Example 2 (2008 - Adapted from China National Training Team Test) If $\frac{x y}{z}+\frac{y z}{x}+\frac{z x}{y}>M \sqrt[3]{x^{3}+y^{3}+z^{3}}$ holds for any positive real numbers $x, y, z$, find the maximum value of the real number $M$.

保持源文本的换行和格式如下：

Example 2 (2008 - Adapted from China National Training Team Test) If $\frac{x y}{z}+\frac{y z}{x}+\frac{z x}{y}>M \sqrt[3]{x^{3}+y^{3}+z^{3}}$ holds for any positive real numbers $x, y, z$, find the maximum value of the real number $M$.
Let $x=y=t>0, z=1$, then $t^{2}+2 > M \sqrt[3]{2 t^{3}+1}$, i.e., $M < \frac{t^{2}+2}{\sqrt[3]{2 t^{3}+1}}$. Let $f(t) = \frac{t^{2}+2}{\sqrt[3]{2 t^{3}+1}}$, it is easy to know that $f(t)$ is increasing on $\left(0, \frac{-1+\sqrt{5}}{2}\right)$, decreasing on $\left(\frac{-1+\sqrt{5}}{2}, 1\right)$, and increasing on $(1,+\infty)$.
$$\text{Also, } f(1)=\sqrt[3]{9}, \lim _{t \rightarrow 0^{+}} f(t)=2 \text{, so } M_{\max } \leqslant 2 \text{,}$$

Next, we prove that $M_{\max }=2$.
That is, we need to prove the inequality $\frac{x y}{z}+\frac{y z}{x}+\frac{z x}{y} > 2 \sqrt[3]{x^{3}+y^{3}+z^{3}}$ holds for any positive real numbers $x, y, z$.

Let $\frac{y z}{x}=X^{2}, \frac{z x}{y}=Y^{2}, \frac{x y}{z}=Z^{2}$, then $x=Y Z$, $y=Z X, z=X Y$,
$$\begin{array}{c}
\frac{x y}{z}+\frac{y z}{x}+\frac{z x}{y} > 2 \sqrt[3]{x^{3}+y^{3}+z^{3}} \Leftrightarrow \\
\left(X^{2}+Y^{2}+Z^{2}\right)^{3} > 8\left(Y^{3} Z^{3}+Z^{3} X^{3}+X^{3} Y^{3}\right) \\
\left(X^{2}+Y^{2}+Z^{2}\right)^{3}=\sum X^{6}+3 \sum X^{2} Y^{2}\left(X^{2}+\right.
\end{array}$$
$$\begin{array}{l}
\left.Y^{2}\right)+6 X^{2} Y^{2} Z^{2} \\
=\sum\left(X^{6}+X^{2} Y^{2} Z^{2}\right)+3 \sum X^{2} Y^{2}\left(X^{2}+Y^{2}\right) \\
+3 X^{2} Y^{2} Z^{2} \\
\geqslant \sum X^{2} Y^{2}\left(X^{2}+Y^{2}\right)+3 \sum X^{2} Y^{2}\left(X^{2}+Y^{2}\right)+
\end{array}$$
$3 X^{2} Y^{2} Z^{2}$ (an equivalent form of Schur's inequality)
$$=4 \sum X^{2} Y^{2}\left(X^{2}+Y^{2}\right)+3 X^{2} Y^{2} Z^{2} (A M-G M \text{ inequality})$$
$\geqslant 8 \sum X^{3} Y^{3}+3 X^{2} Y^{2} Z^{2} > 8 \sum X^{3} Y^{3}$, so
$$M_{\max }=2$$