blademan9999 asked: "At what speed will the required proper acceleration be minimised."
In that example with a circular orbit in the $\theta$ direction, the coordinate acceleration is
$$\rm \frac{d^2 x^{\mu}}{d \tau^2}=\{ \ddot{t}, \ \ddot{r}, \ \ddot{\theta}, \ \ddot{\phi}\}=\{0,\ 0, \ 0, \ 0\}$$
and the 4-velocity
$$\rm \frac{d x^{\mu}}{d \tau}=u^{\mu}=\{ \dot{t}, \ \dot{r}, \ \dot{\theta}, \ \dot{\phi}\}=\{ \gamma \surd g^{tt}, \ 0, \ v \gamma /r, \ 0\}$$
so the force required for that is in natural units
$$\rm F= \surd \ |\sum_{\mu, \nu} g_{\mu \nu} \ a^{\mu} a^{\nu}| = \surd \
|\frac{(r-2) \ v^2-1}{\sqrt{(r-2) \ r^3 \left(v^2-1\right)^2}}|$$
where $\rm v$ is the local velocity relative to a stationary observer, and the proper acceleration
$$ \rm a^{\mu}= \frac{d^2 x^{\mu}}{d \tau^2}+\sum_{\alpha, \beta} \ \Gamma^{\mu}_{\alpha \beta} \ u^{\alpha}u^{\beta}$$
so at $\rm r=4$ we get $\rm F=0$ with $\rm v= \sqrt{1/(r-2)}=\sqrt{1/2}$ as expected. At $\rm r=1.5 \ r_s=3$ the radial force $\rm F=1/\sqrt{27}$
(in units of $\rm c^2 m/M/G$) is independend of the angular velocity, and at $\rm r<3$ the required force to keep a constant $\rm r$ is smallest when $\rm v=0$.
That is due to the $-3 \dot{\theta}^2$ term in the geodesic equation
$$\rm \ddot{r}=-3 \dot{\theta}^2+ r \dot{\theta}^2 -1/r^2$$
which differs from the Newtonian expression and makes the angular velocity contribute negatively below the critical radius $\rm r=3$, and cancel itself at that exact radius.
This is assuming the Schwarzschild metric, for rotating black holes it would be a little bit more complicated. The solution might be shortened into a more elegant form, but since the coordinate acceleration is $0$ on a circular path the calculation is straightforward.
