Triangle free graph with $n$ vertices and maximum degree $k$ has at most $k(n-k)$ edges.

Question

Having a graph $G$ which is simple and non-directed with $n$ vertices and max degree of a vertex $k$. Then show if it does not contain $K_3$ as induced subgraph then that proves $|E(G)|\le (n-k)k$

So,having to find an upper limit to the number of edges. My thoughts so far are:

1)If it does not contain an induced subgraph of $K_3$ then it does not have any kind of bigger $K$ and then the degree of $k$ is $2$.

Extra thought(not complete):Having a vertex ν that is $k(=2)$ degree then compute the sum of degrees $V(G)-N(v)$. [$N(ν)$ is neighbourhood of $v$].I need some analysis if correct to this last statement.

Answer 1

Let $v$ be a vertex of degree $k$. Then every vertex $u \in N_G(v)$ cannot have a neighbor in $N_G(v)$ lest there be a triangle. So each such $u$ can have degree at most $N-k$; and there are $k$ such $u$. Every vertex $w \not \in N_G(v)$ [which includes $v$ itself] can have degree at most $k$ by the hypothesis that $G$ has maximum degree $k$; there are $N-k$ such $w$.

So $$|E(G)| = \frac{1}{2} \times \left(\sum_{u \in N_G(v)} d_G(u) + \sum_{w \not \in N_G(v)} d_G(w) \right)$$

$$\le \frac{1}{2} \times \left(k \cdot (N-k) + (N-k)\cdot k\right)$$

which gives the desired bound.