Why is the uncertainty principle not $\sigma_A^2 \sigma_B^2\geq(\langle A B\rangle +\langle B A\rangle -2 \langle A\rangle\langle B\rangle)^2/4$?

Griffiths' formulation makes it explicit that operators which commute are not restricted by the uncertainty principle. Your boxed expression obscures this physical and mathematical insight.

The most correct relation is the following general relation, that actually contains both terms.

If you omit the inequality $(|z|^2\geq(Im(z))^2)$ from the derivation, the next steps toward the uncertainty relation would be:
$$\sigma_A^2 \sigma_B^2\geq|\langle f\lvert g\rangle|^2$$ $$|\langle f\lvert g\rangle|^2=\langle f\lvert g\rangle\langle g\lvert f\rangle=(\langle A B\rangle-\langle A\rangle \langle B\rangle)(\langle B A\rangle-\langle A\rangle \langle B\rangle)$$ $$|\langle f\lvert g\rangle|^2=\langle A B\rangle \langle B A\rangle+(\langle A\rangle \langle B\rangle)^2-\langle A\rangle \langle B\rangle(\{A,B\})\to$$ $$\boxed{\sigma_A^2 \sigma_B^2\geq\langle A B\rangle \langle B A\rangle+(\langle A\rangle \langle B\rangle)^2-\langle A\rangle \langle B\rangle(\{A,B\})}$$ The above relation can be written better as $$|\langle f|g\rangle|^{2} = \bigg(\frac{\langle f|g\rangle+\langle g|f\rangle}{2}\bigg)^{2} + \bigg(\frac{\langle f|g\rangle-\langle g|f\rangle}{2i}\bigg)^{2}$$ $$\langle f|g\rangle-\langle g|f\rangle =\langle [{A},{B}]\rangle$$ $$\langle f|g\rangle+\langle g|f\rangle = \langle \{{A},{B}\}\rangle -2\langle {A}\rangle\langle {B}\rangle\to$$ $$\boxed{\sigma_A^2 \sigma_B^2\geq\Big(\frac{1}{2}\langle\{{A},{B}\}\rangle - \langle {A} \rangle\langle {B}\rangle\Big)^{2}+ \Big(\frac{1}{2i}\langle[{A},{B}]\rangle\Big)^{2}}$$ So, Griffiths ignores the first term in the above relation. It remains to tell why he does so. The answer is provided in the last paragraph of the Lubos Motl's answer to this question:

Well, in normal cases, the stronger version is not "terribly" useful because the anticommutator term is only nonzero if there is a "correlation" in the distributions of $A,B$ - i.e. if the distribution is "tilted" in the $A,B$ plane rather than similar to a vertical-horizontal ellipse which is usually the case in simple wave packets etc. Maybe this is what you wanted to hear as the physical explanation of the anticommutator term - because $AB+BA$ is just twice the Hermitean part of $AB$, it measures the correlation of $A,B$ in the distribution given by the wave function - although the precise meaning of these words has to be determined by the formula.