Evalution of a function where $t = x + \frac{1}{x}$

Question

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

Answer 1

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 $t\geq2$.

Answer 2

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 $x=1$.

Answer 3

hint: use AM GM inequality ,a+b/2 is greater than or equal to root(ab) for positive a,b