Check my solution to $x^2 + x + 1 > 0$ 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?
 A: $$x^2 + x + 1 > 0 \Leftrightarrow x^2 + x + \frac{1}{4} > -1 + \frac{1}{4} \Leftrightarrow (x+\frac{1}{2})^2 > -\frac{3}{4} $$ And we have the desired result by the positivity of the square. (i.e. $\forall x \in \mathbb{R}: x^2 \geq 0$)
A: First I will show that with small modifications, your approach can be made to work. (It is, however, simpler to use one of the methods in other answers.)  At the end, I suggest an additional simple approach. 
It is clear that $x^2+x+1 \ge 1$ if $x \ge 0$. There is potential trouble only when $x<$, so suppose that $x<0$.
Because it is all too easy to make errors when handling negative numbers, let $w=-x$. Then $x^2+x+1=w^2-w+1$, and $w$ is positive.
If $w \ge 1$ (that is, if $x\le -1$), then $w^2 \ge w$, so $w^2 -w+1\ge 0$. (Here we used the method that you proposed.)
That method doesn't work when $0<w<1$, for then $w^2<w$.  But $w^2>0$ and $w<1$, so $w^2-w>-1$, and therefore $w^2-w+1>0$. 
Another way: Note that $(x-1)(x^2+x+1)=x^3-1$, and that $x-1$ and $x^3-1$ are positive if and only if $x>1$, and are negative if and only if $x<1$. 
If $x>1$, then $x-1$ and $x^3-1$ are both positive, so the ratio $\frac{x^3-1}{x-1}$, that is, $x^2+x+1$, is positive.
If $x-1$ and $x^3-1$ are both negative, then again their ratio $x^2+x+1$ is positive.
And finally if $x=1$, then $x^2+x+1$ is positive. 
Note that in exactly the same way, we can show that $x^n+x^{n-1}+\cdots +x+1$ is always positive if $n$ is even.
A: Hint $\rm\ \ 4\:f(x)\: =\: 4\:(x^2+x+1)\ =\ (2x+1)^2 + 3\: \ge\: 3\ $ thus $\rm\:f(x)\ge 3/4$
More generally one can decide polynomial  (in)equations by partitioning the real line into intervals based on the finite number of roots, and testing on each interval. Here there are no real roots, so it has constant sign, so has the sign of $\rm\:f(0) = 1.\:$
This is a special case of Collin's cylindrical algebraic decomposition algorithm, an effective form of Tarski-Seidenberg quantifier elimination for the reals. For deep generalizations, search on "semialgebraic geometry" and "o-minimal".
A: There's also this answer (but I'm not sure if it qualifies as a proof):

This is a plot of $y=x^2+x+1$, notice it never touches the $x$-axis.
A: I bring x and 1 to the other side to change the problem to $x^2 > -1 - x$.
Note that $x^2$ is always positive for real $x$, so the equality can only be false when $x < -1$.
But, if $x < -1$, $x^2$ is still positive, and so is $- x$. So we need to prove that when $x > 1$, then $x^2 > x - 1$ holds. But since $x^2 > x$ when $x > 1$ this holds. 
A: One more way to solve this (actually a different approach):
Since $1^2+1+1=3>0$ any y for which $y^2+y+1 \leq 0$ holds would yield a root of the polynomial $p(x)=x^2+x+1$ by the intermediate value theorem. Therefore you get a second root (besides $1$) of the polynomial $q(x)=p(x)(x-1)=x^3-1$. But this is a monotonically increasing function and hence has at most one zero.
