Let's suppose we have a quantum system that consists of a free particle on the real line. The Hilbert space associated to this quantum system is $L^2(\mathbb{R})$. Let $x$ and $p$ be the position and momentum operators. The operatorial version of Schrodinger equation states that if the system is in a state $\psi=\psi(x)$ then
$$\sigma_\psi(x)\sigma_\psi(p)\geq \frac12 |\langle \psi,[x,p]\psi\rangle|$$
It's pretty easy to see that $\langle \psi,[x,p]\psi\rangle=0$ iff $\psi=0$, but $\psi$ can't be $0$, so there is no state in which the uncertainty of $(x,p)$ can be reduced to $0$.
What if we try with the couple $(x,p^2)$. This is trickier because:
$$\langle \psi,[x,p^2]\psi\rangle=0 \iff \langle \psi,\psi'\rangle=0$$
I understand that $\psi'$ can't be $0$, but why does this imply that $\langle \psi,\psi'\rangle$ can't be $0$ for any $\psi$? I think that $\psi(x)=e^{-x^2}$ does the job! Right?
I'm not sure how you are misconstruing the uncertainty principle, because I didn't understand your argument. I am a little baffled by why you failed to directly compute the uncertainty of your Gaussian state, (basically the ground state of the oscillator), bypassing generic Robertson inequality jazz. Nondimensionalize ℏ=1, m=1.
For the state $\psi= N e^{-x^2}$ where $ N^2=\sqrt{2/\pi}$, you have, directly, the variances $$ \sigma_\psi ^2 (x)= N^2\int\!\!dx~ x^2 e^{-2x^2}= 1/4, \\ \sigma_\psi ^2 (p^2/2)= N^2\int\!\!dx~ e^{-2x^2}(4x^2-12x^2 +3) -\left ( N^2\int\!\!dx~ e^{-2x^2}(2x^2-1) \right )^2\\ =3/4-1/4=1/2, $$ where I hope you have recognized the Hermite polynomials controlling the problem here, so that $$ \sigma_\psi (x) \sigma_\psi (p^2/2)= \sqrt{2}/4. $$
It is not evident what "trick" you are invoking, but, for that state, the precise uncertainty is nonvanishing.