Find all the points (x, y, z) with x, y, z => 0 which maximizes the function
f(x, y, z) = xy + xz + yz - 4xyz subject to the constraint x + y + z = 1.
Please do NOT "hijack" other peoples threads by asking completely unrelated questions in them! It is extremely rude.
guess, use the "Lagrange Multiplier Method"- the max or min of f(x,y,z) subject to the constraint g(x,y,z)= constant must satisfy $\displaystyle \nabla f= \lambda \nabla g$ for some number $\displaystyle \lambda$.
For this problem, that is $\displaystyle <y+ z- 4yz, x+ z- 4xz, x+ y- 4xy>= \lambda <1, 1, 1>$. That is, you have $\displaystyle y+ z- 4yz= \lambda$, $\displaystyle x+ z- 4xz= \lambda$ and $\displaystyle x+ y- 4xy= \lambda$.
Those fairly easily reduce to y+ z- 4yz= x+ z- 4xz and y+ z- 4yz= x+ y- 4xy. Together with x+ y+ z= 1, you have three equations to solve for x, y, and z.
well , i am not keen on inequalities but i think it is a great attempt !
$\displaystyle xy + yz + zx - 4xyz $
$\displaystyle = \frac{ (1-4x)(1-4y)(1-4z) - 1 + 4(x+y+z) }{16}$
$\displaystyle = \frac{ (1-4x)(1-4y)(1-4z) + 3 }{16} $
$\displaystyle = \frac{1}{16} \left[ 3 + (1-4x)(1-4y)(1-4z) \right] $
use $\displaystyle GM \leq AM $
$\displaystyle \leq \frac{1}{16} \left[ 3 + \left( \frac{ (1-4x ) + (1-4y ) + (1-4z ) }{3} \right )^3\right] $
$\displaystyle = \frac{1}{16} ( 3 - \frac{1}{27} ) = \frac{5}{27} $
the equality holds when $\displaystyle x =y =z = 1/3 $
but it may be wrong because we can find that :
$\displaystyle GM \leq AM \leq 0 $ !!
I hope it will be perfectly corrected by some experts .