$\langle xp+px\rangle|_{t=0}=2\langle p\rangle\langle x\rangle|_{t=0}$ for the free particle? 
Quantum Mechanics, Volume 1 by Claude Cohen-Tannoudji, Bernard Diu and Frank Laloe. Complement L-III, exercise 4 (page 342).

Basically consider a free particle, and calculate the variance(uncertainty) of $(\Delta X)^2$ by using Ehrenfest's theorem.
I got $$\langle X\rangle =\frac{\langle p\rangle }{m}t+K, \langle X^2\rangle =\frac{\langle p^2\rangle }{m^2}t^2+\frac{c_1}{m}t+c_2$$ and $$\langle xp+px\rangle =\frac{2}{m}\langle p^2\rangle t+c_1 ,$$where $c_1,c_2,K$ are arbitrary constant from integration with respect to time $t$.
The answer in the text book is $$(\Delta X)^2 =\frac{1}{m}(\Delta p)_0^2t^2+(\Delta X)_0^2.$$
According to my calculation, there is an extra term $$(\frac{c_1}{m}-\frac{2\langle p\rangle }{m}K)t .$$
If I assume $c_1=K=0$, then it's done. 
However, this does not seem to be the case the textbook suggested, instead, it seemed to be suggesting that $$\langle xp+px\rangle |_0=c_1=2\langle p\rangle K=2\langle p\rangle \langle x\rangle |_0 .$$
(Notice $\langle p\rangle $ and $\langle p^2\rangle $ are independent of time. Further, if one assumes transitional symmetry, then as a term in measurement $$(\frac{c_1}{m}-\frac{2\langle p\rangle }{m}K)t$$ must disappear, so even without the context of the textbook $$\langle xp+px\rangle |_0=c_1=2\langle p\rangle K=2\langle p\rangle \langle x\rangle |_0$$ for a free particle.)
My question is:


*

*How to prove $$\langle xp+px\rangle |_0= 2\langle p\rangle \langle x\rangle |_0$$

*If the expression is not true, what happened to the transitional symmetry?
(* the subscript $0$ meant measure at time $t=0$)
 A: Your text instructs you to heed  "with a suitable choice of the time origin". Certainly not to  nullify the extra term 
$$A_0 t\equiv \frac{t}{m} \langle \{(p-\langle p\rangle_0),(  x-\langle x\rangle_0) \}\rangle_0
$$ you correctly found. Instead, to absorb it into a shifted time variable. 
To wit, from your original 
$$(\Delta X)^2 _t=\frac{1}{m}(\Delta p)_0^2~t^2+(\Delta X)_0^2 +A_0 ~t,$$ 
pick a time $\tau$, so
$$(\Delta X)^2 _\tau=\frac{1}{m}(\Delta p)_0^2~\tau^2+(\Delta X)_0^2 +A_0 ~\tau,$$ 
which you may subtract from the above to get 
$$(\Delta X)^2 _t -(\Delta X)^2 _\tau =\frac{1}{m}(\Delta p)_0^2~(t^2 -\tau^2)+A_0 ~(t-\tau)\\
= \frac{1}{m}(\Delta p)_0^2~(t  -\tau ) ^2   + \left ( \frac{2\tau}{m}(\Delta p)_0^2    +A_0\right ) ~(t-\tau) .   $$ 
You then solve for $\tau$ to nullify the big parenthesis,
$$
 \tau  =-mA_0/2  (\Delta p)_0^2 = \frac{\langle p\rangle \langle x\rangle_0 -\langle xp+px\rangle_0/2}{(\Delta p)^2_0} = \frac{-\langle \{(p-\langle p\rangle_0),(  x-\langle x\rangle_0) \}\rangle_0}{2(\Delta p)^2_0}, 
$$
substituting your initial conditions. Note this does not vanish for all distributions (initial conditions): it it absolutely not constrained by the TDSE, or measurements, or anything else. It is a constant.
You then see that for this origin of time, 
$$
(\Delta X)^2   = (\Delta X)^2 _\tau +
  \frac{1}{m}(\Delta p)_\tau^2~(t  -\tau ) ^2 ,
$$
your book's expression, recalling that $(\Delta p)_0=(\Delta p)_\tau$.  
