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If $a$ and $b$ are positive integers and $a\ge b$ and $b$ is an odd prime, 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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Let $a=nb+k$, with $0\le k < b$. Your inequality simplifies to

$$\left\lfloor\frac{6k-1}{b}\right\rfloor\ge\left\lfloor\frac{2k}{b}\right\rfloor+\left\lfloor\frac{3k-1}{b}\right\rfloor+\left\lfloor\frac{2k+1}{b}\right\rfloor$$

Now let's examine it case by case.

Case 1. $k = 0:$ Both sides of the inequality are $-1$, so it holds.
Case 2. $0< k < \dfrac{b+1}3:$ The left side is at least $0$ and the right side is $0$.
Case 3. $\dfrac{b+1}3\le k <\dfrac b2:$ The left side is at least $2$ and the right side is $1$.
Case 4. $\dfrac b2\le k<\dfrac{2b+1}3:$ The left side is at least $3$ and the right side is at most $3$.
Case 5. $\dfrac{2b+1}3\le k < b:$ The left side is at least $4$ and the right side is at most $4$.

In case 4, note that since $b$ is odd and $k$ is an integer, $\dfrac b2\le k$ implies that $\dfrac{b+1}2\le k$.

As you can see, the inequality holds in every case. I also only used the fact that $b$ is odd (and greater than $1$), not that it is prime, so I believe it holds for all odd $b>1$ and $a\ge b$.

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  • $\begingroup$ Please let me know if you need any clarification or if I've made any mistakes. $\endgroup$ – Regret Apr 29 '15 at 8:16

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