<s>[INST] Given the $ x+y+z =3$ Prove that $ x^2+y^2+z^2 \geq 3xyz$ Given the $ x+y+z =3$ and x, y and z are positive numbers. How to prove that $ x^2+y^2+z^2 \geq 3xyz$. 
I tried many methods but I failed. 
I did the AM-HM in-equality, but failed.
$\frac{\frac{x}{yz} + \frac{y}{zx} + \frac{z}{xy}}{3} \geq \frac{3}{\frac{xy}{z}+\frac{yz}{x}+\frac{zx}{y}} $.
The denominator of RHS of the equation must have, minimum value. So it boils down to 
$\frac{x^2y^2+y^2z^2+z^2x^2}{xyz} \leq 3 $. Don't know how to proceed after that.
Taking partial derivative in range $(0,3)$ for the function $(x^2+y^2+z^2)$ also didn't give me the required insight. 
Thankyou.
 [/INST] Using AM-GM and Cauchy-Schwarz inequalities, For $x,y,z > 0 \Rightarrow 3 = x+y+z \geq 3\sqrt[3]{xyz} \Rightarrow 1 = 1^3 \geq xyz \Rightarrow 3^2 = (x+y+z)^2 \leq 3(x^2+y^2+z^2) \Rightarrow 3xyz \leq 3 \leq x^2+y^2+z^2 \Rightarrow 3xyz \leq x^2+y^2+z^2$
</s>