$(\sqrt{a}+\sqrt{a+1}+\sqrt{a+2})^2 = 3a+3 + 2\sqrt{a(a+1)} + 2\sqrt{(a+1)(a+2)} + 2\sqrt{(a+2)a} < 9a+9$, By AM-GM. $2\sqrt{a(a+1)} + 2\sqrt{(a+1)(a+2)} + 2\sqrt{(a+2)a} > 6a+4$,because $2\sqrt{a(a+1)} + 2\sqrt{(a+2)a} > 4a+2$ and $2\sqrt{(a+1)(a+2)} > 2a+2$ $9a+7<(\sqrt{a}+\sqrt{a+1}+\sqrt{a+2})^2<9a+9$ Notice that $9a+8$ can not be a perfect square and $\lfloor \sqrt{9a+7}\rfloor = \lfloor \sqrt{9a+8}\rfloor$, gives us $\lfloor \sqrt{a}+\sqrt{a+1}+\sqrt{a+2}\rfloor = \lfloor \sqrt{9a+8}\rfloor$

How do we show $9a+8$ can't be a perfect square? (I forgot how to show that, do we take mod 4 or 9)

You can show that $9a+8$ cannot be a perfect square by taking it mod 9. Note that $9a+8\equiv8 \text{ mod } 9$ and $8$ is a quadratic non-residue mod $9$.

Lol I totally forgot about quadratic residues

smartguy888 wrote: Lol I totally forgot about quadratic residues quadratic residues $0,1,4,7$ mod $9$