1
$\begingroup$

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

$\endgroup$
  • 1
    $\begingroup$ You could consider $\frac{a}{b} - \frac{a+\epsilon }{b+\epsilon}$ (which you want to be $ > 0$), and combine the fractions. $\endgroup$ – Minus One-Twelfth Apr 2 at 21:08
  • $\begingroup$ @MinusOne-Twelfth I see, that works. Thank you. $\endgroup$ – user106860 Apr 2 at 21:13
  • 1
    $\begingroup$ Are $a$ and $b$ positive? $\endgroup$ – Bernard Apr 2 at 21:20
  • 1
    $\begingroup$ Note that for all $a,b,c,d>0$ it holds $$\min\big\{\frac{a}{b},\frac{c}{d}\big\}\leq \frac{a+c}{b+d}\leq \max\big\{\frac{a}{b},\frac{c}{d}\big\}.$$ It doesn't bring you straight to the point, but is definitely related and might be useful in the future. $\endgroup$ – Surb Apr 2 at 21:24
  • $\begingroup$ you do not need derivative for this kind of problem $\endgroup$ – qwr Apr 2 at 23:45
1
$\begingroup$

$\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.

$\endgroup$
0
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.