1. determine all solutions in natural numbers of the equation

$$
a b+b c+c a=2(a+b+c)
$$
## 1st solution:

We can assume $a \leq b \leq c$ for reasons of symmetry. First let $a \geq 2$. Then we obtain the estimate for the left-hand side

$$
a b+b c+c a \geq 2 a+2 b+2 c
$$

However, according to the prerequisite, equality applies here, so $a=b=c=2$ must be true. This is actually a solution. Let $a=1$ in the following, the equation then becomes

$$
b c=b+c+2
$$

If $b \geq 3$ applies here, then $b c \geq 3 c \geq b+c+3$, a contradiction. Consequently, $b=1$ or $b=2$. The first case leads to no solution, the second to $c=4$. The solutions are therefore the permutations of the triples

$$
(2,2,2) \quad \text { and } \quad(1,2,4)
$$