Another Algebraic Proof Problem

Let a,ba, b and cc be positive real numbers such that abc=1abc = 1. Prove that

aba5+b5+ab+bcb5+c5+bc+cac5+a5+ca1\frac {ab}{a^5 + b^5 + ab} + \frac {bc}{b^5 + c^5 + bc} + \frac {ca}{c^5 + a^5 + ca} \leq 1

Looking for multiple solutions.

Note by Sharky Kesa
5 years, 5 months ago

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A,b,c can have least value of 3 if. a=b=c=1 Taking a,b,c as 1 we can prove it

Bob Man - 5 years, 5 months ago

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put a=1,b=1/2,c=2 .. we get (16/49)+(2/35)+(1/34) on left hand side.. on solving, (163534)+(24934)+(14935)<=1 .. therefore, 24087 <= 58310

Pinky Sri - 5 years, 5 months ago

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You are supposed to prove for all a,b,ca, b, c, not just certain values. :D

Sharky Kesa - 5 years, 4 months ago

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take the values of (a,b,c) as (1,x,1/x) or (x,1,1/x) or (x,1/x,1) or (1,1,1)....here x is a positive real number..I'm suggesting numbers of this kind because condition was given that abc=1....I kindly request you to check for various values of x..still the condition is satisfied :D

Pinky Sri - 5 years, 4 months ago

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Cool one. At first I tried several standard approaches but failed.

We keep in mind that a,b,ca,b,c are positive reals and abc=1abc=1. First note that, from the inequality

a5a3b2a2b3+b5=(a2b2)(a3b3)=(ab)2(a+b)(a2+ab+b2)0a^5-a^3b^2-a^2b^3+b^5=(a^2-b^2)(a^3-b^3)=(a-b)^2(a+b)(a^2+ab+b^2)\geq 0

follows that

a5+b5+aba2b2(a+b)+ab      1a5+b5+ab1a2b2(a+b)+ab.a^5+b^5+ab\geq a^2b^2(a+b)+ab~~\implies \dfrac{1}{a^5+b^5+ab}\le \dfrac{1}{a^2b^2(a+b)+ab}.

Thus we have

cycaba5+b5+abcycaba2b2(a+b)+ab=cycabc2a2b2c2(a+b)+abc2=cycca+b+c=1.\sum_{\text{cyc}} \dfrac{ab}{a^5+b^5+ab} \le \sum_{\text{cyc}} \dfrac{ab}{a^2 b^2(a+b)+ab} = \sum_{\text{cyc}} \dfrac{abc^2}{a^2b^2c^2(a+b)+abc^2} = \sum_{\text{cyc}} \dfrac{c}{a+b+c} = 1.

Jubayer Nirjhor - 5 years, 5 months ago

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Your first step can also be done by the rearrangement inequality, as a, b>0, so a^3, b^3 and a^2, b^2 are in the same order, so a^5+b^5>a^2b^3+a^3b^2

Neel Nanda - 5 years, 5 months ago

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Nice observation.

Jubayer Nirjhor - 5 years, 5 months ago

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Wow, how did I not think of that? Now it's so obvious! :o

Great Solution!

Daniel Liu - 5 years, 5 months ago

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Did you have your own solution?

Sharky Kesa - 5 years, 5 months ago

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Multiply each fraction by whatever variable isn't used in it (so by cc in the first one) to get

abca5c+b5c+abc+abcb5a+c5a+abc+abcc5b+a5b+abc\dfrac{abc}{a^5c+b^5c+abc}+\dfrac{abc}{b^5a+c^5a+abc}+\dfrac{abc}{c^5b+a^5b+abc}

Now substitute abc=1abc=1:

1a5c+b5c+1+1b5a+c5a+1+1c5b+a5b+1\dfrac{1}{a^5c+b^5c+1}+\dfrac{1}{b^5a+c^5a+1}+\dfrac{1}{c^5b+a^5b+1}

We're set to apply an AM-HM inequality now! Here we go:

31a5c+b5c+1+1b5a+c5a+1+1c5b+a5b+1\dfrac{3}{\dfrac{1}{a^5c+b^5c+1}+\dfrac{1}{b^5a+c^5a+1}+\dfrac{1}{c^5b+a^5b+1}}

a5c+b5c+abc+b5a+c5a+abc+c5b+a5b+abc3\geq \dfrac{a^5c+b^5c+abc+b^5a+c^5a+abc+c^5b+a^5b+abc}{3}

Factor the numerator of the RHS, which is just

a5(b+c)+b5(a+c)+c5(a+b)a^5(b+c)+b^5(a+c)+c^5(a+b)

This expression has its maximum when the LHS has it's minimum and vice versa. By the Intermediate Value theorem, there exists an equality case, which happens to be when a=b=c=1a=b=c=1. Otherwise, the inequality is satisfied and we're done. AWESOME PROBLEM!! :D I'm probably going to write like three more proofs since I like this so much. :D

Finn Hulse - 5 years, 5 months ago

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I don't quite get your solution. When did you prove that cycaba5+b5+ab1\displaystyle\sum_{cyc}\dfrac{ab}{a^5+b^5+ab}\le 1?

In addition, the inequality sign is pointing the wrong way in the last expression. You can't find a maximum of cyc1a5c+b5c+1\sum\limits_{cyc}\dfrac{1}{a^5c+b^5c+1} if you put it in the denominator and have the sign point \le .

Daniel Liu - 5 years, 5 months ago

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I didn't really, I kind of skipped through the last step but the maximum was 1/31/3 for the HM and 1/31/3 also for the AM.

Finn Hulse - 5 years, 5 months ago

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@Finn Hulse Well, whatever you did, you surely did not prove the inequality.

Daniel Liu - 5 years, 5 months ago

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@Daniel Liu Yeah maybe. But I'm going to do a couple more, so at least one of them will be legit! :D

Finn Hulse - 5 years, 5 months ago

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Sharky, I can't find anything smart to do. I've just tried like 3030 approaches and they are all cumbersome and are just a bunch of substitutions. Can you give me a hint?

Finn Hulse - 5 years, 5 months ago

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a5+b5a2b2(a+b)a^5 + b^5 \geq a^2b^2(a + b)

because

(a3b3)(a2b2)0(a^3 - b^3)(a^2 - b^2) \geq 0

with equality if and only if a=ba = b. That was the first bit of my solution.

Sharky Kesa - 5 years, 5 months ago

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Gah dude I would feel bad if I just took your solution but that is a very good start. I'm going to find some way to apply some discrete inequality to this problem. :D

Finn Hulse - 5 years, 5 months ago

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@Finn Hulse No worries. The first part was the easiest part to find. The next steps are harder. It'll be basically a challenge where you got some help to start off.

Sharky Kesa - 5 years, 5 months ago

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