How would I prove $|x + y| \le |x| + |y|$?

Question

How would I write a detailed structured proof for:

for all real numbers $x$ and $y$, $|x + y| \le |x| + |y|$

I'm planning on breaking it up into four cases, where both $x,y < 0$, $x \ge 0$ and $y<0$, $x<0$ and $y \ge0$, and $x,y \ge 0$. But I'm not sure how I'd go about writing it formally.

Thanks!

Answer 1

You are absolutely on the right track. I'll model one case for you, and you can try the other cases on your own.

Case 1: $x,y\geq 0$. Then $x+y\geq 0$, so $|x+y|=x+y$. Similarly, $|x|=x$ because $x\geq 0$, and $|y|=y$ because $y\geq 0$. Thus $|x+y|=x+y=|x|+|y|$.

Answer 2

Here is an alternative proof, without any case distinctions, using the definition $$ |p| = p \max -p $$ so that we can use the properties of $\;\max\;$, which are simpler than those of $\;|\phantom p|\;$. In this case we start at the right hand side of the equation, which seems to be the most complex side, and calculate for every $\;x,y\;$: \begin{align} & |x| + |y| \\ = & \;\;\;\;\;\text{"the above definition, twice"} \\ & (x \max -x) + (y \max -y) \\ = & \;\;\;\;\;\text{"$\;+\;$ distributes over $\;\max\;$"} \\ & (x + (y \max -y)) \max (-x + (y \max -y)) \\ = & \;\;\;\;\;\text{"$\;+\;$ distributes over $\;\max\;$, twice more; $\;\max\;$ is associative"} \\ & (x+y) \max (x-y) \max (-x+y) \max (-x-y) \\ \geq & \;\;\;\;\;\text{"$\;p \max q \geq p\;$, twice"} \\ & (x+y) \max (-x-y) \\ = & \;\;\;\;\;\text{"the above definition"} \\ & |x + y| \\ \end{align}

Answer 3

You might consider the following:

Consider squaring both sides since both sides are non negative (you don't have to worry about sign flips). Canceling yields $$xy \leq |x||y|$$

which is true.