Example 6 Find the maximum real number $k$, such that for any simple graph $G$ of order $n(n \geqslant 3)$, we have $x^{3} \geqslant k y^{2}$, where $x$ is the number of edges in $G$, and $y$ is the number of triangles in $G$.
【Analysis】First consider a special $n$-order complete graph, we have
$\left(\mathrm{C}_{n}^{2}\right)^{3} \geqslant k\left(\mathrm{C}_{n}^{3}\right)^{2}$
$\Rightarrow k \leqslant \frac{9 n(n-1)}{2(n-2)^{2}}$.
Let $n \rightarrow+\infty$, then $k \leqslant \frac{9}{2}$.
Next, use mathematical induction to prove: $x^{3} \geqslant \frac{9}{2} y^{2}$ holds for any $n$-order simple graph $G$.
When $n=3$, it is obviously true.
Assume when $n=m$, the conclusion holds.
When $n=m+1$, take $A$ as the vertex with the smallest degree in the $(m+1)$-order simple graph, let its degree be $a$, and let the remaining $m$ points form a simple graph with $x$ edges and $y$ triangles.

Lemma: Let an $n$-order simple graph have $x$ edges and $y$ triangles. Then $y \leqslant \frac{n-2}{3} x$.

Proof: In an $n$-order simple graph, $y$ triangles have a total of $3 y$ edges, and each edge can appear in at most $n-2$ triangles.
Thus $\frac{3 y}{n-2} \leqslant x \Rightarrow y \leqslant \frac{n-2}{3} x$.
The lemma is proved.
By the induction hypothesis,
$x^{3} \geqslant \frac{9}{2} y^{2}$.
Since vertex $A$ has the smallest degree, then,
$(m+1) a \leqslant$ sum of degrees $=2(x+a)$
$\Rightarrow a \leqslant \frac{2 x}{m-1}$.
Let the $a$ vertices adjacent to $A$ be connected by $b$ edges, each of these $b$ edges forms a triangle with vertex $A$, so we need to prove
$(x+a)^{3} \geqslant \frac{9}{2}(y+b)^{2}$.
It is easy to know, $b \leqslant x, b \leqslant \mathrm{C}_{a}^{2}=\frac{a(a-1)}{2}$.
Thus $9 b y \leqslant 9 \cdot \frac{a(a-1)}{2} y$.
Substituting (1) and (3) into the above equation, we get
$$
\begin{array}{c}
9 b y \leqslant \frac{9(a-1)}{2} \cdot \frac{2 x}{m-1} \cdot \frac{m-2}{3} x \\
=\frac{3(a-1)(m-2) x^{2}}{m-1} \leqslant 3 a x^{2} .
\end{array}
$$

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

From (2), (6), and (7), we know (4) holds, i.e., when $n=m+1$, $x^{3} \geqslant \frac{9}{2} y^{2}$ holds for any $n$-order simple graph $G$. Therefore, for all $n \geqslant 3$, the conclusion holds.
Thus, $k_{\text {max }}=\frac{9}{2}$.