1981 USAMO Problems/Problem 5

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 $nx\geq\sum_{1}^{n}\left(\frac{[kx]}{k}\right)$ . $\blacksquare$

1981 USAMO (Problems • Resources)
Preceded by Problem 4	Followed by Last Question
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1981 USAMO Problems/Problem 5

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