Find $\inf A$ and $\sup A$ for $A=\{x+\frac{4}{x}: x>0\}$
My attempt:
$$x+\frac{4}{x}\geq 2\sqrt{x\cdot \frac{4}{x}}=4$$
$$\Rightarrow \inf A=4$$
Now I'm not sure about supremum. $A$ is not bounded from above so I'd say there doesn't exist $\sup A$ in $\Bbb R$ and I understand that $A$ tends to infinity as $x$ gets bigger and bigger but how do I prove this formally?
Proving that $$ x+\frac{4}{x}\ge4 $$ (by the way, your proof is fine) doesn't by itself show that $4$ is the infimum, but just that it is a lower bound.
However, if $x=2$…
For the supremum, you're on the right track: if $a>0$ were an upper bound, we'd have $$ a+\frac{4}{a}\le a $$ which is a contradiction. Thus $A$ is not upper bounded.