Is Proof method valid So I am studying for an upcoming midterm, and I am practicing my proofs. I found an old test online, that states the following:
$| x + y | + | x − y |≥| x | + | y |, x, y∈R .$
I want to know if my proof or attempt is fair enough to prove it?
my attempt:
let x be any arbitrary number, and since the absolute value of a number is always positive or zero 
then $|x| ≥ x  $  then $ |x+y| ≥ x+y  ≥ |x| ≥ x  $
(given that y is an element of the real number set) --->$|x+y|+|x| ≥ |x|+|y| +x ≥ |x|+|y| ≥ x+y ≥ x  $
and lastly $ |x+y| +|x-y| ≥  |x|+|y|+|x| ≥ |x|+|y|≥ x+y ≥ x   $ 
end proof.
 A: Note that if $s$ and $t$ are non-negative, then $s\ge t$ if and only if $s^2\ge t^2$. Now
$$(|x+y|+|x-y|)^2=2x^2+2y^2+2|x^2-y^2|\ge 2x^2+2y^2\tag{1}$$
and 
$$(|x|+|y|)^2=x^2+y^2+2|xy|.\tag{2}$$
So it is enough to show that $2x^2+2y^2\ge x^2+y^2+2|xy|$, or equivalently that $x^2+y^2\ge 2|xy|$.
But this last inequality  is clear, for $x^2+y^2-2|xy|=(|x|-|y|)^2$.
A: No, for example the step $x+y\ge |x|$ is not true, the counter example would be if $x+y<0$. Then there's more errors further down, but it's hard to tell which are a consequence of the first error and which are new errors, but for example the step $|x|+|y|+x\ge|x|+|y|$ definitely looks odd and the step $x+y\ge x$.
Finally it's not nice to keep irrelevant parts in the proof. For example the last step $|x+y|+|x-y|\ge...\ge|x|+|y|\ge...$ everything to the right of $|x|+|y|\ge$ is irrelevant (you was only to proove that $|x+y|+|x-y|\ge|x|+|y|$ - never forget what you're about to prove).
What you could do if you want to prove it using brute force is to divide it into the cases where the $x$, $y$, $x-y$ and $x+y$ are each positive or negative - there 16 combinations, some of which are impossible (you probably see those). It's a tedious way, but given you don't find any other way it's a possibility.
An easier way would be to use the triangle inequality:
$$|x+y| + |x-y| \ge |x+y+x-y| = |2x|$$
and then use this to prove it if $|x|\ge|y|$. How you us it for the case $|x|\le|y|$ you maybe can figure out yourself?
A: we have for any $x$ and $y$ (using triangle inequality)
$$|x+y| + |x-y| \ge |x+y+x-y| = |2x|$$
and $$|x+y| + |y-x| \ge |x+y-x+y| = |2y|$$
Adding up we get   $$2(|x+y| + |x-y| )\ge |2x|+ |2y| $$
hence  $$|x+y| + |x-y| \ge |x|+ |y| $$
