If $f,g\in \mathscr{L}^2$, then $\|fg\|_1\leq\|f\|_2\|g\|_2$. My textbook says that this is a consequence of Cauchy-Schwarz inequality.
How so? Cauchy-Schwarz says that $|\langle v,w\rangle|\leq\|v\|\|w\|$. Putting in $f$ and $g$, we have $$\left|\int_X f\overline{g}d\mu\right|\leq \|f\|_2\|g\|_2$$
But $$\|fg\|_1=\int_X|fg|d\mu$$
So I don't see how $\|fg\|_1\leq\|f\|_2\|g\|_2$ is a consequence of Cauchy-Schwarz.
Multiply $f$ by $e^{\alpha(x)}$, where $\alpha$ is chosen to make $f\bar g$ real and positive, and use Cauchy's inequality on $e^\alpha f$ and $g$. The right side remains the same, but the left side becomes $\int e^{\alpha}f\bar g = \int |e^{\alpha}f\bar g|=\int |f\bar g|$.
You might want to think about why $e^{\alpha(x)}$ is measurable.