Floor Ceiling Floor Ceiling Floor Ceiling 2

Similar solution as part 1, so just cut and paste solution and change a bit.

Let $\space f(x) = \left \lfloor x + \frac{1}{2} \right \rfloor + \left \lceil x \right \rceil + \left \lfloor x - \frac{1}{2} \right \rfloor$ , $\space g(x) = \left \lceil x + \frac{1}{2} \right \rceil + \left \lfloor x \right \rfloor + \left \lceil x - \frac{1}{2} \right \rceil$ and $\space x = \lfloor x \rfloor + \{x\} = n + \delta$ , where $\lfloor x \rfloor = n$ and $\{x\} = \delta$ are the integral and fractional parts of $x$ respectively; $n$ and $\delta$ are positive for $x \in [1,1000]$ .

There are four cases of $f(x)$ and corresponding $g(x)$ as follows:

$\begin{cases} \delta = 0 & \Rightarrow \begin{cases} f(x) = n+n+(n-1) \\ g(x) = (n+1)+n+n \end{cases} & \Rightarrow \color{#D61F06} {f(x) < g(x)} \\ 0 < \delta < \frac{1}{2} & \Rightarrow \begin{cases} f(x) = n+(n+1)+(n-1) \\ g(x) = (n+1)+n+n \end{cases} & \Rightarrow \color{#D61F06} {f(x) < g(x)} \\ \delta = \frac{1}{2} & \Rightarrow \begin{cases} f(x) = (n+1)+(n+1)+n \\ g(x) = (n+1)+n+n \end{cases} & \Rightarrow \color{#3D99F6} {f(x) > g(x)} \\ \frac{1}{2} < \delta < 1 & \Rightarrow \begin{cases} f(x) = (n+1)+(n+1)+n \\ g(x) = (n+2)+n+(n+1) \end{cases} & \Rightarrow \color{#D61F06} {f(x) < g(x)} \end{cases}$

Therefore, for each $n \in [1,1000)$ , there is a case that $\color{#3D99F6} {f(x) > g(x)}$ or $\boxed{999}x$ 's that satisfy the inequality.

$\left\lfloor x+\dfrac{1}{2}\right\rfloor +\lceil x \rceil+\left\lfloor x-\dfrac{1}{2}\right\rfloor\ge \left\lceil x+\dfrac{1}{2}\right\rceil+\lfloor x \rfloor+\left\lceil x-\dfrac{1}{2}\right\rceil$

For now, assume $x$ is an integer. We have

$x+x+x-1\geq x+1+x+x$

This is obviously false so $x$ can't be an integer. Now, rearrange the equation.

$0\geq \left(\left\lceil x+\dfrac{1}{2}\right\rceil-\left\lfloor x+\dfrac{1}{2} \right\rfloor\right)+\left(\lfloor x \rfloor-\lceil x\rceil \right)+\left(\left\lceil x-\dfrac{1}{2}\right\rceil-\left\lfloor x-\dfrac{1}{2}\right\rfloor \right)$

(The equality sign has been change do, it's now facing the correct way).

We already established $x$ is not an integer, thus $\left(\lfloor x \rfloor-\lceil x\rceil \right)=-1$ . The other two sections are either BOTH equal to 0 or 1. If they're both 1, then the sum is greater than 0 and we can't have that.

For both parts to be equal to 0, $x=n+0.5$ where $n\displaystyle\epsilon \Bbb{N}$ . Since there are 999 of these between 1 and 1000, out answer is 999.

Amazingly, this problem is exactly the same as Floor Ceiling Floor Ceiling Floor Ceiling

See if you can figure out why $\left\lfloor x+\dfrac{1}{2}\right\rfloor +\lceil x \rceil+\left\lfloor x-\dfrac{1}{2}\right\rfloor\ge \left\lceil x+\dfrac{1}{2}\right\rceil+\lfloor x \rfloor+\left\lceil x-\dfrac{1}{2}\right\rceil$ $\iff\left\lfloor x+\dfrac{1}{2}\right\rfloor \lceil x \rceil\left\lfloor x-\dfrac{1}{2}\right\rfloor\ge \left\lceil x+\dfrac{1}{2}\right\rceil\lfloor x \rfloor\left\lceil x-\dfrac{1}{2}\right\rceil$

There is an obvious casework solution, and a not-so-obvious, but much more enjoyable, solution by algebraic manipulation.

We are under the assumption, of course, that $|x|\ge 1$ .