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

• well, going from right to left, note $(x+2)^(1/x)$ is equal to $2$ when $x=2$ but smaller when $x > 2.$ Feb 9 at 23:56
• Very similar to the well-known problem of showing $\sqrt{2{\sqrt{3\sqrt{4 \sqrt{\ldots {\sqrt{n} } } } } } } < 3$. Feb 10 at 0:13
• 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. Feb 10 at 9:19

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!

• 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). Feb 11 at 3:09

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

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

• 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....
– D S
Feb 10 at 16:43
• 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. Feb 11 at 3:11