Croatian Mathematical Competitions

Let $S_n = \sqrt{n + \sqrt{(n-1) + \sqrt{(n-2) + \dots + \sqrt{2 + \sqrt{1}}}}}$.

We will prove by induction on $n$ that $S_n < \sqrt{n} + 1$ for all positive integers $n$.

Base case ($n = 1$):
$S_1 = \sqrt{1} = 1 < \sqrt{1} + 1 = 2$.

Inductive step:
Assume $S_{n-1} < \sqrt{n-1} + 1$.
Consider $S_n = \sqrt{n + S_{n-1}}$.

We want to show:
$$\sqrt{n + S_{n-1}} < \sqrt{n} + 1.$$

It suffices to show that $n + S_{n-1} < (\sqrt{n} + 1)^2 = n + 2\sqrt{n} + 1$.
Subtract $n$ from both sides:
$$S_{n-1} < 2\sqrt{n} + 1.$$

But by the induction hypothesis, $S_{n-1} < \sqrt{n-1} + 1$.
So it is enough to show:
$$\sqrt{n-1} + 1 < 2\sqrt{n} + 1$$
which is equivalent to $\sqrt{n-1} < 2\sqrt{n}$, which is true for all $n \geq 1$.

Therefore, by induction, $S_n < \sqrt{n} + 1$ for all positive integers $n$.