Question

prove that x^2+y^2+z^2-xy-yz-zx is always positive

Answer 1

We know that,

$p (x.y.z)= {x}^{2} + {y}^{2} + {z}^{2} \\ - xy - yz - zx \\ = \frac{1}{2} (2 {x}^{2} + 2 {y}^{2} + 2 {z}^{2} \\ - 2xy - 2yz - 2zx) \\ = \frac{1}{2} ( {x}^{2} - 2xy + {y}^{2} + \\ \: \: \: \: \: \: \: \: \: \: \: {y}^{2} - 2yz + {z}^{2} + \\ \: \: \: \: \: \: {z}^{2} - 2zx + {x}^{2} ) \\ = \frac{1}{2} ( {(x - y) }^{2} + {(y - z)}^{2} + \\ {(z - x)}^{2} )$
Now, this square terms are always non negative.
P(x,y,z) >=0


Hence, if x is not equal to y not equal to z.
p(x,y,z) is always positive.