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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.

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Simply use Hölder's inequality: $$\mu(X)=\int_X \sqrt{1} \mathrm d \mu\leq\int_X \sqrt{f}\sqrt{g} \mathrm d \mu\leq\left(\int_X f \mathrm d \mu\right)^{1/2}\left(\int_X g \mathrm d \mu\right)^{1/2}.$$

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    $\begingroup$ Thanks, I looked at it again, and it made sense... I don't know what I was missing. $\endgroup$
    – sandy
    Commented Mar 12, 2021 at 0:07

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