#System of equations
14 messages · Page 1 of 1 (latest)
add the 3 equations to get:
x^2 + y^2 + z^2 + (x+y+z)^2 + 3 = 4(x+y+z)
then use AM-GM with:
x^2 + 1/4
y^2 + 1/4
z^2 + 1/4
(x+y+z)^2 + 9/4
@heady plover
thats not a solution?
?
whats the solution
The solution is x=y=z=1/2
nvm the problem never states that x, y and z are positive
you should turn this into (x + y + z -3/2)^2 + (x - 1/2)^2 + (y - 1/2)^2 + (z - 1/2)^2 = 0 instead
How did u do that
as in, how did i have the insight to do that?
you should see that this is a cyclic system of equations (if you replace x -> y; y -> z and z -> x you get the exact same system of equation)