32. $[\mathbf{1 7}]$ Let $a, b, c$ be non-negative real numbers such that $a b+b c+c a=3$. Suppose that
$$
a^{3} b+b^{3} c+c^{3} a+2 a b c(a+b+c)=\frac{9}{2} .
$$

What is the minimum possible value of $a b^{3}+b c^{3}+c a^{3}$ ?
Answer:
18

Expanding the inequality $\sum_{\text {cyc }} a b(b+c-2 a)^{2} \geq 0$ gives
$$
\left(\sum_{\text {cyc }} a b^{3}\right)+4\left(\sum_{\text {cyc }} a^{3} b\right)-4\left(\sum_{\text {cyc }} a^{2} b^{2}\right)-a b c(a+b+c) \geq 0
$$

Using $\left(\sum_{\text {cyc }} a^{3} b\right)+2 a b c(a+b+c)=\frac{9}{2}$ in the inequality above yields
$$
\left(\sum_{\text {cyc }} a b^{3}\right)-4(a b+b c+c a)^{2} \geq\left(\sum_{\text {cyc }} a b^{3}\right)-4\left(\sum_{\text {cyc }} a^{2} b^{2}\right)-9 a b c(a+b+c) \geq-18
$$

Since $a b+b c+c a=3$, we have $\sum_{\text {cyc }} a b^{3} \geq 18$ as desired.
The equality occurs when $(a, b, c) \underset{\text { cyc }}{\sim}\left(\sqrt{\frac{3}{2}}, \sqrt{6}, 0\right)$.