<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] 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.
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