Q:

Upper bound on the number of edges of a ten vertexes graph no $C_4, C_3$ subgraph.

We are given a graph $G$ with ten vertices. Any three or four vertices of $G$ don't form a cycle. What is the maximum number of edges $G$ could have?
I know that if the graph is planar and triangle free, it would have 16 edges. It seems that between any 4 vertices, there can be at most 3 edges.
Maybe there is a formula and can be proven by induction on number of vertex. I'm not sure.

A:

Let $A$ be a set of connected pairs and $B$ a set of unconnected pairs of vertices in $G$. 
Connect a pair of vertices $u,v$ with vertex $w$ iff $w$ is connected with both (that is we make a new graph which is bipartite). 
Since there is no triangles we have $d(u,v) =0$ if $\{u,v\}\in A$ and
since there is no 4-cycles we have $d(u,v) \leq 1$ if $\{u,v\}\in B$
So we have $G$ $$0\cdot |A|+1\cdot |B| \geq \sum _{i=1}^{10} {d_i\choose 2}$$
By handshake lemma in starting graph $G$ we have $$\sum _{i=1}^{10} d_i = 2\varepsilon$$
where the number of edges in starting graph $G$ is $\varepsilon$. Since $|A|=\varepsilon$ we have $|B| = {10\choose 2} -\varepsilon$
Now we have by Cauchy inequality $$\sum _{i=1}^{10} {d_i\choose 2}\geq {1\over 2}({1\over 10}4\ \varepsilon^2 -2\varepsilon)$$
so $${10\choose 2} -\varepsilon\geq {1\over 2}({1\over 10}4\ \varepsilon^2 -2\varepsilon)$$
After solwing this quadratic inequality we get $\varepsilon \leq 15$.

This value can not be improved since we have a configuration for $\varepsilon = 15$. Say vertices are $1,2,...,10$ and let $N(v)$ be a set of neighours for $v$. Then if we set:
$$ N(1) = \{2,3,4\}$$
$$ N(2) = \{1,5,6\}$$
$$ N(3) = \{1,7,8\}$$
$$ N(4) = \{1,9,10\}$$
$$ N(5) = \{2,7,9\}$$
$$ N(6) = \{2,8,10\}$$
$$ N(7) = \{3,5,10\}$$
$$ N(8) = \{3,6,9\}$$
$$ N(9) = \{4,5,8\}$$
$$ N(10) = \{4,6,7\}$$
we get a graph with $10$ vertices and $15$ edges.

