Is my proof of $|x|-|y| \le |x-y|$ correct? So I just started Spivak's Calculus and I am working my way through the problem set. However, my proofs don't always aline with Spivaks and I would like to know if they are still acceptable proofs.
Note this is my first time reading a math book like this and I haven't written proofs before.
Here it goes
Prove that: $|x|-|y| \le |x-y|$
$(|x|-|y|)^2 \le (|x-y|)^2$ 
$|x|^2-2|x||y|+|y^2| \le x^2 + y^2 - 2xy $
$x^2-2|x||y|+y^2 \le x^2 + y^2 - 2xy $
$ -2|x||y| \le 2xy $
And we know the last part is true, so $|x|-|y| \le |x-y|$
 A: Before the first line of your proof just write this text. 
If $|x| \lt |y|$ then the inequality is trivially true (since the RHS is $\ge 0$).
If $|x| \ge |y|$, then it suffices to prove that the same inequality holds for the squares of the two sides.  
And then you proceed with your proof.
Between the lines of your proof write equivalence signs. 
Otherwise it's not clear if you knew this and considered it too trivial
to mention, or you simply overlooked it and you're not aware of it.    
A: You shouldn't write proofs in this direction. 
Start what what you know, then continue. 
For example: \begin{align*} |xy| &\geq xy \\ |x||y| &\geq xy \\ -2|x||y| &\leq -2xy \\ x^2-2|x||y|+y^2 &\leq x^2-2xy+y^2 \\ |x|^2-2|x||y|+|y|^2 &\leq x^2-2xy+y^2 \\ (|x|-|y|)^2 &\leq (x-y)^2  \\ (|x|-|y|)^2 &\leq (|x-y|)^2 \end{align*}
However, we now reach an problem when we try to write your last step, because we can't take the square root of inequalities. To fix this, note that we may take the square root of inequalities if we know that  $|x|-|y|\geq0$ and $|x-y|\geq0$.  The latter is true by definition, and if the former isn't true, then we have $|x|-|y|\leq0\leq|x-y|$.
Also note that I fixed a minus sign in this writeup, because from the next to last line to the last line, you are making a sign error: you went form $-2|x||y| \leq 2xy$ to $x^2-2|x||y|+y^2 \leq x^2-2xy+y^2$, but then you change the sign of $2xy$, while you don't change the other signs. Fortunately it is easy to fix. 
