precalculus algebra: show the inequality $ sum x + 6 ge 2 ( sum sqrt {xy}) $

Leave $ x; Y; z in R ^ + $ such that $ x + y + z + 2 = xyz $. Such that $$ x + y + z + 6 ge 2 ( sqrt {xy} + sqrt {yz} + sqrt {xz}) $$


This inequality is not homogeneous and look at the condition that I thought
I would replace the variables $ x; Y; z $ such that

+)$ x ^ 2 + y ^ 2 + z ^ 2 + 2xyz = 1 $. Leave $ x = frac {2a} { sqrt { left (a + b right) left (a + c right)}} $

+)$ xy + yz + xz + xyz = 4 $. Leave $ a = frac {2 sqrt {xy}} { sqrt { left (y + z right) left (x + z right)}} $.Leave $ x = frac {2a} {b + c} $

but failed. Please explain to me how I can obtain this substitution (if I have a solution by substitution)

I also tried to solve it by $ u, v, w $.Leave $ sum_ {cyc} x = 3u; sum_ {cyc} xy; Pi_ {cyc} a = w ^ 3 (3u + 2 = w ^ 3; u, v, w> 0) $ so $ u le w ^ 3-3u $ or $ 4u le w ^ 3 $ but stagnant (I'm very bad at $ uvw $)