. $$
Write
$$ \mathcal{L} :=16\Big((S_{123}S_{234}+S_{134}S_{124}) -(\sqrt{S_{123}S_{134}}-\sqrt{S_{234}S_{124}})^2\Big). $$
Then a direct substitution of the formulas above and routine algebra give the exact identity
$$ \boxed{\quad \mathcal{L} =\left(\underbrace{\mathrm{Pt}}_{\text{Ptolemy poly}}\; \bigl(s_{23}+s_{41}\bigr)^{2} \;+\; \underbrace{\bigl(d_{13}s_{23}-d_{24}s_{41}\bigr)^{2}}_{\text{perfect square}\ge0}\right)\,s_{12}s_{34}. \quad} \tag{$\heartsuit$} $$
Because $\mathrm{Pt}=0$ for concyclic points, the first term vanishes and we obtain
$$ \mathcal{L}=\bigl(d_{13}s_{23}-d_{24}s_{41}\bigr)^{2}\,s_{12}s_{34}\;\ge\;0, $$
which is precisely the desired inequality
$$ (\sqrt{S_{123}S_{134}}-\sqrt{S_{234}S_{124}})^2\;\le\; S_{123}S_{234}+S_{134}S_{124}. $$
How to verify $(\heartsuit)$ quickly
Start from
$$ \sqrt{S_{123}S_{134}}= \frac{d_{13}}4\sqrt{s_{12}s_{23}s_{34}s_{41}},\qquad \sqrt{S_{234}S_{124}}= \frac{d_{24}}4\sqrt{s_{12}s_{23}s_{34}s_{41}}, $$
so
$$ 16\big(\sqrt{S_{123}S_{134}}-\sqrt{S_{234}S_{124}}\big)^2 =(d_{13}-d_{24})^{2}\,s_{12}s_{23}s_{34}s_{41}. $$
Also,
$$ 16(S_{123}S_{234}+S_{134}S_{124}) =d_{13}d_{24}\,s_{12}s_{34}\,(s_{23}^{2}+s_{41}^{2}). $$
Hence
$$ \mathcal{L} =s_{12}s_{34}\Big(d_{13}d_{24}(s_{23}^{2}+s_{41}^{2}) -s_{23}s_{41}(d_{13}-d_{24})^{2}\Big), $$
and one checks (expand both sides) that the bracket equals
$$ \underbrace{(d_{13}d_{24}-s_{12}s_{34}-s_{23}s_{41})}_{\mathrm{Pt}}\,(s_{23}+s_{41})^{2} +\,(d_{13}s_{23}-d_{24}s_{41})^{2}, $$
which yields $(\heartsuit)$ upon multiplying by $s_{12}s_{34}$.
When does equality hold?
On a circle, the Ptolemy factor vanishes; with distinct points $s_{12},s_{34},s_{23},s_{41}>0$. Thus
$$ \mathcal L=0 \ \Longleftrightarrow\ (d_{13}s_{23}-d_{24}s_{41})^{2}=0 \ \Longleftrightarrow\ d_{13}s_{23}=d_{24}s_{41}. $$
Parametrize by consecutive central angles (arc gaps)
$$ a=\angle P_1OP_2,\quad b=\angle P_2OP_3,\quad c=\angle P_3OP_4,\quad d=\angle P_4OP_1,\qquad a,b,c,d\in(0,\pi),\ a+b+c+d=2\pi, $$
so the chord lengths on the unit circle are
$$ s_{12}=2\sin\frac a2,\quad s_{23}=2\sin\frac b2,\quad s_{34}=2\sin\frac c2,\quad s_{41}=2\sin\frac d2,\quad d_{13}=2\sin\frac{a+b}2,\quad d_{24}=2\sin\frac{b+c}2. $$
The equality $d_{13}s_{23}=d_{24}s_{41}$ becomes
$$ \sin\frac{a+b}{2}\,\sin\frac b2 =\sin\frac{b+c}{2}\,\sin\frac d2 \quad\text{with}\quad \frac d2=\pi-\frac{a+b+c}{2}. $$
Use product-to-sum on both sides:
$$ \sin X\sin Y=\tfrac12\big(\cos(X-Y)-\cos(X+Y)\big) $$
to reduce the equality to
$$ \cos\frac a2-\cos\!\Big(\frac a2+b\Big) =\cos\frac a2-\cos\!\Big(\frac a2+d\Big) \quad\Longleftrightarrow\quad \cos\!\Big(\frac a2+b\Big)=\cos\!\Big(\frac a2+d\Big). $$
Hence either $b=d$ or $\frac a2+b=-(\frac a2+d)+2\pi$. The latter would imply $a+b+d=2\pi$, forcing $c=0$ (forbidden since points are distinct). Therefore
$$ d_{13}s_{23}=d_{24}s_{41}\iff b=d \iff P_1P_2\parallel P_3P_4 $$