9. (16 points) Find all positive real numbers $k$, such that for any positive real numbers $a, b, c$, we have
$$
\frac{a}{b+c}+\frac{b}{c+a}+\frac{k c}{a+b} \geqslant 2 \text {. }
$$
9. $k \geqslant 4$.
On one hand, let $a=b=1$, we get
$$
\frac{2}{1+c}+\frac{k c}{2} \geqslant 2 \Leftrightarrow \frac{k c}{2} \geqslant \frac{2 c}{1+c} \Leftrightarrow k \geqslant \frac{4}{1+c} \text {. }
$$

Let $c \rightarrow 0$, we get $k \geqslant 4$.
On the other hand, it suffices to prove that the inequality holds when $k=4$.
By Cauchy-Schwarz inequality, we have
$$
\begin{array}{l}
{[a(b+c)+b(c+a)+c(a+b)] .} \\
\left(\frac{a}{b+c}+\frac{b}{c+a}+\frac{4 c}{a+b}\right) \\
\geqslant(a+b+2 c)^{2} \\
\Rightarrow \frac{a}{b+c}+\frac{b}{c+a}+\frac{4 c}{a+b} \\
\quad \geqslant \frac{(a+b+2 c)^{2}}{2(a b+b c+c a)} \geqslant 2 .
\end{array}
$$
(because $(a+b+2 c)^{2}-4(a b+b c+c a)$
$$
=a^{2}+b^{2}+c^{2}-2 a b \geqslant 0 \text {.) }
$$

Therefore, the given inequality holds.