I know intuitively that this is true, but whenever I try and break apart that intuition to see where it's coming from, I essentially end up re-writing the assumption I'm trying to prove.
I've tried playing around with the denter preformatted text hereefinition that we use for the ceiling function (given $x \in \mathbb{R}$, there exists a integer $n$ such that $n-1 < x \leq n < x+1$ ), but that's been fruitless so far, as well.
Found stuff on Wikipedia/Wolfram pertaining to sums and floor/ceiling functions, but nothing pertaining to products and floor/ceiling functions.
(answered)
For any real numbers x and y, we can let them be equal to the sum of integers k1, k2 and remaining decimal values d1, d2.
The proof will hinge on the fact that the decimal values d1, d2 will always fall in the interval [0,1).
You then look at the cases exhausting the possible combinations of values for d1 and d2, most importantly both = 0 or both > 0 (the boundary cases). In each of these cases, plug in (k1 + d1) for x and (k2 + d2) for y and expand LHS and RHS of your inequalities until they have reached the necessary form.
Bear in mind that in our integer + decimal form, ceiling(k + d) = k if d=0 and = k+1 if d>0.
Best Answer
It is obvious that ceiling of a real number is at least as much as that number, therefore, $Ceiling(x)Ceiling(y)\geq xy$. As $Ceiling(xy)$ is defined to be such minimum integer, we have $Ceiling(x)Ceiling(y)\geq Ceiling(xy)$.