I am unsure on how to prove this statement. I am relatively certain that it is true, any advice on how to start?
Edit: It has occurred to me that this will be significantly easier to show by restricting the simplex to be regular (and may be only true in this case).
On
If the "origin" is arbitrary, then the statement is clearly wrong: if it is sufficiently far enough from the simplex, the vectors from it to the vertices are almost collinear.
Even for a regular triangle and the origin inside it, but close to one of the vertices, the angle between the the vectors pointing to the other two vertices will be acute - almost $\pi/2$.
Even if the origin is the centroid of the simplex, it is still false for an arbitrary simplex because one can arrange the vertices almost along a circle and the angles for "neighboring" vertices will be $2\pi/n$.
However, for a regular simplex and its center all the dot products are the same and it is pretty easy to compute: in $n$-dimensional space, the regular $n-1$-dimensional simplex with $n$ vertices has the standard orthonormal basis $(0,...,0,1,0,...,0)$ for vertices and the centroid is $(1/n,...,1/n)$. The dot product in question is:
$$ 2\times(-\frac1n)(1-\frac1n)+(n-2)\times(-\frac1n)(-\frac1n) = - \frac2n+\frac2{n^2} +\frac{n}{n^2}-\frac2{n^2} =-\frac1n < 0 $$
More generally, given a segment, the locus of points from which the segment is seen under an obtuse angle is the interior of the sphere having the segment as its diameter. Thus the locus of point such that the vectors from it to the vertices of a simplex have all negative dot products is precisely the intersection of spheres that have simplex edges as diameters, IOW, those points whose distances to the edge centers are less than the half edge length.
[Note: Some of my $n$s are off by one because the simplexes I was thinking about weren't the standard simplexes. I don't think that changes the core idea though, and this is only a sketch of an idea.]
I haven't worked this all the way out, but it's got an $n$ in it, so why not try induction?
It feels like you should be able to express the vectors for an $n+1$ simplex as {the vectors for an $n$ simplex} plus (the vector from the $n$ center to the $n+1$ center), along with one new vector for the new vertex. Then you have to show 1) you still have a negative dot product for the old vectors, even after adding on the center-to-center vector, and 2) the vector for the new vertex has a negative dot product with all the previous vectors.
Edit - Adding details on the induction: I think of an n-simplex as lying in $\mathbb{R}^n$, and you build an $(n+1)$-simplex by mapping $\mathbb{R}^n$ into $\mathbb{R}^{n+1}$ as $$(x_1, ... x_n) \mapsto (x_1, ... x_n, 0)$$
and then you add a new vertex that becomes the "top of the pyramid" and the embedded $n$-simplex becomes the base of the $n+1$-simplex. BTW, as I'm explaining this I'm realizing you could either track the vertices or track the vectors from your center to the vertices when doing your induction. I'm not sure which would be cleaner/easier.
I'll confess that I'm still not sure which class of simplexes you want to prove this for, so I'm not even sure that this construction always applies to all the cases you're interested in, but it works for the standard simplexes we used in algebraic topology, and it might let you figure out how to construct counter-examples if you find that you can't carry the induction forward.