I received this interesting problem in an email. It comes from the $1993$ All-Russian Olympiad grade 10, round 4. Prove that:
$$\sqrt{2 + \sqrt[3]{3 + \cdots + \sqrt[1993]{1993}}} < 2$$
I did verify this easily in a spreadsheet, but of course that misses the point and the fun of it all.
I have tried a few methods but not sure how to proceed. I thought about the AM-GM inequality but did not see a good substitution.
I also thought about the infinite nested square root sum of 2s:
$$ \sqrt{2 + \sqrt{2 + \cdots }} = 2 $$
That sum does converge to 2. However, when comparing to this nested root, we have $\sqrt[3]{3} > \sqrt{2}$ so that does not help show this nested sum is less using a term by term comparison.
This is a problem for talented 10th graders (I'm guessing 15 year olds), so there must be some proof that is not too advanced! What's the idea to solve this one?
Best Answer
I suggest a recursive/inductive proof:
We will show, for every $n\in\{2,\dots,1993\}$, that $$f(n):= \sqrt[n]{n+\dots+\sqrt[1993]{1993}}<2.$$
First, for $n=1993$ this is true because $1993<2^{1993}$.
Now we make a step from $n$ to $n-1$: Assume that $f(n)<2$ for some $n\in\{3,\dots,1993\}$. Then $$f(n-1)=\sqrt[n-1]{n-1+f(n)}<\sqrt[n-1]{n+1}\le 2,$$ where the last inequality follows from $n\ge 3$.
This establishes the desired result.
PS: I have seen quite a lot of your videos!