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 $0$.
We note that:

$$x^2+x+1=\frac{x^3-1}{x-1}$$

and the sign of the RHS numerator and denominator are always equal, except for $x=1$.

We handle $x=1$ separately, but this is trivial on the LHS.
Put $x=\dfrac{b}{a}$ 

$x^2+x+1=\dfrac{b^2}{a^2}+\dfrac{b}{a}+1$ 
$=\dfrac{b^2+ab+a^2}{a^2}=\dfrac{\dfrac{1}{2}(a+b)^2+\dfrac{b^2}{2}}{a^2}+\dfrac{1}{2}&gt; 
 0$