# 2013 Canadian MO Problems/Problem 4

## Problem

Let be a positive integer. For any positive integer and positive real number , define where denotes the smallest integer greater than or equal to . Prove that for all positive real numbers .

## Solution

First thing to note on both functions is the following:

and

Thus, we are going to look at two cases:\. When , and when which is the same as when

**Case 1:**

Since in the sum, then

, and the equality holds.

Likewise,

Since is integer we have:

, and the equality holds.

Thus for we have equality as:

**Case:**

Since , then

Therefore,

Since ,

Since , then

Therefore,

, which together with the equality case of proves the left side of the equation:

Now we look at :

Since , then

Therefore,

Since ,

When is a whole number and is divisible by we notice the following:

Then is not divisible by then we add more ceiling terms to the expression. Likewise, when is not a whole number and , the sum is larger.

Therefore,

Hence,

which together with the case where , we have:

and together with we have:

~Tomas Diaz. orders@tomasdiaz.com

*Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.*