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Suppose $f$,$g$ are non-negative functions on a space $X$, with $\mu(X)=1$, such that $fg\ge1$ a.e.

Prove that $(\int f)(\int g)\ge1$.

We know that if both functions are larger or equal to 1, we are done, So it remains to check when one functions is larger than 1, the other is less than 1, and the product is still greater than or equal to 1.

We also know that $\int fg \ge 1$. I would then like to apply the Holder's inequality, since if $f$ or $g$ are each not in $L_p$ for some $p$, we are also done. The problem is then what do I do?

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As $f\cdot g\geq 1$ and $f, g$ are non-negatives, then $\sqrt{f\cdot g}\geq 1$, and then $$1\leq f^{\frac{1}{2}}g^{\frac{1}{2}}\Rightarrow \int 1 d\mu\leq \int f^{\frac{1}{2}}g^{\frac{1}{2}}d\mu. $$

As $\int 1 d\mu=\mu(X)=1$, by Holder's inequality with $p=q=2$, we have $$1=\int d\mu\leq \int f^{\frac{1}{2}}g^{\frac{1}{2}}d\mu\leq \left(\int (f^{\frac{1}{2}} )^2d\mu\right)^{\frac{1}{2}} \left(\int (g^{\frac{1}{2}} )^2d\mu\right)^{\frac{1}{2}}= \left(\int fd\mu\right)^{\frac{1}{2}} \left(\int g d\mu\right)^{\frac{1}{2}}, $$ i.e., $$\left(\int fd\mu\right)^{\frac{1}{2}} \left(\int g d\mu\right)^{\frac{1}{2}}\geq 1. $$ Squaring, $$\left(\int f d\mu\right) \left(\int g d\mu\right)\geq 1.$$

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