<s>[INST] Proving $x^2+x+1\gt0$ I was doing a question recently, and it came down to proving that $x^2+x+1\gt0$. There are of course many different methods for proving it, and I want to ask the people here for as many ways as you can think of.

My methods:


*

*$x^2+x+1=(x+\frac12)^2+\frac34$, which is always greater than $0$.

*Let it be $0$ for some $x=k$. Then $x^2+x+1=0$ has a real solution. But since $1^2\not\gt4$, this has no real solution. Therefore it is more than $0$.
 [/INST] Here a rather geometric way:
$$y = x^2+x+1 = x(x+1) + 1$$
So, $y = x^2+x+1$ is the parabola $y=x(x+1)$ shiftet by $1$ upwards.
$y=x(x+1)$ has its vertex at $x_V = -\frac{1}{2} \Rightarrow y_V = -\frac{1}{4}$
So, the vertex of $y= x(x+1) + 1$ is also at $x_V = -\frac{1}{2}$ with a minimum value of $y_{min}= -\frac{1}{4}+1 = \frac{3}{4}>0$
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