Let $f : (-1, +\infty) \rightarrow \mathbb{R}$ be the function defined by $f(x) = \frac{1}{\sqrt{1+x}}$. Since $f'(x) = -\frac{1}{2(1+x)^{3/2}} < 0$ and $f''(x) = \frac{3}{4(1+x)^2\sqrt{1+x}} > 0$, then $f$ is convex. Applying Jensen's inequality to the function $f$ with $a_1 = \frac{yz}{x}$, $a_2 = \frac{zx}{y}$, $a_3 = \frac{xy}{z}$, we have
$$\frac{1}{3} \sum_{k=1}^{3} f(a_k) \ge f\left(\frac{1}{3} \sum_{k=1}^{3} a_k\right)$$
or
$$\frac{1}{3} \left( \sqrt{\frac{x}{x + yz}} + \sqrt{\frac{y}{y + zx}} + \sqrt{\frac{z}{z + xy}} \right) \ge f(2) = \frac{\sqrt{3}}{3}$$
because from $x^2y^2 + y^2z^2 + z^2x^2 = 6xyz$ immediately follows that
$$\frac{a_1 + a_2 + a_3}{3} = \frac{1}{3} \left( \frac{yz}{x} + \frac{zx}{y} + \frac{xy}{z} \right) = 2$$
Equality holds when $x = y = z = 2$, and we are done. $\square$