Example 10.3 (Han Jingjun) Determine the minimum value of $k$ such that when non-negative real numbers $a, b, c \geqslant k$, and $a + b + c = 9$, we have
$$\sqrt{a b + b c + c a} \leqslant \sqrt{a} + \sqrt{b} + \sqrt{c}$$
Let $a=\frac{13-4 \sqrt{10}}{3}, b=c=\frac{7+2 \sqrt{10}}{3}$, at this time
$$\sqrt{a b+b c+c a}=\sqrt{a}+\sqrt{b}+\sqrt{c}$$

Thus, $k \geqslant \frac{13-4 \sqrt{10}}{3}$. Next, we prove that the inequality holds when $k=\frac{13-4 \sqrt{10}}{3}$.
We keep $a+b+c$ and $a b+b c+c a$ unchanged, which means $a^{2}+b^{2}+c^{2}$ remains constant. At this point, $m=2$. Let $f(x)=\sqrt{x}, g(x)=f^{\prime}(x)$, then $g^{\prime \prime}(x)=\frac{3}{8} x^{-\frac{5}{2}}>0$, which means $g(x)$ is a convex function. Therefore, by Theorem 10.12, the minimum value of $\sqrt{a}+\sqrt{b}+\sqrt{c}$ is achieved when $k \leqslant a \leqslant b=c$.

The original inequality becomes, when $a+2 b=9, b \geqslant a \geqslant k \geqslant 0$,
$$\begin{array}{l}
\sqrt{\frac{(9-a)^{2}}{4}+a(9-a)} \leqslant \sqrt{a}+2 \sqrt{\frac{9-a}{2}} \Leftrightarrow \\
-3 a^{2}+9+22 a \leqslant 8 \sqrt{a(18-2 a)} \Leftrightarrow \\
\left(9 a^{2}-78 a+9\right)(a-3)^{2} \leqslant 0
\end{array}$$

Then, we can solve that $a \geqslant \frac{13-4 \sqrt{10}}{3}$.
Therefore, the minimum value of $k$ is $k=\frac{13-4 \sqrt{10}}{3}$.