Prove the inequality $\sum x+6\ge 2(\sum\sqrt{xy}) $ Let $x;y;z\in R^+$ such that $x+y+z+2=xyz$. Prove that $$x+y+z+6\ge 2(\sqrt{xy}+\sqrt{yz}+\sqrt{xz}) $$

This inequality is not homogeneous and look at the condition i thought that 
 i would substitute the variables $x;y;z$ such as:
+)If i need to solve $x^2+y^2+z^2+2xyz=1$, i will let $x=\frac{2a}{\sqrt{\left(a+b\right)\left(a+c\right)}}$ 
+)If i need to solve $xy+yz+xz+xyz=4$,i will let $x=\frac{2a}{b+c}$
but failed. Please explain for me how can i get this substitution (if have a solution by substitution)
I also tried to solve it by $u,v,w$.Let $\sum_{cyc} x=3u;\sum_{cyc} xy;\Pi_{cyc}a=w^3(3u+2=w^3;u,v,w>0 )$ then $u\le w^3-3u$ or $4u\le w^3$  but stuck (I am really bad at $uvw$)

The condition gives
$$\sum_{cyc}\frac{1}{x+1}=1.$$
Now, let $x=\frac{b+c}{a}$ and $y=\frac{a+c}{b},$ where $a$, $b$ and $c$ are positives.
Thus, $z=\frac{a+b}{c}$ and we need to prove that:
$$\sum_{cyc}\frac{b+c}{a}+6\geq2\sum_{cyc}\sqrt{\frac{(b+c)(a+c)}{ab}}$$ or
$$\sum_{cyc}(a^2b+a^2c+2abc)\geq2\sum_{cyc}a\sqrt{bc(a+b)(a+c)},$$
which is true by AM-GM.
Indeed,
$$2\sum_{cyc}a\sqrt{bc(a+b)(a+c)}=2\sum_{cyc}a\sqrt{(ac+bc)(ab+bc)}\leq$$
$$\leq\sum_{cyc}a(ac+bc+ab+bc)=\sum_{cyc}(a^2b+a^2c+2abc).$$
Done!
