, 2017

Solution:

Expanding the inequality $\sum_{\text{cyc}} ab(b + c - 2a)^2 \geq 0$ gives
$$\left(\sum_{\text{cyc}} ab^3\right) + 4\left(\sum_{\text{cyc}} a^3 b\right) - 4\left(\sum_{\text{cyc}} a^2 b^2\right) - abc(a + b + c) \geq 0$$
Using $\left(\sum_{\text{cyc}} a^3 b\right) + 2abc(a + b + c) = \frac{9}{2}$ in the inequality above yields
$$\left(\sum_{\text{cyc}} ab^3\right) - 4(ab + bc + ca)^2 \geq \left(\sum_{\text{cyc}} ab^3\right) - 4\left(\sum_{\text{cyc}} a^2 b^2\right) - 9abc(a + b + c) \geq -18$$
Since $ab + bc + ca = 3$, we have $\sum_{\text{cyc}} ab^3 \geq 18$ as desired.

The equality occurs when $(a, b, c) \underset{\text{cyc}}{\sim} \left(\sqrt{\frac{3}{2}}, \sqrt{6}, 0\right)$.