18. Let $a, b, c$ be positive real numbers, find the value of $k$ such that
$$\left(k+\frac{a}{b+c}\right)\left(k+\frac{b}{c+a}\right)\left(k+\frac{c}{a+b}\right) \geqslant\left(k+\frac{1}{2}\right)^{3}$$
(2009 Vietnam National Team Selection Exam Problem)
18. When $a=b, c \rightarrow 0$, the original inequality transforms to
$$(k+1)^{2} k \geqslant\left(k+\frac{1}{2}\right)^{3} \Rightarrow 2 k^{2}+k \geqslant \frac{3}{2} k^{2}+\frac{3}{4} k+\frac{1}{8} \Rightarrow 4 k^{2}+2 k-1 \geqslant 0$$

Solving, we get $k \geqslant \frac{\sqrt{5}-1}{4}$ or $k \leqslant-\frac{\sqrt{5}+1}{4}$.
Next, we prove that when $4 k^{2}+2 k-1 \geqslant 0$,
$$\left(k+\frac{a}{b+c}\right)\left(k+\frac{b}{c+a}\right)\left(k+\frac{c}{a+b}\right) \geqslant\left(k+\frac{1}{2}\right)^{3}$$

holds.
$$\begin{aligned}
\text { Equation }(1) \Leftrightarrow & \left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right) k^{2}+\left[\frac{a c}{(a+b)(b+c)}+\right. \\
& \left.\frac{b c}{(b+a)(c+a)}+\frac{a b}{(c+a)(b+c)}\right] k+
\end{aligned}$$
$$\begin{array}{l}
\frac{a b c}{(a+b)(b+c)(c+a)} \geqslant \frac{3}{2} k^{2}+\frac{3}{4} k+\frac{1}{8} \Leftrightarrow \\
{[a(a+b)(a+c)+b(a+b)(b+c)+c(a+c)(b+c)] k^{2}+} \\
{[a c(a+c)+b c(b+c)+a b(a+b)] k+} \\
a b c(a+b)(b+c)(c+a)\left(\frac{3}{2} k^{2}+\frac{3}{4} k+\frac{1}{8}\right) \Leftrightarrow \\
4 k^{2}\left[2\left(a^{3}+b^{3}+c^{3}\right)-\left(a^{2} b+a b^{2}+b^{2} c+b c^{2}+c^{2} a+c a^{2}\right)\right]+ \\
2 k\left[\left(a^{2} b+a b^{2}+b^{2} c+b c^{2}+c^{2} a+c a^{2}\right)-6 a b c\right]- \\
\left(a^{2} b+a b^{2}+b^{2} c+b c^{2}+c^{2} a+c a^{2}\right)-6 a b c \geqslant 0
\end{array}$$

By Schur's inequality,
$$a^{3}+b^{3}+c^{3}-\left(a^{2} b+a b^{2}+b^{2} c+b c^{2}+c^{2} a+c a^{2}\right)+3 a b c \geqslant 0$$

Thus, the left side of the above inequality is greater than or equal to
$$\left[\left(a^{2} b+a b^{2}+b^{2} c+b c^{2}+c^{2} a+c a^{2}\right)-6 a b c\right]\left(4 k^{2}+2 k-1\right) \geqslant 0$$