Let $x,y,z$ be positive real numbers such that $x^2+y^2+z^2=xyz$. Prove the following $xyz\geq27,xy+yz+zx\geq27,x+y+z\geq9$

Question

Let $x,y,z$ be positive real numbers such that $x^2+y^2+z^2=xyz$.
Prove $$xyz\geq27,\\xy+yz+zx\geq27,\\x+y+z\geq9.$$

Here's what I tried:
$$\begin{align} \frac{x^3+y^3+z^3}{3}&\ge\sqrt[3]{x^3y^3z^3}\\ x^3+y^3+z^3&\ge3\sqrt[3]{x^3y^3z^3}\\ x^3+y^3+z^3&\ge3xyz \\ x^3+y^3+z^3&\ge 3(x^2+y^2+z^2)\end{align} $$

Not sure where to go from here, I think we have to use the arithmetic and geometrical means and their relation. ($A\ge G$)

Answer 1

By the AM-GM inequality, $$x^2 + y^2 + z^2 \ge 3 (x^2y^2z^2)^{1/3}$$ but $x^2 + y^2 + z^2 = xyz$, so $$xyz\ge 3(xyz)^{2/3} \implies (xyz)^{1/3} \ge 3 \implies xyz\ge 27$$ By two more applications of the AM-GM inequality and $xyz\ge 27$, you can show that $$x+y+z \ge 3(xyz)^{1/3} \ge 9$$ and $$xy + yz + zx \ge 3(x^2y^2z^2)^{1/3} \ge 3\cdot 9 = 27$$ as required.

Answer 2

(1)

$xyz=x^2+y^2+z^2 \ge 3\sqrt[3]{x^2y^2z^2}$

So $\sqrt[3]{xyz}\ge3$

$xyz\ge27$

(2)

$xy+yz+zx\ge3\sqrt[3]{x^2y^2z^2}\ge3\sqrt[3]{27^2}=27$

(3)

$x^2+y^2\ge2xy$

$x^2+z^2\ge2xz$

$y^2+z^2\ge2yz$

add and divided by 2

$\Rightarrow x^2+y^2+z^2\ge xy+yz+zx$

$\Rightarrow (x+y+z)^2=x^2+y^2+z^2+2xy+2xz+2yz\ge3(xy+yz+zx)\ge3*27$

use (2)

So $x+y+z\ge9$