Quick floor function This isn't true, right?
$$k\left\lfloor\frac n {2k}\right\rfloor\leq \left\lfloor\frac n k\right\rfloor$$
$2<k\leq \left\lfloor\dfrac {n-1} 2\right\rfloor$, $n>4$, $k,n$ are coprime.

Let $n=15$ and $k=4$. Not that $15>4$, $2<4<\left\lfloor\dfrac{15-1}{2}\right\rfloor=7$, and that $gcd(4,15)=1$
Now
$$k\left\lfloor\dfrac {2n} k\right\rfloor=4\left\lfloor\dfrac {30} 4\right\rfloor=28$$
and
$$\left\lfloor\dfrac n k\right\rfloor=\left\lfloor\dfrac {15}4\right\rfloor=3$$ 
Clearly $28>3$, so this provides a counterexample.
