, 2008

The Cauchy-Schwarz inequality gives
$$(a + b + c)^2 \le 3(a^2 + b^2 + c^2).$$
Thus $(a + b + c) < 6$ and hence $\frac{(a + b + c)^3}{9} < 4(a + b + c)$. Now the AM-GM inequality gives
$$(a + b + c)^3 \ge 27abc.$$
Thus
$$3abc \le \frac{(a + b + c)^3}{9} < 4(a + b + c).$$

Alternately, we have $a + b + c < 6$ so that $a^2 + b^2 + c^2 < 2(a + b + c) < 12$. Thus
$$(a^2 + b^2 + c^2)(a + b + c) < 12(a + b + c).$$
Using the AM-GM inequality, we have
$$a^2 + b^2 + c^2 \ge 3(abc)^{2/3}, \quad a + b + c \ge 3(abc)^{1/3}.$$
Thus
$$9abc \le (a^2 + b^2 + c^2)(a + b + c) < 12(a + b + c),$$
giving the required inequality. Thus