I would greatly appreciate it if someone could show me how to solve this problem. The question is this:
Let $D$ be the region in $\mathbb{R}^2$ enclosed by the circle with equation $(x-4)^2 + (y-1/2)^2 = 1/16$.
Show that: $$ \frac{3\pi}{8} \le \iint_D (xy+\frac{8}{x}+\frac{1}{y})\,dA$$
Assume that the global minimum of the integrand does not occur on the boundary of D.
Hint:
Since all points $(x,y)$ that lie in $D$ have a positive $x$ and $y$ value, by the AM-GM inequality we have:
$$\frac{\left(xy+\frac{8}{x}+\frac{1}{y} \right) }{3} \geq \left(xy \cdot \frac{8}{x} \cdot \frac{1}{y}\right)^{1/3}$$
So, what does that say about the integrand in the region $D$?