Prove $[\sqrt{x}+\sqrt{x+1}]=[\sqrt{4x+2}], x \in \Bbb N$

First a kooky Lemma:

For $x > 1$ then $\sqrt{1 - \frac 1 x} + \sqrt{1 + \frac 3 x} > 2$

[$\sqrt{1 - \frac 1 x} + \sqrt{1 + \frac 3 x} > 2$ iff

$1 - \frac 1 x + 1 + \frac 3 x + 2\sqrt{(1 - \frac 1 x)(1 + \frac 3 x)} > 4$ iff

$2 + \frac 2 x + 2\sqrt{(1 + \frac 2 x -\frac 2 {x^2}} > 4$ iff

$1 + \frac 2 x -\frac 2 {x^2} > 1$ iff

$\frac 2 x -\frac 2 {x^2} > 0$ ]

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So now to answer your question:

Let $n^2 \le x < (n+1)^2$ or in other words $n^2 \le x \le n^2 + 2n$.

It's obvious:

$2n < \sqrt x + \sqrt {x + 1} < (n + 1) + (n + 1) = 2n + 2$.

So $\lfloor \sqrt x + \sqrt {x + 1} \rfloor = \{2n, 2n + 1\}$.

Likewise $2n = \sqrt{4n} < \sqrt{4x + 2} \le \sqrt{4n^2 + 8n + 2} < \sqrt{4^2 + 8 + 4} = 2(n + 1) = 2n + 2$.

So $\lfloor \sqrt {4x + 2} \rfloor = \{2n, 2n + 1\}$.

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Case 1:

$n^2 \le x < x + 1 \le n^2 + n < n^2 + n + \frac 1 4 = (n + \frac 1 2)^2$

Then $ \sqrt x + \sqrt {x + 1} < n + \frac 1 2 + n + \frac 1 2 < 2n + 1$

So $\lfloor \sqrt x + \sqrt {x + 1} \rfloor = 2n$.

Likewise $ \sqrt{4x + 2} \le \sqrt{4n^2 + 4(n -1) + 2} < \sqrt{4n^2 + 4n + 1} = 2(n + \frac 1 2) = 2n + 1$.

So $\lfloor \sqrt {4x + 2} \rfloor = 2n = \lfloor \sqrt x + \sqrt {x + 1} \rfloor$.

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Case 2:

$n^2 + 2n + 1 \ge x + 1 > x \ge n^2 + n$.

Then $\sqrt{4x + 2} \ge \sqrt{4n^2 + 4n + 2} = 2(n + \frac 12) = 2n + 1$.

So $\lfloor \sqrt {4x + 2} \rfloor = 2n + 1$.

Now $\sqrt x + \sqrt {x + 1} \ge \sqrt{n^2 + n} + \sqrt{n^2 + n + 1} = \sqrt{n^2 + n + \frac 1 4 - \frac 1 4} + \sqrt{n^2 + n + \frac 1 4 + \frac 3 4}$

$= (n + \frac 1 2)[\sqrt{1 - \frac 1 {4(n + \frac 1 2)^2}} + \sqrt{1 + \frac 3 {4(n + \frac 1 2)^2}}]$

(remember the kooky lemma?)

$ > (n + \frac 1 2)2 = 2n + 1$.

So $\lfloor \sqrt x + \sqrt {x + 1} \rfloor = 2n + 1 = \lfloor \sqrt{4x +2} \rfloor$.