Since $x$, $y > 0$ the given inequality can be written as:
$$\begin{align*} (xy+2)(y+2x) \ge 8xy &\Leftrightarrow xy^2 + 2y + 2x^2y + 4x - 8xy \ge 0 \\ &\Leftrightarrow (xy^2 - 4xy + 4x) + (2x^2y - 4xy + 2y) \ge 0 \\ &\Leftrightarrow x(y^2 - 4y + 4) + 2y(x^2 - 2x + 1) \ge 0 \\ &\Leftrightarrow x(y-2)^2 + 2y(x-1)^2 \ge 0, \text{ which is valid for } x, y > 0. \end{align*}$$
Equality holds, if and only if:
$$x(y-2) = 0 \text{ and } 2y(x-1) = 0 \Leftrightarrow x = 1, y = 2, \text{ since } x, y > 0.$$

Since $x$, $y > 0$ the given inequality can be written as:
$$(xy+2)(y+2x) \ge 8xy \quad (1)$$
We apply the well-known inequality of arithmetic – geometric mean
$$a+b \ge 2\sqrt{ab}, \text{ yea } a,b \ge 0, \text{ (equality when } a=b) \text{ two times, for } a=xy, b=2 \text{ and for } a=y, b=2x, \text{ to receive:}$$
$$xy + 2 \ge 2\sqrt{2xy}, \quad \text{(equality for } xy = 2\text{)} \quad (2)$$
$$y+2x \geq 2\sqrt{2xy}, \quad (\text{equality for } y=2x) \quad (3)$$
From (2) and (3) (multiplication by parts) we get: $(xy+2)(y+2x) \geq 8xy$.
Equality holds, if and only if $xy=2$ and $y=2x$, i.e. $x=1$, $y=2$.