Prove $|x|-|y| \le |x+y|$

Question

I'm trying to prove $|x|-|y| \le |x+y|$. I read a solution that said it follows from $|x-y| \le |x|-|y|$ that $|x|=|y-(y-x)| \le |y|+|y-x|$ so $|x|-|y| \le |x-y|$. That last step seems like a bit of a jump to me. Am I missing something?

Answer 1

By the triangular inequality, $|a \pm b| \le |a| + |b|$. So: $$ |x|-|y|= |x+y-y|-|y|\le |x+y|+|-y|-|y|=|x+y|. $$

Answer 2

You have $$|x| \le |y| + |x - y|,$$ so $$|x| - |y| \le |x - y|.$$ Now reverse the roles of $x$ and $y$ to obtain your conclusion.

Answer 3

Just for variety, here's a different approach: the right side is nonnegative, so if the left side is negative, the claim is true. If both sides are nonnegative, then the claim is equivalent to the inequality resulting from squaring both sides: $$x^2-\left|x\right|\left|y\right|+y^2\leq x^2+y^2$$ which is trivially true.