# Let E = ${x + \frac{4}{x}: x > 0}$. Prove or disprove that E is a bounded set.

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!

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

Consider Positive Values for x alone for now,

AM-GM

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

$$f(x) \geq 4$$

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 $$x + \frac{4}{x} > B$$.

• 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. Commented Feb 25, 2020 at 3:03
• @fancyfawn28971: bounded means bounded above and below. If you only want to prove it is bounded below, you should say that. Commented Feb 25, 2020 at 3:39