Replacing $x=a^{2}, y=b^{2}, z=c^{2}$, where $a, b, c$ are positive numbers, our inequality is equivalent to

$$\frac{a^{2}}{\sqrt{\sqrt{b}+\sqrt{c}}}+\frac{b^{2}}{\sqrt{\sqrt{c}+\sqrt{a}}}+\frac{c^{2}}{\sqrt{\sqrt{a}+\sqrt{b}}} \geq \frac{\sqrt[4]{(a+b+c)^{7}}}{\sqrt{2 \sqrt{27}}}$$

Using the Cauchy-Schwarz inequality for the left hand side we get

$$\frac{a^{2}}{\sqrt{\sqrt{b}+\sqrt{c}}}+\frac{b^{2}}{\sqrt{\sqrt{c}+\sqrt{a}}}+\frac{c^{2}}{\sqrt{\sqrt{a}+\sqrt{b}}} \geq \frac{(a+b+c)^{2}}{\sqrt{\sqrt{b}+\sqrt{c}}+\sqrt{\sqrt{c}+\sqrt{a}}+\sqrt{\sqrt{a}+\sqrt{b}}}$$

Using Cauchy-Schwarz inequality for three positive numbers $\alpha, \beta, \gamma$, we have

$$\sqrt{\alpha}+\sqrt{\beta}+\sqrt{\gamma} \leq \sqrt{3(\alpha+\beta+\gamma)}$$

Using this result twice, we have

$$\begin{aligned} \sqrt{\sqrt{b}+\sqrt{c}}+\sqrt{\sqrt{c}+\sqrt{a}}+\sqrt{\sqrt{a}+\sqrt{b}} & \leq \sqrt{6(\sqrt{a}+\sqrt{b}+\sqrt{c})} \\ & \leq \sqrt{6 \sqrt{3(a+b+c)}} \end{aligned}$$

Combining (1) and (2) we get the desired result.

Alternative solution by PSC. We will use Hölder's inequality in the form

$$\begin{aligned} & \left(a_{11}+a_{12}+a_{13}\right)\left(a_{21}+a_{22}+a_{23}\right)\left(a_{31}+a_{32}+a_{33}\right)\left(a_{41}+a_{42}+a_{43}\right) \\ & \quad \geq\left(\left(a_{11} a_{21} a_{31} a_{41}\right)^{1 / 4}+\left(a_{12} a_{22} a_{32} a_{42}\right)^{1 / 4}+\left(a_{13} a_{23} a_{33} a_{43}\right)^{1 / 4}\right)^{4} \end{aligned}$$

where $a_{ij}$ are positive numbers. Using this appropriately we get

$$\begin{aligned} &(1+1+1)((\sqrt{b}+\sqrt{c})+(\sqrt{c}+\sqrt{a})+(\sqrt{a}+\sqrt{b}))\left(\frac{a^{2}}{\sqrt{\sqrt{b}+\sqrt{c}}}+\frac{b^{2}}{\sqrt{\sqrt{c}+\sqrt{a}}}+\frac{c^{2}}{\sqrt{\sqrt{a}+\sqrt{b}}}\right)^{2} \\ & \geq(a+b+c)^{4} \end{aligned}$$

By the Cauchy-Schwarz inequality we have

$$(\sqrt{b}+\sqrt{c})+(\sqrt{c}+\sqrt{a})+(\sqrt{a}+\sqrt{b})=2(\sqrt{a}+\sqrt{b}+\sqrt{c}) \leq 2 \sqrt{3(a+b+c)}$$

Combining these two inequalities we get the desired result.