Help with proof of $|x+y| \geq |x|-|y|$ I've been working on this problem from Lang's 'A First Course in Calculus'. 
I'll restate the problem in full.
Prove the following inequalities for all numbers x, y. Then, the problem I'm having trouble with under this header is:
$$ |x+y| \ge |x|-|y| $$
It gives as a hint to write x = x + y -y, and apply Theorem 2.3 (If a, b are two numbers then |a+b|<= |a| + |b|), together with the fact that |-y| = |y| but I haven't been able to successfully apply that information to solve it. 
Here is my initial attempt to apply the advice. 
$$ x = x + y - y $$
Then, stating Theorem 2.3:
$$ |x+y| \le |x| + |y| $$
Now, substituting x:
$$ |(x+y-y) + y| \le |x+y-y| + |y| $$
I'm not really sure where to go after this point, but I think these are probably the correct steps so far to reach the answer. I know I'll need to get |x| - |y| on the right side and flip the inequality sign (multiplying by negative one) somehow.
I did restate the inequality in a few simplified possibilities, when x, y are both positive numbers and when x, y are both negative numbers. I know there are other cases though, such as when x is negative and y is positive or vice versa. And I'd also need to do varying cases when x+y >= 0, and x+y <= 0, where x,y could be all possible combinations of positive and negative.
It seems like there might be a solution through tedious case analysis along these lines but I believe there is probably a better way to come about the answer. In any case, the ones I did restate are:
Case 1: 
$$ x + y \ge x - y $$ when x, y are both positive or zero. This simplifies somewhat.
$$ y \ge -y $$
$$ 2y \ge 0 $$
Case 2: 
$$ -x - y \ge -x + y $$ when x, y are both negative or zero.
In the first case, when x and y are positive or equal to zero, it is obvious that it is correct. 
Intuitively speaking, the second case and indeed any case should simplify down to the first case, since I know any negatives attached to the numbers will simply make them positive or stay zero since it is applying the rules of the absolute value function. I'm not sure how to state this in a proof though, and I'd like to be able to apply the technique that the book suggests. 
I would appreciate any guidance on this problem, thanks in advance. 

The hint is trying to get you to think:
$|x| = |x+ y -y| = |x+y+(-y)| \le |x+y| + |-y| = |x+y| + |y|$.
Subtract $|y|$ from both sides and "flip".
$|x|-|y| \le |x+y|$.
....
Another way of putting it is: Let $a = x +y$ and $b = -y$.
Then 
$|a + b| \le |a| + |b|$ so
$|(x+y) + (-y)| \le |x+y| + |-y|$ and......
.....
Yet another way of putting it:
$|x+y| + |-y| \ge |x+y-y| =|x|$
So $|x| -|-y| = |x| - |y| \le |x+y|$.
