From the inequality $2yz \le y^2 + z^2$ we have that $x^2 + 2yz + 2 \le x^2 + y^2 + z^2 + 2$, so
$$\frac{x}{\sqrt{x^2 + 2yz + 2}} \ge \frac{x}{\sqrt{x^2 + y^2 + z^2 + 2}}$$
Working similarly and adding we have that
$$\frac{x}{\sqrt{x^2 + 2yz + 2}} + \frac{y}{\sqrt{y^2 + 2zx + 2}} + \frac{z}{\sqrt{z^2 + 2xy + 2}} \ge \frac{x+y+z}{\sqrt{x^2 + y^2 + z^2 + 2}}$$
Therefore, it suffices to prove that
$$\frac{x+y+z}{\sqrt{x^2+y^2+z^2+2}} \ge 1 \Leftrightarrow (x+y+z)^2 \ge x^2+y^2+z^2+2 \Leftrightarrow xy+yz+zx \ge 1$$
However, from the given condition we have $\frac{1}{xy} + \frac{1}{yz} + \frac{1}{zx} = 9$, and from the Cauchy-Schwarz inequality we have
$$(xy + yz + zx) \left( \frac{1}{xy} + \frac{1}{yz} + \frac{1}{zx} \right) \ge 9, \text{ so } xy + yz + zx \ge 1, \text{ which is the desired result.}$$

2ºς τρόπος: Using Holder's inequality we have:
$$\left( \sum_{cyc} \frac{x}{\sqrt{x^2 + 2yz + 2}} \right) \left( \sum_{cyc} \frac{x}{\sqrt{x^2 + 2yz + 2}} \right) \left( \sum_{cyc} x(x^2 + 2yz + 2) \right) \ge (x+y+z)^3$$
Therefore, it suffices to prove that
$$\begin{aligned} \frac{(x+y+z)^3}{\sum_{cyc} x(x^2+2yz+2)} &\ge 1 \\ &\Leftrightarrow (x+y+z)^3 \ge x^3+y^3+z^3+6xyz+2(x+y+z) \\ &\Leftrightarrow x^3+y^3+z^3+3(x+y)(y+z)(z+x) \ge x^3+y^3+z^3+6xyz+2(9xyz) \\ &\Leftrightarrow (x+y)(y+z)(z+x) \ge 8xyz \end{aligned}$$
but the last one holds since
$$x + y \ge 2\sqrt{xy}, \quad y + z \ge 2\sqrt{yz}, \quad z + x \ge 2\sqrt{zx}.$$