2013 Canadian MO Problems/Problem 4

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Problem

Let $n$ be a positive integer. For any positive integer $j$ and positive real number $r$, define \[f_j(r) =\min (jr, n)+\min\left(\frac{j}{r}, n\right),\text{ and }g_j(r) =\min (\lceil jr\rceil, n)+\min\left(\left\lceil\frac{j}{r}\right\rceil, n\right),\] where $\lceil x\rceil$ denotes the smallest integer greater than or equal to $x$. Prove that \[\sum_{j=1}^n f_j(r)\leq n^2+n\leq \sum_{j=1}^n g_j(r)\] for all positive real numbers $r$.

Solution

First thing to note on both functions is the following:

$f_j\left(\frac{1}{r}\right) =\min\left(\frac{j}{r}, n\right)+\min (jr, n)=f_j(r)$

and $g_j \left( \frac{1}{r} \right) =\min\left(\left\lceil\frac{j}{r}\right\rceil, n\right)+\min (\lceil jr\rceil, n)=g_j(r)$


Case 1: $r=1$

Since $j \le n$ in the sum, the f_j(r) =\min (jr, n)+\min\left(\frac{j}{r}, n\right)

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