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Because $\chi (H)=3$, every bipartite graph is $H$-free. The balanced complete bipartite graph therefore gives

For the reverse bound, fix $\delta >0$ and suppose that $G$ has more than $(1/2+\delta )\left(\genfrac{}{}{0ex}{}{n}{2}\right)$ edges. Apply the Szemerédi regularity lemma with parameters much smaller than $\delta$. Form the reduced graph whose vertices are the regularity classes and whose edges are the regular pairs of density above a small fixed threshold. Edges inside classes, irregular pairs, and regular pairs below the threshold account for $o_{\delta }(n^2)$ edges. The remaining edges force the reduced graph to have more than $m^2/4$ edges.

By the Turan theorem, the reduced graph contains a triangle. The three corresponding regular pairs all have positive density, and the graph embedding lemma for regular pairs embeds every fixed three-colourable graph, in particular $H$, across suitable repeated subclusters of these three classes. Thus every sufficiently large graph of density above $1/2+\delta$ contains $H$. Letting $\delta \downarrow 0$ givesThis is the chromatic-number-three case of the Erdős-Stone theorem.

The Turan theorem states that every $n$-vertex graph with no $K_{r+1}$ has at most $e(T_r(n))$ edges, where $T_r(n)$ is the complete $r$-partite graph with part sizes differing by at most one.

Here is an induction on $n$ and $r$. If a $K_{r+1}$-free graph $G$ contains no $K_r$, induction on $r$ givesOtherwise choose a copy $S$ of $K_r$. Every vertex outside $S$ has at most $r-1$ neighbors in $S$, and induction on $n$ givesThe last identity is obtained by removing one vertex from every part of $T_r(n)$. This proves the theorem; the standard equality analysis forces the balanced complete $r$-partite graph.

Now suppose $G$ is rhombus-free. If it is triangle-free, the $r=2$ case just proved gives
$e(G)\le e(T_2(n))$. Otherwise remove the three vertices of a triangle. No remaining vertex can be adjacent to two vertices of that triangle, since those two triangle vertices and the outside vertex would form a second triangle sharing an edge with the first. Induction therefore givesThis proves the rhombus-free edge bound and hence the requested strict contrapositive.

For equality at $n=6$, take the triangular prism graph: two disjoint triangles joined by a matching. It has $6+3=9=e(T_2(6))$ edges. Every edge belongs to at most one triangle, so there is no rhombus, while the presence of triangles proves that it is not isomorphic to $T_2(6)=K_{3,3}$.