Solving $x^2 + x + 1 > 0$ I'm having a lot of trouble solving the above inequality in the "standard" way. I know by the plot that $x$ can be any value, but I can't seem to get to that answer.
Here are my steps:
$$x^2 + x + 1 > 0$$
$$x^2 + x > -1 $$
$$x (x + 1) > -1 $$
So now i check when both $x$ and $x+1$ are bigger than $-1$ and when both $x$ and $x+1$ are smaller than $-1$
First system of equations:
$$ x > -1$$
$$ x + 1 > -1 $$ turns into: $$ x > -2 $$
So the answer to the first set of equations is : $ x > -1 $
Second system of equations:
$$ x < -1$$
$$ x + 1 < -1 $$ turns into: $$ x < -2 $$
So the answer to the second set of equations is : $ x < -2 $
Which would give us a final answer of 
$$x < -2, x > 1$$, which I know is wrong.
My question is: where is my mistake in all this? Thanks for the help.
 A: Your logic falters at the step $x(x+1)>-1 \Rightarrow x>-1 \wedge x+1>-1$  and the corresponding assuming $x<-1$ and $x+1<-1$. That $x(x+1)>-1$ has two possibilities, \begin{align} -1<&x(x+1)\leq0 \\ 0<&x(x+1)\end{align}In the first case is only true if $x\leq0$ and $x+1\geq0$, so $-1\leq x\leq0$. The second case is possible if $x<0$ and $x+1<0$ or $x>0$ and $x+1>0$. The first case occurs precisely when $x<-1$, the second for $x>0$. From these, we conclude it holds for all $x\in\mathbb{R}$.
Your logic says $xy>z$ implies $x>z \wedge y>z$ or $xy>z$ implies $x<z\wedge y<z$. This is not true! If $x=z=-1, y=0$, then $$0=(-1)0>-1\not\Rightarrow -1>-1 \wedge 0>-1$$ and the same for the other case, which is a contradiction.
A: Knowing $ab > -1$ doesn't let you split into the cases you give. For instance, $a=-1/3$, $b=-3$.  I can't tell for sure, but you seem to be committing the error of lots of beginning algebra students:  They forget that $0$ is special.  If $ab=0$ then one of $a$ or $b$ is zero.  What works for $0$ doesn't work for other numbers, but they'll get excited and write "$x(x-1)=-1$ so either $x=-1$ or $x-1=-1$."  It seems like your error is similar, but with inequalities.  
A: Hint: Your function has no zeros (try p-q-formula). So its a parabola, that doesn't cross the x-axis. So the leading factor $\textbf{1}x^2 + x + 1$ tells you that the function is positive on $\mathbb R$.
A: You are wrong when you say:

now i check when both $x$ and $x+1$ are bigger than $-1$ and when both $x$ and $x+1$ are smaller than $-1$

supposing that this gives the solution of $x(x+1)>-1$. 
This is false. You can test, as an example: $x=-\frac{3}{2}<-1$ and $x+1=-\frac{1}{2} >-1$ that gives:
$x(x+1)=3/4>-1$
A: I don’t see how your method could have worked, since you don’t seem to have multiplied $x$ and $x+1$ anywhere. Following your lead, though, I would split the situation into three exhaustive and mutually exclusive cases: (1) $x\le-1$; (2) $x\ge0$; and (3) $-1<x<0$. My task is to show that in each of these three cases, $x(x+1)>-1$.
\begin{align}
x\le-1&\Longrightarrow x+1\le0\Longrightarrow x(x+1)\ge0>-1\quad\text{(good)}\\
x\ge0&\Longrightarrow x+1>0\Longrightarrow x(x+1)\ge0>-1\quad\text{(good)}\\
-1<x<0&\Longrightarrow0<x+1<1\text{ and }0<-x<1\Longrightarrow-x(x+1)<1
\Longrightarrow x(x+1)>-1\quad\text{(good)}\,,
\end{align}
which does it.
(But I think the right way to do it without Calculus would be to complete the square: $x^2+x+1=\bigl(x+\frac12\bigr)^2+\frac34$, clearly always positive.)
