Problem : Consider finitely many points in $\mathbb{R}^3$. The boundary of the convex hull is $\Sigma$. When $f_i$ is a face and $u_i$ is unit outnormal to $f_i$, then assume that $$(-u_1)\cdot u_i >\eta>0 \ \ast$$ for all $i>1$ where $\cdot $ is an inner product. When $P$ is a shortest path in $\Sigma -f_1$,
then the total curvature of $P$ is smaller than $\frac{\pi}{\eta}$.
I need another proof, because the following well-known proof is not easy.
Proof : If $P$ contains line segments $[z_iz],\ [z z_{i+1}]$ in faces $f_i, \ f_{i+1}$ where $z\in f_i\bigcap f_{i+1}$, then we define $$ x_i =\frac{z-z_i}{|z-z_i|},\ x_{i+1} = \frac{z_{i+1}-z}{|z_{i+1} -z|}$$
Similarly, we have any line segments $[\overline{z}_iz],\ [z \overline{z}_{i+1}]$ in faces $f_i, \ f_{i+1}$ s.t. they are orthogonal to $f_i\bigcap f_{i+1}$. Similarly we have $\overline{x}_i,\ \overline{x}_{i+1}$. Then there is $$ x_i-x_{i+1} = C_i (\overline{x}_i-\overline{x}_{i+1} )=\lambda_i( u_i + u_{i+1}) $$ where $C_i,\ \lambda_i>0$ : Here the second inequality is followed from drawing directly. If we set $x_i=A \overline{x}_i +B u_i\times \overline{x_i}$, then note that $x_{i+1}=A \overline{x}_{i+1} +B u_{i+1}\times \overline{x}_{i+1}$. Hence the first inequality is followed.
Hence $ -u_1\cdot( x_i-x_{i+1} ) \geq \lambda_i (2\eta ) $ Hence \begin{align*} \angle \ (x_i,x_{i+1}) &\leq \frac{\pi}{2}|x_i-x_{i+1}| \\ &=\frac{\pi}{2} |\lambda_i (u_i+u_{i+1}) | \\&\leq \pi \lambda_i \\&\leq \pi\frac{-u_1\cdot ( x_i-x_{i+1}) }{2\eta } \end{align*}
Hence $\sum_{i=1}^{n-1} \angle(x_i,x_{i+1}) \leq \frac{\pi}{2\eta} |u_1\cdot (x_1-x_n)| \leq \frac{\pi}{\eta}$.
Consider the case where $\Sigma$ is smooth Riemannian surface in $\mathbb{R}^3$ of positive Gaussian curvature, homeomorphic to a sphere. Define a region $D =\bigg\{ x\in \Sigma \bigg| N(x)\cdot v >\eta \bigg\}$ for some $|v|=1$ and $\eta>0$ where $N$ is unit out normal to $\Sigma$.
Assume that $c:[0,l]\rightarrow D$ is a $\Sigma$-minimizing geodesic of unit speed in $D$. Hence we have $|w(t)|=1,\ v\perp w(t)$ s.t. $$ N(t)=a(t)w(t) + b(t) v,\ b\geq \eta ,\ a^2+b^2=1$$
If $k$ is a curvature for $c$, then $c''(t)=-k(t)N(t),\ k>0$. Hence \begin{align*} -c''\cdot v&=k b \geq k\eta \\ \int_0^l\ k(t)dt &\leq \int_0^l\ \frac{-c''\cdot v}{\eta} \ dt \\&= -\frac{1}{\eta} c'(t)\cdot v \bigg|_0^l \\&\leq \frac{2}{\eta } \end{align*}