The proposition is proven, when $0\leq | \sqrt{9n+8}-\sqrt{n}-\sqrt{n+1}-\sqrt{n+2} | \leq 1$, i.e. $-1\leq \sqrt{9n+8}-\sqrt{n}-\sqrt{n+1}-\sqrt{n+2} \leq 1$. First, i will show that $-1 \leq \sqrt{9n+8}-\sqrt{n}-\sqrt{n+1}-\sqrt{n+2}$: Therefore, you need estimate 1: $3\sqrt{n+3/2}\geq\sqrt{n}+\sqrt{n+1}+\sqrt{n+2}$ (proof by estimate 2: $\sqrt{n+1}\geq\sqrt{n(n+1)}$ and calculation) So, you have to show that $-1 \leq \sqrt{9n+8}-3\sqrt{n+3/2} \Leftrightarrow 3\sqrt{n+3/2}\leq \sqrt{9n+8}+1 \\ 9n +27/2\leq 1+9n+8+2\sqrt{9n+8} \\ 11/4\leq\sqrt{9n+8} \\ 8 \leq 9n+8 \\ 0 \leq 9n$ , what is true for all $n>0$. Now, you have to show that $\sqrt{9n+8}-\sqrt{n}-\sqrt{n+1}-\sqrt{n+2} \leq 1$. Therefore, you need estimate 3: $3\sqrt{n+1}\leq \sqrt{n}+\sqrt{n+1}+\sqrt{n+2}$ (proof by calculation). So, you have to show that $\sqrt{9n+8}-3\sqrt{n+1}\leq 1 \\ 12n+11-6\sqrt{(9n+8)(n+1)}\leq 1$ Now you need estimate 4: $(2n+2)^2\leq (9n+8)(n+1)$ (again proof by calculation) and you have to show that $12n+11-6(2n+2)\leq 1 \\ 0\leq 5n^2+9n+4$ , what is obviously true for alle $n>0$ Because of that, $-1\leq \sqrt{9n+8}-\sqrt{n}-\sqrt{n+1}-\sqrt{n+2} \leq 1$ is true and hence, $\left\lfloor \sqrt{9n+8} \right\rfloor = \left\lfloor \sqrt{n}+\sqrt{n+1}+\sqrt{n+2} \right\rfloor$ is true.

I think it is not enough to prove that $ 0\leq | \sqrt{9n+8}-\sqrt{n}-\sqrt{n+1}-\sqrt{n+2} | \leq 1$. For example, $ 4,5-3,6<1$ but $ \lfloor 4,5\rfloor\ne\lfloor 3,6\rfloor$. Here is my proof: Since $ 9n+8$ is never a square, there is no integer in the interval $ (9n+8,9n+9)$. I will proof that $ \sqrt{9n+8}<\sqrt{n}+\sqrt{n+1}+\sqrt{n+2}<\sqrt{9n+9}$. The first inequality can be proven by squaring and rearranging terms a few times. It's very boring, so I won't write it here. The second inequality is equivalent to $ \sqrt{n}+\sqrt{n+2}<\sqrt{9n+9}-\sqrt{n+1}=3\sqrt{n+1}-\sqrt{n+1}=2\sqrt{n+1}$, which is obvious after squaring. Hence, proved.

I am unsure about the above proof; the lower bound $\sqrt{9n+8} < \sqrt{n} + \sqrt{n+1} + \sqrt{n+2}$ is not trivial, and actually false for real $n$ in the interval $[0, 0.6]$, so I don't think it's easily proven 'by squaring and rearranging terms a few times'. I agree with the upper bound $\sqrt{n} + \sqrt{n+1} + \sqrt{n+2} < \sqrt{9n+9}$, which follows from the concavity of the square root function, and from this the inequality $\lfloor \sqrt{n} + \sqrt{n + 1} + \sqrt{n + 2} \rfloor \le \lfloor \sqrt{9n+8} \rfloor$ can be shown. To show that we actually have the reverse inequality $\lfloor \sqrt{n} + \sqrt{n + 1} + \sqrt{n + 2} \rfloor \ge \lfloor \sqrt{9n+8} \rfloor$ as well, I want to make use of the fact that $9n + 8$ is never a square, from which we get that $ \lfloor \sqrt{9n+8} \rfloor = \lfloor \sqrt{9n+7} \rfloor$ for all $n \in \mathbb{N}$. It therefore suffices to show $\sqrt{n} + \sqrt{n + 1} + \sqrt{n + 2} > \sqrt{9n+7}$. To prove this inequality, I claim that $\sqrt{n} + \sqrt{n+2} > 2\sqrt{n + 0.8}$ for all $n \ge 1$. Taking this latter claim for granted at the moment, we deduce $\sqrt{n} + \sqrt{n + 1} + \sqrt{n + 2} > 3\sqrt{n + 0.8} = \sqrt{9n + 7.2} > \sqrt{9n+7}$, and we are done. So it comes down to proving $\sqrt{n} + \sqrt{n+2} > 2\sqrt{n + 0.8}$, to finish off the proof. We will show this by starting with the inequality $n > 0.45$ and then repeatedly add or multiply by something positive, factorize both sides, or take square roots. \begin{align*} n &> 0.45 &\text{By assumption} \\ 3.2n & > 1.44 &\text{Multiply both sides by } 3.2\\ 8n &> 4.8n + 1.44 &\text{Add } 4.8n \text{ to both sides} \\ 4n^2 + 8n &> 4n^2 + 4.8n + 1.44 &\text{Add } 4n^2 \\ 4n(n+2) &> (2n + 1.2)^2 &\text{Factorize both sides } \\ 2\sqrt{n}\sqrt{n+2} &> 2n + 1.2 &\text{Take the square root}\\ 2n + 2 + 2\sqrt{n}\sqrt{n+2} &> 4n + 3.2 &\text{Add } 2n + 2\\ (\sqrt{n} + \sqrt{n+2})^2 &> 4(n + 0.8) &\text{Factorize}\\ \sqrt{n} + \sqrt{n+2} &> 2\sqrt{n + 0.8} &\text{Take the square root} \end{align*}