$\epsilon < \frac{a}{b} < \frac{c}{d}$ implies that: $\epsilon \leq \frac{a+c}{b+d}$? Let $a,b,c,d \in \mathbb{N}$ and $\epsilon \in \mathbb{R}$  Let $\epsilon < \frac{a}{b} < \frac{c}{d}$ Does this imply that: $\epsilon \leq \frac{a+c}{b+d}$?
 A: Hint:
$$\epsilon\le \frac{a+c}{b+d}\iff b\epsilon+d\epsilon \le a+c$$
Because $\epsilon < \frac{a}{b}$ and $\epsilon< \frac{c}{d}$,... (you don't actually need that $\frac ab<\frac cd$ this way)
A: Suppose that $\dfrac{a}{b}\lt \dfrac{c}{d}$, where $b$ and $d$ are positive real numbers. Then
$$\frac{a}{b}\lt \frac{a+c}{b+d}\lt \frac{c}{d}.$$
We prove the half of the above result that you do not need, that $\dfrac{a+c}{b+d}\lt \dfrac{c}{d}$.
A natural approach is to consider the difference $\dfrac{c}{d}-\dfrac{a+c}{b+d}$, which simplifies to $\dfrac{bc-ad}{d(b+d)}$. The denominator is positive. And since $\dfrac{c}{d}-\dfrac{a}{b}\gt 0$, the numerator is positive. 
Remark: You might be interested in other properties of the mediant.
A: More generally, you can show that if $\frac a b < \frac c d$ then:$$\frac{a}{b}\leq\frac{a+c}{b+d}\leq \frac c d$$
A: More generally, 
if $\epsilon < a/b$ and
$\epsilon < c/d$,
where $a, b, c, d$ are positive reals,
then
$\epsilon < (r a+s c)/(r b + s d)$,
where $r$ and $s$ are any positive reals.
Proof:
$\epsilon < (r a+s c)/(r b + s d)$
$\iff$
$\epsilon(r b + s d) < r a+s c$
$\iff$
$r(a-\epsilon b) > s(d \epsilon - c)$
which is true because
 the left side is positive
and the right side is negative.
The reverse inequality is also true,
but $\epsilon$ likes being less than other values
and is uncomfortable when asked to be greater.
