×

# 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
2 years, 11 months ago

Sort by:

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

$\sum_{\text{cyc}} \dfrac{ab}{a^5+b^5+ab} \le \sum_{\text{cyc}} \dfrac{ab}{a^2 b^2(a+b)+ab} = \sum_{\text{cyc}} \dfrac{abc^2}{a^2b^2c^2(a+b)+abc^2} = \sum_{\text{cyc}} \dfrac{c}{a+b+c} = 1.$ · 2 years, 11 months ago

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

Great Solution! · 2 years, 11 months ago

Did you have your own solution? · 2 years, 11 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 · 2 years, 11 months ago

Nice observation. · 2 years, 11 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 · 2 years, 11 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 $$\le$$. · 2 years, 11 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. · 2 years, 11 months ago

Well, whatever you did, you surely did not prove the inequality. · 2 years, 11 months ago

Yeah maybe. But I'm going to do a couple more, so at least one of them will be legit! :D · 2 years, 11 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? · 2 years, 11 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. · 2 years, 11 months ago

Comment deleted Jun 20, 2014

Dude it's in the domain $$abc=1$$. · 2 years, 11 months ago

How does that have anything to do with what I said? · 2 years, 11 months ago

What did you say? · 2 years, 11 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 · 2 years, 11 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. · 2 years, 11 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 · 2 years, 11 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 · 2 years, 11 months ago

You are supposed to prove for all $$a, b, c$$, not just certain values. :D · 2 years, 9 months ago