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 $$uvw$$)