The solutions of the following equation

Question

Please consider the following equation:

$$\left\lfloor x+\frac{1}{x}\right\rfloor=\frac{2x}{3}$$

where $\lfloor x\rfloor$ is the largest integer not greater than $x$.

It is clear that it has not a integer solution. Has it a real solution?

Answer 1

Let $\lfloor x\rfloor$ be the largest integer not greater than $x$.

Let $\lfloor x+\frac 1x\rfloor=m$ where $m$ is an integer. Since we have $$m\le x+\frac 1x\lt m+1,$$ having $m=\frac{2x}{3}$ gives us $$\frac{2x}{3}\le x+\frac 1x\lt \frac{2x}{3}+1.$$

If $x\gt 0$, then we have$$x+\frac 1x\lt \frac{2x}{3}+1\Rightarrow 3x\left(x+\frac 1x\right)\lt 3x\left(\frac{2x}{3}+1\right)\Rightarrow \left(x-\frac 32\right)^2+\frac{3}{4}\lt 0.$$ There is no such $x\in\mathbb R$.

If $x\lt 0$, then we have $$\frac{2x}{3}\le x+\frac 1x\Rightarrow 3x\times\frac{2x}{3}\ge 3x\left(x+\frac 1x\right)\Rightarrow x^2\le -3.$$ There is no such $x\in\mathbb R$.

Hence, there is no $x\in\mathbb R$ such that $$\left\lfloor x+\frac 1x\right\rfloor=\frac{2x}{3}.$$

Answer 2

Notice the left side is an integer, we have x = 3t/2 for some integer t. There're two cases,

1). t = 2k. If k > 0, the left is 3k, 3k = 2k, no solution. If k < 0, the left is 3k - 1, 3k - 1 = 2k, or k = 1, which is a contradiction to k < 0.

2). t = 2k + 1. If k >= 1, then 1/x < 1/2, and the left is 3k + 1, 3k + 1 = 2k + 1, no solution. If k <= -2, then -1/2 < 1/x < 0, the left is 3k + 1, 3k + 1 = 2k + 1, no solution. And finally we check k = -1, 0, find no one satisfies the equation.

Thus, no real solution.

Answer 3

$⌊x+\frac{1}{x}⌋$=$(x+\frac{1}{x})$-{$x+\frac{1}{x}$} where {$x+\frac{1}{x}$} represents fractional part which is always positve therefore $(x+\frac{1}{x})$-{$x+\frac{1}{x}$} =$\frac{2x}{3}$ therefore $\frac{x}{3}+\frac{1}{x}$={$x+\frac{1}{x}$} now as fractional part is always positive and less than 1 , $x$ cannot be negetive and by AM-GM you can see that $\frac{x}{3}+\frac{1}{x}$ is greater than $\frac{2}{\sqrt{3}}$ which is greater than 1 which is not possible , thus no solution