4. Determine the smallest possible value of the expression

$$
\frac{a b+1}{a+b}+\frac{b c+1}{b+c}+\frac{c a+1}{c+a}
$$

where $a, b, c \in \mathbb{R}$ satisfy $a+b+c=-1$ and $a b c \leq-3$.
Solution: The minimum is 3 , which is obtained for $(a, b, c)=(1,1,-3)$ and permutations of this triple.

As $a b c$ is negative, the triple $(a, b, c)$ has either exactly one negative number or three negative numbers. Also, since $|a b c| \geq 3$, at least one of the three numbers has absolute value greater than 1 .

If all of $a, b, c$ were negative, the previous statement would contradict $a+b+c=-1$, hence exactly one of $a, b, c$ is negative.

Wlog let $c$ be the unique negative number. So $a, b>0>c$, as the value 0 isn't possible by $|a b c| \geq 3$. Let $S$ be the given sum of fractions. We then have

$$
\begin{aligned}
S+3 & =\sum_{c y c} \frac{a b+1+a+b}{a+b}=\sum_{c y c} \frac{(a+1)(b+1)}{a+b}=\sum_{c y c}-\frac{(a+1)(b+1)}{c+1} \\
& \geq \sum_{c y c}|a+1|=(a+1)+(b+1)-(c+1)=2 a+2 b+2
\end{aligned}
$$

using AM-GM on the three pairs of summands respectively for the inequality. We can do this since $a+1, b+1>0$ and $-(c+1)=a+b>0$, so every summand is positive.

So all we want to do now is show $a+b \geq 2$, to conclude $S \geq 3$. From the two given conditions we have $a b(1+a+b) \geq 3$. If $a+b<2$, then $a b \leq\left(\frac{a+b}{2}\right)^{2}<1$ and thereby $a b(1+a+b)<3$. So the implication $a b(1+a+b) \geq 3 \Rightarrow a+b \geq 2$ is indeed true.