Prove that $x^2 y+ y^2 z + z^2x ≥ 2(x + y + z) − 3$

Question

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.

Answer 1

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!

Answer 2

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!