that equation only works for x,y,z≥0 right?=/
well, how I solved it was
let x≥0
let y=x+a
let z=x+b
let b≥a≥0
substitute into equation, expand, cancel out, you get
3xa^2 + 3xb^2 + a^3 + b^3 ≥ 3xab
Now x, a and b are ≥ so all terms above are ≥ 0,
For 3xb^2 ≥ 3xab
b ≥ a which is true.
Therefore...