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?
 A: HINT + INFORMATION:
If you know about the inequality that exists for all positive real, that is
AM $\ge$ GM $\ge$ HM, then this question will not take much time.
For just information,

AM–GM-HM inequality, states that the arithmetic mean of a list of non-negative real numbers is greater than or equal to the geometric mean of the same list and geometric mean of non-negative real numbers is greater than or equal to harmonic mean of those numbers.

Read more about this here.
This inequality for a,b,c states that:
$$\frac {a+b+c} 3 \ge (abc)^{1/3} \ge\frac 3 {\frac 1 a + \frac 1 b + \frac 1 c}$$
Solving GM, HM part, we get
$ab+bc+ca \ge12$
Thus $ab+4+bc+4+ca+4 \ge 24$
Also solving AM,GM part we get,
$a+b+c\ge6$
Now all you need to do is to set these inequalities in such a way that you can get something to prove about the question inequality.
A: I solved it in the following way:

A: Hint $${ab+4\over a+2} = {2ab+8\over 2a+4} = {2ab+abc\over 2a+4} = {ab\over 2}\cdot{2+c\over a+2}$$
Similary for the rest of terms. Now use AM-GM for 3 terms...
A: 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:
$$\sum_{cyc}\frac{ab+4}{a+2}=\sum_{cyc}\frac{\frac{4z}{x}+4}{\frac{2y}{x}+2}=2\sum_{cyc}\frac{z+x}{x+y}\geq6\sqrt[3]{\prod_{cyc}\frac{z+x}{x+y}}=6.$$
