Show that $f(x,y,z)=6x^2+4y^2+2z^2+4xz-4yz \geq 0$ for all $x, y, z$ except for $x=y=z=0$ When I try to factor the quadratic form, I end up with 
$$6x^2+4y^2+2z^2+4xz-4yz = 2((x+z)^2+2x^2+(y-z)^2-z^2)$$ 
which does not ensure that $f(x,y,z) \geq 0$ for all $x, y, z \geq 0$ since the $z$ term is negative. How should these kinds of problems be tackled?

$$6x^2+4y^2+2z^2+4xz−4yz=(4x^2+z^2+4xz)+(4y^2+z^2−4yz)+2x^2=(2x+z)^2+(2y-z)^2+2x^2\ge 0$$
