Evalution of a function where $t = x + \frac{1}{x}$ Consider a function $$y=(x^3+\frac{1}{x^3})-6(x^2+\frac{1}{x^2})+3(x+\frac{1}{x})$$ defined for real $x>0$. Letting $t=x+\frac{1}{x}$ gives: $$y=t^3-6t^2+12$$
Here it holds that $$t=x+\frac{1}{x}\geq2$$
My question is: how do  I know that $t=x+\frac{1}{x}\geq2$  ?
I want to know how to get to this point without previouly knowing that $t=x+\frac{1}{x}\geq2$

Because by AM-GM $$x+\frac{1}{x}\geq2\sqrt{x\cdot\frac{1}{x}}=2.$$
Your calculation of $y$ is right:
$$y=t^3-3t-6(t^2-2)+3t=t^3-6t^2+12$$ and you got it without using $t\geq2$.
