I know that it is a bounded below set and the infimum is 4, but I'm unsure of going about how to prove that it is indeed bounded. Any help would be greatly appreciated!

  • $\begingroup$ Formatting tip: use $\{$ to get $\{$. What happens to $x+\dfrac4x$ as $x\to\infty$? $\endgroup$ Commented Feb 25, 2020 at 2:58
  • $\begingroup$ You think it's bounded... Hmm... What happens when $x\rightarrow +\infty$? Moreover, the infimum is 4. $\endgroup$ Commented Feb 25, 2020 at 2:58
  • $\begingroup$ Try plotting the function. Can you see an upper bound? $\endgroup$
    – user744868
    Commented Feb 25, 2020 at 2:59

2 Answers 2


Consider Positive Values for x alone for now,


$x + \frac 4x \geq 2 \sqrt {x \frac 4x}$

$f(x) \geq 4$

It is Easy to see that it is a monotonically increasing function beyond x = 2 by differentiating,

Thus it suffices to show that Limit at Infinity Is not Infinity, Easy to see that It is,

Noe generalize this to -ve side as it is an odd function. This gives you the Range $(-\infty,-2] \ U \ [2,\infty]$ and thus it's unbounded.


Hint #1: If you want to investigate intuitively whether or not it is bounded, try evaluating $x + \frac{4}{x}$ for values of $x$ that get larger and larger, such as $x=10$, $x=100$, $x=1000$, $x=100000$, and so on.

Hint #2: Perhaps, if someone tells you "This number $B>0$ is an upper bound", you can figure out rigorously whether or not they are telling the truth by attempting to solve the inequality $x + \frac{4}{x} > B$.

  • $\begingroup$ Thank you for the tips. I can indeed prove that it has no upper bound, but I must prove using the formal definition of a lower bound, (Let A be a subset of the reals, A not empty, we'll say L is a lower bound for A f for all a in A, L is less than or equal to a. $\endgroup$ Commented Feb 25, 2020 at 3:03
  • $\begingroup$ @fancyfawn28971: bounded means bounded above and below. If you only want to prove it is bounded below, you should say that. $\endgroup$ Commented Feb 25, 2020 at 3:39

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