Floor and ceiling functions in a inequality Given $f(x)= x/\lfloor x\rfloor$, find $f(x) \geq  3/2$.
I have been stuck on this for a couple of hours. Thanks in advance.
 A: $$f(x) = \frac{x}{\lfloor x\rfloor}$$
Solve $f(x) \ge \frac32$. Since $f(x) \gt 0$(why?) we can write,
$$\frac{1}{f(x)} = \frac{\lfloor x\rfloor}{x} \le\frac23 \implies \frac{x-\{x\}}{x}\le\frac23$$, where $\{x\}$ is the fractional part of $x$
$$1-\frac{\{x\}}{x} \le \frac{2}{3} \implies \frac{\{x\}}{x} \ge \frac13$$
$$\{x\} \ge \frac{x}{3}$$ But since, $\{x\}\lt 1$,therefore $0\le x\le3$. 
But for $0\le x\lt1$ $f(x)$ is undefined, so $1\le x\lt3$. 
Consider $1\le x\lt2$ and $2\le x\lt3$ separately.
A: Since $\lfloor x\rfloor\le x$, if $x\lt0$,
$$
x\le\frac32\lfloor x\rfloor\le\frac32x
$$
which simplifies to $x\ge0$. Therefore, $x\ge0$.
Since $\lfloor x\rfloor\gt x-1$,
$$
x\ge\frac32\lfloor x\rfloor\gt\frac32(x-1)
$$
which simplifies to $0\le x\lt3$.
If $2\le x\lt3$, then
$$
\frac x2\ge\frac32\implies x\ge3
$$
If $1\le x\lt2$, then
$$
\frac x1\ge\frac32\implies x\ge\frac32
$$
$f$ is not defined when $0\le x\lt 1$, so the solution set is
$$
\frac32\le x\lt2
$$
A: The floor function gives the largest integer less than or equal to x.
So, consider x = 1.7.  The floor of x will be 1, so you will have 1.7/1 > 3/2 since 3/2 = 1.5.
The only possible solutions for this problem are from 1.5 through 2 (exclusive).
A: For such problems, it is better to work in intervals of $[k,k+1]$ where $k \in \mathbb{Z}$.
Now the function is $ \frac{x}{k}$ in this interval. Since you want $f(x) \geq 1.5$, the interval $[1.5,2)$ would be the only interval where this would work.
A: If $x\ge 1$ then $$\frac{3}{2}[x]\le x<[x]+1\Rightarrow [x]<2\Rightarrow [x]=1\Rightarrow x\ge 3/2$$
Then $3/2\le x<2$.
If $x<0$ then $$\frac{3}{2}[x]\ge x\ge[x]\Rightarrow [x]\ge 0\Rightarrow x\ge 0$$
which is absurd.
