Prove by contradiction or contrapositive? If $|x+y|<|x|+|y|$, then $x<0$ or $y<0$. Prove: If $|x+y|<|x|+|y|$, then $x<0$ or $y<0$
This looks as though it's true from the start. Take $x=-4, y=4$.
$|-4+4|<|-4|+|4|$
$0<8$ is true.  

The question is asking for a proof by contradiction or contrapositive.  Which means I am going to negate some part of the ending in order to find a contradiction in the hypothesis.
This is of form: If $ P\implies Q$
So for a proof by contradiction I need:
$P \implies \lnot Q $
If pr the contrapositive: 
$ \lnot Q \implies \lnot P$

Will the following proof work?  Also, is my proof formal enough?  What can be done to improve it's form?
PF. (by contradiction)
If $|x+y|<|x|+|y|, \implies x \geq 0 \lor y \geq 0$
$x \geq 0, y \geq0$
since $x \geq 0$ 
$|x+y|<|x|+|y|$ is false
proof by contradiction
 A: When coming to prove $P\Rightarrow Q$, we can either:


*

*Prove directly, that is assume $P$ and show $Q$;

*Prove by contradiction, that is assume $P$ and $\lnot Q$ and derive contradcition; or

*Prove the contrapositive, that is assume $\lnot Q$ and show $\lnot P$.


So to prove by contrapositive is to assume that $\lnot(x<0\lor y<0)\equiv (x\ge 0\land y\ge 0)$ and show that $|x+y|\nless|x|+|y|$;
and to prove by contradiction is to assume that $|x+y|<|x|+|y|$ but $x>0$ and $y>0$, and thus derive contradiction.

As for the added proof attempt:
This proof is unclear, when I was grading papers in a similar introductory course I would probably mark this question with a nice little X mark and move on.


*

*The statement of the assumption by contradiction is unclear. You should say something such as "assume by contradiction $x\ge 0, y\ge 0$." to make it clear where is the contradiction is going to come from.

*Now you are trying to prove, not to falsify. The next step is to say $x+y\ge 0$, therefore $|x+y|=x+y$ and since $|x|=x, |y|=y$ we have that $x+y=|x+y|<|x|+|y|=x+y$.
Now we have derived a contradiction, since $x+y< x+y$ is clearly a false theorem in the context above.
