Upper bound for the total curvature of a shortest path in the boundary of a convex polyhedron in $mathbb{R}^3$.

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 $$(-u_1)cdot u_i >eta>0$$ 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
: If $P$ contains line segments $[z_iz], [z z_{i+1}]$ in faces
$f_i, f_{i+1}$ where $zin f_ibigcap f_{i+1}$, 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 $f_ibigcap f_{i+1}$. 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 $C_i, lambda_i>0$.

Hence $$ -u_1cdot( x_i-x_{i+1} ) geq lambda_i (2eta ) $$

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 pifrac{-u_1cdot (
x_i-x_{i+1}) }{2eta } end{align*}

Hence $sum_{i=1}^{n-1} angle(x_i,x_{i+1}) leq frac{pi}{2eta} |u_1cdot (x_1-x_n)| leq frac{pi}{eta}$.