The assumption that $G$ is connected is crucial.
Otherwise, consider a graph of $6$ vertices which is composed of $2$ disjoint triangles.
Also, it is not true that $\overline{G}$ needs to be triangle free (all you need is an independent set of size $3$ in $G$).
A proof by induction actually works.
Base cases
$n=3,4$ are easily verified.
Induction step
Suppose we are given a graph $G$ with $n \ge 5$ vertices. Let $e = |E(G)|$.
Now if every vertex of $G$ had degree $\ge 3$, then we are done, as $2e \ge 3n$.
So assume there is a vertex $a$ of degree exactly $2$. (Why not $0$ or $1$?).
Let $b,c$ be the neighbours of $a$. Note that $\{b,c\}$ is an edge in $G$ (Why?)
Consider two cases:
Case 1) The edge $\{b,c\}$ is a part of a triangle other than $abc$.
In which case, we delete $a$ from $G$ to give a new graph G' of $n-1$ vertices, satisfying the induction hypothesis.
Thus we get $e-2 \ge 3(n-2)/2$ and so $e \ge (3n - 2)/2 \gt 3(n-1)/2$ and so we are done.
Case 2) The edge $\{b,c\}$ is not part of any other triangle.
In this case, we delete the vertex $a$ and the edge $\{b,c\}$ to get a new graph G'.
Now it is possible that G' is disconnected due to deletion of edge $\{b,c\}$. Since $G$ was connected, G' has no more than two connected components, and we can try and apply the induction step to each connected component if each component has at least two vertices.
Thus if $n_1$ and $n_2$ (with $n_1 \ge 2$ and $n_2 \ge 2$) are the number of vertices in each connected component, we have that,
$e - 3 \ge 3(n_1 - 1)/2 + 3(n_2 -1)/2 = 3(n-3)/2$
Thus
$e \ge 3(n-1)/2$
and we are done.
If one of $n_1$, $n_2$ is $0$ or $1$, we can ignore that component and the above inequality still holds.