# Math Help - Stuck on a Minimum Value Problem

1. ## 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

2. Originally Posted by catbodak
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 $\nabla f = 0$

3. Ok, I have been working on this, but am still stuck. What am I doing wrong, or what do I need to do next:

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.

4. Let $A = \cos^2(x-y)+\cos^2(y-z)+\cos^2(z-x)$

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

By $AM \geq GM$

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

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

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

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

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

Sub $x = 1/t$ , it becomes

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

sub $x = 1 + y$ , it becomes

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

back substitution , we obtain

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

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

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

so the minimum value of it is $\frac{3}{4}$

5. Here is another way to factorize the equation

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

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

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

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

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

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

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

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

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