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Intuitively, it seems that

$$ \min \frac{f(x)}{g(x)} \geq \frac{\min f(x)}{\max g(x)} $$

if both $f$ and $g$ are positive-valued functions, but is there a more general relationship? Can one also deduce a similar relationship with the $\min$ and $\max$ reversed?

Rahul
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  • I am almost sure there are no other relationships. – Jean Marie Oct 23 '23 at 07:43
  • @JeanMarie I see. So this sort of relationship only works for positive-valued functions? – Rahul Oct 23 '23 at 07:48
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    +1 to this posting, despite the posting not actually showing work. For one thing, although the posting is focusing on a purely mathematical question, in my opinion, this is not the sort of question for which it is reasonable to expect that work is shown. As a second point, this is actually a very relevant question when (for example) constructing an $~\epsilon,\delta~$ proof that $~\displaystyle \lim_{x \to a} \frac{f(x)}{g(x)} = L > 0 ~: ~0 < f(a),g(a),~$ where $~f(x),g(x)~$ are continuous functions. – user2661923 Oct 23 '23 at 09:18

1 Answers1

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You can make the l.h.s. arbitrarily large compared to the r.h.s., so you won't be able to improve that inequality.

For simplicity, we restrict ourselves to a domain consisting of only two elements: $x\in\{0, 1\}$. Define $$f(x) = \begin{cases}1,\quad\text{for } x=0,\\X,\quad\text{for } x=1,\\\end{cases},$$ and $$g(x) = \begin{cases}1/X,\quad\text{for } x=0,\\1,\quad\text{for } x=1,\\\end{cases},$$ where $X>1$ is some parameter. Then $f/g = X$ always, and can be made arbitrarily large by chosing $X$ large, but $$\frac{\min f}{\max g}= 1.$$

Comment: There is an analogous statement for the maximum: $$\max\frac{f(x)}{g(x)} \le \frac{\max f(x)}{\min g(x)},$$ provided that $f,g\ge 0$ and $\min g > 0$.

DominikS
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