Reasoning about numbers close to two other numbers $a,b$ (inequalities) Let $a < b$ and $0 < \varepsilon < (b - a)$ and let $x, y \in \mathbb R$ be such that
$$
 | x - a | < \frac{(b - a) - \varepsilon}{2}, \qquad
 | y - b | < \frac{(b - a) - \varepsilon}{2}
$$
then $|x - y| > \varepsilon$. How to prove this?
 A: Your statement (as written at first) is not true. Let 
$$a=0,b=1,\varepsilon=\frac 14,x=y=\frac 12$$
Then 
$$0<1$$
$$0<\frac 14<(1-0)$$
$$\left|\frac 12-0\right|<(1-0)-\frac 14$$
$$\left|\frac 12-1\right|<(1-0)-\frac 14$$
But
$$\left|\frac 12-\frac 12\right| \not\gt \frac 14$$
A: These kind of inequalities are the alphabet of "real-analysis" and most students/beginners think that these require "reasonable amount of manipulation of basic rules of inequalities" to establish. However such routine inequalities are almost obvious if we visualize them on the real number line. Once a student sees it visually, he is convinced of the trivialities of the such results and does not need to bother with a formal proof based on "manipulation of basic rules of inequalities". BTW another strange difficulty with such results is the use of greek letter $\epsilon$. Had we used letter $c$ in place of $\epsilon$, it would have been less intimidating.
Let's try to visualize the current question on real line.
-------------------$a$---$x$--------$c$---$c'$----$d$----$y$---$b$----------------------
The number $(b - a)$ represents distance between $a$ and $b$ and we want $\epsilon$ less than this distance and hence the number $c = a + \epsilon$ must lie between $a$ and $b$. Also note that distance between $a$ and $c$ is $\epsilon$ and hence the distance between $b$ and $c$ is $(b - a) - \epsilon$. If $d$ lies exactly half-way between $b$ and $c$ then clearly distance of $d$ from both $b$ and $c$ is $p = ((b - a) - \epsilon)/2$. Now we want $|x - a| < p, |y - b| < p$ so that $x$ is near $a$, $y$ is near $b$ and distance of $x$ from $a$ (and that of $y$ from $b$) is less than $p$ i.e. it is less than the distance between $d$ and $b$.
We need to show that $|x - y| > \epsilon$. Now note that since $x$ is near $a$ and $y$ is near $b$ and hence the value $|x - y|$ (which represents distance of $x$ from $y$) will be minimum when $x$ is to the right of $a$ and $y$ to the left of $b$ so that both $x$ and $y$ lie between $a$ and $b$. If that is the case then $|y - b| < p$ implies that $y$ lies somewhere between $d$ and $b$. We have kept $x$ to the right of $a$.
Now we need to shift $c$ to right by the amount equal to distance between $x$ and $a$ (which is less than $p$). This new position of $c$ we call $c'$. This will ensure that distance between $c'$ and $x$ is same as distance between $c$ and $a$ and hence is equal to $\epsilon$. Also note that distance between $c$ and $c'$ is same as distance between $x$ and $a$ and hence it is less than $p$. Thus $c'$ remains somewhere between $c$ and $d$. The conclusion is now obvious. The distance between $x$ and $y$ is obviously greater than that between $x$ and $c'$. Note that this conclusion is based on the fact that $y$ cannot come to the left of $d$ and $c'$ can not go to the right of $d$ and hence distance $|x - y|$ remains greater than $|c' - x| = \epsilon$.
A: After two applications of the reverse triangle inequality you get:
$$|x - y| = |(x - a) - (y - b) - (b - a)| \geq (b - a) - |x - a| - |y - b|.$$
Using your assumption on $|x - a|$ and $|x - b|$, you get:
$$|x - a| + |y - b| < (b - a) - \epsilon.$$
So,
$$|x - y| > (b - a) - (b - a) + \epsilon.$$
