From this link Heisenberg uncertainty principle, It says:
Clearly, when $\Delta p_x$ shrinks, $\Delta x$ has to grow larger and larger in order to satisfy the Heisenberg inequality. For example, a plane wave $\psi(x) = \exp[2\pi i px/h]$ is an eigenfunction of $p = -ih/(2\pi)\ d/dx$, so that $\Delta p_x = 0$; a plane wave particle has a position $x$ that is completely undetermined. Conversely, if the position of the particle is very well determined, its momentum is very uncertain. The p-expansion (Fourier transform) of a well-localized wave packet ($\Delta x \approx 0$) requires eigenstates of many different eigenvalues $p$ and hence gives rise to a large spread in $p$.
so one of the variance from the left side of this equation could be zero (my calculation about measure spin in different basis also implies that one of the hermitian operator can have zero variance), this equation becomes to $0\geq h/4\pi$ which is clearly wrong. Does this mean the inequality does not work for 0 variance?