I use units in which hbar = 1, so xp - px = i.

<xp>* = <psi,xp psi>* = <xp psi,psi> = <p psi,x psi>
      = <psi,px psi> = <px> = <xp - i> = <xp> - i,
so the imaginary part of <xp> is 1/2, meaning |<xp>|^2 >= 1/4.

It is a general rule in Hilbert spaces that
|<phi,psi>| <= ||phi|| ||psi||.  Thus:
1/4 <= |<xp>|^2 = |<psi,xp psi>|^2 = |<x psi,p psi>|^2 
    <= ||x psi||^2 ||p psi||^2 = <x psi,x psi><p psi,p psi> 
    = <psi,xx psi><psi,pp psi> = <x^2><p^2>.

Now, here's the trick.
By "x" and "p", I do *not* mean the position and momentum operators!
If X and P are the position and momentum operators,
then x := X - <X>, and p := P - <P>.
Note that [x,p] = i, just like [X,P];
note that x and p are Hermitian, just like X and P;
in fact, everything I did with x and p was perfectly valid.
So that last inequality, <x^2><p^2> >= 1/4, is just what we want.

BTW, if psi is Gaussian, <x^2><p^2> = 1/4,
so this is the strongest inequality possible.