Difference between revisions of "2013 Canadian MO Problems/Problem 4"
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<math>g_j(r)=\begin{cases} n+\left\lceil \frac{j}{r} \right\rceil\; , & \left\lceil jr \right\rceil > n \\ | <math>g_j(r)=\begin{cases} n+\left\lceil \frac{j}{r} \right\rceil\; , & \left\lceil jr \right\rceil > n \\ | ||
\left\lceil jr \right\rceil+\left\lceil \frac{j}{r} \right\rceil\; , & \left\lceil jr \right\rceil \le n \end{cases}</math> | \left\lceil jr \right\rceil+\left\lceil \frac{j}{r} \right\rceil\; , & \left\lceil jr \right\rceil \le n \end{cases}</math> | ||
+ | |||
+ | <math>\sum_{j=1}^n g_j(r)=\sum_{j=1}^{\left\lceil \frac{n}{r} \right\rceil}\left\lceil jr \right\rceil+\sum_{j=\left\lceil \frac{n}{r} \right\rceil +1}^{n}n+\sum_{j=1}^{n}\left\lceil \frac{j}{r} \right\rceil</math> | ||
+ | |||
+ | <math>\sum_{j=1}^n g_j(r) \ge \sum_{j=1}^{\left\lceil \frac{n}{r} \right\rceil} jr +\sum_{j=\left\lceil \frac{n}{r} \right\rceil +1}^{n}n+\sum_{j=1}^{n}\left\lceil \frac{j}{r} \right\rceil</math> | ||
+ | |||
+ | <math>\sum_{j=1}^n g_j(r) > \frac{r}{2}\left( \left\lceil \frac{n}{r} \right\rceil \right)\left( \left\lceil \frac{n}{r} \right\rceil+1 \right)+n\left( n- \left\lceil \frac{n}{r} \right\rceil\right)+\sum_{j=1}^{n}\left\lceil \frac{j}{r} \right\rceil</math> | ||
+ | |||
+ | Since <math>\left\lceil \frac{n}{r} \right\rceil \ge \frac{n}{r}</math>, | ||
~Tomas Diaz. orders@tomasdiaz.com | ~Tomas Diaz. orders@tomasdiaz.com | ||
{{alternate solutions}} | {{alternate solutions}} |
Revision as of 17:34, 27 November 2023
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 ,
~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.