NMO Selection Tests for the Balkan and International Mathematical Olympiads

Set $a_{n+1}=a_1$ and let $0\le x\le y$. Since $$f(y)-f(x)=(y-x)\sum _{i=1}^n\frac{a_{i+1}-a_i}{(a_{i+1}+x)(a_{i+1}+y)}$$ showing $f$ is decreasing amounts to showing $$\sum _{i=1}^n\frac{a_{i+1}}{(a_{i+1}+x)(a_{i+1}+y)}\le \sum _{i=1}^n\frac{a_i}{(a_{i+1}+x)(a_{i+1}+y)}$$ Noticing that $a_i\le a_j$ if and only if $(a_i+x{)}^{-1}(a_i+y{)}^{-1}\ge (a_j+x{)}^{-1}(a_j+y{)}^{-1}$, the above inequality is a straightforward consequence of the rearrangement inequality for the $a_i$ and the $(a_i+x{)}^{-1}(a_i+y{)}^{-1}$.