In addition to the other good answers directly explaining the meaning, it would be constructive to see how it is derived:
The inequality is the generalized uncertainty principle, which is the most complete form of uncertainty principle.
Given Hermitian operators $A,B$, we define vectora $\left|q\right> \& \left|p\right>$ with an arbitrary wavefunction $\left|\phi \right>$ as:
$$\left|q\right>=(A-\left< A\right>)\left|\phi\right>
\ ;\
\left|p\right>=(B-\left< B\right>)\left|\phi\right>\tag{1}
$$
By Schwarz inequality:
$$\left<q|q\right>\left<p|p\right> \ge \left<q|p\right>\left<p|q\right>\tag{2}
$$
$$(\left<A\right>^2-\left<A^2\right>)(\left<B\right>^2-\left<B^2\right>)\ge \Re^2[\left<p|q\right>]+ \Im^2[\left<p|q\right>]\tag{3}
$$
After some work:
$$(\Delta A)^2(\Delta B)^2\ge [\frac{1}{2}(\left<p|q\right>+\left<q|p\right>)]^2+[\frac{1}{2i}(\left<p|q\right>-\left<q|p\right>)]^2\tag{4}
$$
$$(\Delta A)^2(\Delta B)^2\ge (\left<\phi\right|\frac{1}{2}[AB+BA]\left|\phi\right>+\left<A\right>\left<B\right>)^2+(\left<\phi\right|\frac{1}{2i}[AB-BA]\left|\phi\right>)^2\tag{5}
$$
Simplifying it, we have:
$$(\Delta A)^2(\Delta B)^2\ge (\left<\phi\right|\frac{1}{2}\{A^*,B^*\}\left|\phi\right>)^2+(\left<\phi\right|\frac{1}{2i}[A,B]\left|\phi\right>)^2\tag{6}
$$
Where $\{A,B\}$ is anticommutator: $AB+BA$ ; $A^*$ and $B^*$ are defined as $A-\left<A\right>$ and $B-\left<B\right>$ respectively. In general, the anticommutator part is too complicated to be useful.
As an inequality, the generalized uncertainty principle always suggests:
$$(\Delta A)^2(\Delta B)^2\ge (\left<\phi\right|\frac{1}{2}\{A^*,B^*\}\left|\phi\right>)^2+(\left<\phi\right|\frac{1}{2i}[A,B]\left|\phi\right>)^2 \ge (\left<\phi\right|\frac{1}{2i}[A,B]\left|\phi\right>)^2
$$
This is the usual uncertainty principle.