Prove that if $a,b,c \in \mathbb{R^+}\text{ and } abc=8\text{ then } {ab+4\over a+2}+{bc+4\over b+2}+{ca+4\over c+2}\ge6$

My Approach:

Now we have: $${ab+4\over a+2}={2\times (ab+4)\over2\times (a+2)}={2ab+8\over2(a+2)}={2ab+abc\over2(a+2)}={ab(2+c)\over2(a+2)}$$ Now similarly we can acheive: $${bc+4\over b+2}={bc(2+a)\over2(b+2)};{ca+4\over c+2}={ca(2+b)\over2(c+2)}$$ Using AM-GM we get: $${ab+4\over a+2}+{bc+4\over b+2}+{ca+4\over c+2}$$ $$={ab(2+c)\over2(a+2)}+{bc(2+a)\over2(b+2)}+{ca(2+b)\over2(c+2)}$$ $$\ge\sqrt[3]{{ab(2+c)\over2(a+2)}\times{bc(2+a)\over2(b+2)}\times{ca(2+b)\over2(c+2)}}$$ $$=\sqrt[3]{(abc)^2\over2^3}$$ $$=\sqrt[3]{8^2\over8}=2$$ Therefore, we get: $${ab+4\over a+2}+{bc+4\over b+2}+{ca+4\over c+2}\ge2$$ However, the question wants me to prove that its greater than or equal to $6$ and when I try to plug in I always get a value larger than or equal to $6$. So where did I go wrong and how can I fix my mistake. Thank you in advance.

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    $\begingroup$ You need to divide the left part by 3, hence you got it, well done $\endgroup$ – Marco Bellocchi Dec 14 '18 at 14:43
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    $\begingroup$ $(X+Y+Z)/3\geq...$ $\endgroup$ – Marco Bellocchi Dec 14 '18 at 14:45
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    $\begingroup$ @MarcoBellocchi Oh thank you, I guess that was my mistake. $\endgroup$ – user587054 Dec 14 '18 at 14:46

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