Difference between revisions of "1981 USAMO Problems/Problem 5"

(Solution)
(Solution)
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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> <math>\blacksquare</math>
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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>.
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<math>\blacksquare</math>
  
 
==See Also==
 
==See Also==

Revision as of 11:39, 7 July 2020

Problem

Show that for any positive real $x$, $[nx]\ge \sum_{1}^{n}\left(\frac{[kx]}{k}\right)$

Solution

We know that $x\geq\lfloor x \rfloor$. Also, $nx\geq\lfloor nx \rfloor$, so $x\geq\frac{\lfloor nx \rfloor}{n}$. Thus, each of the terms in the sum is $x\geq$, so the total sum is $nx\geq\sum_{1}^{n}\left(\frac{[kx]}{k}\right)$. $\blacksquare$

See Also

1981 USAMO (ProblemsResources)
Preceded by
Problem 4
Followed by
Last Question
1 2 3 4 5
All USAMO Problems and Solutions

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