Assume that $(X,s,\mu)$ is a finite measure space and $f,g$ are non-negative $s$-measurable functions such that $fg \ge 1$ a.e. Show that $(\int f \,d\mu) (\int g \, d\mu) \ge (\mu(X) )^2$
This inequality is clearly true if either integral is infinite, so I am looking for a solution for the finite case. I have an answer, but it does not depend on anything that we have recently covered, so I am pretty sure I am missing something big. Any help would be appreciated.
Simply use Hölder's inequality: $$\mu(X)=\int_X \sqrt{1} \mathrm d \mu\leq\int_X \sqrt{f}\sqrt{g} \mathrm d \mu\leq\left(\int_X f \mathrm d \mu\right)^{1/2}\left(\int_X g \mathrm d \mu\right)^{1/2}.$$