2. Suppose $\frac{1}{2} \leq x \leq 2$ and $\frac{4}{3} \leq y \leq \frac{3}{2}$. Determine the minimum value of
$$
\frac{x^{3} y^{3}}{x^{6}+3 x^{4} y^{2}+3 x^{3} y^{3}+3 x^{2} y^{4}+y^{6}} \text {. }
$$
Answer: $\frac{27}{1081}$
Solution: Note that
$$
\begin{aligned}
\frac{x^{3} y^{3}}{x^{6}+3 x^{4} y^{2}+3 x^{3} y^{3}+3 x^{2} y^{4}+y^{6}} & =\frac{x^{3} y^{3}}{\left(x^{2}+y^{2}\right)^{3}+3 x^{3} y^{3}} \\
& =\frac{1}{\frac{\left(x^{2}+y^{2}\right)^{3}}{x^{3} y^{3}}+3} \\
& =\frac{1}{\left(\frac{x^{2}+y^{2}}{x y}\right)^{3}+3} \\
& =\frac{1}{\left(\frac{x}{y}+\frac{y}{x}\right)^{3}+3} .
\end{aligned}
$$

Let $u=\frac{x}{y}$. We have $\frac{x}{y}+\frac{y}{x}=u+\frac{1}{u}$. To get the minimum value of the entire expression, we need to make $u+\frac{1}{u}$ as large as possible. We can do this by setting $x=\frac{1}{2}$ and $y=\frac{3}{2}$. The function $f(u)=u+\frac{1}{u}$ is increasing on $(1,+\infty)$. Therefore $\frac{x}{y}+\frac{y}{x}=\frac{1}{3}+3=\frac{10}{3}$ and the minimum is
$$
\frac{1}{\left(\frac{x}{y}+\frac{y}{x}\right)^{3}+3}=\frac{1}{\left(\frac{10}{3}\right)^{3}+3}=\frac{27}{1081}
$$