Let \(a, b\) and \(c\) be positive real numbers such that \(abc = 1\). Prove that

\[\frac {ab}{a^5 + b^5 + ab} + \frac {bc}{b^5 + c^5 + bc} + \frac {ca}{c^5 + a^5 + ca} \leq 1\]

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## Comments

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

We keep in mind that \(a,b,c\) are

positivereals and \(abc=1\). First note that, from the inequality\[a^5-a^3b^2-a^2b^3+b^5=(a^2-b^2)(a^3-b^3)=(a-b)^2(a+b)(a^2+ab+b^2)\geq 0\]

follows that

\[a^5+b^5+ab\geq a^2b^2(a+b)+ab~~\implies \dfrac{1}{a^5+b^5+ab}\le \dfrac{1}{a^2b^2(a+b)+ab}.\]

Thus we have

\[\sum_{\text{cyc}} \dfrac{ab}{a^5+b^5+ab} \le \sum_{\text{cyc}} \dfrac{ab}{a^2 b^2(a+b)+ab} = \sum_{\text{cyc}} \dfrac{abc^2}{a^2b^2c^2(a+b)+abc^2} = \sum_{\text{cyc}} \dfrac{c}{a+b+c} = 1.\]

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

Great Solution!

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Did you have your own solution?

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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

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Nice observation.

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

\[\dfrac{abc}{a^5c+b^5c+abc}+\dfrac{abc}{b^5a+c^5a+abc}+\dfrac{abc}{c^5b+a^5b+abc}\]

Now substitute \(abc=1\):

\[\dfrac{1}{a^5c+b^5c+1}+\dfrac{1}{b^5a+c^5a+1}+\dfrac{1}{c^5b+a^5b+1}\]

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

\[\dfrac{3}{\dfrac{1}{a^5c+b^5c+1}+\dfrac{1}{b^5a+c^5a+1}+\dfrac{1}{c^5b+a^5b+1}}\]

\[\geq \dfrac{a^5c+b^5c+abc+b^5a+c^5a+abc+c^5b+a^5b+abc}{3}\]

Factor the numerator of the RHS, which is just

\[a^5(b+c)+b^5(a+c)+c^5(a+b)\]

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 \(a=b=c=1\). 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

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

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

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

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Sharky, I can't find anything smart to do. I've just tried like \(30\) approaches and they are all cumbersome and are just a bunch of substitutions. Can you give me a hint?

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\[a^5 + b^5 \geq a^2b^2(a + b)\]

because

\[(a^3 - b^3)(a^2 - b^2) \geq 0\]

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

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Comment deleted Jun 20, 2014

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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

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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

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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, (16

3534)+(24934)+(14935)<=1 .. therefore, 24087 <= 58310Log in to reply

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

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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

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