This is a question from an old exam and I don't know how to do c).
Let $n$ be an even number and let $G=(V,E)$ be a graph with $n\geq 4$.
a) Draw a $K_3$ graph with $n^2/4$ edges.
b) Show that $G$ has a $K_3$ if $\delta(G) > n/2$.
c) Show that $G$ has a $K_3$-free subgraph with $|E|/2$ edges.
a) is just a square
For b) we had a theorem that says that the maximum number of edges in a $K_3$-free graph is $\left\lfloor \frac{n^2}{4} \right\rfloor$. The sum of all degrees is equal to twice the number of edges, we get $2\left\lfloor \frac{n^2}{4} \right\rfloor$ = $\left\lfloor \frac{n^2}{2} \right\rfloor$. $\delta(G) > n/2$ is the minimal degree a vertex, so all vertices are $n^2/2$.
Hint: Instead of trying to find any $K_3$-free subgraph with $|E|/2$ edges try to find a bipartite subgraph with $|E|/2$ edges. To do this there are several ways, including an elegant probabilistic one. I suggest you try it yourself.