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

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

Looking for multiple solutions.

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

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

Looking for multiple solutions.

No vote yet

1 vote

×

Problem Loading...

Note Loading...

Set Loading...

## Comments

Sort by:

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

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

Log in to reply

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

Great Solution!

Log in to reply

Did you have your own solution?

Log in to reply

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

Log in to reply

Nice observation.

Log in to reply

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

Log in to reply

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 \(\le \).

Log in to reply

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.

Log in to reply

Log in to reply

Log in to reply

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?

Log in to reply

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

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.

Log in to reply

Comment deleted Jun 20, 2014

Log in to reply

Log in to reply

Log in to reply

Log in to reply

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

Log in to reply

Log in to reply

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

Log in to reply

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

Log in to reply

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

Log in to reply