Example 5 If for any positive real numbers, $\frac{a^{2}}{\sqrt{a^{4}+3 b^{4}+3 c^{4}}}+\frac{k}{a^{3}} \cdot\left(\frac{c^{4}}{b}+\frac{b^{4}}{c}\right) \geqslant \frac{2 \sqrt{2}}{3}$. Always holds, find the minimum value of the real number $k$.


Solve $\frac{a^{2}}{\sqrt{a^{4}+3 b^{4}+3 c^{4}}}=\frac{\sqrt{2} a^{4}}{\sqrt{2 a^{4}\left(a^{4}+3 b^{4}+3 c^{4}\right)}}$.
$$\begin{array}{l}
\geqslant \frac{\sqrt{2} a^{4}}{\frac{1}{2}\left[2 a^{4}+\left(a^{4}+3 b^{4}+3 c^{4}\right)\right]} \\
=\frac{2 \sqrt{2}}{3} \cdot \frac{a^{4}}{a^{4}+b^{4}+c^{4}}
\end{array}$$

From the form, we need to prove $\frac{k}{a^{3}} \cdot\left(\frac{c^{4}}{b}+\frac{b^{4}}{c}\right) \geqslant \frac{2 \sqrt{2}}{3} \cdot \frac{b^{4}+c^{4}}{a^{4}+b^{4}+c^{4}}$, and from the condition for equality $2 a^{4}=a^{4}+3 b^{4}+3 c^{4}$ and the symmetry of $b, c$, we guess that $k$ might take its maximum value when $a^{4}=6 b^{4}=$ $6 c^{4}$, and this value is the minimum value that makes the inequality hold for any $a, b, c$.

Let $a=\sqrt[4]{6}, b=c=1$, then $\frac{\sqrt{2}}{2}+\frac{k}{(\sqrt[4]{6})^{3}} \cdot 2 \geqslant \frac{2 \sqrt{2}}{3} \Rightarrow k \geqslant \frac{1}{\sqrt[4]{24}}$.
We will prove that when $k=\frac{1}{\sqrt[4]{24}}$, $\frac{k}{a^{3}} \cdot\left(\frac{c^{4}}{b}+\frac{b^{4}}{c}\right) \geqslant \frac{2 \sqrt{2}}{3} \cdot \frac{b^{4}+c^{4}}{a^{4}+b^{4}+c^{4}}$.
This is equivalent to proving $\left(a^{4}+b^{4}+c^{4}\right)\left(b^{5}+c^{5}\right) \geqslant \frac{4 \sqrt[3]{6}}{3} a^{3} b c\left(b^{4}+c^{4}\right)$.
Since $\left(b^{9}+c^{9}\right)-\left(b^{5} c^{4}+b^{4} c^{5}\right)=\left(b^{5}-c^{5}\right)\left(b^{4}-c^{4}\right) \geqslant 0$, we have $b^{9}+c^{9} \geqslant$ $b^{5} c^{4}+b^{4} c^{5}$

By the weighted mean inequality, we get:
$$\begin{aligned}
a^{4} b^{5}+2 b^{5} c^{4} & =6 \cdot \frac{a^{4} b^{5}}{6}+2 \cdot b^{5} c^{4} \\
& \geqslant 8 \sqrt[8]{\left(\frac{a^{4} b^{5}}{6}\right)^{6} \cdot\left(b^{5} c^{4}\right)^{2}} \\
& =\frac{4 \sqrt[3]{6}}{3} a^{3} b^{5} c \\
a^{4} c^{5}+2 c^{5} b^{4} & =6 \cdot \frac{a^{4} c^{5}}{6}+2 \cdot c^{5} b^{4} \\
& \geqslant 8 \sqrt[8]{\left(\frac{a^{4} c^{5}}{6}\right)^{6} \cdot\left(c^{5} b^{4}\right)^{2}} \\
& =\frac{4 \sqrt[3]{6}}{3} a^{3} b c^{5}
\end{aligned}$$

Adding the three inequalities and rearranging, we get
$$\left(a^{4}+b^{4}+c^{4}\right)\left(b^{5}+c^{5}\right) \geqslant \frac{4 \sqrt[3]{6}}{3} a^{3} b c\left(b^{4}+c^{4}\right)$$

Thus, the original left side $\geqslant \frac{2 \sqrt{2}}{3} \cdot \frac{a^{4}}{a^{4}+b^{4}+c^{4}}+\frac{2 \sqrt{2}}{3} \cdot \frac{b^{4}+c^{4}}{a^{4}+b^{4}+c^{4}}=\frac{2 \sqrt{2}}{3}=$ the original right side, so the minimum value of $k$ is $\frac{1}{\sqrt[4]{24}}$.