This exercise gives me trouble: Let $F$ denote the collection of measurable functions which are positive $\mu$-a.e. and let $A \in \mathbb X$ satisfy $0 < \mu(A) < \infty$. For $f \in F$ let $T_f := \left(\int_A f \, d\mu\right)\left(\int_A \frac{1}{f} \, d\mu\right)$. Find $\inf_{f \in F} T_f$ and show that the infimum is attained.
Firstly, we show that \begin{align*} \left(\int f\chi_A \, d\mu\right)\left(\int \frac{\chi_A}{f} \, d\mu\right) \ge \int \chi_A \, d\mu. \end{align*} That was my plan. I needed this because I wanted to have something similar to Hoelder's inequality, but that does not work for $p = q = 1$. Because then we have that \begin{align*} T_f &= \left(\int_A f \, d\mu\right)\left(\int_A \frac{1}{f} \, d\mu\right) = \left(\int f\chi_A \, d\mu\right)\left(\int \frac{\chi_A}{f} \, d\mu\right) \ge \int \chi_A \, d\mu = \int_A 1 \, d\mu. \end{align*} So this would be the infimum and is attained because $1 \in F$.
Does this work? The general plan was to estimate the integral to something which is smaller.
After all if that does not work, I would suppose that the infimum is just $0$.
Use the Cauchy-Schwarz inequality on $\sqrt{f}$ and $\dfrac{1}{\sqrt{f}}$. Then your plan works with a slight modification.