<s>[INST] 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 ?
 [/INST] 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 $\lfloor \sqrt n+\sqrt {n+2}+\sqrt{n+4} \rfloor=\lfloor \sqrt {9n
+17}\rfloor$
</s>