, 2011

Let $a_i$ and $b_i$ denote the width and height of each white rectangle, and let $c_i$ and $d_i$ denote the width and height of each red rectangle. Also, let $L$ denote the side length of the original square.

Lemma: Either $\sum a_i \ge L$ or $\sum d_i \ge L$.

Proof of lemma: Suppose there exists a horizontal line across the square that is covered entirely with white rectangles. Then, the total width of these rectangles is at least $L$, and the claim is proven. Otherwise, there is a red rectangle intersecting every horizontal line, and hence the total height of these rectangles is at least $L$. □

Now, let us assume without loss of generality that $\sum a_i \ge L$. By the Cauchy-Schwarz inequality,
$$\left(\sum \frac{a_i}{b_i}\right) \cdot \left(\sum a_i b_i\right) \ge \left(\sum a_i\right)^2 \ge L^2.$$
But we know $\sum a_i b_i = \frac{L^2}{2}$, so it follows that $\sum \frac{a_i}{b_i} \ge 2$. Furthermore, each $c_i \le L$, so
$$\sum \frac{d_i}{c_i} \ge \frac{1}{L^2} \cdot \sum c_i d_i = \frac{1}{2}.$$
Therefore, $x$ is at least $2.5$. Conversely, $x = 2.5$ can be achieved by making the top half of the square one colour, and the bottom half the other colour. □