Proof for an identity (from Ramanujan written) I saw an identity by Ramanujan
$$\forall n \in \mathbb{N} ,n>1 :\lfloor \sqrt n+\sqrt {n+2}+\sqrt{n+4} \rfloor=\lfloor \sqrt {9n
+17}\rfloor$$ I tried to prove it by limit definition .  I post my trial below . If possible check my prove (right , wrong) ?
Then Is there more Idea to proof ?
 A: Assume $(m-1)^2<9n+17<m^2$. (We never get equality because $9n+17\equiv -1\pmod{3}$.) So we get: $$(m-1)^2+1\leq 9n+17\leq m^2-1$$
or $$\frac{(m-1)^2-16}{9}\leq n\leq\frac{m^2-18}{9}.$$
So $$\frac{\sqrt{(m-1)^2-16}}{3}\leq\sqrt{n}\leq\frac{\sqrt{m^2-18}}{3}\\
\frac{\sqrt{(m-1)^2+2}}3\leq\sqrt{n+2}\leq\frac m3\\
\frac{\sqrt{(m-1)^2+20}}3\leq\sqrt{n+4}\leq\frac{\sqrt{m^2+18}}{3}$$
Now by AM/GM, we get, for any real $a\geq 4$ that:
$$\begin{align}\frac{\sqrt{a^2-16}+\sqrt{a^2+2}+\sqrt{a^2+20}}{3}&>\left((a^2-16)(a^2+2)(a^2+20)\right)^{1/6}\\&=\left(a^6 + 6 a^4 - 312 a^2 - 64\right)^{1/6}
\end{align}$$
For $a\geq 8$ this gives a lower bound of $a$, because then:
$$0<6 a^4 - 312 a^2 - 64$$
This means that for $a=m-1$ with $m\geq 9$ we get that $\sqrt{n}+\sqrt{n+2}+\sqrt{n+4}\geq m-1$.
You can hand-check the values when $m<9$.
Finally, we'd like to show that $\sqrt{m^2-18}+m+\sqrt{m^2+18}\leq 3m$. This follows from concavity of $\sqrt{\cdot}$ function - $\sqrt{m^2-18}+\sqrt{m^2+18}\leq 2\sqrt{m^2}=2m$.
This gives that $\sqrt{m^2-18}+m+\sqrt{m^2+18}<3m$.
So we have that $m-1<\sqrt{n}+\sqrt{n+2}+\sqrt{n+4}<m.$ and we are done.

The key step is the AM/GM step, which gives a good approximation of the left sum because when $a$ is large, the interval $\left(\sqrt{a^2-16},\sqrt{a^2+20}\right)$ is small, so the AM/GM inequality is close to equality.
The other key is that $\sqrt{9n+17}$ is never an integer.
It might well be a numeric "accident" that it works when $m\leq 8$ (that is, when $n\leq 5$.) After all, it doesn't work for $n=1$.

More generally, for any odd $d$ such that $-1$ is not a square modulo $d$ and any value $a$, you get that, for large enough $n$:
$$\left\lfloor \sum_{k=0}^{d-1} \sqrt{n+ka}\right\rfloor = \left\lfloor \sqrt{d^2n+\frac{d^2(d-1)a}{2}-1}\right\rfloor$$
(Experimentation seems to indicate this is true with no condition on $d$, but I don't have a proof.)
For example, when $d=7$ and $a=2$ then you get for $n>22$:
$$\left\lfloor \sqrt{n}+\sqrt{n+2}+\cdots+\sqrt{n+12}\right\rfloor = \left\lfloor\sqrt{49n+293}\right\rfloor$$
For $d=3,a=1$ you actually get for all $n\geq 1$ that:
$$\left\lfloor \sqrt n+\sqrt{n+1}+\sqrt{n+2}\right\rfloor=\left\lfloor\sqrt{9n+8}\right\rfloor$$
For $d=7,a=1$ you get equality for all $n\neq 3$.
A: if $$|\sqrt n+\sqrt {n+2}+\sqrt{n+4} - \sqrt {9n
+17}|<\epsilon \\\Rightarrow \lfloor \sqrt n+\sqrt {n+2}+\sqrt{n+4} \rfloor=\lfloor \sqrt {9n
+17}\rfloor $$ 
so ,I will show $|\sqrt n+\sqrt {n+2}+\sqrt{n+4} - \sqrt {9n
+17}|<\epsilon$
$$\quad{|\sqrt n+\sqrt {n+2}+\sqrt{n+4} - \sqrt {9n
+17}|=\\|\sqrt n+\sqrt {n+2}+\sqrt{n+4} - 3\sqrt {n
+\frac{17}{9}}|\leq\\
|\sqrt n-\sqrt {n
+\frac{17}{9}}|+|\sqrt {n+2}-\sqrt {n
+\frac{17}{9}}|+|\sqrt{n+4} - \sqrt {n
+\frac{17}{9}}|\\\leq|\frac{0-\frac{17}{9}}{\sqrt n+\sqrt {n
+\frac{17}{9}}}|+\leq|\frac{2-\frac{17}{9}}{\sqrt {n+2}+\sqrt {n
+\frac{17}{9}}}|+\leq|\frac{4-\frac{17}{9}}{\sqrt {n+4}+\sqrt {n
+\frac{17}{9}}}|\\
\leq|\frac{\frac{17}{9}}{2\sqrt n}|+|\frac{\frac{1}{9}}{2\sqrt n}|+|\frac{\frac{19}{9}}{2\sqrt n}|\\=\frac{37}{18\sqrt n}\\\leq \frac{37}{18.5\sqrt n}\\=\frac{1}{2\sqrt{n}}\\\leq \frac{1}{2}}$$ I checked max $\{\epsilon\}<0.15$ with numerical method ,also graphing..
so $\lfloor \sqrt n+\sqrt {n+2}+\sqrt{n+4} \rfloor=\lfloor \sqrt {9n
+17}\rfloor$
