Floor and ceiling functions in a inequality

Question

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.

Answer 1

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 $$ \frac32\le x\lt2 $$

Answer 2

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).

Answer 3

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.

Answer 4

$$f(x) = \frac{x}{\lfloor x\rfloor}$$

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.

Answer 5

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.