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 $\lvert x\rvert \geq -1$?
On
These are two independent cases. A particular value of $x$ satisfies one of the cases, not both. Therefore combining the inequalities is incorrect.
You have "if $x<0$ then $-x > -1$" and "if $x\geq 0$ then $x > -1$" but no number satisfies both of those conditionals.
On
Don't approach it like that. I mean, don't consider $x \ge 0$ as one case and $x < 0$ as another case. Breaking into two inequalities like you did is fine, but technically they are not "$x \ge 0$" and "$x < 0$".
$|x| > -1$ is really the following compound inequality:
$$x > -1 \text{ OR } x < -(-1)$$
I think you approached it with AND in mind instead of OR.
But since it's "$|x| > \ $" then it should be OR.
The first inequality just gives you $x > -1$. The second one is $x < 1$. So you get:
$$ x > -1 \text{ OR } x < 1$$
And together these two inequalities just mean $x$ can be any real number.
Formal side note: OR is union: $\cup$. AND is intersection: $\cap$.
An easier way to approach this specific problem: $|x| \ge 0$ no matter what $x$ is. And since $0 > -1$, then no matter what $x$ is, you will always have $|x| \ge 0 > -1$. Therefore $|x| > -1$ is true for all real values of $x$.
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.