(15) Find the smallest positive real number $k$, such that the inequality
$$
a b+b c+c a+k\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right) \geqslant 9,
$$

holds for all positive real numbers $a, b, c$.
(15) When $a=b=c=1$, we can get $k \geqslant 2$. Now we prove the inequality
$$
a b+b c+c a+2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right) \geqslant 9
$$

holds for all positive real numbers $a, b, c$.
By the AM-GM inequality, we have
$$
a b+\frac{1}{a}+\frac{1}{b} \geqslant 3 \sqrt[3]{a b \cdot \frac{1}{a} \cdot \frac{1}{b}}=3,
$$

Similarly, we get
$$
\begin{array}{l}
b c+\frac{1}{b}+\frac{1}{c} \geqslant 3, \\
c a+\frac{1}{c}+\frac{1}{a} \geqslant 3 .
\end{array}
$$

Adding these 3 inequalities, we obtain
$$
a b+b c+c a+2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right) \geqslant 9 .
$$

In conclusion, the minimum value of $k$ is 2.