49th Mathematical Olympiad in Ukraine

Using the inequality $a^2+b^2+c^2\ge ab+bc+ca$ we have $$\frac{a}{(a+b)(a+c)}+\frac{b}{(b+c)(b+a)}+\frac{c}{(c+a)(c+b)}\le \frac{9}{4(a+b+c)}$$ , is equivalent to $$8(a+b+c)(ab+bc+ca)\le 9(a+b)(b+c)(c+a)$$ and this equivalent to inequality $$6abc\le a^{2}b+a^{2}c+b^{2}a+b^{2}c+c^{2}a+c^{2}b$$ which follows immediately by the AM-GM inequality applied to the six numbers.