Junior Balkan Mathematical Olympiad

$$\begin{array}{rl}A& =\frac{2-a^3}{a}+\frac{2-b^3}{b}+\frac{2-c^3}{c}=2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)-a^2-b^2-c^2\\ & =2\frac{ab+bc+ca}{abc}-(a^2+b^2+c^2)\\ & =2\frac{ab+bc+ca}{abc}-\left((a+b+c{)}^2-2(ab+bc+ca)\right)\\ & =2\frac{ab+bc+ca}{abc}-(9-2(ab+bc+ca))\\ & =2\frac{ab+bc+ca}{abc}+2(ab+bc+ca)-9\\ & =2(ab+bc+ca)\left(\frac{1}{abc}+1\right)-9\end{array}$$

Recall now the well-known inequality $(x+y+z{)}^2\ge 3(xy+yz+zx)$ and set $x=ab$, $y=bc$, $z=ca$, to obtain $(ab+bc+ca{)}^2\ge 3abc(a+b+c)=9abc$ where we have used $a+b+c=3$. By taking the square roots on both sides of the last one we obtain: $$ab+bc+ca\ge 3\sqrt{abc}.\text{\hspace{1em}(1)}$$ Also by using AM-GM inequality we get that $$\frac{1}{abc}+1\ge 2\sqrt{\frac{1}{abc}}.\text{\hspace{1em}(2)}$$ Multiplication of (1) and (2) gives $$(ab+bc+ca)\left(\frac{1}{abc}+1\right)\ge 3\sqrt{abc}\cdot 2\sqrt{\frac{1}{abc}}=6.$$ So $A\ge 2\cdot 6-9=3$ and the equality holds if and only if $a=b=c=1$, so the minimum value is $3$.