Difference between revisions of "1981 USAMO Problems/Problem 5"
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==Solution== | ==Solution== | ||
| − | We know that <math>x\geq\lfloor x \rfloor</math>. Also, <math>nx\geq\lfloor nx \rfloor</math>, so <math>x\geq\frac{\lfloor nx \rfloor}{n}</math>. Thus, each of the terms in the sum is <math>x\geq</math>, so the total sum is <math>nx\geq\sum_{1}^{n}\left(\frac{[kx]}{k}\right)</math> | + | We know that <math>x\geq\lfloor x \rfloor</math>. Also, <math>nx\geq\lfloor nx \rfloor</math>, so <math>x\geq\frac{\lfloor nx \rfloor}{n}</math>. Thus, each of the terms in the sum is <math>x\geq</math>, so the total sum is <math>nx\geq\sum_{1}^{n}\left(\frac{[kx]}{k}\right)</math>. |
| + | <math>\blacksquare</math> | ||
==See Also== | ==See Also== | ||
Revision as of 10:39, 7 July 2020
Problem
Show that for any positive real 
, ![$[nx]\ge \sum_{1}^{n}\left(\frac{[kx]}{k}\right)$](http://latex.artofproblemsolving.com/8/9/0/890ce23466abfd1bcb2eb253ac571925c0362163.png)
Solution
We know that 
. Also, 
, so 
. Thus, each of the terms in the sum is 
, so the total sum is ![$nx\geq\sum_{1}^{n}\left(\frac{[kx]}{k}\right)$](http://latex.artofproblemsolving.com/b/3/b/b3bd1e3cb9c457742d6efd965ae2841120ade827.png)
.
See Also
| 1981 USAMO (Problems • Resources) | ||
| Preceded by Problem 4 |
Followed by Last Question | |
| 1 • 2 • 3 • 4 • 5 | ||
| All USAMO Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
