Is it true that $\lfloor\sqrt{n}+\sqrt{n+1}+\sqrt{n+2} \rfloor=\lfloor\sqrt{9n+7}\rfloor$? I know it's true that 
$$\lfloor\sqrt{n}+\sqrt{n+1} \rfloor=\lfloor\sqrt{4n+1}\rfloor,\forall n\in \mathbb N. \tag 1$$
Is it true that 
$$\left[\sqrt{n}+\sqrt{n+1}+\sqrt{n+2} \right ]=\left[\sqrt{9n+7}\right], \tag 2$$
$$\left[\sqrt{n} +\sqrt{n+1}+\sqrt{n+2}+\sqrt{n+3}\right ]=\left[\sqrt{16n+20}\right], \tag 3$$
$$\left[\sqrt{n} +\sqrt{n+1}+\sqrt{n+2}+\sqrt{n+3}+\sqrt{n+4}\right ]=\left[\sqrt{25n+49}\right], \tag 4$$
for all $n\in \mathbb N$? I have checked $(2),(3),(4)$ through $10^6.$ 
(I think $(4)$ maybe has counter-examples, but I cannot find it.)
 A: For $0<a<b$ we have $\sqrt b-\sqrt a=\frac{b-a}{\sqrt b+\sqrt a}$ and hence the estimate 
$$\frac{b-a}{2\sqrt b}<\sqrt b-\sqrt a<\frac{b-a}{2\sqrt a}.$$
Therefore 
$$x:=\sqrt {n}+\sqrt {n+1}+\sqrt{n+2} =3\sqrt{n+1}+(\sqrt{n+2}-\sqrt {n+1})-(\sqrt {n+1}-\sqrt n)$$
is estimated by
$$3\sqrt {n+1}+\frac1{2\sqrt{n+2}}-\frac1{2\sqrt n}< x<3\sqrt{n+1}$$
and then (applying the same trick to $a=\frac1{n+2}$, $b=\frac1n$) 
$$ x > 3\sqrt{n+1}-\frac{\frac1n-\frac1{n+2}}{4\sqrt{\frac1{n+2}}}= 3\sqrt{n+1} - \frac{1}{2n\sqrt{n+2}}.$$
Therefore 
$$ 9(n+1)-\frac{3\sqrt{n+1}}{n\sqrt{n+2}}+\frac1{4n^2(n+2)}< x^2 < 9n+9$$
For $n>3$ this gives us $ 9n+8<x^2<9n+9$ and hence
$ x=\sqrt{9n+\theta}$
with $8<\theta<9$. Since $9n+8$ cannot be a perfect square(!), we conclude that  $$\lfloor\sqrt{9n+7}\rfloor = \lfloor\sqrt{9n+\theta}\rfloor = \lfloor \sqrt n+\sqrt{n+1}+\sqrt {n+2}\rfloor$$
(at least for $n>3$, but one checks the rest manually).
The same approach, though with more complicated terms, should work for $(3)$ and $(4)$.
