An inequality with a measure probability space

Question

Let $(X , \mathcal{M} , \mu)$ a measure space with $\mu(X) = 1$ and let $f , g : X \to (0 , \infty)$ two positive measurable functions such that $(f g)(x) \geq 1$ for all $x \in X$. Show that $$ \left(\int_X f d \mu\right) \left(\int_X g d \mu\right) \geq 1\mbox{.} $$ Using the conditions $\mu(x) = 1$ and $f g \geq 1$, the inequality $1 \leq \int_X (f g) d \mu$ is trivial: $$ 1 = \mu(X) = \int_X 1 d \mu \leq \int_X (f g) d \mu\mbox{,} $$ so using that $f , g > 0$, we should obtain the final inequality: $$ \int_X (f g) d \mu \leq \left(\int_X f d \mu\right) \left(\int_X g d \mu\right)\mbox{,} $$ but I can't see that. Any help? Thank you very much.

Answer 1

$X \ge \frac{1}{Y}$. Therefore $E[ X ] \ge E [ \frac{1}{Y}] \underset{Jensen}{\ge} \frac{1}{E[Y]},$