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