Proving $\frac{a}{b} >\frac{a+\epsilon}{b+\epsilon}$ if and only if $b<a$, for $\epsilon >0$, $a,b$ positive.

Question

The way I use to see that this is true is to take the derivative of the LHS w.r.t to $\epsilon$. This derivative is negative if $b<a$.

I am not sure how I can use this to prove the if and only if statement though, or even if this is a good approach.

Would it just be something like the following: For the forward direct -- $\frac{a}{b} > \frac{a+\epsilon}{b+\epsilon} \implies b<a$, -- would I just say that because the LHS of $\frac{a}{b} > \frac{a+\epsilon}{b+\epsilon}$ is the RHS when $\epsilon =0$, then this means that the RHS is decreasing in $\epsilon$.

The RHS being decreasing in $\epsilon$ then means that $\frac{d}{d\epsilon} \left [\frac{a+\epsilon}{b+\epsilon}\right ] <0$ which requires $b<a$?

Using the notation of derivative feels more high powered than necessary though.

So the question is what is a good method to prove the result in the question. And, if possible give some comment or example of how to prove one direction

Answer 1

$\begin{array}\\ \dfrac{a+c}{b+c}-\dfrac{a}{b} &=\dfrac{b(a+c)-a(b+c)}{b(b+c)}\\ &=\dfrac{c(b-a)}{b(b+c)}\\ \end{array} $

The sign of the difference depends on all the expressions in this final fraction.

Answer 2

For a= 3, b= -5, and epsilon= 6, the statement does not hold, so we simply cannot prove it. So I am assuming that a and b are positive too, unless you want to correct me and add some other information that you might have missed:Basically a proof for the "if" statement and the "only if" statement separately. Please correct me if I'm wrong.

Answer 3

Note all quantities $a, b, \epsilon, a+\epsilon, b+\epsilon$ are positive, so we can multiply/divide by denominators and preserve the inequality: $$\frac{a}{b} > \frac{a + \epsilon}{b+\epsilon} \iff a(b+\epsilon) > b(a+\epsilon) \iff ab + a \epsilon > ab + b\epsilon \iff a > b$$