# Find $\inf A$ and $\sup A$ for $A=\{x+\frac{4}{x}: x>0\}$

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?

• $\text{sup}(A) = +\infty$. – DeepSea Jan 8 '17 at 9:36
• Proving that the supremum is $\infty$ could be done by contradiction i.e assuming that a supremum less than $\infty$ exists and showing that it leads to a contradiction. – Lundborg Jan 8 '17 at 9:37

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.