We have that $ (ab + cd)(ac + bd)(ad + bc)\leq(\frac {ab + cd + ac + bd + ad + bc}{3})^3\leq$ $ \frac {1}{2^9}(a + b + c + d)^6\leq 4^3\cdot r^6\,.$ The equality holds if and only if $ a = b = c = d$.

Let $ e,f$ diagonals of this quadrilateral,$ R$ radius of a given circle,$ S$area of this quadrilateral so by Ptolemey $ T=(ab + cd)(ac + bd)(ad + bc) = ef(ab + cd)(ad + bc) = (abe + cde)(adf + cdf) = 16R^2(\frac {abe + cde}{4R})(\frac {abe + cde}{4R}) = 16R^2S^2$ S max when the quadrilateral is a square, so T max the quadrilateral is a square

When the product $ (ab + cd)(ac + bd)(ad + bc)$ acquires its maximum, we can show that $c=d.$