By twice using the inequality between the arithmetical mean and geometrical mean we get
$$\begin{aligned} (16a^2 + 8b + 17) &= (16a^2 + 1 + 8b + 16) \ge 8a + 8b + 16 = 8(a+b+2) = 8(a+1+b+1) \ge \\ &\ge 8 \cdot 2\sqrt{(a+1)(b+1)} = 2^4\sqrt{(a+1)(b+1)}. \end{aligned} \quad (1)$$
Analogously we have
$$(16b^2 + 8c + 17) \ge 2^4 \sqrt{(b+1)(c+1)} \quad (2)$$
$$(16c^2 + 8a + 17) \ge 2^4 \sqrt{(c+1)(a+1)} \quad (3).$$
If we multiply the three inequalities we get
$$(16a^2 + 8b + 17)(16b^2 + 8c + 17)(16c^2 + 8a + 17) \ge 2^{12}(a+1)(b+1)(c+1).$$
In (1) equality is obtained when $16a^2 = 1$ and $a = b$, i.e. $a = b = \frac{1}{4}$. By an analogous argument for (2) and (3) we get $a = b = c = \frac{1}{4}$.