1981 USAMO Problems/Problem 5

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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
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All USAMO Problems and Solutions

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