Iberoamerican Mathematical Olympiad

Solution:

Put $f(a, b, c) = \frac{a-b}{a+b} + \frac{b-c}{b+c} + \frac{c-a}{a+c}$. Let $A, B, C$ be a permutation of $a, b, c$, with $A \leq B \leq C$. If $(A, B, C) = (b, a, c), (a, c, b)$ or $(c, b, a)$, then $f(a, b, c) = X$, where
$$X = \frac{B-A}{B+A} + \frac{C-B}{C+B} - \frac{C-A}{A+C}.$$
If $(A, B, C) = (a, b, c), (b, c, a)$ or $(c, a, b)$, then $f(a, b, c) = -X$.

Put $B = A + h$, $C = B + k = A + h + k$, where $h, k \geq 0$. Since $A, B, C$ are the sides of a triangle, we also have $A + B > C$ or $A > k$. So
$$X = \frac{h}{2A + h} + \frac{k}{2A + 2h + k} - \frac{h + k}{2A + h + k} = \frac{hk(h + k)}{(2A + h)(2A + h + k)(2A + 2h + k)}.$$
This is obviously non-negative. We claim also that it is $< \frac{1}{20}$. That is equivalent to:
$$20h^2k + 20hk^2 < (2A + h)(2A + h + k)(2A + 2h + k).$$
Since $k < A$ it is sufficient to show that
$$20h^2k + 20hk^2 \leq (2k + h)(2k + h + k)(2k + 2h + k)$$
$$= 18k^3 + 27hk^2 + 13h^2k + 2h^3$$
or
$$18k^3 + 7hk^2 - 7h^2k + 2h^3 \geq 0.$$
But $7k^2 - 7hk + 2h^2 = 7(k - h/2)^2 + h^2/4 \geq 0$ and $h$ and $k$ are non-negative, so
$$18k^3 + h\left(7k^2 - 7hk + 2h^2\right) \geq 0.$$
Thus we have established that $0 \leq X < \frac{1}{20}$, which shows that $f(a, b, c) < \frac{1}{20}$, which is slightly stronger than the required result.