Absolute Value Brackets Proof Prove that if $|x-y| < c$ then $|x| < |y| + c$
Workings:
I have to ideas of a proof
Proof 1:
$$|x| =|x-y+y|\leq |x-y|+|y|\leq c+|y|$$
Thus $|x| < |y| + c$
Proof 2:
From the triangle inequality:
$$-|x-y| \leq |x| - |y| \leq |x-y| < c$$
So $$|x| - |y| < c$$
Split this into two cases:
Case 1: When $|x| > |y|$
Case 2: When $|y| > |x|$
When $$|x| > |y|, |x-y| > 0$$ Meaning $c > 0$
So by (< preserved by addition) $|x|$ < $|y| + c$
When $|y|$ > $x$, $x-y < 0$, though $|x-y| > 0$. So $x-y < 0 < |x-y|$
And $c > |x-y|$
So $c > 0$
So $|x| < |y| + c$
I'm not sure if what one, both or neither of my proofs are correct. So any help will be appreciated.
 A: Both of your proofs look perfectly fine for me. Except for where you added cases. You'll notice you get exactly the result you were trying to prove before you had to consider cases. Hence, there is no need for it. You also don't need to worry about whether or not $c>0$. (Although by your assumption that $|x-y|<c$ you were already implicitly given that $c>0$.) To be slightly more thorough, you might want to add a line to the beginning of your second proof where you begin with $||x|-|y||\leq |x-y|$ before moving to $-|x-y|\leq |x|-|y|\leq |x-y|$. But that depends on whoever will be reading your proof, and may be clear without the extra line.
A: Proof 1 seems perfect, except for one small detail: you have written $|x - y| + |y| \leq c + |y|$, whereas you need to write (and presumably meant to write) $|x - y| + |y| < c + |y|$.
I agree with the other answer and comments about Proof 2, in that it's already essentially complete almost at the start, without any need for the subsequent division into cases.
Also, because of this, Proof 2 is virtually equivalent to Proof 1, because, in quoting the triangle inequality as $|x| - |y| \leq |x - y|$, you are essentially making use of the same, more immediate deduction from the triangle inequality you made in Proof 1, viz. $|x| = |x - y + y| \leq |x - y| + |y|$.
