, 2010

Let $ABCD$ be a cyclic quadrilateral with $AB = a$, $BC = b$, $CD = c$, $DA = d$, $AC = e$ and $BD = f$. Because the midpoints of the cyclic quadrilateral $ABCD$ form another cyclic quadrilateral, which is a parallelogram, we deduce that this parallelogram is a rectangle and $ABCD$ is orthogonal.

It follows that
$$S = \sigma(ABCD) = \frac{ef}{2}$$
and
$$a^2 + c^2 = b^2 + d^2$$
Let $R$ and $R_1$ be the circumradii of the cyclic quadrilateral $ABCD$ and the rectangle respectively. We obtain
$$4R_1^2 = \frac{e^2 + f^2}{4}$$
and
$$16R^2 S^2 = (ac + bd)(ab + cd)(ad + bc) = ef[(ac(b^2 + d^2) + bd(a^2 + c^2))] = (ef)^2(a^2 + c^2),$$
so
$$4R^2 = a^2 + c^2$$
Note that our inequality $R^2 \geq 2R_1^2$ is equivalent to
$$2(a^2 + c^2) \geq e^2 + f^2$$
$$a^2 + b^2 + c^2 + d^2 - e^2 - f^2 \geq 0,$$
hence $2(a^2 + c^2) \geq (e^2 + f^2)$, and we are done. $\square$