Is it ok? Lot of calculations $ 1/6 + \frac {1}{108n} < n$ $ 1/6 n + 1/108 < n^2$ $ 1/12 n + 1/216 < n^2/2$ $ n^3 + n^2/2 + 1/12 n + 1/216 < n^3 + n^2$ $ (n + 1/6)^3 < ((n^3 + n^2)^{1/3})^3$ $ n + 1/6 < (n^3 + n^2)^{1/3}$ $ 1/8 < n^2 /2 + n/4$ $ n^3 + 3/2 n^2 + 3/4 n + 1/8 < n^3 + 2n^2 + n$ $ (n + 1/2)^3 < ((n^3 + 2n^2 + n)^{1/3})^3$ $ n + 1/2 < (n^3 + 2n^2 + n)^{1/3}$ $ 2n + 2/3 < (n^3 + n^2)^{1/3} + (n^3 + 2n^2 + n)^{1/3}$ $ 6n + 2 < 3 (n^3 + n^2)^{1/3} + 3 (n^3 + 2n^2 + n)^{1/3}$ $ 8n + 3 < 2n + 1 + 3 (n^3 + n^2)^{1/3} + 3 (n^3 + 2n^2 + n)^{1/3}$ $ ((8n + 3)^{1/3})^3 < (n^{1/3} + (n + 1)^ {1/3})^{3}$ $ (8n + 3)^{1/3} < n^{1/3} + (n + 1)^ {1/3}$ $ 0 < n^{1/3} + (n + 1)^ {1/3} - (8n + 3)^{1/3}$ $ n^3 + 2n^2 + n < n^3 + 8n^2 + 48n + 9/8$ $ n^3 + 2n^2 + n < n^3 + (64n^2 + 48n + 9)/8$ $ ((n^3 + 2n^2 + n)^{1/3})^3 < (n + (64n^2 + 48n + 9)^{1/3}/2)^3$ $ (n^3 + 2n^2 + n)^{1/3} < n + (64n^2 + 48n + 9)^{1/3}/2$ $ n^3 + n^2 < n^3 + 8n^2 + 6n + 9/8$ $ n^3 + n^2 < n^3 + (64n^2 + 48n + 9)/8$ $ ((n^3 + n^2)^{1/3})^3 < (n + (64n^2 + 48n + 9)^{1/3}/2)^3$ $ (n^3 + n^2)^{1/3} < n + (64n^2 + 48n + 9)^{1/3}/2$ $ (n^3 + n^2)^{1/3} + (n^3 + 2n^2 + n)^{1/3} < 2n + 1 + (64n^2 + 48n + 9)^{1/3} + (8n + 3)^{1/3}$ $ 3 (n^3 + n^2)^{1/3} + 3 (n^3 + 2n^2 + n)^{1/3} < 6n + 3 + 3 (8n + 3)^{2/3} + 3 (8n + 3)^{1/3}$ $ 2n + 1 + 3 (n^3 + n^2)^{1/3} + 3 (n^3 + 2n^2 + n)^{1/3} < 8n + 3 + 1 + 3 (8n + 3)^{2/3} + 3 (8n + 3)^{1/3}$ $ (n^{1/3} + (n + 1)^ {1/3})^{3} < ((8n + 3)^{1/3} + 1)^3$ $ n^{1/3} + (n + 1)^ {1/3} < (8n + 3)^{1/3} + 1$ $ n^{1/3} + (n + 1)^ {1/3} - (8n + 3)^{1/3} < 1$

I work on the principle "a few words are worth a paragraph of $ \text{\LaTeX}$", so let's try at least to give some explanations. First of all, let's notice that it suffices to prove the inequalities $ \sqrt[3]{8n + 3} < \sqrt[3]{n} + \sqrt[3]{n+1} < \sqrt[3]{8n + 4}$ $ (\star)$, for all $ n \in \mathbb{N}$. Hence $ \lfloor \sqrt[3]{8n + 3} \rfloor \le \lfloor \sqrt[3]{n} + \sqrt[3]{n+1} \rfloor \le \lfloor \sqrt[3]{8n + 4} \rfloor$. Now it's a well-known fact that if $ \sqrt[3]{8n + 4}$ is irrational, $ \lfloor \sqrt[3]{8n + 3} \rfloor = \lfloor \sqrt[3]{8n + 4} \rfloor$, which proves our identity. Otherwise, $ \sqrt[3]{n} + \sqrt[3]{n+1} < \sqrt[3]{8n + 4}$ yields, in the view of $ \sqrt[3]{8n + 4} \in \mathbb{N}$, $ \lfloor \sqrt[3]{n} + \sqrt[3]{n+1} \rfloor \le \lfloor \sqrt[3]{8n + 4} \rfloor - 1 = \lfloor \sqrt[3]{8n + 3} \rfloor \le \lfloor \sqrt[3]{n} + \sqrt[3]{n+1} \rfloor$, which was to be proved. Now $ (\star)$ is equivalent to $ 2n + \frac{2}{3} < \sqrt[3]{n^2(n+1)} + \sqrt[3]{n(n+1)^2} < 2n + 1$. At first, we prove the leftmost part. This is implied by $ n + \frac{1}{5} < \sqrt[3]{n^2(n+1)}$ and $ n + \frac{7}{15} < \sqrt[3]{n(n+1)^2}$. The first reduces to $ 15n + 1 < 50n^2$, which is obvious, and the second to $ 343 < 2025n^2 + 1170n$, again clear. The rightmost is broken even more straightforward into $ \sqrt[3]{n^2(n+1)} < n + \frac{1}{3}$ and $ \sqrt[3]{n(n+1)^2} < n + \frac{2}{3}$. The first reduces clearly to $ 9n + 1 > 0$, and the second to $ 9n + 8 > 0$. The inequalities have been proved. I have used above a well-known Lemma. Let $ n, k$ be positive integers, $ k \ge 2$. Then, $ \sqrt[k]{n} \in \mathbb{R} \backslash \mathbb{Q} \Rightarrow \lfloor \sqrt[k]{n-1} \rfloor = \lfloor \sqrt[k]{n} \rfloor$. $ \sqrt[k]{n} \in \mathbb{N} \Rightarrow \lfloor \sqrt[k]{n-1} \rfloor + 1 = \lfloor \sqrt[k]{n} \rfloor$. The proof is standard as it is left to the reader