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 $$\frac{2.5}{1.5}=\frac{5}{3}<2$$
To give you a feel for the geometry involved, consider $\displaystyle y={p\over q}$ and $\displaystyle z={p\over q}-{p+c\over q+c}={c(p-q)\over q(q+c)}$ where $0<c<1$. Without loss of generality, take $p=1$. In that case, the surface below on the left is $z(q,c)$ and on the right is the region in the $q$-$c$ plane where $z(q,c)>0$.

![enter image description here](https://i.stack.imgur.com/FOnj3.png)

One can quickly see (graphically and analytically) the role that $0<q<1$ plays for any $c$ as well as the role of $c<-q$ for $q<0$. That is, assuming $c>0$, you need precisely two of these to hold:
$$
1-q>0, \quad q>0, \quad q+c>0.
$$

A quick analysis shows that the first two holding while the third fails is impossible. The first and third holding while the second fails leads to $-c<q<0$. The second and third holding while the first fails leads to $0<q<1$. 

Thus, $z(q,c)>0$ for $c>0$ when 
$$
\{-c<q<0\}\cup \{0<q<1\},
$$
just as the figure on the right shows.

PS I just realized you wanted to know when the new fraction was *larger*, not *smaller*. However, this can be easily deduced from the answer I gave (when the new fraction is smaller).