How can I find the number of different triangles in $n$-vertex graph? First of all, I reveal this problem stems from below statement.


*

*a graph $G = (V, E)$ with $n$ vertices is extremal for $K_3$ if it contains "no triangles" and has $\left\lfloor\frac{n^2}{4}\right\rfloor$ edges.

*Thm : For every $n$, every extremal $n$-vertex graph for $K_3$ is isomorphic to the graph $K_{a, b}$ with $a=\left\lfloor\frac{n}{2}\right\rfloor$, $b = n -\left\lfloor\frac{n}{2}\right\rfloor$.
So, conversely saying, if the number of edges of any $n$-vertex graph is greater than $\left\lfloor\frac{n^2}{4}\right\rfloor$, We can find a triangle(or triangles), that is, can find at least "one" triangle. 
Likewise, if average degree of the n-vertex graph is greater than $\frac{n}{2}$, so the number of edges is the graph is greater than $\frac{n^2}{4}$, thus we can find at least one triangle.
Back to main problem, I can't reach how can I find "different $\frac{1}{10}$ ${n\choose 3}$ triangles" if the average degree of the graph ($n$-vertex) is greater than $\frac{3n}{5}$.
I don't just want to find answers. 
If you're willing to help me, Give me hints, please.
thanks for reading.
 A: 
[APMO 1989] Show that a graph with $n$ vertices and $k$ edges has at least $ \frac{k(4k-n^2) } { 3n } $ triangles. 

Corollary: With $k = \frac{3}{10}n^2$, we get $\frac{k(4k-n^2) } { 3n }  = \frac{ \frac{3}{10} n^2 \times \frac{2}{10} n^2 } { 3n } =  \frac{1}{50} n^3 > \frac{1}{10} { n \choose 3} $

This is a very useful result, which often makes olympiad problems with a similar setup trivial. In my set of notes, I encourage people to remember the result, and how to prove it. 
Sketch of proof: Double Counting
Hint: Each edge appears in 3 triangles.   

 Given an edge $v_i v_j$, it is in at least $d(v_i ) + d(v_j) + C$ triangles. Find the constant $C$.   

$ $

Hence, the number of triangles is
$ \displaystyle \geq \frac{1}{3} \left( \sum_{\text{edge}} d(v_i ) + d(v_j) + C \right) $
$ \displaystyle = \frac{1}{3} \left( Ck+ \frac{1}{2}\sum_{\text{vertex}}  \text{some function of the degree} \right)$
$= \frac{k(4k-n^2) } { 3n }.$ 


A: I have a weaker inequality (that is, I got $\displaystyle \frac1{25}\binom{n}{3}$ instead of $\displaystyle \frac1{10}\binom{n}{3}$).  Here is a proof.  Maybe somebody can tweak it and get the bound you need.
A branch of a graph is a pair $\big(v,\{e,f\}\big)$, where $e$ and $f$ are two distinct edges with the same endpoint $v$ such that $e$ and $f$ are not edges of a triangle.  Let $b(G)$ and $t(G)$ denote the number of branches and triangles in a simple graph $G$ on $n$ vertices with $m$ edges.
Observe that
$$b(G)+3\,t(G)=\sum_{v\in V(G)}\,\binom{\deg_G(v)}{2}\,,$$
where $V(G)$ is the vertex set of $G$ and $\deg_G(v)$ is the degree of each $v\in V(G)$ in $G$.  By, for example, the AM-QM Inequality, we can show that
$$\sum_{v\in V(G)}\,\binom{\deg_G(v)}{2}\geq n\,\binom{\frac{1}{n}\,\sum\limits_{v\in V(G)}\,\deg_G(v)}{2}=n\,\binom{\left(\frac{2m}{n}\right)}{2}\,.$$
This proves that
$$b(G)+3\,t(G)\geq \frac{n}{2}\,\left(\frac{2m}{n}\right)\,\left(\frac{2m}{n}-1\right)=\frac{m\,(2m-n)}{n}\,.\tag{*}$$
However, it is easy to see that
$$b(G)+t(G)\leq \binom{n}{3}\,.$$
Therefore,
$$t(G)\geq \frac{1}{2}\,\left(\frac{m\,(2m-n)}{n}-\binom{n}{3}\right)\,.$$
This means
$$t(G)\geq \frac{6m(2m-n)-n^2(n-1)(n-2)}{12n}\,.$$
If the average degree is greater than or equal to $\dfrac{3n}{5}$, then 
$$m\geq \frac{n}{2}\left(\dfrac{3n}{10}\right)=\frac{3n^2}{10}\,.$$
Therefore,
$$t(G)\geq \frac{n(n^2+15n-25)}{150}=\frac{(n^2+15n-15)}{25(n-1)(n-2)}\,\binom{n}{3}\,.$$
It can be easily seen that
$$\frac{(n^2+15n-15)}{25(n-1)(n-2)}>\frac{1}{25}$$
for all $n\geq 3$.  Hence, if $n\geq 3$ and $m\geq \dfrac{3n^2}{10}$, we indeed have
$$t(G)>\frac{1}{25}\,\binom{n}{3}\,.$$
We can also count anti-branches and anti-triangles to improve the bound, I think.  I am too exhausted to think about them.

Edit.  Let $b'(G)$ and $t'(G)$ denote the number of branches and triangles, respectively, of the complementary graph $G'$ of $G$ (that is, they count anti-branches and anti-triangles of $G$, respectively).  In other words, $b'(G)=b(G')$ and $t'(G)=t(G')$.  Write $\deg'_G(v)$ for $\deg_{G'}(v)$ for each $v\in V(G)=V(G')$.
Then, we also have
$$b'(G)+3\,t'(G)=\sum_{v\in V(G)}\,\binom{\deg'_G(v)}{2}\,.$$
However,
$$\sum_{v\in V(G)}\,\deg'_G(v)=n(n-1)-2m\,.$$
Therefore,
$$b'(G)+3\,t'(G)\geq n\,\binom{\frac{1}{n}\,\sum\limits_{v\in V(G)}\,\deg'_G(v)}{2}=n\,\binom{n-1-\frac{2m}{n}}{2}\,.$$
Thus,
$$b'(G)+t'(G)\geq \frac{1}{3}\,\big(b'(G)+3\,t'(G)\big)\geq\frac{n}{3}\,\binom{n-1-\frac{2m}{n}}{2}\,.$$
That is
$$b'(G)+t'(G)\geq \frac{(n^2-n-2m)(n^2-2n-2m)}{6n}\,.$$
Because
$$b(G)+t(G)+b'(G)+t'(G)=\binom{n}{3}\,,$$
we get
$$b(G)+t(G)\leq \binom{n}{3}-\frac{(n^2-n-2m)(n^2-2n-2m)}{6n}\,.$$
From (*), we obtain a better bound
$$t(G)\geq \frac{1}{2}\,\left(\frac{m\,(2m-n)}{n}-\binom{n}{3}+\frac{(n^2-n-2m)(n^2-2n-2m)}{6n}\right)\,,$$ 
leading to
$$t(G)\geq \frac{m(4m-n^2)}{3n}\,.\tag{#}$$
If $m\geq \dfrac{3n^2}{10}$, then
$$t(G)\geq \frac{n^3}{50}=\frac{3n^2}{25(n-1)(n-2)}\,\binom{n}{3}\,.$$
It can easily be seen that
$$\frac{3n^2}{25(n-1)(n-2)}>\frac{3}{25}$$
for all $n\geq 3$.  Therefore,
$$t(G)>\frac{3}{25}\,\binom{n}{3}$$
when $n\geq 3$ and $m\geq \dfrac{3n^2}{10}$. 
In fact, for any $\alpha>\dfrac14$, if $n\geq 3$ and $m\geq \alpha\,n^2$, then
$$t(G)\geq\frac{\alpha\,(4\alpha-1)}{3}\,n^3>2\alpha\,(4\alpha-1)\,\binom{n}{3}\,.$$ 
In terms of the average degree $d(G)$ of the graph, we have
$$t(G)\geq\frac{d(G)\,\big(2\,d(G)-n\big)\,n}{6} >\frac{d(G)\,\big(2\,d(G)-n\big)}{n^2}\,\binom{n}{3}$$
if $n\geq 3$ and $d(G)>\dfrac{n}2$.
Incidentally, (#) also shows that, if $m> \dfrac{n^2}{4}$, then the graph has a triangle.  This is quite an unexpected result because it is sharp, but I thought my bounds were weak.
