Difference between revisions of "2013 Canadian MO Problems/Problem 4"
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and <math>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)</math> | and <math>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)</math> | ||
| + | Thus, we are going to look at two cases:\. When <math>r=1</math>, and when <math>r>1</math> which is the same as when <math>0<r<1</math> | ||
Case 1: <math>r=1</math> | Case 1: <math>r=1</math> | ||
| − | Since <math>j \le n</math> in the sum, | + | Since <math>j \le n</math> in the sum, then |
| − | f_j( | + | |
| + | <math>f_j(1) =\min (j, n)+\min (j, n)=2j</math> | ||
| + | |||
| + | $sum_{j=1}^n f_j(r)=2sum_{j=1}^n=n^2+n and the equality holds. | ||
| + | |||
~Tomas Diaz. orders@tomasdiaz.com | ~Tomas Diaz. orders@tomasdiaz.com | ||
{{alternate solutions}} | {{alternate solutions}} | ||
Revision as of 16:44, 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 
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
$sum_{j=1}^n f_j(r)=2sum_{j=1}^n=n^2+n and the equality holds.
~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.
