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# Reverse Rearrangement Inequality

Finally posted the paper on AoPS here!

It is on a new inequality I discovered while writing a Proofathon problem a while back.

Note by Daniel Liu
3 years ago

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That you're amazing might be sort of an understatement!

- 3 years ago

This is a very interesting read. I especially like the examples which you have chosen to illustrate the technique, as they are otherwise non-trivial. The lemma used in your proof is reminiscent to that of the Rearrangement Inequality.

Can you add it to the Reverse Rearrangement Inequality Wiki page?

P.S. I think it could have a much better name. What you need to do, is find a bunch of "hard" inequalities of which this approach simplifies it, and someday we can call it Daniel's Lemma. I've definitely seen this approach used in several olympiad problems. The 2 variable case is often overlooked by simply expanding terms and canceling, while the 3 variable case starts to show it's usefulness.

Staff - 3 years ago

That's where I would need to have some sort of spark of ingenuity. Currently, most of the inequalities in the problems section can be solved using Holder's Inequality (the symmetric ones usually) and I heard that some of the others can be solved using clever applications of AM-GM. But then, most problems can be solved with AM-GM used in some way or the other, so I'm not sure if that qualifies as my problems being bad.

Currently, I'm thinking of doing something based on the corollaries, which seem to give the problem maker relative freedom, as we just need to plug in any arbitrary increasing/decreasing function in some domain (which will be specified in the problem) in order to create a problem. However, in most cases the application is pretty obvious. I'm not sure if it is also obvious using other inequalities though. I also want to use the "any permutation" condition to my advantage, as not every inequality allows that.

As an example of an inequality trivial by Reverse Rearrangement: $\prod_{cyc}\big((y+1)^2+(x+y)(x-y)\big)\ge \big((x+1)(y+1)(z+1)\big)^2$

- 3 years ago

I created the Wiki page. I also went along and created the Holder's Inequality page too (for basic holders)

- 3 years ago

Thanks. If you add more examples to it, I can add it to the featured Wiki list. I will try and add a section on motivation / explanation too.

Staff - 3 years ago

Well, now there are three examples. Sorry, I was too lazy to make up new problems, so I just took problems from my paper.

- 2 years, 11 months ago

Thanks. I've made some further edits to it, can you take a look and do a sanity check?

Comment:

Observe that the base case of $$n=1, 2$$ are trivially true for similarly ordered sequences, and we do not require the condition of non-negativity.

I believe the condition on non-negative sequences, can be relaxed to having $$a_3 + b_3$$ be positive (where $$a_1, b_1$$ are the smallest terms in their sequence). I tried tracking down where non-negative sequence was used in the proof, and the only place that I see it is in the induction hypothesis where we have to multiply by $$(a_{k+1} + b_{k+1} )$$. We apply it when $$k = 2$$, hence we just require $$a_3 + b_3$$ to be positive, which makes $$a_ i + b_i$$ positive for $$i \geq 3$$.

If this were true, it could lead to very interesting results! We could have $$a_1, a_2, a_3, b_1, b_2$$ as negative values. Thoughts?

Staff - 2 years, 11 months ago

That's interesting... I was too lazy myself to check where it stopped working when some of the terms were negative. If what you said is true, wouldn't it mean we could have that the entire sequence $$b_1,b_2,\ldots b_n$$ can be negative, as long as $$a_k\ge b_k$$ for all $$k=1\to n$$?

I'll ask Cody Johnson do a quick check of this on Mathematica to see if what you said is actually true.

- 2 years, 11 months ago

- 3 years ago

wow man that was really great. :o kudos!! God..you guys are geniuses.. what am I even doing here among y'all :/ anyways....again..brilliant job :D

- 3 years ago

Don't worry, you will get there with persistence and ardor. As for the document, it is very intriguing indeed. Daniel, you have certainly cracked upon a new discovery that will be talked about for a while. I will spread the word immediately to my school and to my friends.

- 3 years ago

The Reverse Rearrangement Inequality Lower Bound is proven in the Rearrangement Inequality section in Math Olympiad Treasures by Titu Andreescu.

- 3 years ago

@Daniel Liu Please can you post it on Brilliant. No matter what i tried I have mot been able to find your inequality. P.s. Is it in the paper .If so where?

- 3 years ago

It should be right under section 2.

- 3 years ago

Your status says they didn't accept your submission. Can you tell what was wrong?

- 3 years ago

I'm wondering too. I sent an email asking why right now and am waiting for their reply.

EDIT: Dr. Andreescu said that they were not interested in articles about inequalities to publish right now. Bad timing, I guess.

- 3 years ago

@Daniel Liu how did you wrote in pdf ( can you explain me , i too want to create a pdf) , i know how to write in latex , can it be converted into pdf?

- 3 years ago

I used a program called TexWorks. It creates PDF's with the nice latex font that is characteristic in mathematical research papers.

- 3 years ago

Amazing. There you are creating (i.e-discovering) new concepts, here I am struggling to understand even the basic ones :/ (My status)

- 3 years ago

Hi Daniel, Can you remember what you were thinking when you discovered the inequality? In other words, What was going through your mind when you created something new?

- 3 years ago

I was creating an Algebra problem. I created the problem, then proceeded to solve it. However, I ended up having to prove that $$(m+n)(2m+2n)\cdots (km+kn)\le (m+\sigma(1)n)(2m+\sigma(2)n)\cdots (km+\sigma(k)n)$$ where $$\sigma(1),\ldots \sigma(k)$$ is a permutation of $$1,\ldots k$$.

I couldn't prove it, so I just gave up on the problem.

A month later, I came back, wondering if I could make a more generalized version of that inequality. It looked so much like the Rearrangement Inequality, it just had to be true. So I created the inequality as it is now, and then managed to prove it.

- 3 years ago

For this particular inequality, you can make use of

$\prod ( 1 + a_i ) \geq [ 1 + \sqrt[k]{ \prod{a_i} } ] ^k \quad - (1)$

Substitute in $$a_i = \frac{ \sigma(i) } { i } \times \frac{ n}{m}$$, clear out denominators and the result follows.

Note: The simplest approach that I know to prove inequality (1) is to take logarithms and apply Jensens to $$f(x) = \ln (1 + x )$$.

Staff - 3 years ago

Have you considered presenting it to the AMS journal?

- 3 years ago

I tried submitting it to AwesomeMath journal, but they declined because they weren't looking for inequality articles at the time. And now I already made it public, so I think it's too late already. These types of topics I doubt the AMS journal cares about, since it is purely competition-math style.

- 3 years ago

Where is the inequality?

- 3 years ago

- 3 years ago

- 3 years ago

If you're on mobile, then it doesn't show up. You must be on a computer for the attachment to show up.

As for the inequality, I also posed it on Brilliant Wiki, you can find it here.

- 3 years ago