Min of Product with Fraction

Opening brackets,
$\displaystyle 4 + 4xy + \frac{1}{xy}$

Using AM-GM inequality,
$\displaystyle \frac{4xy+\frac{1}{xy}}{2} \geq \sqrt{4}$

Thus,
$\displaystyle 4xy + \frac{1}{xy} \geq 4$

Hence,
the minimum value is $\displaystyle 4 + 4 = \boxed{8}$

Let's break precedence and use calculus here. Let $z = xy$ ( $z \in \mathbb{R^{+}}$ ) such that $f(z) = 4z + 4 + \frac{1}{z}$ . Setting the first derivative equal to zero gives $f'(z) = 4 - \frac{1}{z^2} = 0 \Rightarrow z^2 = \frac{1}{4} \Rightarrow z = \pm \frac{1}{2}$ . Since we require $z > 0$ we only admit the positive root. Evaluating the second derivative at $z = \frac{1}{2}$ produces $f''(\frac{1}{2}) = \frac{2}{(1/2)^3} = 16 > 0$ , hence a global minimum over $z \in \mathbb{R^{+}}.$ Our required minimum value is just $f(\frac{1}{2}) = 4(\frac{1}{2}) + 4 + 2 = \boxed{8}.$