Equality using floor function 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.
 A: 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.
