Special properties of subgraphs I have two statements I am not sure how to prove them:
(1) Every graph with $n=|V(G)|\ge 6$ and size $\lfloor n^2/4\rfloor+1$ contains a 5-cycle $C_5$
(2) Every graph with $n=|V(G)|\ge 5$ and size $\lfloor n^2/4\rfloor+2$ contains two trinagles with exactly one vertex in common. 
I think the best way is prove it by induction. For the simple cases n=6 or n=5 it is obvious for both cases (can be verfied geometrically.)
I do not know how to proceed with the induction step, may you can help me with that. If there is an alternative way to prove this it would be also very interesting. 

Induction step for first problem (everything else seems to be settled),
so we assume that the statement has been proven for all graphs with less than $n$ vertices.
Now let $G$ be a graph with $n$ vertices and size $e\geq\lfloor\frac{n^2}{4}\rfloor+1$.
The average degree is $\frac{2e}{n}$, so there is a vertex $v$ with degree $d(v)\leq\frac{2e}{n}$.
Now remove this vertex for the induction step.
The size of the new graph is $e'=e-d(v)$.
We need to prove that $e'=e-d(v)\geq \lfloor\frac{(n-1)^2}{4}\rfloor+1$.
Then the induction hypothesis gives us a 5-cycle in the new graph, which certainly is a 5-cycle in the larger graph.
Case 1: $n$ is even.
The righthand side evaluates to:
$\frac{(n-1)^2-1}{4}+1=\frac{n^2-2n+4}{4}=\frac{n^3-2n^2+4n}{4n}$.
For the lefthand side we have
$e'=e-d(v)\geq e-\frac{2e}{n}=e\frac{n-2}{n}\geq \frac{n-2}{n}(\lfloor\frac{n^2}{4}\rfloor+1)
=\frac{n-2}{n}(\frac{n^2}{4}+1)=\frac{n-2}{n}\frac{n^2+4}{4}
=\frac{n^3-2n^2+4n-8}{4n}=\frac{n^3-2n^2+4n}{4n}-\frac{8}{4n}$
$\frac{8}{4n}<1$ since $n>2$, and because the number of edges always is an integer
we may conclude that $e'\geq\frac{n^3-2n^2+4n}{4n}$.
Case 2: $n$ is odd.
I leave that one to you.
