Example 8.17 (2009 Vietnam) Determine the minimum value of $k$ such that the following inequality holds for all positive real numbers $a, b, c$
$$\left(k+\frac{a}{b+c}\right)\left(k+\frac{b}{c+a}\right)\left(k+\frac{c}{a+b}\right) \geqslant\left(k+\frac{1}{2}\right)^{3}$$
Prove that by setting $a=b=1, c=0$, we can obtain $k \geqslant \frac{\sqrt{5}-1}{4}$.
Let $m=2k, x=\frac{2a}{b+c}, y=\frac{2b}{c+a}, z=\frac{2c}{a+b}$. It is not hard to verify that $xy+yz+zx+xyz=4$.
First, we need to prove that under this condition, we have $x+y+z \geqslant xy+yz+zx$.
Clearly, if $x+y+z > 4$, then it holds. If $x+y+z \leqslant 4$.
By the third-degree Schur's inequality combined with the condition, we can get
$$xy+yz+zx \leqslant \frac{36+(x+y+z)^{3}}{4(x+y+z)+9}$$

We only need to prove
$$\frac{36+(x+y+z)^{3}}{4(x+y+z)+9} \leqslant x+y+z \Leftrightarrow x+y+z \geqslant 3$$

And
$$4=xy+yz+zx+xyz \geqslant 4 \sqrt[4]{(xyz)^{3}} \Rightarrow xyz \leqslant 1$$

That is $\square$
$$xy+yz+zx \geqslant 3 \Rightarrow x+y+z \geqslant \sqrt{3(xy+yz+zx)} \geqslant 3$$

Returning to the original inequality, we need to prove $(m+x)(m+y)(m+z) \geqslant (m+1)^{3}$, and
$$\begin{aligned}
(m+x)(m+y)(m+z)= & m^{3}+m^{2}(x+y+z)+m(xy+yz+zx)+xyz \geqslant \\
& m^{3}+\left(m^{2}+m-1\right)(xy+yz+zx)+4
\end{aligned}$$

Then we need
$$\begin{array}{l}
m^{3}+\left(m^{2}+m-1\right)(xy+yz+zx)+4 \geqslant (m+1)^{3} \\
\left(m^{2}+m-1\right)(xy+yz+zx-3) \geqslant 0 \Rightarrow m \geqslant \frac{\sqrt{5}-1}{2}
\end{array}$$

Therefore, the minimum value of $k$ is $\frac{\sqrt{5}-1}{4}$.