A logical Math question and am bruding too much over it.Help

[Vishalrox]

Given a,b,c>0 satisfying a^2+b^2+c^2=2; find the maximum value of
(a^5+b^5)(b^5+c^5)(c^5+a^5)

Note : Do not use calculus...use AM-GM or anything...

 

[HallsofIvy]

So you post a question in "Linear and Abstract Algebra" that has nothing to do with either Linear Algebra or Abstract Algebra and challenge us to answer it without using anything? Sort of a Zen thing- the only correct response is to delete this thread!

 

[spamiam]

From what I understood, he meant don't use calculus, but we may use AM-GM or anything else. I admit, I would probably solve this using Lagrange multipliers if he hadn't said that...

 

[HallsofIvy]

Yes, I see that now. So instead of deleting, I will move it to "General Mathematics".

 

[mathman]

Given a,b,c>0 satisfying a^2+b^2+c^2=2; find the maximum value of
(a^5+b^5)(b^5+c^5)(c^5+a^5)

Note : Do not use calculus...use AM-GM or anything...

I haven't solved it directly, but from past experience with questions like this, there are most likely two possibilities: (1) a=b=c, (2) two of the unknowns are 0.

 

[SteveL27]

I haven't solved it directly, but from past experience with questions like this, there are most likely two possibilities: (1) a=b=c, (2) two of the unknowns are 0.

Two of the unknowns can't be 0, because then the product of the three expressions would be 0.

On the other hand if you take $a = b = c = \frac{\sqrt{2}}{\sqrt{3}}$, then

$(a^5+b^5)(b^5+c^5)(c^5+a^5)$

$= (2 (\frac{\sqrt{2}}{\sqrt{3}})^5)^3$

$= 8 (\frac{\sqrt{2}}{\sqrt{3}})^{15} = 8 (\frac{2}{3})^7 (\frac{\sqrt{2}}{\sqrt{3}}) > 0$

Can anyone show that this is the max?

(edit)
Hmmm, doesn't look like that's max. If you take

a = b = 1, c = 0

then the product is (1+1)(1+0)(0+1) = 2, and that's greater than what I got above for a = b = c. ((2/3)^7 is way less than 1/8). Unless I made an algebra mistake, a = b = c is not the max.

 

[SpectraCat]

Two of the unknowns can't be 0, because then the product of the three expressions would be 0.

On the other hand if you take $a = b = c = \frac{\sqrt{2}}{\sqrt{3}}$, then

$(a^5+b^5)(b^5+c^5)(c^5+a^5)$

$= (2 (\frac{\sqrt{2}}{\sqrt{3}})^5)^3$

$= 8 (\frac{\sqrt{2}}{\sqrt{3}})^{15} = 8 (\frac{2}{3})^7 (\frac{\sqrt{2}}{\sqrt{3}}) > 0$

Can anyone show that this is the max?

(edit)
Hmmm, doesn't look like that's max. If you take

a = b = 1, c = 0

then the product is 2 * 1 * 1 = 2, and that's greater than what I got above for a = b = c. ((2/3)^7 is way less than 1/8). Unless I made an algebra mistake, a = b = c is not the max.

It certainly isn't the max ... the set $\{1,\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\}$ gives a larger result, and $\{1,\frac{\sqrt{3}}{2},\frac{1}{2}\}$ is even larger than that.

My intuition is that that maximum is given by the set {1,1,0}, which gives a result of 2, but I haven't been able to figure out how to prove it yet.

[EDIT] you must have submitted your edit just as I was clicking to respond ... it made it into my quote, but I didn't even notice.

 

[agentredlum]

What are you people doing? Look at the post again, NONE of the unknowns can be zero.

 

[SpectraCat]

What are you people doing? Look at the post again, NONE of the unknowns can be zero.

Hey, when you can't figure it out, try to change the rules !

Seriously, I did miss that. However in that case, I am not sure you can even express the answer without calculus, because the maximum value can only be expressed as a limit. That is, the following condition holds:

$\lim_{\epsilon\rightarrow 0}\left[(1+\epsilon^5)(1+\sqrt{1-\epsilon^2}^5) (\epsilon^5+\sqrt{1-\epsilon^2}^5)\right]=2$

so the set $\{1,1-\epsilon,\epsilon\}$ gives a result of 2 in the limit where epsilon goes to zero. I don't know how to express that result without using calculus.

Of course, I am assuming that 2 is the theoretical maximum value of the expression. I know this is true when one of {a,b,c} is equal to 1. My intuition tells me that having one of {a,b,c} greater than 1, or having all of them less than 1 (which are the only other possibilities), will always yield a smaller result. However, once again I can't prove it.

 

[agentredlum]

Hey, when you can't figure it out, try to change the rules !

Seriously, I did miss that. However in that case, I am not sure you can even express the answer without calculus, because the maximum value can only be expressed as a limit. That is, the following condition holds:

$\lim_{\epsilon\rightarrow 0}\left[(1+\epsilon^5)(1+\sqrt{1-\epsilon^2}^5) (\epsilon^5+\sqrt{1-\epsilon^2}^5)\right]=2$

so the set $\{1,1-\epsilon,\epsilon\}$ gives a result of 2 in the limit where epsilon goes to zero. I don't know how to express that result without using calculus.

Of course, I am assuming that 2 is the theoretical maximum value of the expression. I know this is true when one of {a,b,c} is equal to 1. My intuition tells me that having one of {a,b,c} greater than 1, or having all of them less than 1 (which are the only other possibilities), will always yield a smaller result. However, once again I can't prove it.

I have an idea...why don't you get the right answer using calculus and then derive it without calculus? You would know for sure what the right answer SHOULD be, this will untie your hands for the moment, then you could concentrate on proving it.

 

[SteveL27]

What are you people doing? Look at the post again, NONE of the unknowns can be zero.

Oh yes, right you are ... but when a and b are close to 1 and c is small, the expression to be maximized gets close to 2. It does look like a limit is involved.

 

[pmsrw3]

I have an idea...why don't you get the right answer using calculus and then derive it without calculus? You would know for sure what the right answer SHOULD be, this will untie your hands for the moment, then you could concentrate on proving it.

The right answer is in fact 2. (I cheated LONG ago.)

 

[SteveL27]

[EDIT] you must have submitted your edit just as I was clicking to respond ... it made it into my quote, but I didn't even notice.

Oh that's funny. If we were moving relative to one another would that cause a simultaneity paradox?

 

[SteveL27]

The right answer is in fact 2. (I cheated LONG ago.)

It's pretty clear that there is no solution to the problem as stated, since the expression does not attain a max with a,b,c>0.

The remaining question is now whether there's a non-calculus solution if we take a,b,c $\geq$ 0.

 

[agentredlum]

It's pretty clear that there is no solution to the problem as stated, since the expression does not attain a max with a,b,c>0.

The remaining question is now whether there's a non-calculus solution if we take a,b,c $\geq$ 0.

a, b, c must all be positive numbers less than sqrt(2) so clearly the expression must have a maximum.

Now there may be many combinations of a, b, c that give the same maximum.

 

[pmsrw3]

a, b, c must all be positive numbers less than sqrt(2) so clearly the expression must have a maximum.

The proposal (which I think is correct) is that it has a supremum, but no max that satisfies the inequality constraints a,b,c > 0.

 

[agentredlum]

As a matter of fact given the symmetry of (a^5+b^5)(b^5+c^5)(c^5+a^5) there will be at least 6 triples giving a maximum.

If {x1,x2,x3} solve this problem, so will all permutatations of it. So finding 1 solution will automatically give you 5 more simply by taking symmetry into account.

As long as all xi are distinct.

 

[pmsrw3]

As a matter of fact given the symmetry of (a^5+b^5)(b^5+c^5)(c^5+a^5) there will be at least 6 different triples giving a maximum.

If {x1,x2,x3} solve this problem, so will all permutatations of it. So finding 1 solution will automatically give you 5 more simply by taking symmetry into account.

Only if all three are distinct.

 

[agentredlum]

Only if all three are distinct.

very good sir.

what if 2 are the same and 1 different, then they are not all distinct.

Even if they are all the same, permute the solutions, you still get 6 solutions all equal.

BTW...oh forget it.

 

[agentredlum]

My point is that a symmetry argument can provide more solutions without doing any more work. I find that useful.

 

[pmsrw3]

very good sir.

what if 2 are the same and 1 different, then they are not all distinct.

No, you get 3 distinct solutions in that case.

Even if they are all the same, permute the solutions, you still get 6 solutions all equal.

If they are all the same, you get no additional solutions by permuting them.

 

[agentredlum]

No, you get 3 distinct solutions in that case.

I agree. I edited the post to account for your wonderful observation.

 

[agentredlum]

It is easy to show they are not all the same, cause if they were a^2 + a^2 + a^2 = 2

3a^2 = 2 so a, b, c would all be sqrt(2/3)

Also a^5 would equal b^5 would equal c^5 so...

(a^5+b^5)(b^5+c^5)(c^5+a^5) = 8a^15 every binomial in parenthesis would be 2a^5

since a is about .8165 raising it to the 15th power gives about .04779 and multiplying by 8 gives about .382

a counterexample is enough to show .382 is not the max.

let a^2 = 1.1, let b^2 = .89, let c^2 = .01 the sum is 2

a^5 is about 1.269, b^5 is about .7472, c^5 is about zero.

(1.269 + .7472)(.7472 + 0)(0 + 1.269) = (2.0162)(.7472)(1.269) = 1.912

 

[SpectraCat]

It is easy to show they are not all the same, cause if they were a^2 + a^2 + a^2 = 2

3a^2 = 2 so a, b, c would all be sqrt(2/3)

Also a^5 would equal b^5 would equal c^5 so...

(a^5+b^5)(b^5+c^5)(c^5+a^5) = 8a^15 every binomial in parenthesis would be 2a^5

since a is about .8165 raising it to the 15th power gives about .04779 and multiplying by 8 gives about .382

a counterexample is enough to show .382 is not the max.

let a^2 = 1.1, let b^2 = .89, let c^2 = .01 the sum is 2

a^5 is about 1.269, b^5 is about .7472, c^5 is about zero.

(1.269 + .7472)(.7472 + 0)(0 + 1.269) = (2.0162)(.7472)(1.269) = 1.912

Yeah .. we already worked this out back in posts 6 & 7

 

[agentredlum]

Yeah .. we already worked this out back in posts 6 & 7

Oh...sorry. My browser does not decode TeX so most of the time i see gibberish instead of formulas.

 

[pmsrw3]

Greetings SpectraCat!

If you can prove that the maximum is given by {1, 1, 0} then this has a very interesting consequence.
The consequence is that the exponents DON'T matter in (a^5+b^5)(b^5+c^5)(c^5+a^5) no matter what natural number exponent you put on a, b, or c you still get the same maximum, 2.

This would mean that the supremum of (a^n+b^n)(b^n+c^n)(c^n+a^n) is 'locked in' by the constraint
a^2 + b^2 + c^2 = 2

No, this is clearly wrong. For instance, let n = 2. Then with a = b = c = (2/3)^(1/2), you have (a^2+b^2)(b^2+c^2)(c^2+a^2) = (4/3)^3 = 64/27 > 2.

 

[agentredlum]

No, this is clearly wrong. For instance, let n = 2. Then with a = b = c = (2/3)^(1/2), you have (a^2+b^2)(b^2+c^2)(c^2+a^2) = (4/3)^3 = 64/27 > 2.

Good point, however your counterexample doesn't work for integer n>2