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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.

user601846
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1 Answers1

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$x^2+y^2+z^2<xy+xz+yz\implies$

$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 $\implies$

$x+y+z > 0$

fleablood
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