$x, y$ and $z$ - 2.0

I got this wrong at first. Here is my solution by Lagrange Multipliers: Define the Lagrangian $\mathcal{L}=xy+xz+yz-3xyz-\lambda(x+y+z-1)$ We set the derivatives with respect to each variable to 0: $\frac{\partial \mathcal{L}}{\partial x}=y+z-3yz-\lambda=0\\\frac{\partial \mathcal{L}}{\partial y}=x+z-3xz-\lambda=0\\\frac{\partial \mathcal{L}}{\partial z}=x+y-3xy-\lambda=0\\\frac{\partial \mathcal{L}}{\partial \lambda}=1-x-y-z=0$

Then consider the equation $y+z-3yz-\lambda=x+z-3xz-\lambda\Rightarrow y-x=3z(y-x)\Rightarrow x=y \text{ or } z=0$ Similarly, $y=z$ or $x=0$ and $x=z$ or $y=0$ . This implies that for extremal cases $(x,y,z)=(1,0,0),(0.5,0.5,0),(1/3,1/3,1/3)$ . We try all three and (0.5, 0.5, 0) results in the maximum value of 0.25.