Firstly I have seen that the $\Gamma$ function clearly has a minimum point at about $(1.462,0.886)$ and that this cannot be completely calculated using calculus, as shown below: $$\Gamma(z)=\int_{0}^{\infty}t^{z-1}e^{-t}dt$$ and since the simplified Liebniz integral rule states that: $$\frac{d}{dx}\int_{a}^{b}{f(x,t)}dt=\int_{a}^{b}{\frac{\partial}{\partial_{x}}f(x,t)dt}$$ so if we rewrite the gamma function as: $$\Gamma(z)=\int_{0}^{\infty}{e^{(z-1)\ln(t)}\,e^{-t}}\,dt$$ then we can deduce that: $$\Gamma'(z)=\int_{0}^{\infty}{\ln(t)\,t^{z-1}\,e^{-t}}\,dt$$ and since we want to find the minimum point we can let $\Gamma'(z)=0$
In addition to this, since it is a minimum point, we know that the second differential is greater than zero, so: $\Gamma''(z)>0$
However, after this point I have got stuck on how people have got from this result to a numerical answer, my only thought was: $$\frac{d}{dt}\Gamma'(z)=[\ln(t)\,t^{z-1}\,e^{-t}]_{0}^{\infty}$$ so maybe: $$\lim_{t\to\infty}[\ln(t)\,t^{z-1}\,e^{-t}]-\lim_{t\to0}[\ln(t)\,t^{z-1}\,e^{-t}]=0$$ But I am not sure if this is correct and how this would help find a solution.
As far as I remember, Ramanujan wrote (I cannot find it again) $$\psi(x)\sim \log \left(x-\frac{1}{2}\right)$$ which is a very good approximation as soon as $x > 4$.
So, the zero of $\psi(x)$ is "close" to $\frac 3 2$ (using any numerical method, it is $1.46163$). Using the approximation, the minimum would be $$\approx\Gamma \left(\frac{3}{2}\right)=\frac{\sqrt{\pi }}{2}\approx 0.886227$$ while the numerical value would be $\approx 0.885603$.