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Let $n\in \mathbb{N}$. How we can show this : $$\lfloor \sqrt{n}+\sqrt{n+1}+\sqrt{n+2}+\sqrt{n+3}\rfloor =\lfloor \sqrt{16n+20}\rfloor$$ by using the concavity of $x\longmapsto \sqrt{x}$.

I read an article a few years ago about this but I can not find it on the internet.

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Let's first show the $\leq$ direction: By concavity, $$\frac{\sqrt{n} + \sqrt{n+1}}{2} < \sqrt{n + \frac{1}{2}}$$ so $$\sqrt{n} + \sqrt{n+1} < \sqrt{4n + 2}\,\,\,\,\, (1)$$ Similarly $$\sqrt{n+2} + \sqrt{n+3} < \sqrt{4n + 10} \,\,\,\,\, (2)$$ And using the same principle again: $$\sqrt{4n + 2} + \sqrt{4n + 10} < \sqrt{16n + 24}$$ So, adding $(1)$ and $(2)$: $$ \sqrt{n} + \sqrt{n+1} + \sqrt{n+2} + \sqrt{n+3} < \sqrt{16n + 24} \,\,\,\,\, (\ast)$$ To prove the $\leq$ direction, it's enough to show that there is no integer between $\sqrt{16n+20}$ (excluded) and $\sqrt{16n + 24}$ (included). The four critical numbers,

i) $8(2n+3)$,

ii) $16n + 23$,

iii) $ 2(8n + 11)$ and

iv) $16n + 21$

are never squares:

i) and iii) because they have an odd multiplicity of prime factor 2, ii) because it's $\equiv 3 \text{ mod } 4$ and iv) because it's $\equiv 5 \text{ mod } 16$.

Therefore $\lfloor \sqrt{16n + 20} \rfloor = \lfloor \sqrt{16n + 24} \rfloor $ and by $(\ast)$, the inequality with $\leq$ holds.

Now let's show the $\geq$: By the arithmetic-geometric mean inequality:

$$ \frac{\sqrt{n} + \sqrt{n + 1} +\sqrt{n+2} + \sqrt{n+3}}{4} > (n(n+1)(n+2)(n+3))^{1/8} \,\,\, (\ast \ast) $$ Note that $(n(n+1)(n+2)(n+3))=((n+\frac{3}{2})^2 - \frac{1}{4})((n+\frac{3}{2})^2 - \frac{9}{4}) > (n+\frac{5}{4})^4$ for all $n > 3$ since $n^2 +3n > n^2 + \frac{5}{2}n + \frac{25}{16}$ if $ n > 3$ (so it remains at the end to verify the identiy for 1,2,3 by hand). Now substituting this back into $(\ast \ast)$ gives $$ \frac{\sqrt{n} + \sqrt{n + 1} +\sqrt{n+2} + \sqrt{n+3}}{4} > \sqrt{n+\frac{5}{4}}$$ for all $n>3$ and multiplying through by $4$ finally $$ \sqrt{n} + \sqrt{n + 1} +\sqrt{n+2} + \sqrt{n+3} > \sqrt{16n+20}$$ completing the proof for $n>3$. For $n = 1,2,3$, the original identity is easy to check by hand.

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    but how we can show :$$\lfloor \sqrt{n}+\sqrt{n+1}+\sqrt{n+2}+\sqrt{n+3}\rfloor>\lfloor\sqrt{4n+20}\rfloor$$2017-02-19
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    done. I've also checked 1,2,3. They hold as expected.2017-02-19