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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?

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    $\begingroup$ $\text{sup}(A) = +\infty$. $\endgroup$ – DeepSea Jan 8 '17 at 9:36
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    $\begingroup$ 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. $\endgroup$ – Lundborg Jan 8 '17 at 9:37
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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.

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