Recently there have been some interesting questions on standard QM and especially on uncertainty principle and I enjoyed reviewing these basic concepts. And I came to realize I have an interesting question of my own. I guess the answer should be known but I wasn't able to resolve the problem myself so I hope it's not entirely trivial.

So, what do we know about the error of simultaneous measurement under time evolution? More precisely, is it always true that for $t \geq 0$
$$\left<x(t)^2\right>\left<p(t)^2\right> \geq \left<x(0)^2\right>\left<p(0)^2\right>$$
(here argument $(t)$ denotes expectation in evolved state $\psi(t)$, or equivalently for operator in Heisenberg picture).

I tried to get general bounds from Schrodinger equation and decomposition into energy eigenstates, etc. but I don't see any way of proving this. I know this statement is true for a free Gaussian wave packet. In this case, we obtain equality, in fact (because the packet stays Gaussian and because it minimizes HUP). I believe this is, in fact, the best we can get and for other distributions, we would obtain strict inequality.

So, to summarize the questions


  
  Is the statement true?
  If so, how does one prove it? And is there an intuitive way to see it is true?
  


The question asks about the time dependence of the function

$$f(t) := \langle\psi(t)|(\Delta \hat{x})^2|\psi(t)\rangle
 \langle\psi(t)|(\Delta \hat{p})^2|\psi(t)\rangle,$$

where 

$$\Delta \hat{x} := \hat{x} - \langle\psi(t)|\hat{x}|\psi(t)\rangle, \qquad
\Delta \hat{p} := \hat{p} - \langle\psi(t)|\hat{p}|\psi(t)\rangle, \qquad
\langle\psi(t)|\psi(t)\rangle=1.$$

We will here use the Schrödinger picture where operators are constant in time, while the kets and bras are evolving. 

Edit: Spurred by remarks of Moshe R. and Ted Bunn let us add that (under assumption (1) below) the Schroedinger equation itself is invariant under the time reversal operator $\hat{T}$, which is a conjugated linear operator, so that 

$$\hat{T} t = - t \hat{T}, \qquad \hat{T}\hat{x} = \hat{x}\hat{T}, \qquad \hat{T}\hat{p} = -\hat{p}\hat{T}, \qquad  \hat{T}^2=1.$$

Here we are restricting ourselves to Hamiltonians $\hat{H}$ so that

$$[\hat{T},\hat{H}]=0.\qquad (1)$$

Moreover, if 

$$|\psi(t)\rangle = \sum_n\psi_n(t) |n\rangle$$

is a solution to the Schrödinger equation in a certain basis $|n\rangle$, then 

$$\hat{T}|\psi(t)\rangle := \sum_n\psi^{*}_n(-t) |n\rangle$$

will also be a solution to the Schrödinger equation with a time reflected function $f(-t)$.

Thus if $f(t)$ is non-constant in time, then we may assume (possibly after a time reversal operation) that there exist two times $t_1<t_2$ with $f(t_1)>f(t_2)$. This would contradict the statement in the original question. To finish the argument, we provide below an example of a non-constant function $f(t)$.

Consider a simple harmonic oscillator Hamiltonian with the zero point energy $\frac{1}{2}\hbar\omega$ subtracted for later convenience.

$$\hat{H}:=\frac{\hat{p}^2}{2m}+\frac{1}{2}m\omega^{2}\hat{x}^2
-\frac{1}{2}\hbar\omega=\hbar\omega\hat{N},$$

where $\hat{N}:=\hat{a}^{\dagger}\hat{a}$ is the number operator.

Let us put the constants $m=\hbar=\omega=1$ to one for simplicity. Then the annihilation and creation operators are

$$\hat{a}=\frac{1}{\sqrt{2}}(\hat{x} + i \hat{p}), \qquad
\hat{a}^{\dagger}=\frac{1}{\sqrt{2}}(\hat{x} - i \hat{p}), \qquad
[\hat{a},\hat{a}^{\dagger}]=1,$$

or conversely,

$$\hat{x}=\frac{1}{\sqrt{2}}(\hat{a}^{\dagger}+\hat{a}), \qquad
\hat{p}=\frac{i}{\sqrt{2}}(\hat{a}^{\dagger}-\hat{a}), \qquad 
[\hat{x},\hat{p}]=i,$$

$$\hat{x}^2=\hat{N}+\frac{1}{2}\left(1+\hat{a}^2+(\hat{a}^{\dagger})^2\right), \qquad 
\hat{p}^2=\hat{N}+\frac{1}{2}\left(1-\hat{a}^2-(\hat{a}^{\dagger})^2\right).$$

Consider Fock space $|n\rangle := \frac{1}{\sqrt{n!}}(\hat{a}^{\dagger})^n |0\rangle$ 
such that $\hat{a}|0\rangle = 0$. Consider initial state 

$$|\psi(0)\rangle := \frac{1}{\sqrt{2}}\left(|0\rangle+|2\rangle\right), \qquad 
\langle \psi(0)|  = \frac{1}{\sqrt{2}}\left(\langle 0|+\langle 2|\right).$$

Then 

$$|\psi(t)\rangle = e^{-i\hat{H}t}|\psi(0)\rangle
 = \frac{1}{\sqrt{2}}\left(|0\rangle+e^{-2it}|2\rangle\right),$$

$$\langle \psi(t)| = \langle\psi(0)|e^{i\hat{H}t}
 = \frac{1}{\sqrt{2}}\left(\langle 0|+\langle 2|e^{2it}\right),$$

$$\langle\psi(t)|\hat{x}|\psi(t)\rangle=0, \qquad 
\langle\psi(t)|\hat{p}|\psi(t)\rangle=0.$$

Moreover,

$$\langle\psi(t)|\hat{x}^2|\psi(t)\rangle=\frac{3}{2}+\frac{1}{\sqrt{2}}\cos(2t), \qquad 
\langle\psi(t)|\hat{p}^2|\psi(t)\rangle=\frac{3}{2}-\frac{1}{\sqrt{2}}\cos(2t),$$

because $\hat{a}^2|2\rangle=\sqrt{2}|0\rangle$. Therefore,

$$f(t) = \frac{9}{4} - \frac{1}{2}\cos^2(2t),$$

which is non-constant in time, and we are done. Or alternatively, we can complete the counter-example without the use of above time reversal argument by simply performing an appropriate time translation $t\to t-t_0$.
