Question: If I add a constant $\varepsilon < 1$ to the numerator and denominator of a fraction, is the new fraction always greater than the original?
That is, do I have 
$$
\frac{a}{b} \leq \frac{a+\varepsilon}{b+\varepsilon},\ \forall a,b\in\mathbb{R}
$$

Answer: 
With calculus, define the function $$f(a,b)=\frac{a}{b}$$ Then the change resulting from the addition of a small $\epsilon$ will be approximately $$df\approx f\_ada+f\_bdb={1\over b}da-{a\over b^2}db={\epsilon\over b^2}(b-a)$$ This change will be positive if $\epsilon$ and $b-a$ have the same sign.

So the answer is no. For example, take $\epsilon=0.5$ and $a=2,b=1$. Then the original fraction is $2$ and the new one is $$\frac{2.5}{1.5}=\frac{5}{3}<2$$
By high school long division (only one step is needed) we get $\frac{a}{b}$ with a remainder of $\varepsilon - \left(\frac{a}{b}\right)\varepsilon,$ which gives
$$\frac{a+\varepsilon}{b+\varepsilon} \;\; = \;\; \frac{a}{b} \; + \; \frac{\varepsilon \left(1 - \frac{a}{b}\right)}{b + \varepsilon}$$
Now it simply becomes a matter of whether $\frac{\varepsilon \left(1 - \frac{a}{b}\right)}{b + \varepsilon}$ is positive for the values you're interested in. Equivalently, look at the sign of $\frac{\varepsilon}{b}(b + \varepsilon)(b-a)$ or at the sign of $\varepsilon b(b + \varepsilon)(b-a).$