Proof: According to the Cauchy-Schwarz inequality, we have
$$\begin{array}{l} \sum_{c y c} \sqrt{\frac{x}{1+2 y z}}=\sum_{c y c} \frac{x^{2}}{\sqrt{x} \sqrt{x^{2}+2 x^{2} y z}} \\ \geq \frac{(x+y+z)^{2}}{\sqrt{x} \sqrt{x^{2}+2 x^{2} y z}+\sqrt{y} \sqrt{y^{2}+2 y^{2} z x}+\sqrt{z} \sqrt{z^{2}+2 z^{2} x y}} \\ \geq \frac{(x+y+z)^{2}}{\sqrt{(x+y+z)\left[x^{2}+y^{2}+z^{2}+2 x y z(x+y+z)\right]}} \end{array}$$

Therefore, it suffices to prove
$$\begin{array}{l} \left(\sum_{c y c} x\right)^{3} \geq 3\left(\sum_{c y c} x^{2}\right)+6 x y z\left(\sum_{c y c} x\right) \Leftrightarrow\left(\sum_{c y c} x\right)^{3} \geq\left(\sum_{c y c} x\right)\left(\sum_{c y c} x^{2}\right)+6 x y z\left(\sum_{c y c} x\right) \\ \Leftrightarrow 3 \sum_{c y c} x(y-z)^{2} \geq 0 \end{array}$$

This is obviously true. The equality holds when $a=b=c=1$ and $a=3, b=c=0$ and their permutations.