Solve for $x>0, \sqrt{x+\lfloor x\rfloor}+\sqrt{x+\{x\}}=\sqrt{x+\lfloor x\rfloor\{x\}}+\sqrt{x+1}$

Question

Solve for $x>0, \sqrt{x+\lfloor x\rfloor}+\sqrt{x+\{x\}}=\sqrt{x+\lfloor x\rfloor\{x\}}+\sqrt{x+1}$

My attempt:

$x\in (0,1)\implies \sqrt {2x}=\sqrt{x+1}\implies x=1;\,$ Rejected!

let $\lfloor x\rfloor =k,k\in\mathbb{N}$

$\implies \sqrt{x+k}+\sqrt{2x-k}=\sqrt{(k+1)x-k^2}+\sqrt{x+1}$

$\implies \sqrt{x+k}-\sqrt{x+1}=\sqrt{(k+1)x-k^2}-\sqrt{2x-k}$

$\implies (\sqrt{(k+1)x-k^2}+\sqrt{2x-k})(k-1)=(\sqrt{x+k}+\sqrt{x+1})(k-1)(x-k)$

Case 1: $k=1\implies x\in[1,2);\,\,$ this set statisfies the equation.

Case 2: $k\ge 2\implies \sqrt{(k+1)x-k^2}+\sqrt{2x-k}=(\sqrt{x+k}+\sqrt{x+1})(x-k)$

Answer 1

Let's think about only Case 2 $$ x \ge 2$$

Then

$$\sqrt{x + \lfloor{x}\rfloor} + \sqrt{x + \{x}\} = \sqrt{x + \lfloor{x}\rfloor\{x}\} + \sqrt{x + 1}$$

This equal to

$$\sqrt{x + \lfloor{x}\rfloor} - \sqrt{x + \lfloor{x}\rfloor\{x}\} = \sqrt{x + 1} - \sqrt{x + \{x}\}$$

$$ \frac{\lfloor{x}\rfloor * (1 - \{x\})}{\sqrt{x + \lfloor{x}\rfloor} + \sqrt{x + \lfloor{x}\rfloor\{x}\}} = \frac{1 - \{x\}}{\sqrt{x + 1} + \sqrt{x + \{x}\} }$$

In here $$1 - \{x\} \neq 0$$

so It equals
$$ \lfloor{x}\rfloor * (\sqrt{x + 1} + \sqrt{x + \{x\} )} = \sqrt{x + \lfloor{x}\rfloor} + \sqrt{x + \lfloor{x}\rfloor\{x}\} $$

In here $$ \lfloor{x}\rfloor \ge 2$$

so it is converted like this

$$ 2 * ( \sqrt{x + 1} + \sqrt{x + \{x\} )} \le \sqrt{x + \lfloor{x}\rfloor} + \sqrt{x + \lfloor{x}\rfloor\{x}\}$$

But in here. $$2 * \sqrt{x + \{x\}} \ge \sqrt{4x} \gt \sqrt{2x} \ge \sqrt{x + \lfloor{x}\rfloor\{x}\}$$

And

$$2 * \sqrt{x + 1} = \sqrt{4x + 4} \gt \sqrt{2x} \ge \sqrt{x + \lfloor{x}\rfloor}$$

This is contradiction.

so the solution is $$ x \in [1, 2) $$