Belarusian Mathematical Olympiad

(Solution by Y. Dubovik, U. Kazlouski.) By the Cosine Law for the triangles $DAB$ and $DCB$,
$$y^2 = a^2 + d^2 - 2ad \cos A, \quad (1)$$
$$y^2 = b^2 + c^2 + 2bc \cos A. \quad (2)$$
Multiplying (1) and (2) by $bc$ and $ad$, respectively, and summing the obtained equalities, we get
$$(bc + ad)y^2 = (a^2 + d^2)bc + (b^2 + c^2)ad$$

$$\geq 2adbc + 2bcad = 4abcd.$$

$$\frac{8}{y^2} \le \frac{2(bc + ad)}{abcd} = \frac{2}{ad} + \frac{2}{bc}. \quad (3)$$
In the same way we can obtain
$$\frac{8}{x^2} \le \frac{2}{ab} + \frac{2}{cd}. \quad (4)$$
So,
$$\begin{aligned} 8 \left( \frac{1}{x^2} + \frac{1}{y^2} \right) &\le \frac{2}{ad} + \frac{2}{bc} + \frac{2}{ab} + \frac{2}{cd} = 2 \left( \frac{1}{a} + \frac{1}{c} \right) \left( \frac{1}{b} + \frac{1}{d} \right) \\ &\le \left( \frac{1}{a} + \frac{1}{c} \right)^2 + \left( \frac{1}{b} + \frac{1}{d} \right)^2, \end{aligned}$$
as required.