Solution:
Let $x = abc$. Multiplying all equations together and simplifying gives
$$(abc)^2 (a + b)(b + c)(c + a) = 10,$$
$$(abc)^2 \left(a^2(b + c) + b^2(c + a) + c^2(a + b) + 2abc\right) = 10,$$
$$x^2 (1 + 2 + 5 + 2x) = 10,$$
$$x^2 (x + 4) = 5.$$
The resulting cubic factors as $(x - 1)(x^2 + 5x + 5) = 0$. Therefore, the smallest possible value of $abc$ is $\frac{-5 - \sqrt{5}}{2}$.