last week I had such exercise on one of my courses: Prove there exists a constant $C$ such that every simple planar graph on $n$ vertices has $\leqslant C·n$ cliques.
I proved by induction easily that $C=4$ is ok. But as Steve Jobs said, I stay hungry, stay foolish. Do you know what is the best known constant $C$ for that problem?
By a clique I shall mean any clique on $3$ or more vertices. I will prove that the constant $C = 4$ is optimal.
A plane graph is a drawing of a graph in the plane with no crossing edges. Let us say that a plane graph is a plane triangulation (or maximally planar) if every face (including the outer face) is bounded by a triangle. The following recursive construction gives for every $n \geq 3$ a plane graph with $n$ vertices and $4n - 11$ cliques:
Start with $G_3 := C_3$, the cycle in $3$ vertices, which has an essentially unique drawing in the plane. This is a plane triangulation with $3$ vertices and $1$ clique.
Suppose that $G_n$ is a plane triangulation with $n$ vertices and $4n - 11$ cliques. Choose a face $F$ of $G_n$ and let $a,b,c$ be the vertices bounding $F$. Add a vertex $d$ in the interior of the face $F$, and connect it to each of the vertices $a$, $b$ and $c$. This adds four new cliques to the graph, namely $\{a,b,d\}$, $\{a,c,d\}$, $\{b,c,d\}$ and $\{a,b,c,d\}$. Therefore the graph $G_{n+1}$ thus obtained has $n + 1$ vertices and $4(n+1) - 11$ cliques.
This shows that $C = 4$ is optimal.