Example 10 Let $a, b, c$ be positive real numbers, satisfying
$$a+b+c+3 \sqrt[3]{a b c} \geqslant k(\sqrt{a b}+\sqrt{b c}+\sqrt{c a}),$$

Find the maximum value of $k$.
Given that when $a=b=c$, from $6 \geqslant 3 k$, we get $k \leqslant 2$. Now we prove
$$\begin{array}{c}
a+b+c+3 \sqrt[3]{a b c} \geqslant 2(\sqrt{a b}+\sqrt{b c}+\sqrt{c a}) . \\
\text { Let } f(a, b, c)=a+b+c+3 \sqrt[3]{a b c}-2(\sqrt{a b}+\sqrt{b c}+\sqrt{c a}) .
\end{array}$$

Without loss of generality, assume $a \leqslant b \leqslant c$, and make the following adjustments,

then
$$a=a^{\prime}, b=b^{\prime}=c^{\prime}=\sqrt{b c}=A$$
$$\begin{aligned}
f\left(a^{\prime}, b^{\prime}, c^{\prime}\right) & =\left(a+2 A+3 \cdot a^{\frac{1}{3}} \cdot A^{\frac{2}{3}}\right)-2\left[A+(2 a A)^{\frac{1}{2}}\right] \\
& =a+3 \cdot a^{\frac{1}{3}} \cdot A^{\frac{2}{3}}-4(a A)^{\frac{1}{2}} \geqslant 0
\end{aligned}$$

Equality holds if and only if $a=0$ or $a=b=c$.
Next, we prove $f(a, b, c) \geqslant f\left(a^{\prime}, b^{\prime}, c^{\prime}\right)$. Because
$$f(a, b, c)-f\left(a^{\prime}, b^{\prime}, c^{\prime}\right)=b+c-2 a^{\frac{1}{2}}\left(b^{\frac{1}{2}}+c^{\frac{1}{2}}-2 A^{\frac{1}{2}}\right)-2 A$$

Since
$$a \leqslant A, b^{\frac{1}{2}}+c^{\frac{1}{2}} \geqslant 2 A^{\frac{1}{2}}$$

Therefore
$$f(a, b, c)-f\left(a^{\prime}, b^{\prime}, c^{\prime}\right)$$
$$\begin{array}{l}
\geqslant b+c-2 \sqrt{A}(\sqrt{b}+\sqrt{c}-2 \sqrt{A})-2 A \\
=b+c-2(\sqrt{b}+\sqrt{c}) \sqrt{A}+2 A \\
=(\sqrt{b}-\sqrt{A})^{2}+(\sqrt{c}-\sqrt{A})^{2} \\
\geqslant 0
\end{array}$$

Thus
$$f(a, b, c) \geqslant 0$$

Therefore, the maximum value of $k$ is 2.
Note: This solution was provided by Lei Fu, a gold medalist at the 48th IMO.