Floor inequality: $\lfloor \frac{6a-1}{b}\rfloor+\lfloor\frac{a}{b}\rfloor\ge \lfloor \frac{2a}{b}\rfloor+\lfloor \frac{3a-1}{b}\rfloor+\cdots$

Question

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!

Answer 1

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 $7n+3$.