Prove the inequality $4+xy+yz+zx \ge 7xyz$ Let $x,y,z$ be non negative real numbers such that $x+y+z=3$, prove the following inequality:
$$4+xy+yz+zx \ge 7xyz$$
I tried MV and taking out one variable but I got nothing

Alternative approach:
We have that $1=\frac{x+y+z}{3}\geq\sqrt[3]{xyz}$, therefore $xyz\leq 1$. 
Since $xyz\leq 1$, we have $4 \geq 4xyz$. 
Also, by AM-HM, we have:
$$1=\frac{x+y+z}{3} \geq \frac{3}{\frac{1}{x}+\frac{1}{y}+\frac{1}{z}}$$
Therefore 
$$\frac{1}{3} \geq \frac{1}{\frac{1}{x}+\frac{1}{y}+\frac{1}{z}}$$
$$3 \leq \frac{1}{x}+\frac{1}{y}+\frac{1}{z}$$
$$3xyz \leq yz+xz+xy$$
Adding the latter inequality and $4 \geq 4xyz$ gives the desired inequality.
