Is this result about the floor function true?

Question

I want to know if the following result is true or not.

Prove that for all positive integer $n$, $$\left\lfloor\sqrt{n}+\frac{1}{\sqrt{n}+\sqrt{n+2}}\right\rfloor \leq \sqrt{n}$$

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I drew a graph of $$f(x)=\sqrt{n}-\left\lfloor\sqrt{n}+\frac{1}{\sqrt{n}+\sqrt{n+2}}\right\rfloor$$ on desmos, and it seems that not all of $f(x)$ is positive, but at integer $x$ it indeed seems to be non negative. However I'm not sure if it's true for all integer $x$ and I haven't been able to prove it.

Answer 1

Hint: $\sqrt{n } + \frac{1}{\sqrt{n} + \sqrt{n+2} } = \frac{\sqrt{n} + \sqrt{n+2}}{2}$.
While this is easy to verify, it might not have been an obvious guess/observation.

Another hint: If $a < b$, then $ \lfloor a \rfloor \leq \lceil b \rceil - 1$.
This follows because $\lceil b \rceil - \lfloor a \rfloor$ is an integer and is positive, hence is $ \geq 1$.

Corollary: The result follows

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

$ \lfloor \sqrt{n } + \frac{1}{\sqrt{n} + \sqrt{n+2} } \rfloor \leq \lceil \sqrt{n+1} \rceil - 1 \leq \sqrt{n}$
This uses $n$ is an integer. It is not true for all real $n$.

Equality holds when $n$ is a perfect square.
(This follows from the above, but shouldn't be a surprise. It is necessary as the RHS is an integer. It is also easy to check that this is sufficient.)