Is there any relation between $\lfloor n/k\rfloor$ and $\lfloor n/(k+1)\rfloor$? Given two integers $n$ and $k$ such that $n\geq k+1$. 
Can we find any relation between $\left\lfloor\dfrac{n}{k}\right\rfloor$ and $\left\lfloor \dfrac{n}{k+1}\right\rfloor$?
At first, I thought that $\left\lfloor \dfrac{n}{k+1}\right\rfloor=\left\lfloor\dfrac{n}{k}\right\rfloor-1,$ but then I found that, for $k=1$, $$\left\lfloor\dfrac{n}{2}\right\rfloor\neq n-1$$. So, I guess that $$\left\lfloor \dfrac{n}{k+1}\right\rfloor\leq\left\lfloor\dfrac{n}{k}\right\rfloor-1.$$ How can we prove this? Is this the best bound?
 A: The floor function satisfies

  
*
  
*$\lfloor x\rfloor\leq x$
  
*$\lfloor\lfloor x\rfloor\rfloor=\lfloor x\rfloor$
  
*From those two one can easily prove that
  $$
\lfloor x\rfloor +\lfloor y\rfloor\leq \lfloor x+y\rfloor
$$
  
*If we subtract $\lfloor y\rfloor$ from both sides and consider $x=a$ and $y=b-a$ we then get
  $$
\lfloor a\rfloor \leq \lfloor b\rfloor-\lfloor b-a\rfloor
$$
  

To apply the last rule to your question, we must consider the difference between the two fractions inside the floor functions you are asking about:
$$
\begin{align}
a &= \frac n{k+1}\\
b &= \frac nk\\
b-a &= \frac nk-\frac n{k+1}\\
&=\frac n{k(k+1)}
\end{align}
$$
so applying the rule $\lfloor a\rfloor \leq \lfloor b\rfloor-\lfloor b-a\rfloor$ this gets us
$$
\left\lfloor\frac n{k+1}\right\rfloor\leq \left\lfloor\frac n{k}\right\rfloor-m
$$
where
$$
m=\left\lfloor\frac n{k(k+1)}\right\rfloor
$$
depends entirely on the size of $n$ relative to $k(k+1)$. If $n<k(k+1)$ we even have $m=0$ so the only constant you may subtract is in fact just zero, not $1$. So unfortunately we only have the very unsurprising
$$
\left\lfloor\frac n{k+1}\right\rfloor\leq \left\lfloor\frac n{k}\right\rfloor
$$
which appears to be not helpful at all.
A: First let's consider that
$$
\eqalign{
  & \left\lfloor {x + y} \right\rfloor  =   \cr 
  &  = \left\lfloor x \right\rfloor  + \left\lfloor y \right\rfloor  + \left\lfloor {\left\{ x \right\} + \left\{ y \right\}} \right\rfloor  =   \cr 
  &  = \left\lfloor x \right\rfloor  + \left\lfloor y \right\rfloor  + \left[ {1 - \left\{ x \right\} \le \left\{ y \right\}} \right] \cr} 
$$
where the curly brackets denotes the fractional part and where
in the last line $[P]$ denotes the Iverson bracket.
Then we have
$$
\eqalign{
  & \left\lfloor {{n \over k}} \right\rfloor  = \left\lfloor {{n \over {k + 1}}{{k + 1} \over k}} \right\rfloor
  = \left\lfloor {{n \over {k + 1}} + {n \over {k\left( {k + 1} \right)}}} \right\rfloor  =   \cr 
  &  = \left\lfloor {{n \over {k + 1}}} \right\rfloor  + \left\lfloor {{n \over {k\left( {k + 1} \right)}}} \right\rfloor  
+ \left[ {1 - \left\{ {{n \over {k + 1}}} \right\} \le \left\{ {{n \over {k\left( {k + 1} \right)}}} \right\}} \right] \cr} 
$$
which tells much about the difference between the two floors.
