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I am having trouble proving this:

"Let $P$ be a convex polyhedron with $V$ vertices, $E$ edges and $F$ faces. Furthermore, let $F_m$ be the number of faces surrounded by $m$ edges, and let $V_m$ be the number of vertices from which $m$ edges emanate.

Show that every convex polyhedron contains a face with either 3, 4 or 5 edges.

Show that every convex polyhedron contains a vertex from which either 3, 4 or 5 edges emanate."

There is apparently a proof by contradiction for these but I have no idea how or where to start with it.

Thanks

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HINTS: the relation of $V$, $E$ & $F$ for a convex polyhedron is given by Euler's formula $$F+V=E+2$$ the sum of interior angles of all the faces meeting at one vertex of a convex polyhedron is always less than $2\pi$ or $360^\circ$

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