Question 131
Let $a$, $b$, $c$, $d$, $x$, $y$ denote the lengths of the sides $AB$, $BC$, $CD$, $DA$ and the diagonals $AC$, $BD$ of a cyclic quadrilateral $ABCD$, respectively. Prove that
$$
\left(\frac{1}{a} + \frac{1}{c}\right)^2 + \left(\frac{1}{b} + \frac{1}{d}\right)^2 \ge 8 \left(\frac{1}{x^2} + \frac{1}{y^2}\right).
$$
$$
\left(\frac{1}{a} + \frac{1}{c}\right)^2 + \left(\frac{1}{b} + \frac{1}{d}\right)^2 \ge 8 \left(\frac{1}{x^2} + \frac{1}{y^2}\right).
$$