Interacting with Floors

$\begin{aligned} \frac 1{\lfloor x \rfloor} + \frac 1{\lfloor 2x \rfloor} & = {\color{#3D99F6}\{ x \}} + \frac 13 & \small \color{#3D99F6} \text{Note that }0 \le \{x\} < 1 \\ \implies \frac 13 \le \frac 1{\lfloor x \rfloor} & + \frac 1{\lfloor 2x \rfloor} < \frac 43 \end{aligned}$

To satisfy the inequality, $x$ must be positive and $\lfloor x \rfloor > 0$ . Note that for $x>0$ , $\lfloor 2x \rfloor = 2\lfloor x \rfloor + \lfloor 2\{x\}\rfloor = \begin{cases} 2\lfloor x \rfloor & \text{if }\{x\} < \frac 12 \\ 2\lfloor x \rfloor + 1 & \text{if }\{x\} \ge \frac 12 \end{cases}$ and $\dfrac 1{\lfloor x \rfloor} + \dfrac 1{\lfloor 2x \rfloor} = \begin{cases} \dfrac 1{\lfloor x \rfloor} + \dfrac 1{2 \lfloor x \rfloor} & \text{if }\{x\} < \frac 12 \\ \dfrac 1{\lfloor x \rfloor} + \dfrac 1{2 \lfloor x \rfloor+1} & \text{if }\{x\} \ge \frac 12 \end{cases}$ . Then we have:

For $1 \le x < 2\implies \dfrac 1{\lfloor x \rfloor} + \dfrac 1{\lfloor 2x \rfloor} = \begin{cases} \color{#D61F06} \frac 32 & \text{if }\{x\} < \frac 12 \\ \color{#D61F06} \frac 43 & \text{if }\{x\} \ge \frac 12 \end{cases} \color{#D61F06} \ \ge \frac 43 \text{ No solution.}$

For $2 \le x < 3 \implies \dfrac 1{\lfloor x \rfloor} + \dfrac 1{\lfloor 2x \rfloor} = \begin{cases} \frac 34 & \text{if }\{x\} < \frac 12 \\ \frac 7{10} & \text{if }\{x\} \ge \frac 12 \end{cases}$ $\implies \{x\} = \begin{cases} \frac 34 - \frac 13 = \color{#3D99F6} \frac 5{12} < \frac 12 & \color{#3D99F6} \implies x = 2\frac 5{12} \\ \frac 7{10} - \frac 13 = \color{#D61F06} \frac {11}{30} < \frac 12 & \color{#D61F06} \text{Rejected} \end{cases}$

For $3 \le x < 4 \implies \dfrac 1{\lfloor x \rfloor} + \dfrac 1{\lfloor 2x \rfloor} = \begin{cases} \frac 12 & \text{if }\{x\} < \frac 12 \\ \frac {10}{21} & \text{if }\{x\} \ge \frac 12 \end{cases}$ $\implies \{x\} = \begin{cases} \frac 12 - \frac 13 = \color{#3D99F6} \frac 16 < \frac 12 & \color{#3D99F6} \implies x = 3\frac 16 \\ \frac {10}{21} - \frac 13 = \color{#D61F06} \frac 17 < \frac 12 & \color{#D61F06} \text{Rejected} \end{cases}$

For $4 \le x < 5 \implies \dfrac 1{\lfloor x \rfloor} + \dfrac 1{\lfloor 2x \rfloor} = \begin{cases} \frac 38 & \text{if }\{x\} < \frac 12 \\ \frac {13}{36} & \text{if }\{x\} \ge \frac 12 \end{cases}$ $\implies \{x\} = \begin{cases} \frac 38 - \frac 13 = \color{#3D99F6} \frac 1{24} < \frac 12 & \color{#3D99F6} \implies x = 4\frac 1{24} \\ \frac {13}{36} - \frac 13 = \color{#D61F06} \frac 1{36} < \frac 12 & \color{#D61F06} \text{Rejected} \end{cases}$

For $5 \le x < 6 \implies \dfrac 1{\lfloor x \rfloor} + \dfrac 1{\lfloor 2x \rfloor} = \begin{cases} \color{#D61F06} \frac 3{10} & \text{if }\{x\} < \frac 12 \\ \color{#D61F06} \frac {16}{55} & \text{if }\{x\} \ge \frac 12 \end{cases} \color{#D61F06} \ < \frac 13 \text{ No solution.}$ As $\dfrac 1{\lfloor x \rfloor} + \dfrac 1{\lfloor 2x \rfloor}$ decreases with $x$ , this implies that there is no more solution for $x \ge 5$ .

Therefore, the sum of solution is $2\frac 5{12}+3\frac 16 + 4\frac 1{24} = 9\frac 58 = \boxed{9.625}$ .