Hong Kong competition problems, 2008

The minimum value of $xy$ is $\frac{1}{16120225}$.

Note that the left-hand side is positive. Therefore, by rewriting the right-hand side as
$$(x - y + 1) + \frac{1}{x - y + 1},$$
we know that $x - y + 1 > 0$. Thus, we can apply the AM-GM inequality to obtain
$$(x - y + 1) + \frac{1}{x - y + 1} \ge 2.$$
But then $1 + \cos^2(2007x + 2008y - 1) \le 2$. This shows equality should hold, and hence
$$x - y + 1 = 1 \quad \text{and} \quad \cos(2007x + 2008y - 1) = \pm 1.$$
The first relation gives $x = y$. The second relation gives $4015x - 1 = k\pi$ for some $k \in \mathbb{Z}$, and so
$$x = y = \frac{1 + k\pi}{4015}.$$
Since $k$ is an integer, it is obvious that $|1 + k\pi| \ge 1$. This yields
$$xy = \frac{(1 + k\pi)^2}{4015^2} \ge \frac{1}{16120225}.$$
Equality holds when $x = y = \frac{1}{4015}$.