Show that $\int_0^2 \int_0^2 \frac{x}{1+\ln(x^2y^2)} \,\mathrm{d}x \,\mathrm{d}y \leq 4$
About thinking outside the box: one of the main tricks in Math is to exploit the exchange of sums/limits/integrals. This trick has many forms: double counting, Feynman's trick, Wilf-Zeilberger couples, Fubini's theorem, creative telescoping... they all boil down to the same powerful idea. In our case we may try to write $\frac{1}{1+2\log(xy)}$ as $\int_{0}^{+\infty}e^{-t} x^{-2t} y^{-2t}\,dt $ to get...
$$ \iint_{(1,2)^2}\frac{x}{1+2\log(xy)}\,dx\,dy = \int_{0}^{+\infty} e^{-t}\iint_{(1,2)^2} x^{1-2t} y^{-2t}\,dx\,dy\,dt $$
(the situation gets temporarily worse) then
$$ \iint_{(1,2)^2}\frac{x}{1+2\log(xy)}\,dx\,dy = \int_{0}^{+\infty}\frac{(4^t-2)(4^t-4)}{(2t-2)(2t-1)} 4^{-2t} e^{-t}\,dt $$
(which is much better, since we have an integral in a single variable). Despite its algebraic nastyness the function $\frac{(4^t-2)(4^t-4)}{(2t-2)(2t-1)} 4^{-2t}$ has a very simple behaviour on $\mathbb{R}^+$: it is positive, decreasing and bounded above by $\frac{3}{2}\,\exp\left(-\frac{7}{5}t\right)$. In particular the original integral over $(1,2)^2$ does not exceed $\frac{5}{8}$. I leave to you to adapt this approach for dealing with the contribution over $(0,1)\times(1,2)$, $(1,2)\times(0,1)$ and $(0,1)^2$.
$4$ is actually a very loose upper bound for the original integral: a sharper bound is $I\leq \frac{17}{14}$.