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Homework Help: How to prove this?

  1. Feb 19, 2012 #1
    1. The problem statement, all variables and given/known data
    How to prove (a+1/a)^2 + (b+1/b)^2 >= 25/2 given that a+b=1 and a,b positive

    2. Relevant equations

    3. The attempt at a solution
    I tried replacing b with 1-a but I get 6th degree a which I dont know how to find inequality for.
    Since a and b are perfect symmetric in this problem then a=b=0.5 is of interest. I get that its the minimum.
    I tried Lagrange multiplier but I get an equation with 3rd degree. By trial and error, I got solution.
    Is there any other way of finding solution? By expanding it?
  2. jcsd
  3. Feb 19, 2012 #2


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    hi topengonzo! :wink:

    have you tried differentiating? :smile:
  4. Feb 19, 2012 #3


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    Can you show what kind of equations you got from the Lagrange multiplier technique? I think you can argue that a critical point only occurs where a=b, even if you can't explicitly solve the equations.
  5. Feb 19, 2012 #4
    I differentiated it with respect to a and i get the equation

    a - 1/(a^3) - (1-a) + 1/(1-a)^3 = 0 which I cant solve :S
  6. Feb 19, 2012 #5
    the equations are:

    2(a^4) - λ (a^3) -2 = 0
    2(b^4) - λ (b^3) -2 = 0

    I can solve by trial and error to get a=b=0.5 but what if there is another root for these equations
  7. Feb 19, 2012 #6
    I just have to prove ab + 2/ab -4 >= 25/2 when a+b=1
    How to do it without Lagrange
  8. Feb 19, 2012 #7


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    If (a+1/a)^2 + (b+1/b)^2 has a critical point, then if you change the sign of a or b you still have a critical point. Let's just look for critical points where a>0 and b>0. If you solve for λ you have λ=2*(a^4-1)/a^3 and λ=(b^4-1)/b^3. You want to prove f(x)=2*(x^4-1)/x^3 is a 1-1 function when x>0, so if f(a)=f(b) shows a=b. Take the derivative of f(x) and show it's always positive. That would show it's a 1-1 function.
  9. Feb 20, 2012 #8
    If you multiply through by the lcd, it will become

    [tex]a^3(2a - 1)(1 - a)^3 - (1 - a)^3 + a^3 = 0[/tex]

    You may be able to see a solution, or expand it and use the

    Rational Root theorem.

    Then, in part, you can test values near that value less than and greater
    than it to see if it corresponds to a relative minimum.
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