Problem 1533

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

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:
$$ab + bc + ca \geq 3\sqrt{abc}. \tag{1}$$
Also by using AM-GM inequality we get that
$$\frac{1}{abc} + 1 \ge 2\sqrt{\frac{1}{abc}}. \tag{2}$$
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 $3$.