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

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.

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 -- $$\frac{a}{b} > \frac{a+\epsilon}{b+\epsilon} \implies b, -- would I just say that because the LHS of $$\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?

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

• You could consider $\frac{a}{b} - \frac{a+\epsilon }{b+\epsilon}$ (which you want to be $> 0$), and combine the fractions. – Minus One-Twelfth Apr 2 at 21:08
• @MinusOne-Twelfth I see, that works. Thank you. – user106860 Apr 2 at 21:13
• Are $a$ and $b$ positive? – Bernard Apr 2 at 21:20
• 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. – Surb Apr 2 at 21:24
• you do not need derivative for this kind of problem – qwr Apr 2 at 23:45

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