In Griffiths' QM, he uses two inequalities (here numbered as $(1)$ and $(2)$) to prove the following general uncertainty principle: $$\sigma_A^2 \sigma_B^2\geq\left(\frac{1}{2i}\langle [\hat A ,\hat B]\rangle \right)^2$$
defining $\lvert f\rangle=(\hat A-\langle A\rangle)\lvert \Psi\rangle$ and $\lvert g\rangle=(\hat B-\langle B\rangle)\lvert \Psi\rangle$, he uses
Schwarz inequality: $$\langle f\lvert f\rangle\langle g\lvert g\rangle\geq|\langle f\lvert g\rangle|^2\tag{1}$$ and with $\sigma_B^2=\langle g\lvert g\rangle$ and $\sigma_A^2=\langle f\lvert f\rangle$, he arrives at $\sigma_A^2 \sigma_B^2\geq|\langle f\lvert g\rangle|^2$.
the fact that for any complex number $z$ we have $$|z|^2\geq(\mathrm{Im}(z))^2=[\frac{1}{2i}(z-z^*)]^2\tag{2}$$ here $z=\langle f\lvert g\rangle$ and so $z^*=\langle g\lvert f\rangle$ , and we find that $\langle f\lvert g\rangle=\langle\hat A \hat B\rangle-\langle A \rangle\langle B \rangle$ and $\langle g\lvert f\rangle=\langle\hat B \hat A\rangle-\langle A \rangle\langle B \rangle$, so $\langle f\lvert g\rangle-\langle g\lvert f\rangle=\langle [\hat A,\hat B]\rangle$. Replacing this into $(1)$ gives the uncertainty principle.
Why he doesn't use $$|z|^2\geq(\mathrm{Re}(z))^2=[\frac{1}{2}(z+z^*)]^2$$ instead of $(2)$? This could give us a (correct though different) relation between the $\sigma_A$ and $\sigma_B$ too: $$\boxed{\sigma_A^2 \sigma_B^2\geq\frac{1}{4}\left(\langle \hat A \hat B\rangle +\langle \hat B \hat A\rangle -2 \langle \hat A\rangle\langle \hat B\rangle\right)^2}$$