Balkan Mathematical Olympiad

By clearing denominators (brute force), the inequality becomes $$a^3{b}^3+b^3{c}^3+c^3{a}^3\ge a^{2}b{c}^3+b^{2}c{a}^3+c^{2}a{b}^3.(1)$$ In order to justify (1), use the AM-GM inequality. Thus $$a^3{b}^3+b^3{c}^3+c^3{a}^3\ge 3a{b}^3{c}^2.$$ By summation with the two other analogous inequalities, one gets the desired result.

Alternative Solution: Make use of the rearrangement inequality (in various settings). From the start we may suppose either $a\ge b\ge c$, or $a\ge c\ge b$. For example, write the inequality as the equivalent form $$\frac{ab}{c(a+b)}+\frac{bc}{a(b+c)}+\frac{ca}{b(a+c)}\ge \frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a},$$ and use same-order triplets $(ab,ac,bc)$, $(\frac{1}{c(a+b)},\frac{1}{a(b+c)},\frac{1}{b(a+c)})$ when $a\ge b\ge c$, and same-order triplets $(ac,ab,bc)$, $(\frac{1}{b(a+c)},\frac{1}{c(a+b)},\frac{1}{a(b+c)})$ when $a\ge c\ge b$.

Alternative Solution: (Official Jury Solution) Divide by $abc$, to obtain $$\frac{a(b-c)}{c(a+b)}+\frac{b(c-a)}{a(b+c)}+\frac{c(a-b)}{b(c+a)}\ge 0.$$ Adding 1 to each fraction leads to $$\frac{b(c+a)}{c(a+b)}+\frac{c(a+b)}{a(b+c)}+\frac{a(b+c)}{b(c+a)}\ge 3,$$ which immediately follows from the AM-GM inequality $$\frac{b(c+a)}{c(a+b)}+\frac{c(a+b)}{a(b+c)}+\frac{a(b+c)}{b(c+a)}\ge 3\sqrt[3]{\frac{b(c+a)}{c(a+b)}\cdot \frac{c(a+b)}{a(b+c)}\cdot \frac{a(b+c)}{b(c+a)}}=3.$$