Irish Mathematical Olympiad

We factorise $a^3 + b^3 + c^3 - (a+b+c)^3$ as follows:
$$\begin{aligned} & a^3 + b^3 + c^3 - ((a+b)^3 + 3(a+b)^2c + 3(a+b)c^2 + c^3) \\ &= a^3 + b^3 - (a^3 + b^3 + 3a^2b + 3ab^2 + b^3 + 3(a+b)^2c + 3(a+b)c^2) \\ &= -3(a+b)(ab + (a+b)c + c^2) \\ &= -3(a+b)(b+c)(c+a). \end{aligned}$$
Hence $|a^3+b^3+c^3-(a+b+c)^3| = 3(a+b)(b+c)(c+a)$. Now, by the AM-GM inequality,
$$(a+b)(b+c)(c+a) \le \left( \frac{(a+b) + (b+c) + (c+a)}{3} \right)^3 = \frac{8}{27}(a+b+c)^3,$$
with equality iff $a+b=b+c=c+a$, equivalently $a=b=c$. It follows that
$$|a^3 + b^3 + c^3 - (a + b + c)^3| \le \frac{8}{9}(a + b + c)^3,$$
with equality iff $a = b = c$.
But, if $x, y > 0$, then $x + y \ge 2\sqrt{xy}$, with equality iff $x = y$. Hence
$$(a+b)(b+c)(c+a) \ge 8\sqrt{ab}\sqrt{bc}\sqrt{ca},$$
and the inequality is strict unless $a = b = c$. It follows that
$$|a^3 + b^3 + c^3 - (a + b + c)^3| \ge 24abc,$$
with equality iff $a = b = c$.