<s>[INST] Let $x,y,z$ be positive real numbers such that $x^2+y^2+z^2=xyz$. Prove the following $xyz\geq27,xy+yz+zx\geq27,x+y+z\geq9$ 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. ($A\ge G$)
 [/INST] (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 $x+y+z\ge9$
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