$f, 1/f$ integrable implies that $f^2$, $1/f^2$ integrable?

Question

on the measure space $(X,A,m)$ let $f: X \to \mathbb R$ be Borel-measurable, it is given that $f$, $1/f$ are well-defined and integrable.

I would like to prove that this implies that $f^2$ and $1/f^2$ are integrable but am stuck on this part. I thought about using Hölder's inequality, do you guys have any hints?

Answer 1

$\sqrt{x}$ and $1/\sqrt{x}$ are integrable on $(0,1]$ with Lebesgue measure, but $1/x$ is not.

Answer 2

You might prefer trying to prove something true.

Consider $\bigl((0,1), \mathcal B((0,1)), \mathcal L^1\bigr)$ and $f(x)=\sqrt x$. Then $f$ and $1/f$ are integrable, but $1/f^2$ is not.