Difference between revisions of "2013 Canadian MO Problems/Problem 4"
| Line 12: | Line 12: | ||
<math>f_j(1/r) =\min (\frac{j}{r}, n)+\min\left(jr, n\right)=f_j(1/r)</math>, | <math>f_j(1/r) =\min (\frac{j}{r}, n)+\min\left(jr, n\right)=f_j(1/r)</math>, | ||
| − | and <math>g_j(1/r) =\min (\lceil frac{j}{r}\rceil, n)+\min\left(\left\lceil | + | and <math>g_j(1/r) =\min (\lceil \frac{j}{r}\rceil, n)+\min\left(\left\lceil jr\right\rceil, n\right) = g_j(r)</math> |
Revision as of 17:34, 27 November 2023
Problem
Let 
be a positive integer. For any positive integer 
and positive real number 
, define
![\[f_j(r) =\min (jr, n)+\min\left(\frac{j}{r}, n\right),\text{ and }g_j(r) =\min (\lceil jr\rceil, n)+\min\left(\left\lceil\frac{j}{r}\right\rceil, n\right),\]](http://latex.artofproblemsolving.com/5/9/8/59838efe579d7932f02643d88449c449d93f7d59.png)
where 
denotes the smallest integer greater than or equal to 
. Prove that
![\[\sum_{j=1}^n f_j(r)\leq n^2+n\leq \sum_{j=1}^n g_j(r)\]](http://latex.artofproblemsolving.com/6/9/3/69378627bf482631c8a7157b50157ae7a86d3b97.png)
for all positive real numbers .
Solution
First thing to note on both functions is the following:
,
and 
Case 1: 
Since 
in the sum, the
f_j(r) =\min (jr, n)+\min\left(\frac{j}{r}, n\right)
~Tomas Diaz. orders@tomasdiaz.com Template:Olution
