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If $a$ and $b$ are positive integers and $a\ge b$, show that: $$\left\lfloor \frac{6a-1}{b}\right\rfloor+\left\lfloor\frac{a}{b}\right\rfloor\ge \left\lfloor \frac{2a}{b}\right\rfloor+\left\lfloor \frac{3a-1}{b}\right\rfloor+\left\lfloor \frac{2a+1}{b}\right\rfloor$$

I tested some numbers and it looks to be true but I haven't been able to prove it.

Please help, thanks!

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No, it's not true.

For $(a,b)=(2n+1,2)$ where $n\ge 1\in\mathbb N$, the LHS equals $7n+2$ and the RHS equals $7n+3$.

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  • $\begingroup$ Hello, sorry I edited. $\endgroup$ – user45220 Apr 20 '15 at 17:29
  • $\begingroup$ @user45220: I edited my answer. It's not true. $\endgroup$ – mathlove Apr 20 '15 at 17:34
  • $\begingroup$ Thanks, would it work if $b$ is an odd prime? $\endgroup$ – user45220 Apr 20 '15 at 17:36

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