2. let $a, b$ and $c$ be natural numbers. Find the smallest value that the following expression can take:

$$
\frac{a}{\operatorname{ggT}(a+b, a-c)}+\frac{b}{\operatorname{ggT}(b+c, b-a)}+\frac{c}{\operatorname{ggT}(c+a, c-b)}
$$
1st solution (Paul): First we notice that

$$
\operatorname{ggT}(a+b, a-c)=\operatorname{ggT}(a+b-(a-c), a-c)=\operatorname{ggT}(b+c, a-c) \leq b+c
$$

applies. It then follows that

$$
\frac{a}{\operatorname{ggT}(a+b, a-c)}+\frac{b}{\operatorname{ggT}(b+c, b-a)}+\frac{c}{\operatorname{ggT}(c+a, c-b)} \geq \frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b} \geq \frac{3}{2}
$$

where the last inequality is Nesbitt. By inserting $a=b=c$ we see that $\frac{3}{2}$ can actually be achieved.