How to solve these inequalities with floors and ceilings?

Question

I am trying to solve these two inequalities for $x$, as a function of the positive integer $n$:

$$ \newcommand{\floor}[1]{\left\lfloor #1 \right\rfloor} \newcommand{\ceil}[1]{\left\lceil #1 \right\rceil} \ceil{\frac{2 x n}{\ceil{2 x}+2}} \geq n \\ \ceil{\frac{\floor{2 x} n }{\floor{2 x}+2}} \geq n $$

So far, the only progress I could made is remove the outer ceiling, since the ceiling is at least $n$ iff the thing inside the ceiling is larger than $n-1$. So I got:

$$ \newcommand{\floor}[1]{\left\lfloor #1 \right\rfloor} \newcommand{\ceil}[1]{\left\lceil #1 \right\rceil} \frac{2 x n}{\ceil{2 x}+2} > n-1 \\ \frac{\floor{2 x} n }{\floor{2 x}+2} > n-1 $$

But I have no idea, how to make further progress on this?

Answer 1

You are on the right track. Now for the second inequality you can solve for $\lfloor 2x \rfloor$ as a whole, as you normally would for a fractional inequality. For the first, set $m=\lceil 2x \rceil$, and solve $x$ in terms of $n$ and $m$, then impose the condition that $m-1<2x\le m$, which should give you the range of $x$ expressed in terms of $n,m$. Now take the union of all the ranges for each $m\in\mathbb Z$.

To give the reason why I suggested this complicated way for solving #1, let's look at an example where $n=3$. In fact, if $n=3$ then $\frac73 < x \le \frac52 $ or $\frac83 < x$. And "$\frac73 < x \le \frac52 $ or $\frac83 < x$" as a whole serves as the correct answer. For $n=4$ the answer splits into 3 parts, etc. Therefore, it is essential that you analyze every possible $m=\lceil 2x \rceil$, and then piece up all the possibilities.

Answer 2

For the second one ,

split the two cases $\newcommand{\floor}[1]{\left\lfloor #1 \right\rfloor} \newcommand{\ceil}[1]{\left\lceil #1 \right\rceil} \floor{2 x}+2 > 0 \iff x\geq-\frac{1}{2} $ and $\floor{2 x}+2 < 0 \iff x \leq -\frac{1}{2} $.

Taking first case :

$$ \frac{\floor{2 x} n }{\floor{2 x}+2} > n-1 \iff \floor{2 x} >2(n-1) \iff x \geq n-\frac{1}{2} $$

So we have $x \geq n-\frac{1}{2} $.

For the other case we have $ x \leq n-\frac{1}{2} $.

So the solution is $x \leq- \frac{1}{2} \cup x \geq n-\frac{1}{2} $.

The first one can be solved with the same reasoning, splitting it in subcases.

Answer 3

HINT.-Let $f_n(x)$ your first function and consider $g(x)=\dfrac{2 x}{\lceil 2x \rceil+2}$.

What you want is $f_n(x)\ge n$ and for this you need $g(x)\ge1$. Solving $g(x)\gt1$ because it is clear that $g(x)$ cannot be equal to $1$ for $x$ real. Try to find $\color{red}{x\le{-1.5}}$.

Similarly for your second inequation.