Show that for measurable functions: $f,g:[1,\infty]\to [0,\infty]$, $$\left|\int_{[1,\infty)^2}e^{-xy}f(x)g(y)\ dx\,dy \right|\leq \frac{1}{2e} \left(\int_1^\infty |f(x)|^2\ dx\right)^{1/2} \left(\int_1^\infty |g(x)|^2\ dx\right)^{1/2}$$ where $dx$ is the Lebesgue measure on $[1,\infty)$ and $dx\,dy$ denotes the Lebesgue measure on $[1,\infty)^2$
Attempt:
$$\int_{[1,\infty)^2}|e^{-xy}f(x)g(y)|\ dx\,dy \leq \left(\int_{[1,\infty)^2} e^{-2xy}\ dx\,dy\right)^{1/2} \left(\int_{[1,\infty)^2}|f(x)|^2|g(y)|^2\ dx\,dy\right)^{1/2}$$
$$\left(\int_{[1,\infty)^2}|f(x)|^2|g(y)|^2\ dx\,dy\right)^{1/2} = \|f\|_2\|g\|_2$$
$$\left(\int_{[1,\infty)^2} e^{-2xy}\ dx\,dy\right)^{1/2}\leq \sqrt{\frac{1}{2e}}$$ where the last inequality arises from Tonelli's theorem and $\frac{1}{2x}e^{-x}\leq \frac{1}{2}e^{-x}$ on $[1,\infty)$
You made a mistake at the end, $\int_1^\infty \frac{1}{2} e^{-2y} dy = \frac{1}{4e^2}$ so taking the square root at the end gives you exactly what you want. (Also, interchanging the limits of integration in the last step didn't really do anything.)