Just AM-GM.

Take c=1/ab. a+b+c=ab+ac+bc.=> a+b+1/ab=ab+1/b+1/a. Then a²b+ab²+1=a²b²+a+b. (ab-1)(a+b)=(ab-1)(ab+1). First case is ab=1. So c=1. Second case is a+b=ab+1 . (a-1)(b-1)=0. Then a or b is equal to 1.Since everything is symmetric in that 2 equations, we get c=1. Then we have ab=1. By using AM≥GM, we get a+b≥2. So since c=1, a+b+c≥3. Equality holds when a=b=c=1. => Kostas is right!

By Am_GM. (a+b+c)/3>=(abc)^1/3 replace abc by 1. a+b+c=ab+bc+ca>=3. smallest possible sum is 3 so Kostas is right.

Technically, one should also show that Kostas really might not have thought of $a=b=c=1$, i.e. give another example of $(a,b,c)$ with $abc = 1$ and $a+b+c = ab+bc+ca$. But in fact, any triple $(a,\frac{1}{a},1)$ works here and one can even show that these are all possible triples (and then the main problem is solved by noting that $a + \frac{1}{a}$ is minimal for $a=1$).

AM-GM. Also this SL has some of the easiest combo problems i've seen.