If we could get rid of then everything would appear like an arithmetic mean and that would have been pleasant (so far no words about how the result would have appeared). But life is not so pleasant as we have a and that can’t be $0$. Now, note that . So, if we can show the existence of some unique integer satisfying the given inequality, we are done (in simple words, we don't have to make our hands dirty to show that such a integer must positive). Now, can the left most inequality be true for every positive integer . Let's see : multiplying out both sides of the leftmost inequality by and then subtracting what remains on the left side from what remains on the right side gives us the equivalent inequality and a quick observation is that this is and since we know that the sequence is increasing so life's good as the above quantity can't always be positive (since everything is an integer and after subtraction things are decreasing). At some point it has to hit some non negative number and that point will be determined by what $a_0$ is. Then immediately everything in the remaining solution was quite visible. We will take the largest such for which the above inequality holds. And then all that remains to be seen is that once it hits a non negative number, it can't become positive anymore, as after the subtraction everything was decreasing here. So we get a largest for which the Left inequality holds. And if there is any truth in the world, then the right inequality must hold for some $k \le n$. And this largest gives everything that we want, because the right inequality is no different than what we were trying to avoid while ensuring that the left inequality gets satisfied. Clear denominators and add and subtract and the RHS of the right inequality from the LHS of the right inequality and we will get everything we want.

If I am not wrong then this was also one of the official solutions (but I had no connections with the jury, and obviously didn't know about the official solution or any other solution at the time of solving)……

Actually I believe that my solution was along these lines mainly for two reasons : first, my brain somehow always thinks of trading terms (like you give one value to one side and you have to compensate it with adding another value and so on……) and secondly because at that time I was probably very obsessed with the Well Ordering principle and so somehow something along the lines of finding the first point where something goes wrong or where something is satisfied was occupying my thinking procedure and fortunately it worked.

(PS : I did not know about the existence of this topic until today and so I am sorry if I am bumping an old topic).

]]>My process could be described as “micro-monomath” since it involves a step not in the ordinary rules.

– (1 minute) scan the post and comments quickly to see if there is a one-line “aha!” solution or keywords linking the problem to some more theoretical concepts beyond the IMO. No, the symbols and every-tenth-word look more like a bunch of algebra and calculation, so I decided to attempt to solve this myself.

– (8 minutes) look for a formulation with averages. This didn’t work, these are not quite averages. Had the indexing started at a_1 instead of a_0 this would have been easier to notice, but I slowly came around to it.

– (5 minutes) Forget about averages and work with integers, since the statement looks suspicious in the generality of real numbers. We need an increasing integer function g(n) that starts negative and therefore crosses 0 at a unique point. The first candidate that comes to mind is g(n)= n a(n) – (a(0) + a(1) + … + a(n) ) and this works.

Writing everything up might have doubled or tripled the time. I used pencil and paper only at the last step, to verify that g(n) works after having decided to try it.

The key heuristic guiding the failed first step and the successful last step was that there should be a statement that (except for the sign of the inequality) is the same for a_n and a_(n+1). For example the same function g(n) is used at points n and (n+1). And in the averaging formulation, the idea was that a_n should be below average for the first n or n+1 terms or something, and a_(n+1) should be above average in the exact same sense. Although averages seemed more intuitive at first, that path runs in circles and it was easier to junk it and switch to “multiplying out” everything so as to work with integers.

]]>Isn’t that basically what is said to have happened when George Dantzig arrived late for a lecture by Jerzy Neyman on mathematical statistics (see the Wikipedia page for George Dantzig, the Mathematical statistics subsection)?

]]>The question of how it affects one’s approach to a problem if one knows that it has been set in an Olympiad is a very interesting one. It definitely gives a big clue about the nature of the solution, but making precise what the clue is is not easy. But it often seems to involve finding a “key” that suddenly unlocks the problem, whereas in a “real” problem there often isn’t such a key and what is required is more like hard grind. But of course, sometimes even research problems can be unlocked in a beautiful way: it’s just that that is not as common as it is in Olympiads.

]]>What would happen if a professor gave as homework to his students some open problem without telling them it’s open? (and assuming they don’t know it, and a problem that doesn’t look hard) I think they would become frustrated for not being able to solve it much more than if they already knew it was a very hard problem.

Sorry for my bad English. I was never able to learn this crazy language.

]]>I noticed the first inequality always holds for n = 1.

I worked on the uniqueness requirement for the special n (as opposed to its existence) first. (In retrospect, I’m not sure if that was because it seemed weirder, or seemed easier, or both.)

I multiplied each inequality through by n (just to make them easier to work with), and later noticed that the resulting second inequality for n, after adding a_(n+1) to both sides, becomes exactly inverse to the first inequality for n+1. I also noticed that (for any given n) if the first inequality is false, the 2nd one is true (due to ordering of the a’s). (This was natural to notice, since I was trying to understand all the ways those middle terms could insert themselves in between some pair of successive a_i’s.)

This proves uniqueness, since looking at the state of the two inequalities for successive n, “yes yes” must be followed by “no yes”, which must also be followed by “no yes”. (I think getting this far took 10-20 minutes.)

Then I worked on “existence of n” in some ways I won’t record here, except for what eventually either worked or almost worked. It’s clear from the info above that if the special n never exists, the inequality-pairs for successive n are all “yes no”. This means the almost-average (middle term) is bigger than the sequence itself (last term), which is weird given that the sequence is increasing. So let’s look at how fast the middle term might be increasing. Asymptotically, either the sequence of a’s is linear with some slope, or superlinear (this is the part of my solution I am only claiming “almost works” since I never did formalize it as much as I’d like to), and clearly a_n >= n + 1 (from original problem statement) so that slope is at least 1. But if it was actually linear, then for large enough n the average would be about 1/2 the last term, and the almost-average used here (which sets an upper limit for the new term under our assumptions) is only (n+1)/n times the average, a contradiction since (n+1)/n << 2.

I convinced myself "morally" that argument was valid (even for sequences a_i which go above some line infinitely often, but more and more rarely as n increases), but not formally enough to consider it solved. I also tried to come up with a non-asymptotic version, but I can't remember if I really did, or if I lost motivation to continue at some point. (I had stopped for dinner and then was no longer writing things down after that.)

]]>I think you meant, or should have meant, that first a_n to be a_(n-1).

> But since $ a_2<(a_0+a_1+a_2)/2$, we can simply insert $ a_3$ in between the two.

It took me awhile to understand what "two" you meant here — the two terms of that inequality.

]]>I see, thanks. Your entire thought-flow is now clear, but I’ll point out two minor issues anyway:

> In particular, it suggests rearranging the first inequality in the general case, to

>

> .

I think you meant, or should have meant, latex a_2<(a_0+a_1+a_2)/2$, we can simply insert in between the two.

It took me awhile to understand what "two" you meant here — the two terms of that inequality.

]]>The answer to your question becomes clear if you read on, which you are deliberately not doing. But I don’t think it’s spoiling much if I tell you that at that point I had not noticed that the have to be positive integers, rather than positive real numbers. If they’re reals, then it really is obvious, since multiplying everything by a positive constant doesn’t affect the inequalities.

]]>I lose you at this point: why is the following obvious? Or more precisely, what thought did you have, but fail to record, which makes it obvious?

It is clear that WLOG a_0=1.

(I don’t see any obvious way to transform a vector of a’s to one with a_0 = 1 without changing the truth of any inequality in the problem, which is what I’d think you’d need to see to make this obvious.)

I have not read ahead in your answer or anyone else’s, or tried hard to answer this subquestion myself, since I want to help make your record of thoughts complete. In other words, I hope you will remember or somehow reconstruct what you thought then, rather than answering it anew.

]]>The result is true if and only if . The case of the $a_i$ integers is just an example.

]]>Oops, there now – thanks.

]]>It follows from by simple algebraic deductions ðŸ™‚

]]>You are quite right to imply that my attempt to write down my thoughts was not a complete success, and unfortunately the point where I failed, thanks to a moment of laziness that I later regretted, was an important one.

I do, however, have some idea of the source of the trick. It came in the section near the beginning labelled “Question”, where I took , and then found that everything else was . That suggested that was more important than either or individually.

]]>Why does it immediately follow that ???

We are at the last point before we cross the river but we do not know that do we?

Thank you

]]>Also, you were forced to record your thoughts to a linear transcript, whereas surely you are having multiple ideas interconnecting in various ways. This inadequacy of a linear medium is probably what led you to initially not record the idea that ended up playing a key role, as you yourself mention in the second red comment. This is not to say that everything else was a waste of time. There were many constructive insights before you chose to revisit the idea that on the surface appears only to be a slight notational change.

This dichotomy between linearity (of how we learn and communicate mathematics) and interconnection (of how we deeply understand areas of mathematics) plays a key role in the difficulty in teaching mathematics.

]]>It should be:

**with** **if** and if **for**

I watched this great and inspiring talk a few weeks ago so it would be a pleasure to share my experience with the problem. Unfortunatley my approach is not to different to the ones alleady mentioned. But since i already wrote it down, iâ€™ll post it anyway :-).

**Hereâ€™s how i approached problem:**

**Insight 1:**

I started by focussing on the first inequality and trying to build the proof up from there. If you decompose the right side of the inequality into , you get and the arithmetic mean of . I noticed that the inequality holds for but has to break eventually as the arithmetic mean declines in proportion to and you have as gets large.

This insight led me to propose a helping **Lemma 1:**

**for** : **with** **if** and if

Itâ€™s relatively straightforward to see, that the proof of Lemma 1 can be divided into 3 parts:

**(1) The first inequality holds for n = 1**

obvious

**(2) The first inequality breaks for some n**

This can be seen by setting . For the arithmetic mean, you have:

as . As the inequality has integers on both sides and it follows that:

**(3)**

A simple transformation of the inequality is sufficient:

=

=

**qed**

**Insight 2:**

Itâ€™s now clear that the left inequality holds for all n below a certain bound and fails otherwise. It would make sense to assume that the right inequality holds for all n above a certain bound and fails otherwise. As the solution that satisfies both inequalities has to be unique, it also makes sense to assume . These assumptions leads to the following **case analysis:**

**Case 1)** :

The left inequality fails.

**Case 2)** :

We know that , therefore we can follow like in step **(3) in Lemma 1** that: which violates the right inequality.

**Case 3)** :

The left inequality is guaranteed to be true, as for the right inequality we have:

=

=

0. The middle term looks sort of like average. Try multiplying both sides by n. Unclear what to do with that…

1. Hmm. We have two inequalities. Let’s figure out whether we can get at least one of them to hold for some n and then worry about getting both. [= start by proving a weaker statement]

2. A silly guess – maybe LHS simply implies RHS? Quick computation shows that not… but as a by-product of this computation, I got: if RHS hold for n, then LHS doesn’t hold for n+1 ! This sounds like progress.

3. Carrying this line of thought further, I immediately obtain that LHS and RHS are both “monotone” (hold up to some point and then don’t or vice versa).

4. By Observation 2, the only scenario we have to worry about is when neither LHS nor RHS hold. But then we immediately arrive at contradiction with the assumption that a_n is increasing!

]]>Something like: ah – the problem has to do with sums of arrays and some kind of average – seen that before … I have some techniques I could use,

It would be interesting to know if that intuition is 100% reliable (well I suppose Turing forbids 100% – 90% then).

I have an counter example of failing intuition: the problem is how many license plates numbered 000000-999999 can be used if each plate must differ from any other in at least 2 places (rather than 1)? Should be easy, right? But I got stuck. I guess I need a flash of inside like in the dominoes-on-a-chessboard problem.

PS I am not a mathematician, but an engineer who reads your blog with great interest.

]]>