Let $A = \{ x \in \mathbb{R}^+ : x \lfloor \frac{1}{x} \rfloor \leq \frac{3}{4} \}$, with $\lfloor x \rfloor=\max \{d \in \mathbb{Z}: d\leq x \}$.
Find, $\sup{A}$ and $\inf{A}$.
If $x>1$, $\frac{1}{x} < 1$, then $x \lfloor \frac{1}{x}\rfloor =0 \leq \frac{3}{4}$. We see that does not exist $\sup{A}$.
On the other hand, $0<x\leq 1 \implies \frac{1}{x} \geq 1 \implies \lfloor \frac{1}{x}\rfloor \geq 1$. Then, $x \lfloor \frac{1}{x} \rfloor \geq x$...
How do I find $\inf{A}$?
Case 1: $\lfloor \frac{1}{x}\rfloor =1 \iff 1 \leq \frac{1}{x} <2 \iff 1 \geq x >\frac{1}{2} $. Then, $$x \cdot 1 \leq \frac{3}{4} \iff S_1= \left ]\frac{1}{2} , \frac{3}{4} \right ] $$
Case 2: $\lfloor \frac{1}{x}\rfloor =2 \iff 2 \leq \frac{1}{x} <3 \iff \frac{1}{2} \geq x >\frac{1}{3} $. Then, $$x \cdot 2 \leq \frac{3}{4} \iff S_2= \left ]\frac{1}{3} , \frac{3}{8} \right ] $$
Case 3: $\lfloor \frac{1}{x}\rfloor =3 \iff 3 \leq \frac{1}{x} <4 \iff \frac{1}{3} \geq x >\frac{1}{4} $. Then, $$x \cdot 3 \leq \frac{3}{4} \iff S_3= \emptyset $$
General Case: $\lfloor \frac{1}{x}\rfloor =n \iff n \leq \frac{1}{x} < n+1 \iff \frac{1}{n} \geq x >\frac{1}{n+1} $. Then, $$x \cdot n \leq \frac{3}{4} \iff x \leq \frac{3}{4n} \iff S_n= \left ]\frac{1}{n+1} , \frac{3}{4n} \right ]= \emptyset \text{ if } n \geq 3.$$
Therefore, $\inf{A}=\frac{1}{3}$.