A conjecture from Cauchy–Schwarz inequality we know the Cauchy–Schwarz inequality in $R^3$:

$$(x_1^2+x_2^2+x_3^2)(y_1^2+y_2^2+y_3^2)\geq(x_1y_1+x_2y_2+x_3y_3)^2$$

$$(x_1+x_2+x_3)(y_1+y_2+y_3)\geq(\sqrt{x_1y_1}+\sqrt{x_2y_2}+\sqrt{x_3y_3})^2$$

i guess the following inequlity exists too:

$$(x_1+x_2+x_3)(y_1+y_2+y_3)(z_1+z_2+z_3)\geq(\sqrt[3]{x_1y_1z_1}+\sqrt[3]{x_2y_2z_2}+\sqrt[3]{x_3y_3z_3})^3$$

is it true? how to prove?
Lets look at a simple proof of Cauchy Schwarz inequality and see if we can extend in the direction you want.

Let $A^2 = \sum_{i=1}^n {a_i ^2}, \quad B^2 = \sum_{i=1}^n  {b_i^2}$  

Then $\sum_{i=1}^n  \dfrac{a_i b_i}{AB} \le \sum_{i=1}^n \dfrac{1}{2}\left( \dfrac{a_i^2}{A^2} + \dfrac{b_i^2}{B^2} \right) = 1$
So $\sum_{i=1}^n  {a_i b_i} \le AB$, or if you prefer $ A^2 B^2 \ge \left(\sum_{i=1}^n  {a_i b_i} \right)^2$ which is Cauchy Schwarz.

Extending this, we have:  

Let $A^3 = \sum_{i=1}^n {a_i ^3}, \quad B^3 = \sum_{i=1}^n  {b_i^3}, \quad C^3 = \sum_{i=1}^n  {c_i^3}$  

Then $\sum_{i=1}^n  \dfrac{a_i b_i c_i}{ABC} \le \sum_{i=1}^n \dfrac{1}{3} \left( \dfrac{a_i^3}{A^3} + \dfrac{b_i^3}{B^3} + \dfrac{c_i^3}{C^3} \right) = 1$ 

So $\sum_{i=1}^n  {a_i b_i c_i} \le ABC$, or if you prefer, $A^3B^3C^3 \ge \left(\sum_{i=1}^n  {a_i b_i c_i} \right)^3$ is the extended version.  Essentially your RHS needs to be cubed, not squared.
It is false.
$x_{1}=x_{2}=x_{3}=y_{1}=y_{2}=y_{3}=z_{1}=z_{2}=z_{3}=0.1$ Then LHS is $0.3*0.3*0.3=0.027$
RHS is $(0.1+0.1+0.1)^{2}=0.09$