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Thread: Stuck on a Minimum Value Problem

  1. #1
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    Stuck on a Minimum Value Problem

    f(x,y,z)=cos^2(x-y)+cos^2(y-z)+cos^2(z-x)

    x,y,z is in the set of all real numbers

    Find the minimum value
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  2. #2
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    Quote Originally Posted by catbodak View Post
    f(x,y,z)=cos^2(x-y)+cos^2(y-z)+cos^2(z-x)

    x,y,z is in the set of all real numbers

    Find the minimum value
    You need to find $\displaystyle \nabla f = 0$
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  3. #3
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    Ok, I have been working on this, but am still stuck. What am I doing wrong, or what do I need to do next:

    GradF= Fx+F y +Fz
    Fx= -sin(2(x-y))+sin(2(z-x))
    Ff = sin(2(x-y)) -sin (2(y-z))
    Fz = -sin(2(z-x)) +sin(2(y-z))

    But it seems to me like all of these terms cancel out if I add them together. So I do not know how to proceed.
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  4. #4
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    Let $\displaystyle A = \cos^2(x-y)+\cos^2(y-z)+\cos^2(z-x) $

    Note that $\displaystyle A = \cos^2(x-y)+\cos^2(y-z)+\cos^2(z-x) = 1 + 2 \cos(x-y)\cos(y-z)\cos(z-x)$

    By $\displaystyle AM \geq GM $

    $\displaystyle \frac{A}{3} \geq \left( \cos(x-y)\cos(y-z)\cos(z-x) \right)^{2/3} ~~ \implies $

    $\displaystyle \left (\frac{A}{3} \right)^3 \geq (\cos(x-y)\cos(y-z)\cos(z-x))^2 $

    $\displaystyle \frac{A^3}{27} \geq ( \frac{A-1}{2} )^2 $

    $\displaystyle 4A^3 - 27(A-1)^2 \geq 0 $

    For the cubic equation $\displaystyle 4x^3 - 27(x-1)^2 = 0 $

    Sub $\displaystyle x = 1/t $ , it becomes

    $\displaystyle 4 - 27x(x-1)^2 = 0 $

    sub $\displaystyle x = 1 + y $ , it becomes

    $\displaystyle 4 - 27(y+1)y^2 = 0 $
    $\displaystyle 27y^3 + 27y^2 - 4 = 0$
    $\displaystyle (27y^3 - 1 ) + 3(9y^2 - 1) = 0$
    $\displaystyle (3y-1)(9y^2 + 3y + 1) + 3(3y-1)(3y+1) = 0 $
    $\displaystyle (3y-1)[ 9y^2 + 12y + 4) =0$
    $\displaystyle (3y-1)(3y+2)^2 = 0 $

    back substitution , we obtain

    $\displaystyle (4x-3)(x-3)^2 = 0 $

    so the inequality , $\displaystyle 4A^3 - 27(A-1)^2 \geq 0 $ we have

    $\displaystyle (4A-3)(A-3)^2 \geq 0 $
    $\displaystyle 4A -3 \geq 0 ~\implies A \geq \frac{3}{4} $

    so the minimum value of it is $\displaystyle \frac{3}{4} $
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  5. #5
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    Here is another way to factorize the equation

    $\displaystyle 4x^3 - 27(x-1)^2 = 0 $

    $\displaystyle \frac{x^3}{27} - \frac{(x-1)^2}{4} = 0 $

    $\displaystyle \left( \frac{x^3}{27} - 1 \right ) - \left( \frac{(x-1)^2}{4} -1 \right) = 0$

    $\displaystyle \frac{ (x-3)(x^2+3x+9) }{27} - \frac{ (x-1-2)(x-1+2)}{4} = 0$

    $\displaystyle \frac{ (x-3)(x^2+3x+9) }{27} - \frac{ (x-3)(x+1)}{4} = 0 $

    $\displaystyle (x-3)( 4x^2 + 12x + 36 - 27x -27 )= 0$

    $\displaystyle (x-3) ( 4x^2 - 15x + 9 ) = 0 $

    $\displaystyle (x-3) ( 4x-3)(x-3) = 0 $

    $\displaystyle (4x-3)(x-3)^2 = 0$
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