If x, y, z are real and distinct, then u=x2+4y2+9z2-6yz-3zx-zxy is always

x,y,z∈R and distinct.Now, u=x2+4y2+9z2-6yz-3zx-2xy=12(2x2+8y2+18z2-12yz-6zx-4xy)=12{(x2-4xy+4y2)+(x2-6zx+9z2)+(4y2 -12yz+9z2)} =12{(x-2y)2+(x-3z)2+(2y-3z)2} Since it is sum of squares. So u is always non-negative.