# Another Algebraic Proof Problem

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.

Note by Sharky Kesa
5 years, 9 months ago

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

We keep in mind that $a,b,c$ are positive reals 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

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

- 5 years, 9 months ago

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

Great Solution!

- 5 years, 9 months ago

Did you have your own solution?

- 5 years, 9 months ago

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

- 5 years, 9 months ago

Nice observation.

- 5 years, 9 months ago

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?

- 5 years, 9 months ago

$a^5 + b^5 \geq a^2b^2(a + b)$

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.

- 5 years, 9 months ago

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

- 5 years, 9 months ago

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.

- 5 years, 9 months ago

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

- 5 years, 9 months ago

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 $\le$.

- 5 years, 9 months ago

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.

- 5 years, 9 months ago

Well, whatever you did, you surely did not prove the inequality.

- 5 years, 9 months ago

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

- 5 years, 9 months ago

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

- 5 years, 9 months ago

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

- 5 years, 8 months ago

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

- 5 years, 8 months ago

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

- 5 years, 9 months ago