, 2011

For each non-negative integer $n$,
$$2\sqrt{n+1} > \sqrt{n} + \sqrt{n+2}.$$
Indeed,
$$\begin{aligned} \sqrt{n+1} - \sqrt{n} &= \frac{1}{\sqrt{n+1} + \sqrt{n}} \\ &> \frac{1}{\sqrt{n+2} + \sqrt{n+1}} \\ &= \sqrt{n+2} - \sqrt{n+1}, \end{aligned}$$
from which the result follows.

We now prove the required equality. It is true for $n = 0$ and $n = 1$, by inspection. Let $n \ge 2$. By the above result,
$$\sqrt{n} + \sqrt{n+1} + \sqrt{n+2} < 3\sqrt{n+1} = \sqrt{9n+9}.$$
We now show that $\sqrt{n} + \sqrt{n+1} + \sqrt{n+2} > \sqrt{9n+8}$. First note that, for $n \ge 2$,
$$n(n+2) - \left(n + \frac{7}{9}\right)^2 = \frac{4}{9}n - \frac{49}{81} \ge \frac{8}{9} - \frac{49}{81} > 0$$
which implies that
$$\sqrt{n(n+2)} > n + \frac{7}{9}.$$
Hence,
$$\begin{aligned} [\sqrt{n} + \sqrt{n+1} + \sqrt{n+2}]^2 &> \left[\frac{3}{2}(\sqrt{n} + \sqrt{n+2})\right]^2 \\ &= \frac{9}{4}\left[n + (n+2) + 2\sqrt{n(n+2)}\right] \\ &> \frac{9}{4}\left[2n + 2 + 2\left(n + \frac{7}{9}\right)\right] \\ &= \frac{9}{4}\left[4n + \frac{32}{9}\right] \\ &= 9n + 8. \end{aligned}$$
Hence we have
$$\sqrt{9n+8} < \sqrt{n} + \sqrt{n+1} + \sqrt{n+2} < \sqrt{9n+9}.$$
Since $9n+8$ and $9n+9$ are consecutive, no perfect square lies between them, and so no integer lies strictly between $\sqrt{9n+8}$ and $\sqrt{9n+9}$. This is enough to conclude the required equality.