Is $| \lceil \frac{a}{2} \rceil - \lceil \frac{b}{2} \rceil |\geq \lfloor |\frac{a - b}{2}| \rfloor $? Let $a$ and $b$ be integers. Is it true that
$$
 \left | \left \lceil \frac{a}{2} \right \rceil -  \left \lceil \frac{b}{2} \right \rceil \right |\geq  \left \lfloor \left | \frac{a - b}{2} \right |\right \rfloor
$$
Where $\lceil \cdot \rceil$ is the ceiling function, $\lfloor \cdot \rfloor$ the floor function and $|\cdot|$ is the absolute function.
The inequality seems to be true when I check it programatically but I would like to get a proof (or disproof) for this inequality.
 A: Yes, it is true.
$$
 \left | \left \lceil \frac{a}{2} \right \rceil -  \left \lceil \frac{b}{2} \right \rceil \right |\geq  \left \lfloor \left | \frac{a - b}{2} \right |\right \rfloor
\tag1$$
In the following, $m,n$ are integers.
Case 1 : If $a=2m,b=2n$, then both sides of $(1)$ equal $|m-n|$.
Case 2 : If $a=2m,b=2n+1$, then 
$$(1)\iff |m-n-1|\ge \left\lfloor\left |m-n-\frac 12\right|\right\rfloor\tag2$$
If $m-n-\frac 12\ge 0$, then $m-n-1\ge 0$, so$$(2)\iff m-n-1\ge m-n-1$$which is true.
If $m-n-\frac 12\lt 0$, then $m-n-1\lt 0$, so$$(2)\iff -m+n+1\ge -m+n$$which is true.
Case 3 : If $a=2m+1, b=2n$, then
$$(1)\iff |m-n+1|\ge \left\lfloor\left|m-n+\frac 12\right|\right\rfloor\tag3$$
If $m-n+\frac 12\ge 0$, then $m-n+1\ge 0$, so$$(3)\iff m-n+1\ge m-n$$which is true.
If $m-n+\frac 12\lt 0$, then $m-n+1\lt 0$, so$$(3)\iff -m+n-1\ge -m+n-1$$which is true.
Case 4 : If $a=2m+1,b=2n+1$, then both sides of $(1)$ equal $|m-n|$.
A: There is no need for the assumption that $a$ and $b$ are integers. You just need to prove that
$$|\lceil x\rceil-\lceil y\rceil|\ge\lfloor|x-y|\rfloor$$
for any real numbers $x$ and $y$. By symmetry, we may assume $x\ge y$, in which case we can remove the absolute value signs. If, moreover, we write $x=y+u$ with $u\ge0$, we are trying to prove
$$\lceil y+u\rceil\ge\lceil y\rceil+\lfloor u\rfloor$$
But $u=\lfloor u\rfloor+r$ for some $0\le r\lt1$, and $\lceil y+\lfloor u\rfloor +r\rceil=\lceil y+r\rceil+\lfloor u\rfloor$, so the inequality to prove is simply
$$\lceil y+r\rceil\ge\lceil y\rceil$$
which is clearly true, since the ceiling function is never decreasing and $r\ge0$.
A: Assume without loss of generality that $a\ge b$. Then the inequality is
$$
\left\lceil \frac a2 \right\rceil - \left\lceil \frac b2 \right\rceil
\ge \left\lfloor \frac {a-b}2 \right\rfloor
$$ 
If either $a$ or $b$ is an even integer, then we can pull the whole number $\frac a2$ or $\frac b2$ out of the floor function, and the inequality reduces to $\left\lceil \frac a2 \right\rceil \ge \left\lfloor \frac {a}2 \right\rfloor$ or $-\left\lceil \frac b2 \right\rceil \ge \left\lfloor -\frac {b}2 \right\rfloor$ (where the first is trivial and the second is actually an equality). 
Assume therefore that neither of $a$ and $b$ is an even integer. Let $2m<a<2(m+1)$ and $2n<b<2(n+1)$, for some $m,n\in \mathbb Z$. Then
$$
\left\lceil \frac a2 \right\rceil - \left\lceil \frac b2 \right\rceil
= (m+1)-(n+1)
= m-n
$$
On the other hand
$$
m-n-1<\frac a2 - \frac b2 < m-n+1
$$
which means that 
$$
\left\lfloor \frac {a-b}2 \right\rfloor
\le m-n
= \left\lceil \frac a2 \right\rceil - \left\lceil \frac b2 \right\rceil
$$
so we are done.
EDIT: I didn't notice you assumed $a$ and $b$ to be integers. Well, my answer works for all real numbers.
