Romanian Mathematical Olympiad

We use straightforward computation: if $x \le y$ are real numbers, then

$$\begin{align*} f(y) - f(x) &= \sum_{\text{cyc}} \frac{a^y(b^x + c^x) - a^x(b^y + c^y)}{(b^x + c^x)(b^y + c^y)} \ &= \sum_{\text{cyc}} (a^y b^x - a^x b^y) \left( \frac{1}{(b^x + c^x)(b^y + c^y)} - \frac{1}{(a^x + c^x)(a^y + c^y)} \right) \\ &= \sum_{\text{cyc}} a^x b^x (a^{y-x} - b^{y-x}) \cdot \\ &\qquad \frac{((a^{x+y} - b^{x+y}) + c^x(a^y - b^y) + c^y(a^x - b^x))}{(b^x + c^x)(b^y + c^y)(a^x + c^x)(a^y + c^y)}, \end{align*}$$

$$(a^p - b^p)(a^q - b^q) = b^{p+q} \left( \left( \frac{a}{b} \right)^p - 1 \right) \left( \left( \frac{a}{b} \right)^q - 1 \right) \begin{cases} \ge 0 & \text{if } q \ge 0 \\ \le 0 & \text{if } q \le 0 \end{cases}$$
hence $f(y) - f(x) \ge 0$ if $y \ge x \ge 0$ and $f(y) - f(x) \le 0$ if $0 \ge y \ge x$.