Four, (50 points) Find the maximum real number $k$, such that for any simple graph $G$ of order $(n \geqslant 3)$, the inequality
$$
x^{3} \geqslant k y^{2} \text {, }
$$

holds, where $x$ is the number of edges in the graph $G$, and $y$ is the number of triangles in the graph $G$.
First, prove a lemma.
Lemma: Let a simple graph of order $n$ have $x$ edges and $y$ triangles. Then
$$
y \leqslant \frac{n-2}{3} x \text {. }
$$

Proof: In a simple graph of order $n$, $y$ triangles have a total of $3 y$ edges, and each edge can be part of at most $n-2$ triangles, so,
$$
\frac{3 y}{n-2} \leqslant x \Rightarrow y \leqslant \frac{n-2}{3} x \text {. }
$$

Returning to the original problem.
First, consider a special complete graph of order $n$, we have
$$
\left(\mathrm{C}_{n}^{2}\right)^{3} \geqslant k\left(\mathrm{C}_{n}^{3}\right)^{2} \Rightarrow k \leqslant \frac{9 n(n-1)}{2(n-2)^{2}} \text {. }
$$

Let $n \rightarrow+\infty$, then $k \leqslant \frac{9}{2}$.
We will prove by mathematical induction that $x^{3} \geqslant \frac{9}{2} y^{2}$ holds for any simple graph $G$ of order $n$.
When $n=3$, the conclusion is obviously true.
Assume that when $n=m$, the conclusion holds.
When $n=m+1$, take $A$ as the vertex with the smallest degree in the simple graph of order $m+1$, let its degree be $a$, and let the remaining $m$ points form a simple graph with $x$ edges and $y$ triangles. Then by the lemma,
$$
y \leqslant \frac{m-2}{3} x \text {. }
$$

By the induction hypothesis,
$$
x^{3} \geqslant \frac{9}{2} y^{2} \text {. }
$$

Since vertex $A$ has the smallest degree,
$$
a \leqslant \frac{2 x+a}{m+1} \Rightarrow a \leqslant \frac{2 x}{m} \text {. }
$$

Let the $a$ vertices adjacent to vertex $A$ be connected by $b$ edges, and each of these $b$ edges forms a triangle with vertex $A$. Therefore, we need to prove
$$
(x+a)^{3} \geqslant \frac{9}{2}(y+b)^{2} .
$$

It is easy to see that $b \leqslant x, b \leqslant \mathrm{C}_{a}^{2}=\frac{a(a-1)}{2}$.
Thus, $9 b y \leqslant 9 \cdot \frac{a(a-1)}{2} y$.
Substituting equations (1) and (3) into the above inequality, we get
$$
\begin{array}{l}
9 b y \leqslant \frac{9(a-1)}{2} \cdot \frac{2 x}{m} \cdot \frac{m-2}{3} x \\
=\frac{3(a-1)(m-2) x^{2}}{m} \leqslant 3 a x^{2} .
\end{array}
$$

Also, by equation (5),
$$
\frac{9}{2} b^{2} \leqslant \frac{9}{2} \cdot \frac{a(a-1)}{2} x \leqslant \frac{9}{4} a^{2} x \leqslant 3 a^{2} x \text {. }
$$

By equations (2), (6), and (7), equation (4) holds, i.e., when $n=m+1$, the conclusion holds.
Therefore, for all $n \geqslant 3$, the conclusion holds.
Thus, $k_{\max }=\frac{9}{2}$.
(Fang Mudan, No. 1 High School Affiliated to Central China Normal University, 430223)