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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?

Source

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    $\begingroup$ well, going from right to left, note $(x+2)^(1/x)$ is equal to $2$ when $x=2$ but smaller when $x > 2.$ $\endgroup$
    – Will Jagy
    Feb 9 at 23:56
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    $\begingroup$ Very similar to the well-known problem of showing $ \sqrt{2{\sqrt{3\sqrt{4 \sqrt{\ldots {\sqrt{n} } } } } } } < 3$. $\endgroup$
    – Calvin Lin
    Feb 10 at 0:13
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    $\begingroup$ While $\sqrt[3]{3} > \sqrt{2}$, for the next term we have equality and for all subsequent terms the inequality is in the desired direction so maybe this identity can be used with a little care for the last few terms. $\endgroup$
    – quarague
    Feb 10 at 9:19

3 Answers 3

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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!

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  • $\begingroup$ Thanks! I love it when induction can be used. (I have accepted the answer, though I will more carefully work through the details when I have more time). $\endgroup$
    – Presh
    Feb 11 at 3:09
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Your idea of using a “term by term” comparison with the $\sqrt{2}$ infinite nested radical also works without considerably more effort and gives a tighter bound.

Claim: $(x+y)^2 \le (2+y)^x$ for $x \geq 4$ and $y\geq 0$ and $x,y \in \mathbb Z$.

First we show that $P(x): 2^x \geq x^2$ for $x \geq 4$. The proposition holds for $x=4$. Suppose $k>4$ is the least $k$ where $P(k)$ doesn’t hold, meaning that $2^k < k^2$ and $2^{k-1} \geq (k-1)^2$. From the second inequality it follows that $2^k \geq 2(k-1)^2$. But $k^2>2(k-1)^2$ means $k<4$, which is a contradiction. Hence we have proved $P(x)$ for $x \geq 4$.

Now to prove the claim we expand both sides via binomial theorem:

$x^2+y^2+2xy \le 2^x + y^x +2^{x-1}xy + \underbrace{\dots}_{\geq0}$, which is true. It is also clear that equality is attained only at $(x,y)=(4,0)$.

Denote $t = \sqrt{2 +\sqrt[3]{3+\sqrt[4]{4+\dots+\sqrt[1998]{1998}}}} $

From the claim it follows that $t < \sqrt{2+\sqrt[3]{3+\sqrt{2+\dots+\sqrt{2}}}} < \sqrt{2+\sqrt[3]{3+\sqrt{2+\sqrt{2+\dots\infty}}}} = \sqrt{2+\sqrt[3]{3+\color{blue}{2}}}$.

Hence $$t<\sqrt{2+\sqrt[3]{5}}<2$$

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Here's my solution (almost the same as the first answer, however, it also explains the intuition behind it):

We define $(a_n)^{1992}_2$ as the leftover, that is, what must be added inside the $(1/n)$th power to get 2: $$a_2 = 2 \text{ and } a_{n+1} = a_n^{n+1}-(n+1) \text{ for }n\ge 2$$ It is easy to see that this recurrence satisfies $$\sqrt{2+\sqrt[3]{3+\ldots \sqrt[n]{n+a_n}}} = 2$$

Now, let's define $(b_n)^{1992}_2$ as what's actually there, i.e: $$b_n = \sqrt[n+1]{n+1+\sqrt[n+2]{\ldots \sqrt[1993]{1993}}}$$ For example: $b_2 = \sqrt[3]{3+\ldots \sqrt[1993]{1993}}$ and so on. Our goal is to show $a_2>b_2$.

Observe that $b_n$ satisfies the same recurrence as $a_n$ (this is the key idea), as in: $$b_{n+1} = b_n^{n+1}-(n+1)$$ Note that $f(x)= x^{n+1} - (n+1)$ is increasing in $x$ for $n\ge2$ and $x> 0$, and is injective.

Since $b_{1992} = \sqrt[1993]{1993}<2<a_{1992}$, we are done. $\tag*{$\blacksquare$}$

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  • $\begingroup$ I think this is not as good as the first answer, but these were my first thoughts after seeing the question, and I'm 15 btw.... $\endgroup$
    – D S
    Feb 10 at 16:43
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    $\begingroup$ I was thinking in terms of this "leftover" idea, so I am very glad to see this approach and how it relates to the first answer. $\endgroup$
    – Presh
    Feb 11 at 3:11

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