Let $a, b, c \in \mathbb{R^+}$ and $abc=8$ Prove that $\frac {ab+4}{a+2} + \frac {bc+4}{b+2} + \frac {ca+4}{c+2} \ge 6$ 
Let $a, b, c \in \mathbb{R^+}$ and $abc=8$ Prove that $$\frac {ab+4}{a+2} + \frac {bc+4}{b+2} + \frac {ca+4}{c+2} \ge 6$$

I have attempted multiple times in this question and the only method that I can think of is by adding up the fractions and expanding, but took too long and I eventually got stuck. Is there any other method I can use to help me in this question?

Let $a=\frac{2y}{x}$ and $b=\frac{2z}{y}$, where $x$, $y$ and $z$ are positives.
Thus, $c=\frac{2x}{z}$ and by AM-GM we obtain:
$$\sum_{cyc}\frac{ab+4}{a+2}=\sum_{cyc}\frac{\frac{4z}{x}+4}{\frac{2y}{x}+2}=2\sum_{cyc}\frac{z+x}{x+y}\geq6\sqrt[3]{\prod_{cyc}\frac{z+x}{x+y}}=6.$$
