How to prove $ \left(\sum\limits_{cyc}{xy}\right)^2 \ge3xyz(x+y+z)$ with $x,y,z$ being positive real numbers I have tried to improve the inequality by AM-GM. Here is what I have done:
Since
$$\sum_{cyc}{xy}\ge3\sqrt[3]{x^2y^2z^2}$$
$$\Rightarrow \left(\sum_{cyc}{xy}\right)^2 \ge3xyz\sqrt[3]{xyz}$$  That means we got to prove$\sqrt[3]{xyz} \ge x+y+z$.
However that is false in the case $x=1$,$y=2$,$z=3$.
Any help?  

Simplify the inequality to 
$$x^2y^2+y^2z^2+x^2z^2\ge xyz(x+y+z)$$
If some of $x,y,z$ equals $0$, then the inequality is obvious.
So assume $xyz\neq 0$ and divide by $x^2y^2z^2>0$ both sides to get
$$\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\ge \frac{1}{yz}+\frac{1}{xz}+\frac{1}{xy}$$
This is true by the power of Cauchy Schwarz  $$\frac{1}{y}\cdot\frac{1}{z}+\frac{1}{z}\cdot\frac{1}{x}+\frac{1}{x}\cdot\frac{1}{y}\leq \sqrt{\frac{1}{y^2}+\frac{1}{z^2}+\frac{1}{x^2}}\cdot\sqrt{\frac{1}{z^2}+\frac{1}{x^2}+\frac{1}{y^2}}$$
