Sum of Ceilings not less than Ceiling of Sum

Theorem
Let $\left \lceil {x} \right \rceil$ be the ceiling function.

Then:
 * $\left \lceil {x} \right \rceil + \left \lceil {y} \right \rceil \ge \left \lceil {x + y} \right \rceil$

The equality holds:
 * $\left \lceil {x} \right \rceil + \left \lceil {y} \right \rceil = \left \lceil {x + y} \right \rceil$

iff either:
 * $x \in \Z$ or $y \in \Z$

or:
 * $x \,\bmod\, 1 + y \,\bmod\, 1 > 1$

where $x \,\bmod\, 1$ denotes the modulo operation.

Proof
From the definition of the modulo operation, we have that:
 * $x = \left \lfloor {x}\right \rfloor + \left({x \, \bmod \, 1}\right)$

from which we obtain:
 * $x = \left \lceil {x}\right \rceil - \left[{x \notin \Z}\right] + \left({x \, \bmod \, 1}\right)$

where $\left[{x \notin \Z}\right]$ uses Iverson's convention.

It is clear that $x \notin \Z \implies x \, \bmod \, 1$.

As $0 \le x \, \bmod \, 1 < 1$ it follows that $\left[{x \notin \Z}\right] \ge x \, \bmod \, 1$.

Hence the inequality.

The equality holds iff:
 * $\left \lceil {\left({x \, \bmod \, 1}\right) + \left({y \, \bmod \, 1}\right)} \right \rceil = \left[{x \notin \Z}\right] + \left[{y \notin \Z}\right]$

that is, iff one of the following holds:
 * $x \in \Z$, in which case $x \, \bmod \, 1 = 0$
 * $y \in \Z$, in which case $y \, \bmod \, 1 = 0$
 * both $x, y \in \Z$, in which case $\left({x \, \bmod \, 1}\right) + \left({y \, \bmod \, 1}\right) = 0$
 * both $x, y \notin \Z$ and $\left({x \, \bmod \, 1}\right) + \left({y \, \bmod \, 1}\right) > 1$.