Convex polyhedron and its Gauß-curvature I have asked this question on MathSE and no one could give me an answer. So I'll post my question here.
What I am trying to prove:

A convex polyhedron has positive Gauß-Curvature at every vertex.

What we know:


*

*Gauß-Curvature at every vertex is given by $K(p) = 2\pi - \sum\limits_{\text{angle } \alpha_i \text{ around } p} \alpha_i$.

*Gauß-Bonnet state $\sum\limits_{\text{every vertex }p} K(p) = 2\pi\chi(S)$, where $S$ the convex polyhedron. Therefore we can also write$\sum\limits_{\text{every vertex }p} K(p) = 4\pi$.


My attempt:
I plugged the Gauß-Curvature in the Gauß-Bonnet-Formula and obtained
$$ 
\sum\limits_{\text{every vertex }p} K(p) = \sum\limits_{\text{every vertex }p} \left(2\pi - \sum\limits_{\text{angle } \alpha_i \text{ around } p} \alpha_i\right) = 4\pi.
$$
Assume now that there exists a $\tilde{p}$ such that $2\pi - \sum\limits_{\tilde{p}} \alpha_i < 0$. Then $2\pi < \sum\limits_{\tilde{p}} \alpha_i$...
From here i do not know how to get further on. Does anyone have an alternative of proving this statement or at least can somebody tell me how to complete my proof?

My attempt is wrong, since the Gauß-formula can also applied for non-convex polyhedra!
Can someone complete this proof? 
Thanks in advance!
 A: Let me try to sketch a proof of "the evident fact that sum of angles around a vertex of a convex polytope is less than $2\pi$," as phrased by Alexandre Eremenko.
This proof below is more complex than Otis Chodosh's, but perhaps has the advantage
of using convexity explicitly.
Let $v$ be a vertex of the convex polyhedron $P$.
Intersect $P$ with a small sphere $S$ centered on $v$.
The intersection forms a convex spherical polygon $Q$ (blue below)
lying within one hemisphere of $S$.
(The hemisphere rim (red below) corresponds to a plane supporting $P$ at $v$.)
We can take the radius of $S$ to be $1$ by appropriately scaling $P$ prior to
intersecting.
Note that the angles of the faces of $P$ incident to $v$ become the arcs of $Q$.
Now I would like to claim that the perimeter of a convex spherical polygon $Q$
lying within a hemisphere of a unit sphere $S$ is at most $2\pi$.
I will mimic a proof in a
Noam Elkies posting for 
a planar analog of the claim.
The proof is by induction.
Let $Q_0$ be the hemisphere rim; its length is $2\pi$.
$Q_0$ contains $Q$: $Q_0 \supseteq Q$.
We will form a succession of more tightly enclosing convex
spherical polygons, $Q_0 \supseteq Q_1 \supseteq Q_2 \ldots \supseteq Q$,
with decreasing perimeters (more accurately: non-increasing perimeters).
Let $k$ be the number of edges of $Q$ that do not lie along
the boundary of $Q_i$. If $k=0$, then $Q_i=Q$ and we are finished.
Let $e$ be some edge of $Q$ that does not lie on the boundary
of $Q_i$.
Extend $e$ to a great-circle arc that cuts $Q_i$ at points $a$ and $b$.
Define $Q_{i+1}$ as following $Q_i$ but then the "shortcut" $ab$.
Because $ab$ is a geodesic, it is a shortest path between $a$ and $b$,
and therefore cannot be longer than the portion of $Q_i$ cut off.
So the perimeter of $Q_{i+1}$ is at most the perimeter of $Q_i$.
Continuing in this manner produces the nested sequence of
enclosing spherical polygons, ending with $k=0$.
If no edge of $Q$ lies along the original $Q_0$, then
the first cut will be a semicircle (green below), and
the perimeter of $Q_1$ is $\pi+\pi$, i.e., still $2 \pi$.

         


So the perimeter of $Q$ is
at most $2\pi$, the perimeter of $Q_0$. 
Therefore the sum of the angles of the faces of $P$ incident to $v$
is at most $2\pi$.
Therefore, $P$ has positive angle deficit at $v$, i.e., it has positive Gaussian
curvature there.

Note convexity is used crucially, for a nonconvex spherical polygon could have
an arbitrarily large perimeter, still fitting inside a hemisphere.

Added (2Jul2015).
The example below is intended to show it is not correct
to project the angles incident to $v$ to a plane and argue that
the angles only enlarge under projection.
Below, one angle is $68^\circ$ incident to $v$ in 3D,
but that angle projects to $45^\circ$ in 2D.

          


As you can see, some angles grow, some shrink, under projection.
