Reducing an inquality with fractions

Question

can you help me reduce the following inequality (i need to get a relation between x and y -- express x in terms of y)

$\frac{n}{2x} < \frac{n}{(4+\epsilon)y}+1$

I would like to show somehow that $x > (1+\epsilon) y$ OR that $x>2y$

The assumption is that $0 < \epsilon < 1$ and that $n>4x, n>4y$, x and y are positive integers.

I did not know what to do with the 1, so I simply said that if the numerator is the same and the first fraction is less, then we have: 2x > (4+\epsilon)y, thus x > (2+\epsilon/2)y — user118462, Apr 28 '14 at 17:35

score 0 · Answer 1 · answered Apr 28 '14 at 17:53

0

You already have what you want. For $x>0$, $y>0$, $\epsilon>0$, and $n>0$, you have: $$\dfrac{(4+\epsilon)}{n}\dfrac{n-2x}{2x}<y.$$ Hence:

$$y>\dfrac{a}{x}+b.$$ Where $a=\dfrac{4+\epsilon}{2}$ and $b=-\dfrac{4+\epsilon}{n}$

answered Apr 28 '14 at 17:53

npisinp

Well, what I really want is to use this to show that x has to be at least double y or alternatively, at least (1+ \epsilon)y. – user118462 Apr 28 '14 at 18:05
you want to show that $x>(1+\epsilon)y$? – npisinp Apr 28 '14 at 18:08
yes, that would be great. – user118462 Apr 28 '14 at 18:17

1 Answers1