Proof of $\lceil x \rceil + \lceil y \rceil \geq \lceil x + y \rceil$

Question

I have been trying to prove

$$ \lceil x \rceil + \lceil y \rceil \geq \lceil x + y \rceil $$

by using

$$ x \leq \lceil x \rceil < x+1,\\ y \leq \lceil y \rceil < y+1,\\ x + y \leq \lceil x+y \rceil < x + y + 1,\\ x + y \leq \lceil x \rceil + \lceil y \rceil < x + y + 2 $$

where $\lceil x \rceil, \lceil y \rceil, \lceil x + y \rceil$, and $\lceil x \rceil + \lceil y \rceil$ are unique integers satisfying their respective conditions. I have been playing around with these inequalities, but have not made much progress towards a complete proof.

Any help is greatly appreciated.

Answer 1

Adding $\lceil x \rceil \ge x$ to $\lceil y \rceil \ge y$ we get $$\lceil x \rceil + \lceil y \rceil \ge x + y.$$ This shows that the LHS is an integer greater than $x + y$. Since by the definition $ \lceil x + y\rceil$ is the smallest integer greater than $x + y$, then necessarily $$ \lceil x \rceil + \lceil y \rceil \ge \lceil x + y\rceil.$$