Absolute value $\lvert x \rvert >-1$ (precalculus) How can $\lvert x \rvert >-1$ be true for all real $x$?
If $x\geq 0$: $x>-1$. 
If $x<0$: $-x>-1 \iff x<1$. 
So $-1<x<1$.
But if I for instance take $x=-5$ I get $\lvert -5 \rvert >-1 \iff -(-5)>-1 \iff 5>-1$. This is true but contradicts $-1<x<1$. 
What is wrong here?
Update: 
Would it be any difference if I instead had $\lvert x\rvert \geq -1$?

Your analysis is flawed. Your conclusion "$x>-1$" on the second line only follows under your condition that "$x\geq 0$". But the conclusion "$x<1$" on the third line only follows under your condition that "$x<0$".
These are mutually exclusive conditions -- there is no $x$ that simultaneously satisfies them both. You are combining conclusions from different assumptions.
