Solution:
Note that the inequality is symmetric in $a, b, c$ so that we may assume that $a \geq b \geq c$. Since $a b c = 1$, it follows that $a \geq 1$ and $c \leq 1$. Using $b = 1 / (a c)$, we get
$$a^{b+c} b^{c+a} c^{a+b} = \frac{a^{b+c} c^{a+b}}{a^{c+a} c^{c+a}} = \frac{c^{b-c}}{a^{a-b}} \leq 1$$
because $c \leq 1$, $b \geq c$, $a \geq 1$ and $a \geq b$.