Vietnamese Mathematical Olympiad

Note that $x \neq \frac{k\pi}{2}$, $y \neq \frac{m\pi}{2}$ ($k, m \in \mathbb{Z}$) and $xy > 0$. From the given condition, one can get
$$A = 20 \sqrt{\frac{xy}{(x+y)^2}},$$
where $A$ equals to the product of two left hand sides of the given system. Using the Cauchy-Schwarz and AM-GM inequality
$$\begin{aligned} & (\sin^2 x + \frac{1}{\sin^2 x})(\cos^2 x + \frac{1}{\cos^2 x}) \\ & \geq (|\sin x \cos x| + \frac{1}{|\sin x \cos x|})^2 \\ & = (\frac{|\sin 2x|}{2} + \frac{1}{2|\sin 2x|} + \frac{3}{2|\sin 2x|})^2 \\ & \geq (1 + \frac{3}{2})^2 = (\frac{5}{2})^2. \end{aligned}$$
Similarly, we have
$$(\sin^2 y + \frac{1}{\sin^2 y})(\cos^2 y + \frac{1}{\cos^2 y}) \geq (\frac{5}{2})^2$$
Therefore, applying AM-GM inequality, we obtain
$$A \geq 4\sqrt[4]{\left(\frac{5}{2}\right)^4} = 10 \geq 20\sqrt{\frac{xy}{(x+y)^2}}$$
The equality occurs if and only if $|\sin 2x| = 1$ and $x = y = \frac{\pi}{4} + \frac{k\pi}{2}$ where $k \in \mathbb{Z}$. It is easy to check that these solutions satisfy the given system. Thus, $x = y = \frac{\pi}{4} + \frac{k\pi}{2}$ where $k \in \mathbb{Z}$ are all solutions of the given system. $\square$