HMMT February

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