# Proving inequality with multiple gaussian brackets

I'm having trouble proving some inequality. I checked multiple values for $$b,h$$ with Excel and it seems to be true. Do you have some idea on how to prove it?

Let $$b,h$$ be natural numbers with:

• $$b\geq h\geq 6$$
• $$n=hb$$

$$\bigg\lfloor\frac{n}{3}\bigg\rfloor+\bigg\lceil\frac{n}{3}\bigg\rceil-\bigg\lceil\frac{n}{2}\bigg\rceil\leq\bigg\lfloor\frac{n}{4}-\frac{h}{2}\bigg\rfloor$$

Let $$a,b$$ be integers with $$a,b\ge 6$$, and let $$n=ab$$.

Claim: $$\bigg\lfloor\frac{n}{3}\bigg\rfloor + \bigg\lceil\frac{n}{3}\bigg\rceil - \bigg\lceil\frac{n}{2}\bigg\rceil \le \bigg\lfloor\frac{n}{4}-\frac{a}{2}\bigg\rfloor$$ Proof:

Defining $$P,Q$$ by \begin{align*} P&= \bigg\lfloor\frac{n}{3}\bigg\rfloor + \bigg\lceil\frac{n}{3}\bigg\rceil - \bigg\lceil\frac{n}{2}\bigg\rceil \\[4pt] Q&= \bigg\lfloor\frac{n}{4}-\frac{a}{2}\bigg\rfloor \\[4pt] \end{align*} our goal is to show that$$\;P\le Q$$.

Since $$a,b\ge 6$$, we can write \begin{align*} a&=\;6+x\\[4pt] b&=\;6+y\\[4pt] \end{align*} where $$x,y$$ are nonnegative integers.

Considering $$xy$$,$$\;\text{mod}\;12$$, we can write $$xy\,=12k+r$$ where $$k$$ is a nonnegative integer and $$r\in\{0,...,11\}$$.

Expanding $$n$$, we get \begin{align*} n&=\;\, ab \\[4pt] &=\; (6+x)(6+y) \\[4pt] &=\; 36+6x+6y+xy \\[4pt] &=\; 36+6x+6y+12k+r \\[4pt] \end{align*} so for $$P$$ we get \begin{align*} P\;\;\;\;{\mathbf{=}}&\;\;\;\;\; \bigg\lfloor\frac{n}{3}\bigg\rfloor + \bigg\lceil\frac{n}{3}\bigg\rceil - \bigg\lceil\frac{n}{2}\bigg\rceil \\[8pt] {\mathbf{=}}&\;\;\;\;\left(12+2x+2y+4k+\bigg\lfloor\frac{r}{3}\bigg\rfloor\right)\\[2pt] &{\mathbf{+}}\left(12+2x+2y+4k+\bigg\lceil\frac{r}{3}\bigg\rceil\right)\\[2pt] &{\mathbf{-}}\left(18+3x+3y+6k+\bigg\lceil\frac{r}{2}\bigg\rceil\right)\\[8pt] {\mathbf{=}}&\;\;\;\;\;\;\;6+x+y+2k +\bigg\lfloor\frac{r}{3}\bigg\rfloor +\bigg\lceil\frac{r}{3}\bigg\rceil -\bigg\lceil\frac{r}{2}\bigg\rceil\\[4pt] \end{align*} and for $$Q$$ we get \begin{align*} Q\;\;\;\;{\mathbf{=}}&\;\;\;\;\; \bigg\lfloor\frac{n}{4}-\frac{a}{2}\bigg\rfloor \\[4pt] {\mathbf{=}}&\;\;\;\;\; \bigg\lfloor\frac{36+6x+6y+12k+r}{4}-\frac{6+x}{2}\bigg\rfloor \\[4pt] {\mathbf{=}}&\;\;\;\;\; \bigg\lfloor 6+x+{\small{\frac{3}{2}}}y+3k+\frac{r}{4} \bigg\rfloor \\[4pt] {\mathbf{\large{\ge}}}&\;\;\;\;\; 6+x+y+3k+\bigg\lfloor\frac{r}{4}\bigg\rfloor \\[4pt] \end{align*} hence we get \begin{align*} Q-P\;\;\;\;{\mathbf{\large{\ge}}} &\;\;\;\;\;\left(6+x+y+3k+\bigg\lfloor\frac{r}{4}\bigg\rfloor\right)\\[2pt] &{\mathbf{-}}\left(6+x+y+2k +\bigg\lfloor\frac{r}{3}\bigg\rfloor +\bigg\lceil\frac{r}{3}\bigg\rceil -\bigg\lceil\frac{r}{2}\bigg\rceil \right)\\[6pt] {\mathbf{=}} &\;\;\;\;\;\;\;k+\bigg\lfloor\frac{r}{4}\bigg\rfloor -\bigg\lfloor\frac{r}{3}\bigg\rfloor -\bigg\lceil\frac{r}{3}\bigg\rceil +\bigg\lceil\frac{r}{2}\bigg\rceil \\[6pt] {\mathbf{\large{\ge}}} &\;\;\;\;\;\;\bigg\lfloor\frac{r}{4}\bigg\rfloor -\bigg\lfloor\frac{r}{3}\bigg\rfloor -\bigg\lceil\frac{r}{3}\bigg\rceil +\bigg\lceil\frac{r}{2}\bigg\rceil \\[4pt] {\mathbf{\large{\ge}}} &\;\;\;\;\;\;\; 0 \\[4pt] \end{align*} by direct evaluation for $$r=0,...,11$$.

Thus we have$$\;Q-P\ge 0$$, so $$P\le Q$$, as was to be shown.