Simple inequality over positive reals: $2(x+y+z) \geq 3xyz + xy+yz+zx$ for $xyz=1$ Problem
Let $x,y,z$ be real positive numbers with $xyz=1$. Prove: 
$$ 2(x+y+z) \geq 3xyz + xy+yz+zx$$
Note : I don't know whether the inequality is true or not. I couldn't find a prove in the place  found it nor a solution to it.
My try
I firstly took $a = \frac{x}{y}$, $b = \frac{y}{z}$ and $c = \frac{z}{x}$ so that the expression becomes $2\sum_{cyc} a^2c \geq 3abc + \sum_{cyc}a^2b$.
Then I tried to think of it as a function $f(a,b,c)$ and say that if we let two variable swap, we would come from one to another, letting us see that the order of the variables has nothing to do with the result. But I couldn't get anything here.
Any hint or solution would be really apreciated.
PD: I tried using the search function, but I couldn't find anything. so, sorry if this is a duplicate 

HINT
Let's observe that:
$2(x+y+z) \geq 3xyz + xy+yz+zx \geq xyz \left( 3 + \frac1x+\frac1y+\frac1z\right)\geq \left( 3 + \frac1x+\frac1y+\frac1z\right)$
and use AM-HM inequality.
