Floor Ceiling Floor Ceiling Floor Ceiling

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.

It will be proven that the above inequality will be satisfied if and only if $x$ is in the form of $\frac{k}{2}$ for every $\frac{k}{2} \in (0, 1000)$ .

To prove the above statement, we will divide the prove into 3 cases.

(i) $x = m + \frac{a}{b}$ where $m$ is a positive integer and $\frac{a}{b} \in (\frac{1}{2}, 1)$ .

It is obvious that $\lfloor x + \frac{1}{2} \rfloor = (m + 1)$ and $\lfloor x \rfloor = \lfloor x - \frac{1}{2} \rfloor = m$ .

It is also obvious that $\lceil x \rceil = \lceil x - \frac{1}{2} \rceil = (m+1)$ and $\lceil x + \frac{1}{2} \rceil = (m+2).$

Therefore there is no $x \in R$ that satisfies the above inequality with the above property.

(ii) $x = m + \frac{a}{b}$ where $m$ is a positive integer and $\frac{a}{b} \in [0, \frac{1}{2})$ .

It is obvious that $\lfloor x - \frac{1}{2} \rfloor = (m-1)$ and $\lfloor x \rfloor = \lfloor x + \frac{1}{2} \rfloor = m$ .

It is also obvious that $\lceil x \rceil = \lceil x + \frac{1}{2} \rceil = (m+1)$ and $\lceil x - \frac{1}{2} \rceil = m$ .

Therefore there is no $x \in R$ that satisfies the above inequality that satisfies the above property.

(iii) $x = \frac{k}{2}$ with $k$ an odd positive integer.

It is obvious that $\lfloor x + \frac{1}{2} \rfloor = \lceil x + \frac{1}{2} \rceil$ and $\lfloor x - \frac{1}{2} \rfloor = \lceil x - \frac{1}{2} \rceil$ .

It is also obvious that $\lceil x \rceil \geq \lfloor x \rfloor$ for every real number $x$ .

Therefore every $x$ in the form above satisfies the inequality. It implies that there is exactly 999 real number $x$ in the range that satisfies the inequality.