A convex polyhedron has more faces than vertices. What is the minimum number of triangles among the faces?
Let the number of faces, edges, and vertices be denoted by $l, e$, and $c$ respectively. Since the polyhedron is convex, Euler's theorem applies:

$$
l+c=e+2
$$

According to the other condition, $l>c$, so there exists a positive integer $k$ such that

$$
l=c+k
$$

We will show that among the faces of the polyhedron, there are at least 6 triangles. Let $m$ be the number of triangular faces and $n$ be the number of faces with more than three sides.

With this notation,

$$
l=m+n
$$

Since every edge belongs to two faces, the number of edges of the polyhedron is at least $\frac{3 m+4 n}{2}$, that is,

$$
e \geqslant \frac{3 m+4 n}{2}
$$

Substituting the value of $c$ from (2) into (1): $2 l-k=e+2$, hence from (3) and (4): $2(m+n)-k \geqslant \frac{3 m+4 n}{2}+2$. This implies $m \geqslant 4+2 k$, which shows that the number of triangular faces is at least 6. The statement cannot be sharpened, as if we glue two congruent regular tetrahedra along one of their faces, then $m=l=6$ and $c=5<l$, so the case $m=6$ does indeed occur.