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?
 A: 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.
A: Search for Lohwater's "Introduction to inequalities" (unpublished manuscript, was released by his wife). It is almost 200 pages full of all sorts of elementary techniques for proving inequalities.
A: Found in Hewitt & Stromberg, Real and Abstract Analysis, (13.26) Exercise, Generalized Hölder's Inequality  
Let $\alpha_1, \alpha_2, \dots \alpha_n$ be positive real numbers such that $\sum_{j=1}^n \alpha_j=1$.  For nonnegative functions $f_1, f_2, \dots, f_n \in L_1(X,\mu)$, we have
$$
f_1^{\alpha_1}f_2^{\alpha_2}\cdots f_n^{\alpha_n} \in L_1(X,\mu),
$$
and
$$
\int_X\big(f_1^{\alpha_1}f_2^{\alpha_2}\cdots f_n^{\alpha_n}\big) d\mu \le 
\|f_1\|_1^{\alpha_1}\|f_2\|_1^{\alpha_2}\cdots \|f_n\|_1^{\alpha_n} .
$$
[Let $X$ have three points, each of measure $1$.  Let $n=3$ and $\alpha_1 = \alpha_2 = \alpha_3 = 1/3$.]
A: It's just another form of Holder's inequality, which generalises Cauchy-Schwarz.
A: 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 $(0.1+0.1+0.1)^{2}=0.09$
