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I spent an hour or so yesterday trying to solve the inequality $x^2 + x + 1 > 0$. Since I'd spent so long on a problem didn't seem like it should be that difficult, I decided I'd call it a day and try it again later.

I just had another look at it and this solution became immediately obvious:

$$x^2 + x + 1 > 0 \ \ \forall \ \ x \in \mathbb{R}$$

I'd justify this by stating that $x^2 > x \ \ \forall \ \ x \in \mathbb{R}$. Because of this, even if $x < 0$, the right hand side of the inequality will always be positive. Am I correct?

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    actually $x^2 < x $ for $0 .2012-03-09
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    Your claim is false $(1/2)^2< 1/2$.2012-03-09
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    Didn't think about that ..2012-03-09
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    It is true that $x^2 + x + 1 > 0$ for all $x \in \mathbb{R},$ but not for the resons you give. If $0 < x < 1,$ we have $x^2 < x.$2012-03-09

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