<s>[INST] Prove that $x^2 y+ y^2 z + z^2x ≥ 2(x + y + z) − 3$ 
Let $x$, $y$ and $z$ be positive real numbers such that $xy + yz + zx = 3xyz$. Prove that $x^2 y+ y^2 z + z^2x ≥ 2(x + y + z) − 3$     

This question is from BMO 2014.
My working:
By AM-GM inequaluity,
  $$x^2 y+ y^2 z + z^2x ≥ 3xyz=xy+yz+zx$$
So now I need to prove that
$$xy+yz+zx\ge2(x+y+z)-3$$
To prove this do I need the relation $xy + yz + zx = 3xyz$?
Need help.
 [/INST] Given, $xy+yz+zx \ge 3xyz$
$$\Rightarrow 3=\frac{1}{x}+\frac{1}{y} + \frac{1}{z}$$ 
Now $$x^2y+y^2z+z^2x\ge2(x+y+z)-3$$
$$\Rightarrow x^2y+y^2z+z^2x+3\ge2(x+y+z)$$
Now
$$x^2y+y^2z+z^2x+3=x^2y+y^2z+z^2x+\frac{1}{x}+\frac{1}{y} + \frac{1}{z}$$
$$=\sum_{cyc} \frac{x^2y^2+1}{y}$$
$$\ge \sum_{cyc} 2x$$
Done!
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