Find an inequality with more precise bound? In an other topic, i found that : $x-1\le\lfloor x\rfloor\le x$ which really easy to show.
However, i want to found an inequality with more precise bound.
Like i found that : $$\forall x\in \mathbb R\text{,  }x-1\le x-1+\frac{1}{\pi}\vert\sin{\pi x}\vert\le \lfloor x\rfloor\le x -\frac{1}{\pi}\vert\sin{\pi x}\vert\le x$$
So in the end, we have a more precise inequality :$$x-1+\frac{1}{\pi}\vert\sin{\pi x}\vert\le \lfloor x\rfloor\le x -\frac{1}{\pi}\vert\sin{\pi x}\vert$$
But now i just want to know what more precise inequality you could find out !
Just two things :

*

*The boundaries must not have any floor or ceil function in them

*The inequality must work for all x in the real numbers

I am interested what we could find out !
 A: Let us prove that
$$x-1+\frac{1 + \frac{1}{6}(\sin \pi x)^2}{\pi}\vert\sin{\pi x}\vert\le 
\lfloor x\rfloor\le x -\frac{1 + \frac{1}{6}(\sin \pi x)^2}{\pi}\vert\sin{\pi x}\vert .$$
By letting $y = \pi(x - \lfloor x\rfloor)\in [0, \pi)$ and $z = \pi (1 - (x - \lfloor x\rfloor))\in (0, \pi]$, it suffices to prove that
$$(1 + \tfrac{1}{6}(\sin y)^2)\sin y \le y$$
and
$$(1 + \tfrac{1}{6}(\sin z)^2)\sin z \le z.$$
Thus, it suffices to prove that, for $u \in [0, \pi]$,
$$(1 + \tfrac{1}{6}(\sin u)^2)\sin u \le u.$$
It is easy to prove that $\sin u \le u$ and $\sin u \le u - \frac{1}{6}u^3 + \frac{1}{120}u^5$ for $u\in [0, \pi]$.
Thus, it suffices to prove that
$$\left(1 + \tfrac{1}{6}u^2\right)(u - \tfrac{1}{6}u^3 + \tfrac{1}{120}u^5) \le u$$
or
$$\frac{1}{720}u^5(14 - u^2) \ge 0.$$
It is true.
Remark: We may find the bounds of the form
$$x-1+\frac{f(\sin \pi x)}{\pi}\vert\sin{\pi x}\vert\le \lfloor x\rfloor\le x -\frac{f(\sin \pi x)}{\pi}\vert\sin{\pi x}\vert.$$
A better one is the following:
$$x-1+\frac{\frac{3}{2 + \sqrt{1 - (\sin \pi x)^2}}}{\pi}\vert\sin{\pi x}\vert\le
\lfloor x\rfloor\le x -\frac{\frac{3}{2 + \sqrt{1 - (\sin \pi x)^2}}}{\pi}\vert\sin{\pi x}\vert .$$
Here we have used the Shafer-Fink’s inequality for $\arcsin x$:
$$\frac{3x}{2 + \sqrt{1-x^2}} \le \arcsin x \le \frac{\pi x}{2 + \sqrt{1-x^2}}, \ x\in [0, 1]$$
