WLOG, assume $ABCD$ is convex. One can reduce this inequality to $$2 \cdot R_{(OXY)}^2 \ge [ABCD]$$with Brocard's and POP. It appears that the stronger inequality $$2 \cdot R_{(OXY)}^2 \ge \frac{AC \cdot BD}{2}$$holds as well. "Progress" made at MOP involves a degree $28$ polynomial the rest is left as an exercise for the reader.

In equality, $ABCD$ is a kite with diameter $AC$ such that $\angle XYZ=135^\circ$: [asy][asy] size(10cm); pair O, A, B, C, D, X, Y, Z; O=(0,0); A=dir(0); B=dir(24.4698); C=dir(180); D=dir(335.5302); X=extension(A,B,C,D); Y=extension(A,C,B,D); Z=extension(A,D,B,C); draw(unitcircle,magenta); draw(circumcircle(X,Y,Z),pink); draw(B--C--D--A--B,blue+linewidth(1)); draw(Z--B,lightred+dashed); draw(X--D,lightred+dashed); draw(Z--A--X,lightred+dashed); draw(Z--Y--X--Z,orange+linewidth(0.5)); draw(A--C,lightred+dashed); draw(B--D,lightred+dashed); dot("$O$",O,dir(270)); dot("$A$",A,dir(295)); dot("$B$",B,dir(60)); dot("$C$",C,C); dot("$D$",D,dir(300)); dot("$X$",X,X); dot("$Y$",Y,dir(225)); dot("$Z$",Z,Z); [/asy][/asy] Have fun!

To be specific, the reduction in #2 is as follows: by Brokard's Theorem, $(ABCD)$ is the polar circle of $\triangle XYZ$, so if $\triangle XYZ$ has circumradius $\rho$ and angles $\alpha$, $\beta$, and $\gamma$, then \[ OX^2 + OY^2 + OZ^2 - 2R^2 = 4\rho^2\cos^2 \alpha +4\rho^2\cos^2 \beta + 4\rho^2\cos^2 \gamma + 4\rho^2\cos \alpha\cos \beta\cos \gamma.\]But it is well known that $\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma + 2\cos \alpha \cos\beta \cos \gamma = 1$ if $\alpha + \beta + \gamma = \tau / 2$, so this is just $4\rho^2$. It suffices to show that $2\rho^2 \geqslant [ABCD]$. We show the following (much) stronger statement: Theorem: Let $ABCD$ be a convex cyclic quadrilateral, and let $X = AB \cap CD$, $Y = AC \cap BD$, and $Z = AD \cap BC$. Then, we have $AC \cdot BD \leqslant 2 \cdot OY \cdot XZ$. Proof: Proceed with length bash: let $AB = a$, $BC = b$, $CD = c$, and $DA = d$. WLOG assume $a > c$ and $b > d$. [asy][asy]import geometry; size(8cm); defaultpen(fontsize(9pt)); pair A = dir(100); pair B = dir(195); pair C = dir(345); pair D = dir(30); pair X = extension(A, B, C, D); pair Y = extension(A, C, B, D); pair Z = extension(A, D, B, C); pair O = origin; pair a = ((A + B) / 2); pair b = ((B + C) / 2); pair c = ((C + D) / 2); pair d = ((D + A) / 2); pair x = ((X + D) / 2); pair z = ((Z + D) / 2); draw(circle(A, B, C)); draw((B -- X -- C)); draw((A -- C)); draw((B -- D)); draw((A -- Z -- B)); dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$D$", D, dir(D)); dot("$X$", X, plain.N); dot("$Y$", Y, plain.N); dot("$Z$", Z, plain.E); dot("$O$", O, plain.S); label("$a$", a, dir(a)); label("$b$", b, dir(b)); label("$c$", c, dir(c)); label("$d$", d, dir(d)); label("$x$", x, plain.E); label("$z$", z, plain.N);[/asy][/asy] First, we compute $XZ$. Let $x = XD$ and $z = ZD$. Since $\triangle XAD \sim \triangle XCB$, we have \[ \frac{XA}{x + c} = \frac{d}{b} = \frac{x}{XA + a}. \]Thus, \begin{align*} b \cdot XA - d \cdot x &= cd, \\ d \cdot XA - b \cdot x &= -ad \end{align*}By Cramer's rule, \[ x = \frac{-b \cdot ad - d \cdot cd}{-b^2 + d^2} = \frac{d(ab + cd)}{b^2 - d^2}.\]Similarly, we can calculate that $y = \frac{c(ab + cd)}{a^2 - c^2}$. Finally, by Law of Cosines on $\triangle ADC$ and $\triangle ABC$, \[ c^2 + d^2 - 2cd \cos \angle ADC = a^2 + b^2 + 2ab \cos \angle ADC, \]so $\cos XDC = \cos ADC = \frac{c^2 + d^2 - a^2 - b^2}{2(ab + cd)}$. Write $\alpha := a^2 - c^2$ and $\beta := b^2 - d^2$. We have, \begin{align*} XZ^2 &= x^2 + z^2 - 2xz \cos XDC \\ &= \frac{d^2(ab+cd)^2}{\beta^2} + \frac{c^2(ab+cd)^2}{\alpha^2} - \frac{2cd(ab+cd)^2(-\alpha - \beta)}{\alpha \beta \cdot 2(ab+cd)} \\ &= (ab + cd) \cdot \frac{(d^2 \alpha^2 + c^2 \beta^2) (ab + cd) + cd(\alpha + \beta)\alpha \beta}{\alpha^2 \beta^2}. \end{align*}But note that \[ \alpha^2 (d^2 (ab + cd) + cd \beta) = \alpha^2 (abd^2 + cd^3 + b^2cd - cd^3) = \alpha^2 bd (ad + bc).\]Similarly, $\beta^2 (c^2 (ab + cd) + cd \alpha) = \beta^2 ac (ad + bc)$. Thus, \[ XZ^2 = \frac{(ab + cd)(ad + bc)(bd \alpha^2 + ac \beta^2)}{\alpha^2 \beta^2}.\]Now, we compute $OY$. Let $M$ be the midpoint of $AC$, and let $N$ be the midpoint of $BD$. By Euler's quadrilateral theorem, we have \[ 4MN^2 = a^2 + b^2 + c^2 + d^2 - AC^2 - BD^2.\]On the other hand, $OMYN$ is cyclic with diameter $OY$, so by Extended Law of Sines, \[ OY = \frac{MN}{\sin \angle MYN} = \frac{MN}{\sin \angle (AC, BD)} = \frac{MN \cdot AC \cdot BD}{2 [ABCD]}.\]Finally, by Ptolemy's two theorems, we have \[ AC^2 = \sqrt{\frac{(ac + bd)(ad + bc)}{(ab + cd)}}, \quad BD^2 = \sqrt{\frac{(ac + bd)(ab + cd)}{(ad + bc)}}, \quad \text{and} \quad AC \cdot BD = ac + bd.\]Thus, \[ 4 MN^2 = \frac{(a^2+b^2+c^2+d^2) (ad + bc) (ab + cd) - (ac+bd)((ad+bc)^2 + (ab+cd)^2)} {(ad + bc)(ab + cd)}.\]The numerator expands as \begin{align*} \sum_{\text{cyc}} (a^4 bd + a^3 b^2c + a^3cd^2 + a^2bc^2d) - (ac + bd) (a^2b^2 + b^2c^2 + c^2d^2 + d^2a^2 + 4abcd). \end{align*}But the first term is just \begin{align*} &ac(b^4 + d^4) + bd(a^4 + c^4) + ac(a^2b^2 + c^2d^2) + bd(b^2c^2 + d^2a^2) \\ +& \, ac(d^2a^2 + b^2c^2) + bd(a^2b^2 + c^2d^2) + 2ac \cdot b^2d^2 + 2bd \cdot a^2c^2 \\ =& \, ac\left((b^2 + d^2)^2 + \sum_{\text{cyc}} a^2b^2 \right) + bd\left((a^2 + c^2)^2 + \sum_{\text{cyc}} a^2b^2 \right) \end{align*}So the numerator evaluates to $ac(b^2 - d^2)^2 + bd(a^2 - c^2)$. Therefore, \[ OY^2 = \frac{(ac(b^2 - d^2)^2 + bd(a^2 - c^2))(ac + bd)^2} {16[ABCD]^2 \cdot (ad + bc)(ab + cd)}.\]Now, we have, with $\alpha = a^2 - c^2$ and $\beta = b^2 - d^2$, \[ OY^2 \cdot XZ^2 = \frac{(bd \alpha^2 + ac \beta^2)^2 (ac + bd)^2} {\alpha^2 \beta^2 \cdot 16[ABCD]^2}.\]Finally, note that $ac + bd = AC \cdot BD$ by Ptolemy's Theorem, so it suffices to show that \[ (bd \alpha^2 + ac\beta^2)^2 \ge 4 \alpha^2 \beta^2 [ABCD]^2 \qquad\qquad (\star). \]By Brahmagupta's Formula, we have \[ 16 [ABCD]^2 = (a^2+b^2+c^2+d^2)^2 - 2(a^4+b^4+c^4+d^4) + 8abcd, \]so by multiplying $(\star)$ by $4$ and expanding the LHS, we wish to show that \begin{align*} &\;\;\;\;4b^2d^2 \alpha^4 + 4a^2c^2 \beta^4 + 8abcd\alpha^2\beta^2 \\ &\geqslant \alpha^2\beta^2((a^2+b^2+c^2+d^2)^2 - 2(a^4+b^4+c^4+d^4) + 8abcd). \end{align*}Clearly a $8abcd\alpha^2\beta^2$ term cancels. Furthermore, note that $(a^2+b^2+c^2+d^2)^2 - (a^4+b^4+c^4+d^4)$ can be grouped as $2(a^2+c^2)(b^2+d^2) + 2(a^2c^2 + b^2d^2) $. Thus, it suffices to show that, \begin{align*} &\;\;\;\;4b^2d^2 \alpha^4 + 4a^2c^2 \beta^4 + \alpha^2\beta^2(a^4+b^4+c^4+d^4) \\ &\geqslant 2\alpha^2\beta^2(a^2+c^2)(b^2+d^2) + 2\alpha^2\beta^2(a^2c^2 + b^2d^2). \end{align*}Now, observe that $(a^4+b^4+c^4+d^4) - 2(a^2c^2 + b^2d^2) = (a^2 - c^2)^2 + (b^2 - d^2)^2 = \alpha^2 + \beta^2$. Thus, it suffices to show that \[ 4b^2d^2 \alpha^4 + 4a^2c^2 \beta^4 + \alpha^2 \beta^2 (\alpha^2 + \beta^2) \geqslant 2\alpha^2\beta^2(a^2+c^2)(b^2+d^2).\]But $4b^2d^2 \alpha^4 + \alpha^4 \beta^2 = \alpha^4 (4b^2d^2 + (b^2 - d^2)^2) = \alpha^4 (b^2+d^2)^2$, so we wish to show that \[ \alpha^4 (b^2+d^2)^2 + \beta^4 (a^2+c^2)^2 \geqslant 2\alpha^2\beta^2(a^2+c^2)(b^2+d^2),\]which is AM-GM. $\blacksquare$ Finally, note that if we inscribe a rectangle in $(XYZ)$ with one side as $XZ$, then it has area exactly $OY \cdot XZ$ by orthocenter facts. But this rectangle has area at most $2 \rho^2$ since that is the area of a square inscribed in the same circle. Finally, $2[ABCD] = AC \cdot BD \sin \angle (AC, BD) \leqslant AC \cdot BD$. The conclusion follows. $\blacksquare$

Congratulations on making significant progress on this beautiful proposal! Indeed, because $O$ is the orthocenter of $XYZ$ due to Brocard's, we have $R_{(OXY)} = R_{(XYZ)}$. Update: The RELMO organizers commend @cosmicgenius for solving this marvelous problem!