Problem using Hölder inequality

Question

Suppose we have $(X, \mathcal A, \mu )$ a measure space with $\mu (X)=1$ and we have measurable functions $f,g:X \rightarrow (0, \infty ]$ with $f(x)g(x)\geq 1 $ for all $x \in X $.

How do I show that $\int_X f \ d\mu . \int_X g \ d\mu \geq 1$?

I know that since $\mu(X)$=1 that $\int_X fg \ d\mu \geq 1$ but how do I show this is less than the product of the integrals?

Answer 1

\begin{align*} \mu(X)=\int_{X}1d\mu\leq\int_{X}\sqrt{f(x)g(x)}d\mu\leq\left(\int_{X}fd\mu\right)^{1/2}\left(\int_{X}gd\mu\right)^{1/2}. \end{align*}