By Holder and AM-GM we obtain: $$\sum_{cyc}\frac{x}{\sqrt{x^2+2yz+2}}=\sqrt{\frac{\left(\sum\limits_{cyc}\frac{x}{\sqrt{x^2+2yz+2}}\right)^2\sum\limits_{cyc}x(x^2+2yz+2)}{\sum\limits_{cyc}(x^3+2xyz+2x)}}\geq$$$$\geq\sqrt{\frac{(x+y+z)^3}{\sum\limits_{cyc}(x^3+2xyz+6xyz)}}\geq\sqrt{\frac{(x+y+z)^3}{\sum\limits_{cyc}(x^3+2xyz+3x^2y+3x^2z)}}=1.$$

arqady wrote: By Holder and AM-GM we obtain: $$\sum_{cyc}\frac{x}{\sqrt{x^2+2yz+2}}=\sqrt{\frac{\left(\sum\limits_{cyc}\frac{x}{\sqrt{x^2+2yz+2}}\right)^2\sum\limits_{cyc}x(x^2+2yz+2)}{\sum\limits_{cyc}(x^3+2xyz+2x)}}\geq$$$$\geq\sqrt{\frac{(x+y+z)^3}{\sum\limits_{cyc}(x^3+2xyz+6xyz)}}\geq\sqrt{\frac{(x+y+z)^3}{\sum\limits_{cyc}(x^3+2xyz+3x^2y+3x^2z)}}=1.$$ Thank arqady. $$\frac {x} {\sqrt {x^2+2yz+2}}+\frac {y} {\sqrt {y^2+2zx+2}}+\frac {z} {\sqrt {z^2+2xy+2}}\ge \frac {x+y+z} {\sqrt {x^2+y^2+z^2+2}}\ge1.$$

From the initial condition, $$ x+y+z=9xyz \Longrightarrow \frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}=9 \Longrightarrow \frac{1}{3}=\frac{3}{\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}} $$And by AM-HM we get $$ \frac{xy+yz+zx}{3}\geq \frac{1}{3} \Longrightarrow xy+yz+zx\geq 1 $$Hence $$ (x+y+z)^2\geq x^2+y^2+z^2+2 \Longrightarrow \frac{x+y+z}{\sqrt{x^2+y^2+z^2+2}}\geq 1 \quad (\star) $$Also, by GM-QM, it holds $$ yz\leq \frac{y^2+z^2}{2} $$Therefore $$ \sum_{cyc}\frac{x}{\sqrt{x^2+2yz+2}}\geq \sum_{cyc} \frac{x}{\sqrt{x^2+y^2+z^2+2}} \stackrel{(\star)} \geq 1 $$The inequality is proved. By the inequality GM-QM that we have used, we get that the equality case only holds for $x=y=z=\frac{\sqrt{3}}{3}$.

Very easy $\sum{\frac{x} {\sqrt{x^2+2yz+2}}} \ge 1(?)$ By $AM-GM$ $\sum{\frac{x} {\sqrt{x^2+y^2+z^2+2}}} \ge 1(?)$ $\implies$ $xy+yz+zx \geq 1(?) $ $(x+y+z) /9(\sum{\frac{1}{x}}) \geq 1(?)$ By $AM-HM$ it is obvious