prove that $\sum_{cyc}\frac{{a^2}{b}}{c}\ge a^2+b^2+c^2$

Question

prove $$\sum_{cyc}\frac{{a^2}{b}}{c}\ge a^2+b^2+c^2$$ where $a,b,c>0$ and $a\ge b\ge c$

My try it seemed qite simple but i couldnt apply the rearrangement inequality directly. so i tried manipulating the inequality.

the inequality can be written as $$a^2(b-c)+b^2(c-a)+c^2(a-b)\ge 0$$ .It seemed like 'schurs' inequality could be uused but i couldn't procceed.Also i tried using the weighted a-m g-m method. Could anyone give me a hint (i want to solve the problem myself).

source: Excursion in mathematics(Modak)

(Not related to question) Good luck for JEE advanced, although if protesters have their way, you could be looking at a long wait. I gave it the first year they changed format, but now things must be very different! (If you wish to know more, call me to chat). — Sarvesh Ravichandran Iyer, Aug 27 '20 at 15:56
@TeresaLisbon thank you! i have just in 12 th so i guess things would have cleared up a bit by next april. — Albus Dumbledore, Aug 27 '20 at 15:59
@TeresaLisbon you gave JEE Advanced in 2006? What is your rank? I would like to know more. I gave JEE mains in January and have got a fairly good score. Also, I see you are a fan of Mentalist. Pardon me for chatting in the chat. — Vishnu Kadiri, Aug 27 '20 at 16:23
Also, I would like to which method @Quantum used to rewrite the inequality in that manner. — Vishnu Kadiri, Aug 27 '20 at 16:24
@VishnuKadiri I gave advanced in 2013, without any coaching center's help. I did not do too bad, in top 10000. JEE mains score was 200, I think. Also, I love Lisbon, she's the best — Sarvesh Ravichandran Iyer, Aug 27 '20 at 16:28
@TeresaLisbon if it was 2013 that would be quite difficult as the paper pattern changed from subjective to objective — Albus Dumbledore, Aug 27 '20 at 16:33
@Quantum Yes, you are correct. Look, I was not actually too keen on that. I gave the KVPY with more eagerness, and did well there! But now everything is an industry, it wasn't so(at least as much) back then. I do sympathise with you and Vishnu here. Just dug it out, I got 145/360 and rank 9741. — Sarvesh Ravichandran Iyer, Aug 27 '20 at 16:34
@TeresaLisbon what was your rank in KVPY. I wrote that in 11th and bungled up because i ddnt study 12th portions in 11th — Albus Dumbledore, Aug 27 '20 at 16:38
@Quantum I wrote it in $11$th, and bungled up. Then I wrote it in $12$th : rank $515$, but "only interview" rank was in top $10$, I've heard (but never confirmed : one guy told me this). That was pretty good, again given that I did no coaching of any kind, but just past papers, past papers, past papers. (Having a picture of Lisbon on your laptop helps!) — Sarvesh Ravichandran Iyer, Aug 27 '20 at 16:39
@VishnuKadiri i am yet to write it. (i have just started 12) what about yours — Albus Dumbledore, Aug 28 '20 at 04:48
@VishnuKadiri that is quite good. when you are free you can call me to chat.Would like to know more about your preperation strategies.good luck for jee adv.!! — Albus Dumbledore, Aug 28 '20 at 05:48

Michael Rozenberg · Accepted Answer · 2020-08-27T17:17:02.757

3

Your first step is wrong:

We need to prove that: $$\sum_{cyc}(a^3b^2-a^3bc)\geq0$$ and it indeed gives a proof: $$\sum_{cyc}(2a^3b^2-2a^3bc)\geq0$$ or $$\sum_{cyc}(a^3b^2+a^3c^2-2a^3bc)\geq\sum_{cyc}(a^3c^2-a^3b^2)$$ or $$\sum_{cyc}a^3(b-c)^2\geq(ab+ac+bc)(a-b)(b-c)(c-a),$$ which is obvious.

We can get $$\sum_{cyc}(a^3c^2-a^3b^2)=(ab+ac+bc)(a-b)(b-c)(c-a)$$ by the following way.

For $a=b$, $a=c$ and $b=c$ we obtain identity, which says that $$\sum_{cyc}(a^3c^2-a^3b^2)=P(a,b,c)(a-b)(b-c)(c-a),$$ where $P$ is a cyclic homogeneous polynomial of second degree.

Id est $$P(a,b,c)=\sum_{cyc}(ka^2+mab).$$ Now, $k=0$ because, otherwise there is a problem on $\infty$.

Also, let $c=0$.

We obtain: $$b^3a^2-a^3b^2=mab(a-b)b(-a)$$ or $$a^2b^2(b-a)=ma^2b^2(b-a),$$ which gives $m=1$.

Factoring of some Schur's polynomials: $$\sum_{cyc}(a^2b-a^2c)=(a-b)(a-c)(b-c),$$ $$\sum_{cyc}(a^3b-a^3c)=(a+b+c)(a-b)(a-c)(b-c),$$ $$\sum_{cyc}(a^4b-a^4c)=(a^2+b^2+c^2+ab+ac+bc)(a-b)(a-c)(b-c),$$ $$\sum_{cyc}(a^3b^2-a^3c^2)=(ab+ac+bc)(a-b)(a-c)(b-c),$$ $$\sum_{cyc}(a^5b-a^5c)=$$ $$=(a^3+b^3+c^3+a^2b+a^2c+b^2a+b^2c+c^2a+c^2b+abc)(a-b)(a-c)(b-c),$$ $$\sum_{cyc}(a^4b^2-a^4c^2)=(a^2b+a^2c+b^2a+b^2c+c^2a+c^2b+2abc)(a-b)(a-c)(b-c),...$$

edited Aug 27 '20 at 17:17

answered Aug 27 '20 at 16:27

Michael Rozenberg

194,933

1

Nice realization. – cr001 Aug 27 '20 at 16:31
one question : how did you factorise the RHS as ($ab+bc+ca)(a-b).(..) $ did you know it before or does it take time to factorise it – Albus Dumbledore Aug 27 '20 at 16:47
@Quantum Yes, of course. I knew it before. I got this factorization, when was a student of a school. This polynomial named as the Schur's polynomial. It's so useful that if you want I can write by memory factoring of any Schur's polynomials of sixth degree and less. – Michael Rozenberg Aug 27 '20 at 16:57
interesting, is there a link that shows such factorisations. i will be very grateful if you can provide that . i tried researching it a bit but most of them are too complicated and i dont see any similarity btw the ones you have used – Albus Dumbledore Aug 27 '20 at 16:59
thank you so much! i am going to memorize them – Albus Dumbledore Aug 27 '20 at 17:27
1

@MichaelRozenberg Thank you for giving such helpful results. Now I can understand the first step for this problem https://artofproblemsolving.com/wiki/index.php/2006_IMO_Problems/Problem_3 – Book Of Flames Aug 29 '20 at 06:24

score 2 · Answer 2 · answered Aug 27 '20 at 16:19

2

I will continue from the latest inequality presented by @Quantum. $\ {a}^{2}(b-c)+{b}^{2}(c-a)+{c}^{2}(a-b)={a}^{2}(b-c)+{b}^{2}(c-b+b-a)+{c}^{2}(a-b)$ $$\ ={a}^{2}(b-c)-{b}^{2}(b-c)-{b}^{2}(a-b)+{c}^{2}(a-b)=({a}^{2}-{b}^{2})(b-c)-({b}^{2}-{c}^{2})(a-b)$$ $$\ =(a-b)(b-c)(a+b-(b+c))=(a-b)(b-c)(a-c)\geq0$$

I hope it is simple.

answered Aug 27 '20 at 16:19

Vishnu Kadiri

1,516

This is not correct because quantum's representation is not correct – Math Lover Aug 27 '20 at 16:38
Yes i get my mistake. my simplification is wrong and so the whole proof is gone! – Albus Dumbledore Aug 27 '20 at 16:39
Hahaha, I should have simplified the original problem. No worries! – Vishnu Kadiri Aug 27 '20 at 16:42

score 1 · Answer 3 · answered Aug 27 '20 at 16:04

1

A simple solution is to say that for $b=a$ and $b=c$ with $a\ge b\ge c$ we have equality. Now differentiate

$$a^2b+b^2c+c^2a-a^2c-b^2a-c^2b$$ as a polynomial in $b$ it gives $(a-c)(a+c-2b)$ so it should be clear the behaviour as $b\in [c,a]$

answered Aug 27 '20 at 16:04

Toni Mhax

1,252

i understand the intuition behind and due to symmetry a=b=c is an extreme value however i am looking for a solid proof that does not take that assumption. – Albus Dumbledore Aug 27 '20 at 16:08
Your inequality in $b$ for $a$, $c$ fixed as you expand it takes $0$ for $b=c$ and $b=a$ and positive otherwise you know the derivative and the function variation... – Toni Mhax Aug 27 '20 at 16:13

prove that $\sum_{cyc}\frac{{a^2}{b}}{c}\ge a^2+b^2+c^2$

3 Answers3