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

  1. Jul 16, 2011 #1
    Given a,b,c>0 satisfying a^2+b^2+c^2=2; find the maximum value of

    Note : Do not use calculus....use AM-GM or anything...
  2. jcsd
  3. Jul 16, 2011 #2


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    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!
  4. Jul 16, 2011 #3
    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...
  5. Jul 17, 2011 #4


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    Yes, I see that now. So instead of deleting, I will move it to "General Mathematics".
  6. Jul 17, 2011 #5


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    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.
  7. Jul 17, 2011 #6
    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 [itex]a = b = c = \frac{\sqrt{2}}{\sqrt{3}}[/itex], then


    [itex]= (2 (\frac{\sqrt{2}}{\sqrt{3}})^5)^3 [/itex]

    [itex]= 8 (\frac{\sqrt{2}}{\sqrt{3}})^{15}

    = 8 (\frac{2}{3})^7 (\frac{\sqrt{2}}{\sqrt{3}})
    > 0 [/itex]

    Can anyone show that this is the max?

    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.
    Last edited: Jul 17, 2011
  8. Jul 17, 2011 #7


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    It certainly isn't the max ... the set [itex]\{1,\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\}[/itex] gives a larger result, and [itex]\{1,\frac{\sqrt{3}}{2},\frac{1}{2}\}[/itex] 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] :rofl: 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.
  9. Jul 17, 2011 #8
    What are you people doing? Look at the post again, NONE of the unknowns can be zero.:smile:
  10. Jul 17, 2011 #9


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

    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:

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

    so the set [itex]\{1,1-\epsilon,\epsilon\}[/itex] 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.
  11. Jul 17, 2011 #10
    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. :smile:
  12. Jul 17, 2011 #11
    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.
  13. Jul 17, 2011 #12
    The right answer is in fact 2. (I cheated LONG ago.)
  14. Jul 17, 2011 #13
    Oh that's funny. If we were moving relative to one another would that cause a simultaneity paradox?
  15. Jul 17, 2011 #14
    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 [itex]\geq[/itex] 0.
  16. Jul 17, 2011 #15
    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.:smile:
  17. Jul 17, 2011 #16
    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.
  18. Jul 17, 2011 #17
    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.:smile:

    As long as all xi are distinct.
    Last edited: Jul 17, 2011
  19. Jul 17, 2011 #18
    Only if all three are distinct.
  20. Jul 17, 2011 #19
    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. :smile:
  21. Jul 17, 2011 #20
    My point is that a symmetry argument can provide more solutions without doing any more work. I find that useful.:smile:
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