= Solution
For a normalized state and an <observable> $A$, define $\langle A\rangle=\langle\psi,A\psi\rangle$ and $\Delta_\psi A=\|(A-\langle A\rangle)\psi\|$. Suppose first that the position and momentum <variances> are finite and the state has the regularity needed for the usual position-momentum quadratic forms. Put $a=\langle x\rangle$, $b=\langle p\rangle$, $u=(x-a)\psi$ and $v=(-i\hbar\partial_x-b)\psi$. The <Cauchy-Schwarz inequality> gives
$$
(\Delta_\psi x)^2(\Delta_\psi p)^2\ge|\langle u,v\rangle|^2\ge(\Im\langle u,v\rangle)^2.
$$
The term involving $b$ has zero imaginary part. Integration by parts gives
$$
2\Re\int_{\mathbb R}(x-a)\bar\psi\psi'\,dx
=\int_{\mathbb R}(x-a)(|\psi|^2)'\,dx=-1.
$$
Consequently $\Im\langle u,v\rangle=\hbar/2$, and
$$
\boxed{(\Delta_\psi x)(\Delta_\psi p)\ge\frac\hbar2.}
$$
This proves the <Heisenberg uncertainty relation>. For smooth rapidly decaying states the boundary terms vanish directly. The same proof extends to $x\psi\in L^2$ and weak derivative $\psi'\in L^2$ using smooth cutoffs and taking their radius to infinity; the necessary mixed products are integrable by Cauchy-Schwarz. Normalization alone need not give finite <variances> or an operator-domain momentum expectation. A state with infinite spread is interpreted through the corresponding extended bound, rather than assigning a finite uncertainty to an undefined expression.
For the specified <Gaussian wave packet>, write $s=\hbar t$ and $a_t=1+is$. Its <probability density> is
$$
|\psi(x,t)|^2=\frac1{\sqrt{2\pi}\sqrt{1+s^2}}
\exp\left[-\frac{x^2}{2(1+s^2)}\right].
$$
It is a normalized centered <normal distribution> with <variance> $1+s^2$. Symmetry gives $\langle x\rangle=0$, and hence
$$
\boxed{\Delta_\psi x=\sqrt{1+\hbar^2t^2}.}
$$
Also $\partial_x\psi=-x\psi/(2a_t)$, so $p\psi=i\hbar x\psi/(2a_t)$ and $\langle p\rangle=0$ by oddness. Integration by parts, or directly taking the squared norm, gives
$$
\langle p^2\rangle=\hbar^2\int|\psi'|^2\,dx
=\frac{\hbar^2}{4|a_t|^2}\langle x^2\rangle=\frac{\hbar^2}{4}.
$$
Thus
$$
\boxed{\Delta_\psi p=\frac\hbar2,\qquad
(\Delta_\psi x)(\Delta_\psi p)=\frac\hbar2\sqrt{1+\hbar^2t^2}\ge\frac\hbar2.}
$$
The packet saturates the bound at $t=0$ and has a larger position-momentum uncertainty product at every nonzero time. Its spatial spreading does not change the momentum distribution.
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