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.
 A: 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!
A: Let $x=\frac{1}{a}$, $y=\frac{1}{b}$ and $z=\frac{1}{c}.$
Thus, $a+b+c=3$ and we need to prove that
$$a^2c+b^2a+c^2b\geq2abc(ab+ac+bc)-3a^2b^2c^2$$ opr
$$\frac{(a+b+c)^3(a^2c+b^2a+c^2b)}{27}\geq\frac{2abc(ab+ac+bc)(a+b+c)}{3}-3a^2b^2c^2$$ or
$$\sum_{cyc}(a^5c+a^4b^2+3a^4c^2+3a^3b^3+3a^4bc-12a^3b^2c-11a^3c^2b+12a^2b^2c^2)\geq0,$$
which is true by Schur and AM-GM.
The Schur inequality is the following.
Let $x$, $y$ and $z$ be positive numbers.
Prove that:
$$\sum_{cyc}(x^3-x^2y-x^2z+xyz)\geq0.$$
Now, let $x=bc$, $y=ac$ and $z=ab$.
Thus, $$\sum_{cyc}(a^3b^3-a^3b^2c-a^3c^2b+a^2b^2c^2)\geq0$$ or
$$3\sum_{cyc}(a^3b^3-a^3b^2c-a^3c^2b+a^2b^2c^2)\geq0.$$
Also, by AM-GM $$\sum_{cyc}a^5c\geq\sum_{cyc}a^4bc$$,
$$\sum_{cyc}(a^4b^2+a^4c^2)\geq2\sum_{cyc}a^4bc.$$
Thus, by Schur again
$$6\sum_{cyc}(a^4bc-a^3b^2c-a^3c^2b+a^2b^2c^2)\geq0.$$
Hence, it remains to prove that
$$\sum_{cyc}(2a^4c^2-3a^3b^2c-2a^3c^2b+3a^2b^2c^2)\geq0,$$
which is obvious by the $uvw$ method.
Another way.
By C-S
$$\sum_{cyc}x^2y=\frac{1}{3}\sum_{cyc}\frac{1}{y}\sum_{cyc}x^2y\geq\frac{1}{3}(x+y+z)^2\geq2(x+y+z)-3,$$
where the last inequality it's just $$(x+y+z-3)^2\geq0.$$
Done!
