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 $t=x+\frac{1}{x}\geq2$

We have that for $x>0$
$$x+\frac{1}{x}\geq2 \iff x\cdot x+x\cdot \frac{1}{x}\geq x\cdot 2 \iff x^2-2x+1\ge 0 \iff (x-1)^2 \ge 0$$
and the equality holds if and only if $x=1$.
