Balkan Mathematical Olympiad

Dividing both sides by $a$, $b$, $c$, the inequality is equivalent to $$\frac{a(b-c)}{c(a+b)}+\frac{b(c-a)}{a(b+c)}+\frac{c(a-b)}{b(c+a)}\ge 0.$$ By adding $1+1+1$ to both sides, the inequality becomes $$\frac{b(c+a)}{c(a+b)}+\frac{c(a+b)}{a(b+c)}+\frac{a(b+c)}{b(c+a)}\ge 3.$$ By applying AM-GM inequality we have $$\frac{b(c+a)}{c(a+b)}+\frac{c(a+b)}{a(b+c)}+\frac{a(b+c)}{b(c+a)}\ge 3\sqrt{\frac{b(c+a)\cdot c(a+b)\cdot a(b+c)}{c(a+b)\cdot a(b+c)\cdot b(c+a)}}=3.$$ The inequality is proved.