Let $\mu$ be a probability measure on a set $X$, i.e. $\mu(X)=1$, and let $f$ and $g$ be positive measurable functions on $X$. Show that if $fg\geq1$, then the integral of $f$ times the integral of $g$ is greater than or equal to $1$.
I'm not sure how to approach this. Since $f$ and $g$ are positive, the integral of $f$ and the integral of $g$ are equal to the $L_1$ norm of $f$ and the $L_1$ norm of $g$, but I'm not sure why $fg\geq 1$ implies that the product of their $L_1$ norms is greater than or equal to $1$. Considering that $fg$ is involved, I was thinking of using Holder's inequality, but I'm not sure what $p$ and $q$ to use.
This relies more on an arithmetic trick than on analysis. Since $fg \ge 1$ you have also that $\sqrt{fg} \ge 1$ so that $$1 \le \int_X \sqrt{fg} \le \left(\int_X f \right)^{1/2} \left( \int_X g \right)^{1/2}$$ by Holder's inequality. Now square both sides.