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 $$\left({x \, \bmod \, 1}\right) + \left({y \, \bmod \, 1}\right) > 1$$.