= Solution
For a connected cellular embedding on the sphere, the <Euler formula for a connected planar graph> is
$$
\boxed{V-E+F=2}.
$$
The boundary graph of a convex polyhedron has such an embedding. Counting edge ends at vertices and edge sides at faces gives $\sum_{m\ge3}mV_m=2E=\sum_{n\ge3}nF_n$. Therefore
$$
\begin{aligned}
\sum_{n\ge3}(6-n)F_n&=6F-2E\\
&=6(2-V+E)-2E\\
&=12+\sum_{m\ge3}(2m-6)V_m\ge12.
\end{aligned}
$$
Faces of size six contribute zero, and larger faces contribute negatively. The contribution of each triangle, quadrilateral or pentagon is at most three, so
$$
12\le3F_3+2F_4+F_5\le3(F_3+F_4+F_5).
$$
Thus the <small faces of a spherical polyhedral graph> bound is
$$
\boxed{F_3+F_4+F_5\ge4}.
$$
The tetrahedron shows that four is attainable. Connectedness and cellular faces are the standard hypotheses in this Euler formula; an arbitrary disconnected embedded graph has a different count.
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