I'm working on the following proof:
Prove that if x, y, and z are three real numbers such that $x^2+y^2+z^2<xy+xz+yz$, then $x+y+z>0$.
I know that the proof is meant to be either vacuously or trivially true, and since $\exists x,y,z \in\Bbb R$ such that $x+y+z>0$ is false, the proof is not trivially true and it must be that the antecedent is false for $\forall a,b,c \in \Bbb R$.
However, I cannot see how to prove this.
Any help would be appreciated. Thanks.
$x^2+y^2+z^2<xy+xz+yz\implies$
$2x^2 + 2y^2 + 2z^2 - 2xy -2xz -2yz < 0\implies $
$(x^2 -2xy + y^2) + (x^2-2xz + z^2) + (y^2-2yz + z^2) < 0\implies$
$(x-y)^2 + (x-z)^2 + (x-z)^2 < 0\implies$
Pink Elephants rule the world and every number is greater than 50 billion $\implies$
$x+y+z > 0$