Heisenberg uncertainty made stronger by ommiting anticommutator

Question

The following reasoning appears in Sakurai/Napolitano - Modern Quantum Mechanics, 3rd ed, ch 1:

Using a Schwarz inequality, it is proved that $$ | \langle \Delta A \Delta B \rangle |^2 \geq \frac{1}{4} |\left \langle \right[A, B] \rangle |^2 + \frac{1}{4} |\left \langle \right \{ \Delta A, \Delta B \} \rangle|^2 $$ Where $\Delta A = A - \langle A\rangle$ is some sort of dispersion/difference operator, $\langle A \rangle$ is expectation value, $[A,B] = AB-BA$ is the commutator and $\left\{ P, Q \right\} = PQ + QP$ is the anti-commutator. Assume $A$, $B$, $\Delta A$, $\Delta B$ are Hermitian (last two might be proved, but let's just assume).

Both terms are non-negative, therefore it is also true that: $$ | \langle \Delta A \Delta B \rangle |^2 \geq \frac{1}{4} |\left \langle \right[A, B] \rangle |^2 $$ and the author says this second version is stronger than the first.

My question is: how so? Isn't a met condition stronger depending on how restricting it is? The first equation is clearly more restrictive. Also, why don't we take the anti-commutator as the stronger version? Also, if the second equation is stronger than the first, why isn't $| \langle \Delta A \Delta B \rangle |^2 \geq 0$, which is trivially true taken as a even stronger condition (it isn't even useful, then)?

Edit: just found this What is the meaning of the anti-commutator term in the uncertainty principle?, but the direction is this question is opposite, since this book says literally: "The proof of is now complete because the omission of the second (the anticommutator) term of can only make the inequality relation stronger"

Answer 1

$$ | \langle \Delta A \Delta B \rangle |^2 \geq \frac{1}{4} |\left \langle \right[A, B] \rangle |^2 + \frac{1}{4} |\left \langle \right \{ \Delta A, \Delta B \} \rangle|^2 \geq \frac{1}{4} |\left \langle \right[A, B] \rangle |^2 \geq 0. $$

Stronger means "less likely to be saturated" (the equal sign instead of >). That is, less tight or restrictive. A fortiori. The imagery is of a tall guy on the left looking down on possibly progressively shorter ones who might or might not be equal in height to each other...

(You have a point that, confusingly, "more restrictive", a tighter lower bound, is a "stronger statement" in customary use. This is the uncertainty to cherish.)

In the last analysis, the usual one is a nice practical compromise: it can be saturated by Schroedinger wave packets (coherent states), but just those. The commutator is a number easy to calculate oftentimes, unlike the anticommutator. The zero right-hand-side one is terminally strong, as it can normally (e.g. for x and p) never be saturated, so it has basically no-nontrivial information.