<s>[INST] 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
 [/INST] Note that because $x+y+z=3$ we have $1\geq xyz$ and $4\geq 4xyz$. 
In addition, $xy+yz+xz \geq 3(xyz)^\frac{2}{3}$ but because $1\geq xyz$ we have $1\geq xyz^\frac{1}{3}$ and $3(xyz)^\frac{2}{3}\geq3xyz$. Hence, $xy+yz+xz \geq 3xyz$.
Adding the two inequalities we have $4+xy+yz+xz \geq 7xyz$.
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