Kanada 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. □