9. (40 points) For real numbers $a, b$ and $c$ it is known that $a b + b c + c a = 3$. What values can the expression $\frac{a\left(b^{2}+3\right)}{a+b}+\frac{b\left(c^{2}+3\right)}{b+c}+\frac{c\left(a^{2}+3\right)}{c+a}$ take?
Answer: 6.

Solution. Consider the first term of the desired expression. Using the condition that $a b+b c+c a=3$. Then

$$
b^{2}+3=b^{2}+a b+b c+c a=(b+a)(b+c)
$$

Therefore,

$$
\frac{a\left(b^{2}+3\right)}{a+b}=\frac{a(b+a)(b+c)}{a+b}=a(b+c) \text {. }
$$

Similarly for the second and third terms, we get

$$
\frac{b\left(c^{2}+3\right)}{b+c}=b(c+a), \quad \frac{c\left(a^{2}+3\right)}{c+a}=c(a+b)
$$

Thus,

$$
\begin{gathered}
\frac{a\left(b^{2}+3\right)}{a+b}+\frac{b\left(c^{2}+3\right)}{b+c}+\frac{c\left(a^{2}+3\right)}{c+a}=a(b+c)+b(c+a)+c(a+b)= \\
\quad=a b+a c+b c+b a+c a+c b=2(a b+b c+c a)=2 \times 3=6
\end{gathered}
$$